Chapter 2FLUID STATICS

The science of fluid statics the study of pressure and its variation throughout a fluid the study of pressure forces on finite surfaces

Special cases of fluids moving as solids are included in the treatment of statics because of the similarity of forces involved

Since there is no motion of a fluid layer relative to an adjacent layer there are no shear stresses in the fluid

all free bodies in fluid statics have only normal pressure forces acting on their surfaces

21 PRESSURE AT A POINT

Average pressure dividing the normal force pushing against a plane area by the area

Pressure at a point the limit of the ratio of normal force to area as the area approaches zero size at the point

At a point a fluid at rest has the same pressure in all directions an element δA of very small area free to rotate about its center when submerged in a fluid at rest will have a force of constant magnitude acting on either side of it regardless of its orientation

To demonstrate this a small wedge-shaped free body of unit width is taken at the point (x y) in a fluid at rest (Fig21)

Figure 21 Free-body diagram of wedge-shaped particle

There can be no shear forces the only forces are the normal surface forces and gravity the equations of motion in the x and y directions

px py ps are the average pressures on the three faces γ is the unit gravity force of the fluid ρ is its density and ax ay are the accelerations

When the limit is taken as the free body is reduced to zero size by allowing the inclined face to approach (x y) while maintaining the same angle θ and using

the equations simplify to

Last term of the second equation ndash infinitestimal of higher of smallness may be neglected

When divided by δy and δx respectively the equations can be combined

θ is any arbitrary angle this equation proves that the pressure is the same in all directions at a point in a static fluid

Although the proof was carried out for a two-dimensional case it may be demonstrated for the three-dimensional case with the equilibrium equations for a small tetrahedron of fluid with three faces in the coordinate planes and the fourth face inclined arbitrarily

If the fluid is in motion (one layer moves relative lo an adjacent layer) shear stresses occur and the normal stresses are no longer the same in all directions at a point the pressure is defined as the average of any three mutually perpendicular normal compressive stresses at a point

Fictitious fluid of zero viscosity (frictionless fluid) no shear stresses can occur at a point the pressure is the same in all directions

22 BASIC EQUATION OF FLUID STATICS

Pressure Variation in a Static Fluid

Force balance The forces acting on an element of fluid at rest (Fig 22) surface force

s and body forces With gravity the only body force acting and by taking the y axis vertic

ally upward it is -γ δx δy δz in the y direction With pressure p at its center (x y z) the approximate force exerted on

the side normal to the y axis closest to the origin and the opposite e side are approximately

δy2 ndash the distance from center to a face normal to y

Figure 22

Rectangular parallelepiped element of fluid at rest

Summing the forces acting on the element in the y direction

For the x and z directions since no body forces act

The elemental force vector δF

If the element is reduced to zero size alter dividing through by δx δy δz = δV the expression becomes exact

This is the resultant force per unit volume at a point which must be equated to zero for a fluid at rest

The gradient isnabla

- p is the vector field f or the surface pressure force per unit volume nabla

The fluid static law or variation of pressure is then

For an inviscid fluid in motion or a fluid so moving that the shear stress is everywhere zero Newtons second law takes the form

a is the acceleration of the fluid element f - jγ is the resultant fluid force when gravity is the only body force acting

In component form Eq (224) becomes

The partials for variation in horizontal directions are one form of Pascals law they state that two points at the same elevation in the same continuous mass or fluid at rest have the same pressure

Since p is a function of y only

relates the change of pressure to unit gravity force and change of elevation and holds for both compressible and incompressible fluids

For fluids that may be considered homogeneous and incompressible γ is constant and the above equation when integrated becomes

in which c is the constant of integration The hydrostatic law of variation of pressure is frequently written in the form

h = -y p is the increase in pressure from that at the free surface

Example 21 An oceanographer is to design a sea lab 5 m high to withstand submersion to 100 m measured from sea level to the top of the sea lab Find the pressure variation on a side of the container and the pressure on the top if the relative density of salt water is 1020

At the top h = 100 m and

If y is measured from the top of the sea lab downward the pressure variation is

Pressure Variation in a Compressible Fluid

When the fluid is a perfect gas at rest at constant temperature

When the value of γ in Eq (227) is replaced by ρg and ρ is eliminated between Eqs (227) and (229)

If P = P0 when ρ = ρ0 integration between limits

- the equation for variation of pressure with elevation in an isothermal gas

- constant temperature gradient of atmosphere

Example 22 Assuming isothermal conditions to prevail in the atmosphere compute the pressure and density at 2000 m elevation if P = 105Pa ρ = 124 kgm3 at sea level

From Eq (2212)

Then from Eq (229)

23 UNITS AND SCALES OF PRESSURE MEASUREMENT

Pressure may be expressed with reference to any arbitrary datum absolute zero local atmospheric pressure

Absolute pressure difference between its value and a complete vacuum

Gage pressure difference between its value and the local atmospheric pressure

Figure 23 Bourdon gage

The bourdon gage (Fig 23) typical of the devices used for measuring gage pressures pressure element is a hollow curved flat metallic tube closed at one e

nd the other end is connected to the pressure to be measured when the internal pressure is increased the tube tends to straighten

pulling on a linkage to which is attached a pointer and causing the pointer to move

the dial reads zero when the inside and outside of the tube are at the same pressure regardless of its particular value

the gage measures pressure relative to the pressure of the medium surrounding the tube which is the local atmosphere

Figure 24 Units and scales for pressure measurement

Figure 24 the data and the relations of the common units of pressure measurement

Standard atmospheric pressure is the mean pressure at sea level 760 mm Hg

A pressure expressed in terms of the length of a column of liquid is equivalent to the force per unit area at the base of the column The relation for variation of pressure with altitude in a liquid p = γh [Eq (228)] (p is in pascals γ in newtons per cubic metre and h in metres)

With the unit gravity force of any liquid expressed as its relative density S times the unit gravity force of water

Water γ may be taken as 9806 Nm3

Figure 25 Mercury barometer

Local atmospheric pressure is measured by mercury barometer aneroid barometer (measures the difference in

pressure between the atmosphere and an evacuated box or tube in a manner analogous to the bourdon gage except that the tube is evacuated and sealed)

Mercury barometer glass tube closed at one end filled with mercury and inverted so that the open end is submerged in mercury

It has a scale the height of column R can be determined

The space above the mercury contains mercury vapor If the pressure of the mercury vapor hv is given in millimetres of mercury and R is measured in the same units the pressure at A may be expressed as (mm Hg)

Figure 24 a pressure may be located vertically on the chart which indicates its relation to absolute zero and to local atmospheric pressure

If the point is below the local-atmospheric-pressure line and is referred to gage datum it is called negative suction or vacuum

Example the pressure 460 mm Hg abs as at 1 with barometer reading 720 mm may be expressed as -260 mm Hg 260 mm Hg suction or 260 mm Hg vacuum

Note

Pabs = pbar + pgage

Absolute pressures P gage pressures p

Example 23 The rate of temperature change in the atmosphere with change in elevation is called its lapse rate The motion of a parcel of air depends on the density of the parcel relative to the density of the surrounding (ambient) air However as the parcel ascends through the atmosphere the air pressure decreases the parcel expands and its temperature decreases at a rate known as the dry adiabatic lapse rate A firm wants lo burn a large quantity of refuse It is estimated that the temperature of the smoke plume at 10 m above the ground will be 11oC greater than that of the ambient air For the following conditions determine what will happen to the smoke

(a) At standard atmospheric lapse rate β = -000651oC per meter and t0 =20oC

(b) At an inverted lapse rate β = 000365oC per meter

By combining Eqs (227) and (2214)

The relation between pressure and temperature for a mass of gas expanding without heat transfer (isentropic relation Sec 61) is

in which T1 is the initial smoke absolute temperature and P0 the initial absolute pressure k is the specific heat ratio 14 for air and other diatomic gases

Eliminating PP0 in the last two equations

Since the gas will rise until its temperature is equal to the ambient temperature

the last two equations may be solved for y Let

Then

For β = -000651oC per metre R = 287 mN(kgK) a = 2002 and y = 3201 m For the atmospheric temperature inversion β = -000365oC per metre a = -02721 and y = 8092 m

24 MANOMETERS

Manometers are devices that employ liquid columns for determining differences in pressure

Figure 26a the most elementary manometer ndash piezometer It measures the pressure in a liquid when it is above zero gage Glass tube is mounted vertically so that it is connected to the space

within the container Liquid rises in the tube until equilibrium is reached The pressure is then given by the vertical distance h from the menis

cus (liquid surface) to the point where the pressure is to be measured expressed in units of length of the liquid in the container

Piezometer would not work for negative gage pressures because air would flow into the container through the tube

Figure 26 Simple manometers

Figure 26b for small negative or positive gage pressures in a liquid With this arrangement the meniscus may come to rest below A as sho

wn Since the pressure at the meniscus is zero gage and since pressure decreases with elevation

units of length H2O Figure 26c for greater negative or positive gage pressures (a second li

quid of greater relative density employed) It must be immiscible in the first fluid which may now be a gas If the relative density of the fluid at A is S1 (based on water) and the re

lative density of the manometer liquid is S2 the equation for pressure at A

hA - the unknown pressure expressed in length units of water

h1 h2 - in length units

A general procedure in working all manometer problems

1 Start at one end (or any meniscus if the circuit is continuous) and write the pressure there in an appropriate unit (say pascals) or in an appropriate symbol if it is unknown

2 Add to this the change in pressure in the same unit from one meniscus to the next (plus if the next meniscus is lower minus if higher) (For pascals this is the product of the difference in elevation in metres and the unit gravity force of the fluid in newtons per cubic metre)

3 Continue until the other end of the gage (or the starting meniscus) is reached and equate the expression to the pressure at that point known or unknown

The expression will contain one unknown for a simple manometer or will give a difference in pressures for the differential manometer In equation form

Figure 27 Differential manometers

A differential manometer (Fig 27) determines the difference in pressures at two points A and B when the actual pressure at any point in the system cannot be determined

Application of the procedure outlined above to Fig 27a produces

For Fig 27b

If the pressures at A and B are expressed in length of the water column the above results can be written for Fig 27a

For Fig 27b

Example 24 In Fig 27a the liquids at A and B are water and the manometer liquid is oil S = 080 h1 = 300 mm h2 = 200 mm and h3 = 600 mm (a) Determine pA - pB in pacals (b) If pB = 50kPa and the barometer reading is 730 mm Hg find the pressure at A in meters of water absolute (a)

(b)

(a)

Micromanometers

For determining very small differences in pressure or determining large pressure differences precisely ndash several types of manometers

One type very accurately measures the differences in elevation of two menisci of a manometer

By means of small telescopes with horizontal cross hairs mounted along the tubes on a rack which is raised and lowered by a pinion and slow motion screw so that the cross hairs can be set accurately the difference in elevation of menisci (the gage difference) can be read with verniers

Figure 28 Micromanometer using two gage liquids

Fig 28 two gage liquids immiscible in each other and in the fluid to be measured a large gage difference R can be produced for a small pressure difference

The heavier gage liquid fills the lower U tube up to 0-0 then the lighter gage liquid is added to both sides filling the larger reservoirs up to 1-1

The gas or liquid in the system fills the space above 1-1 When the pressure at C is slightly greater than at D the menisci move as indicated in Fig 28

The volume of liquid displaced in each reservoir equals the displacement in the U tube

Manometer equation

γ1 γ2 and γ3 are the unit gravity force

Example 25 In the micromanometer of Fig 28 the pressure difference is wanted in pascals when air is in the system S2 = 10 S3 = 110 aA = 001 R = 5 mm t = 20oC and the barometer reads 760 mm Hg

The term γ1(aA) may be neglected Substituting into Eq (241) gives

Figure 29

Inclined manometer

The inclined manometer frequently used for measuring small differences in gas pressures

Adjusted to read zero by moving the inclined scale when A and B are open Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube it affords greater accuracy in reading the scale

Surface tension causes a capillary rise in small tubes If a U tube is used with a meniscus in each leg the surface-tension effects cancel

25 FORCES ON PLANE AREAS

In the preceding sections variations oF pressure throughout a fluid have been considered

The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force insofar as external reactions to the force system are concerned

In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration by formula and by use of the concept of the pressure prism

Horizontal Surfaces

A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure

The magnitude of the force acting on one side of the surface is

The elemental forces pdA acting on A are all parallel and in the same sense a scalar summation of all such elements yields the magnitude of the resultant force Its direction is normal to the surface and toward the surface if p is positive

Fig 210 arbitrary xy axes - to find the line of action of the resultant ie the point in the area where the moment of the distributed force about any axis through the point is zero

Then since the moment of the resultant must equal the moment of the distributed force system about any axis say the y axis

xrsquo ndash the distance from the y axis to the resultant

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

The science of fluid statics the study of pressure and its variation throughout a fluid the study of pressure forces on finite surfaces

Special cases of fluids moving as solids are included in the treatment of statics because of the similarity of forces involved

Since there is no motion of a fluid layer relative to an adjacent layer there are no shear stresses in the fluid

all free bodies in fluid statics have only normal pressure forces acting on their surfaces

21 PRESSURE AT A POINT

Average pressure dividing the normal force pushing against a plane area by the area

Pressure at a point the limit of the ratio of normal force to area as the area approaches zero size at the point

At a point a fluid at rest has the same pressure in all directions an element δA of very small area free to rotate about its center when submerged in a fluid at rest will have a force of constant magnitude acting on either side of it regardless of its orientation

To demonstrate this a small wedge-shaped free body of unit width is taken at the point (x y) in a fluid at rest (Fig21)

Figure 21 Free-body diagram of wedge-shaped particle

There can be no shear forces the only forces are the normal surface forces and gravity the equations of motion in the x and y directions

px py ps are the average pressures on the three faces γ is the unit gravity force of the fluid ρ is its density and ax ay are the accelerations

When the limit is taken as the free body is reduced to zero size by allowing the inclined face to approach (x y) while maintaining the same angle θ and using

the equations simplify to

Last term of the second equation ndash infinitestimal of higher of smallness may be neglected

When divided by δy and δx respectively the equations can be combined

θ is any arbitrary angle this equation proves that the pressure is the same in all directions at a point in a static fluid

Although the proof was carried out for a two-dimensional case it may be demonstrated for the three-dimensional case with the equilibrium equations for a small tetrahedron of fluid with three faces in the coordinate planes and the fourth face inclined arbitrarily

If the fluid is in motion (one layer moves relative lo an adjacent layer) shear stresses occur and the normal stresses are no longer the same in all directions at a point the pressure is defined as the average of any three mutually perpendicular normal compressive stresses at a point

Fictitious fluid of zero viscosity (frictionless fluid) no shear stresses can occur at a point the pressure is the same in all directions

22 BASIC EQUATION OF FLUID STATICS

Pressure Variation in a Static Fluid

Force balance The forces acting on an element of fluid at rest (Fig 22) surface force

s and body forces With gravity the only body force acting and by taking the y axis vertic

ally upward it is -γ δx δy δz in the y direction With pressure p at its center (x y z) the approximate force exerted on

the side normal to the y axis closest to the origin and the opposite e side are approximately

δy2 ndash the distance from center to a face normal to y

Figure 22

Rectangular parallelepiped element of fluid at rest

Summing the forces acting on the element in the y direction

For the x and z directions since no body forces act

The elemental force vector δF

If the element is reduced to zero size alter dividing through by δx δy δz = δV the expression becomes exact

This is the resultant force per unit volume at a point which must be equated to zero for a fluid at rest

The gradient isnabla

- p is the vector field f or the surface pressure force per unit volume nabla

The fluid static law or variation of pressure is then

For an inviscid fluid in motion or a fluid so moving that the shear stress is everywhere zero Newtons second law takes the form

a is the acceleration of the fluid element f - jγ is the resultant fluid force when gravity is the only body force acting

In component form Eq (224) becomes

The partials for variation in horizontal directions are one form of Pascals law they state that two points at the same elevation in the same continuous mass or fluid at rest have the same pressure

Since p is a function of y only

relates the change of pressure to unit gravity force and change of elevation and holds for both compressible and incompressible fluids

For fluids that may be considered homogeneous and incompressible γ is constant and the above equation when integrated becomes

in which c is the constant of integration The hydrostatic law of variation of pressure is frequently written in the form

h = -y p is the increase in pressure from that at the free surface

Example 21 An oceanographer is to design a sea lab 5 m high to withstand submersion to 100 m measured from sea level to the top of the sea lab Find the pressure variation on a side of the container and the pressure on the top if the relative density of salt water is 1020

At the top h = 100 m and

If y is measured from the top of the sea lab downward the pressure variation is

Pressure Variation in a Compressible Fluid

When the fluid is a perfect gas at rest at constant temperature

When the value of γ in Eq (227) is replaced by ρg and ρ is eliminated between Eqs (227) and (229)

If P = P0 when ρ = ρ0 integration between limits

- the equation for variation of pressure with elevation in an isothermal gas

- constant temperature gradient of atmosphere

Example 22 Assuming isothermal conditions to prevail in the atmosphere compute the pressure and density at 2000 m elevation if P = 105Pa ρ = 124 kgm3 at sea level

From Eq (2212)

Then from Eq (229)

23 UNITS AND SCALES OF PRESSURE MEASUREMENT

Pressure may be expressed with reference to any arbitrary datum absolute zero local atmospheric pressure

Absolute pressure difference between its value and a complete vacuum

Gage pressure difference between its value and the local atmospheric pressure

Figure 23 Bourdon gage

The bourdon gage (Fig 23) typical of the devices used for measuring gage pressures pressure element is a hollow curved flat metallic tube closed at one e

nd the other end is connected to the pressure to be measured when the internal pressure is increased the tube tends to straighten

pulling on a linkage to which is attached a pointer and causing the pointer to move

the dial reads zero when the inside and outside of the tube are at the same pressure regardless of its particular value

the gage measures pressure relative to the pressure of the medium surrounding the tube which is the local atmosphere

Figure 24 Units and scales for pressure measurement

Figure 24 the data and the relations of the common units of pressure measurement

Standard atmospheric pressure is the mean pressure at sea level 760 mm Hg

A pressure expressed in terms of the length of a column of liquid is equivalent to the force per unit area at the base of the column The relation for variation of pressure with altitude in a liquid p = γh [Eq (228)] (p is in pascals γ in newtons per cubic metre and h in metres)

With the unit gravity force of any liquid expressed as its relative density S times the unit gravity force of water

Water γ may be taken as 9806 Nm3

Figure 25 Mercury barometer

Local atmospheric pressure is measured by mercury barometer aneroid barometer (measures the difference in

pressure between the atmosphere and an evacuated box or tube in a manner analogous to the bourdon gage except that the tube is evacuated and sealed)

Mercury barometer glass tube closed at one end filled with mercury and inverted so that the open end is submerged in mercury

It has a scale the height of column R can be determined

The space above the mercury contains mercury vapor If the pressure of the mercury vapor hv is given in millimetres of mercury and R is measured in the same units the pressure at A may be expressed as (mm Hg)

Figure 24 a pressure may be located vertically on the chart which indicates its relation to absolute zero and to local atmospheric pressure

If the point is below the local-atmospheric-pressure line and is referred to gage datum it is called negative suction or vacuum

Example the pressure 460 mm Hg abs as at 1 with barometer reading 720 mm may be expressed as -260 mm Hg 260 mm Hg suction or 260 mm Hg vacuum

Note

Pabs = pbar + pgage

Absolute pressures P gage pressures p

Example 23 The rate of temperature change in the atmosphere with change in elevation is called its lapse rate The motion of a parcel of air depends on the density of the parcel relative to the density of the surrounding (ambient) air However as the parcel ascends through the atmosphere the air pressure decreases the parcel expands and its temperature decreases at a rate known as the dry adiabatic lapse rate A firm wants lo burn a large quantity of refuse It is estimated that the temperature of the smoke plume at 10 m above the ground will be 11oC greater than that of the ambient air For the following conditions determine what will happen to the smoke

(a) At standard atmospheric lapse rate β = -000651oC per meter and t0 =20oC

(b) At an inverted lapse rate β = 000365oC per meter

By combining Eqs (227) and (2214)

The relation between pressure and temperature for a mass of gas expanding without heat transfer (isentropic relation Sec 61) is

in which T1 is the initial smoke absolute temperature and P0 the initial absolute pressure k is the specific heat ratio 14 for air and other diatomic gases

Eliminating PP0 in the last two equations

Since the gas will rise until its temperature is equal to the ambient temperature

the last two equations may be solved for y Let

Then

For β = -000651oC per metre R = 287 mN(kgK) a = 2002 and y = 3201 m For the atmospheric temperature inversion β = -000365oC per metre a = -02721 and y = 8092 m

24 MANOMETERS

Manometers are devices that employ liquid columns for determining differences in pressure

Figure 26a the most elementary manometer ndash piezometer It measures the pressure in a liquid when it is above zero gage Glass tube is mounted vertically so that it is connected to the space

within the container Liquid rises in the tube until equilibrium is reached The pressure is then given by the vertical distance h from the menis

cus (liquid surface) to the point where the pressure is to be measured expressed in units of length of the liquid in the container

Piezometer would not work for negative gage pressures because air would flow into the container through the tube

Figure 26 Simple manometers

Figure 26b for small negative or positive gage pressures in a liquid With this arrangement the meniscus may come to rest below A as sho

wn Since the pressure at the meniscus is zero gage and since pressure decreases with elevation

units of length H2O Figure 26c for greater negative or positive gage pressures (a second li

quid of greater relative density employed) It must be immiscible in the first fluid which may now be a gas If the relative density of the fluid at A is S1 (based on water) and the re

lative density of the manometer liquid is S2 the equation for pressure at A

hA - the unknown pressure expressed in length units of water

h1 h2 - in length units

A general procedure in working all manometer problems

1 Start at one end (or any meniscus if the circuit is continuous) and write the pressure there in an appropriate unit (say pascals) or in an appropriate symbol if it is unknown

2 Add to this the change in pressure in the same unit from one meniscus to the next (plus if the next meniscus is lower minus if higher) (For pascals this is the product of the difference in elevation in metres and the unit gravity force of the fluid in newtons per cubic metre)

3 Continue until the other end of the gage (or the starting meniscus) is reached and equate the expression to the pressure at that point known or unknown

The expression will contain one unknown for a simple manometer or will give a difference in pressures for the differential manometer In equation form

Figure 27 Differential manometers

A differential manometer (Fig 27) determines the difference in pressures at two points A and B when the actual pressure at any point in the system cannot be determined

Application of the procedure outlined above to Fig 27a produces

For Fig 27b

If the pressures at A and B are expressed in length of the water column the above results can be written for Fig 27a

For Fig 27b

Example 24 In Fig 27a the liquids at A and B are water and the manometer liquid is oil S = 080 h1 = 300 mm h2 = 200 mm and h3 = 600 mm (a) Determine pA - pB in pacals (b) If pB = 50kPa and the barometer reading is 730 mm Hg find the pressure at A in meters of water absolute (a)

(b)

(a)

Micromanometers

For determining very small differences in pressure or determining large pressure differences precisely ndash several types of manometers

One type very accurately measures the differences in elevation of two menisci of a manometer

By means of small telescopes with horizontal cross hairs mounted along the tubes on a rack which is raised and lowered by a pinion and slow motion screw so that the cross hairs can be set accurately the difference in elevation of menisci (the gage difference) can be read with verniers

Figure 28 Micromanometer using two gage liquids

Fig 28 two gage liquids immiscible in each other and in the fluid to be measured a large gage difference R can be produced for a small pressure difference

The heavier gage liquid fills the lower U tube up to 0-0 then the lighter gage liquid is added to both sides filling the larger reservoirs up to 1-1

The gas or liquid in the system fills the space above 1-1 When the pressure at C is slightly greater than at D the menisci move as indicated in Fig 28

The volume of liquid displaced in each reservoir equals the displacement in the U tube

Manometer equation

γ1 γ2 and γ3 are the unit gravity force

Example 25 In the micromanometer of Fig 28 the pressure difference is wanted in pascals when air is in the system S2 = 10 S3 = 110 aA = 001 R = 5 mm t = 20oC and the barometer reads 760 mm Hg

The term γ1(aA) may be neglected Substituting into Eq (241) gives

Figure 29

Inclined manometer

The inclined manometer frequently used for measuring small differences in gas pressures

Adjusted to read zero by moving the inclined scale when A and B are open Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube it affords greater accuracy in reading the scale

Surface tension causes a capillary rise in small tubes If a U tube is used with a meniscus in each leg the surface-tension effects cancel

25 FORCES ON PLANE AREAS

In the preceding sections variations oF pressure throughout a fluid have been considered

The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force insofar as external reactions to the force system are concerned

In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration by formula and by use of the concept of the pressure prism

Horizontal Surfaces

A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure

The magnitude of the force acting on one side of the surface is

The elemental forces pdA acting on A are all parallel and in the same sense a scalar summation of all such elements yields the magnitude of the resultant force Its direction is normal to the surface and toward the surface if p is positive

Fig 210 arbitrary xy axes - to find the line of action of the resultant ie the point in the area where the moment of the distributed force about any axis through the point is zero

Then since the moment of the resultant must equal the moment of the distributed force system about any axis say the y axis

xrsquo ndash the distance from the y axis to the resultant

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

21 PRESSURE AT A POINT

Average pressure dividing the normal force pushing against a plane area by the area

Pressure at a point the limit of the ratio of normal force to area as the area approaches zero size at the point

At a point a fluid at rest has the same pressure in all directions an element δA of very small area free to rotate about its center when submerged in a fluid at rest will have a force of constant magnitude acting on either side of it regardless of its orientation

To demonstrate this a small wedge-shaped free body of unit width is taken at the point (x y) in a fluid at rest (Fig21)

Figure 21 Free-body diagram of wedge-shaped particle

There can be no shear forces the only forces are the normal surface forces and gravity the equations of motion in the x and y directions

px py ps are the average pressures on the three faces γ is the unit gravity force of the fluid ρ is its density and ax ay are the accelerations

When the limit is taken as the free body is reduced to zero size by allowing the inclined face to approach (x y) while maintaining the same angle θ and using

the equations simplify to

Last term of the second equation ndash infinitestimal of higher of smallness may be neglected

When divided by δy and δx respectively the equations can be combined

θ is any arbitrary angle this equation proves that the pressure is the same in all directions at a point in a static fluid

Although the proof was carried out for a two-dimensional case it may be demonstrated for the three-dimensional case with the equilibrium equations for a small tetrahedron of fluid with three faces in the coordinate planes and the fourth face inclined arbitrarily

If the fluid is in motion (one layer moves relative lo an adjacent layer) shear stresses occur and the normal stresses are no longer the same in all directions at a point the pressure is defined as the average of any three mutually perpendicular normal compressive stresses at a point

Fictitious fluid of zero viscosity (frictionless fluid) no shear stresses can occur at a point the pressure is the same in all directions

22 BASIC EQUATION OF FLUID STATICS

Pressure Variation in a Static Fluid

Force balance The forces acting on an element of fluid at rest (Fig 22) surface force

s and body forces With gravity the only body force acting and by taking the y axis vertic

ally upward it is -γ δx δy δz in the y direction With pressure p at its center (x y z) the approximate force exerted on

the side normal to the y axis closest to the origin and the opposite e side are approximately

δy2 ndash the distance from center to a face normal to y

Figure 22

Rectangular parallelepiped element of fluid at rest

Summing the forces acting on the element in the y direction

For the x and z directions since no body forces act

The elemental force vector δF

If the element is reduced to zero size alter dividing through by δx δy δz = δV the expression becomes exact

This is the resultant force per unit volume at a point which must be equated to zero for a fluid at rest

The gradient isnabla

- p is the vector field f or the surface pressure force per unit volume nabla

The fluid static law or variation of pressure is then

For an inviscid fluid in motion or a fluid so moving that the shear stress is everywhere zero Newtons second law takes the form

a is the acceleration of the fluid element f - jγ is the resultant fluid force when gravity is the only body force acting

In component form Eq (224) becomes

The partials for variation in horizontal directions are one form of Pascals law they state that two points at the same elevation in the same continuous mass or fluid at rest have the same pressure

Since p is a function of y only

relates the change of pressure to unit gravity force and change of elevation and holds for both compressible and incompressible fluids

For fluids that may be considered homogeneous and incompressible γ is constant and the above equation when integrated becomes

in which c is the constant of integration The hydrostatic law of variation of pressure is frequently written in the form

h = -y p is the increase in pressure from that at the free surface

Example 21 An oceanographer is to design a sea lab 5 m high to withstand submersion to 100 m measured from sea level to the top of the sea lab Find the pressure variation on a side of the container and the pressure on the top if the relative density of salt water is 1020

At the top h = 100 m and

If y is measured from the top of the sea lab downward the pressure variation is

Pressure Variation in a Compressible Fluid

When the fluid is a perfect gas at rest at constant temperature

When the value of γ in Eq (227) is replaced by ρg and ρ is eliminated between Eqs (227) and (229)

If P = P0 when ρ = ρ0 integration between limits

- the equation for variation of pressure with elevation in an isothermal gas

- constant temperature gradient of atmosphere

Example 22 Assuming isothermal conditions to prevail in the atmosphere compute the pressure and density at 2000 m elevation if P = 105Pa ρ = 124 kgm3 at sea level

From Eq (2212)

Then from Eq (229)

23 UNITS AND SCALES OF PRESSURE MEASUREMENT

Pressure may be expressed with reference to any arbitrary datum absolute zero local atmospheric pressure

Absolute pressure difference between its value and a complete vacuum

Gage pressure difference between its value and the local atmospheric pressure

Figure 23 Bourdon gage

The bourdon gage (Fig 23) typical of the devices used for measuring gage pressures pressure element is a hollow curved flat metallic tube closed at one e

nd the other end is connected to the pressure to be measured when the internal pressure is increased the tube tends to straighten

pulling on a linkage to which is attached a pointer and causing the pointer to move

the dial reads zero when the inside and outside of the tube are at the same pressure regardless of its particular value

the gage measures pressure relative to the pressure of the medium surrounding the tube which is the local atmosphere

Figure 24 Units and scales for pressure measurement

Figure 24 the data and the relations of the common units of pressure measurement

Standard atmospheric pressure is the mean pressure at sea level 760 mm Hg

A pressure expressed in terms of the length of a column of liquid is equivalent to the force per unit area at the base of the column The relation for variation of pressure with altitude in a liquid p = γh [Eq (228)] (p is in pascals γ in newtons per cubic metre and h in metres)

With the unit gravity force of any liquid expressed as its relative density S times the unit gravity force of water

Water γ may be taken as 9806 Nm3

Figure 25 Mercury barometer

Local atmospheric pressure is measured by mercury barometer aneroid barometer (measures the difference in

pressure between the atmosphere and an evacuated box or tube in a manner analogous to the bourdon gage except that the tube is evacuated and sealed)

Mercury barometer glass tube closed at one end filled with mercury and inverted so that the open end is submerged in mercury

It has a scale the height of column R can be determined

The space above the mercury contains mercury vapor If the pressure of the mercury vapor hv is given in millimetres of mercury and R is measured in the same units the pressure at A may be expressed as (mm Hg)

Figure 24 a pressure may be located vertically on the chart which indicates its relation to absolute zero and to local atmospheric pressure

If the point is below the local-atmospheric-pressure line and is referred to gage datum it is called negative suction or vacuum

Example the pressure 460 mm Hg abs as at 1 with barometer reading 720 mm may be expressed as -260 mm Hg 260 mm Hg suction or 260 mm Hg vacuum

Note

Pabs = pbar + pgage

Absolute pressures P gage pressures p

Example 23 The rate of temperature change in the atmosphere with change in elevation is called its lapse rate The motion of a parcel of air depends on the density of the parcel relative to the density of the surrounding (ambient) air However as the parcel ascends through the atmosphere the air pressure decreases the parcel expands and its temperature decreases at a rate known as the dry adiabatic lapse rate A firm wants lo burn a large quantity of refuse It is estimated that the temperature of the smoke plume at 10 m above the ground will be 11oC greater than that of the ambient air For the following conditions determine what will happen to the smoke

(a) At standard atmospheric lapse rate β = -000651oC per meter and t0 =20oC

(b) At an inverted lapse rate β = 000365oC per meter

By combining Eqs (227) and (2214)

The relation between pressure and temperature for a mass of gas expanding without heat transfer (isentropic relation Sec 61) is

in which T1 is the initial smoke absolute temperature and P0 the initial absolute pressure k is the specific heat ratio 14 for air and other diatomic gases

Eliminating PP0 in the last two equations

Since the gas will rise until its temperature is equal to the ambient temperature

the last two equations may be solved for y Let

Then

For β = -000651oC per metre R = 287 mN(kgK) a = 2002 and y = 3201 m For the atmospheric temperature inversion β = -000365oC per metre a = -02721 and y = 8092 m

24 MANOMETERS

Manometers are devices that employ liquid columns for determining differences in pressure

Figure 26a the most elementary manometer ndash piezometer It measures the pressure in a liquid when it is above zero gage Glass tube is mounted vertically so that it is connected to the space

within the container Liquid rises in the tube until equilibrium is reached The pressure is then given by the vertical distance h from the menis

cus (liquid surface) to the point where the pressure is to be measured expressed in units of length of the liquid in the container

Piezometer would not work for negative gage pressures because air would flow into the container through the tube

Figure 26 Simple manometers

Figure 26b for small negative or positive gage pressures in a liquid With this arrangement the meniscus may come to rest below A as sho

wn Since the pressure at the meniscus is zero gage and since pressure decreases with elevation

units of length H2O Figure 26c for greater negative or positive gage pressures (a second li

quid of greater relative density employed) It must be immiscible in the first fluid which may now be a gas If the relative density of the fluid at A is S1 (based on water) and the re

lative density of the manometer liquid is S2 the equation for pressure at A

hA - the unknown pressure expressed in length units of water

h1 h2 - in length units

A general procedure in working all manometer problems

1 Start at one end (or any meniscus if the circuit is continuous) and write the pressure there in an appropriate unit (say pascals) or in an appropriate symbol if it is unknown

2 Add to this the change in pressure in the same unit from one meniscus to the next (plus if the next meniscus is lower minus if higher) (For pascals this is the product of the difference in elevation in metres and the unit gravity force of the fluid in newtons per cubic metre)

3 Continue until the other end of the gage (or the starting meniscus) is reached and equate the expression to the pressure at that point known or unknown

The expression will contain one unknown for a simple manometer or will give a difference in pressures for the differential manometer In equation form

Figure 27 Differential manometers

A differential manometer (Fig 27) determines the difference in pressures at two points A and B when the actual pressure at any point in the system cannot be determined

Application of the procedure outlined above to Fig 27a produces

For Fig 27b

If the pressures at A and B are expressed in length of the water column the above results can be written for Fig 27a

For Fig 27b

Example 24 In Fig 27a the liquids at A and B are water and the manometer liquid is oil S = 080 h1 = 300 mm h2 = 200 mm and h3 = 600 mm (a) Determine pA - pB in pacals (b) If pB = 50kPa and the barometer reading is 730 mm Hg find the pressure at A in meters of water absolute (a)

(b)

(a)

Micromanometers

For determining very small differences in pressure or determining large pressure differences precisely ndash several types of manometers

One type very accurately measures the differences in elevation of two menisci of a manometer

By means of small telescopes with horizontal cross hairs mounted along the tubes on a rack which is raised and lowered by a pinion and slow motion screw so that the cross hairs can be set accurately the difference in elevation of menisci (the gage difference) can be read with verniers

Figure 28 Micromanometer using two gage liquids

Fig 28 two gage liquids immiscible in each other and in the fluid to be measured a large gage difference R can be produced for a small pressure difference

The heavier gage liquid fills the lower U tube up to 0-0 then the lighter gage liquid is added to both sides filling the larger reservoirs up to 1-1

The gas or liquid in the system fills the space above 1-1 When the pressure at C is slightly greater than at D the menisci move as indicated in Fig 28

The volume of liquid displaced in each reservoir equals the displacement in the U tube

Manometer equation

γ1 γ2 and γ3 are the unit gravity force

Example 25 In the micromanometer of Fig 28 the pressure difference is wanted in pascals when air is in the system S2 = 10 S3 = 110 aA = 001 R = 5 mm t = 20oC and the barometer reads 760 mm Hg

The term γ1(aA) may be neglected Substituting into Eq (241) gives

Figure 29

Inclined manometer

The inclined manometer frequently used for measuring small differences in gas pressures

Adjusted to read zero by moving the inclined scale when A and B are open Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube it affords greater accuracy in reading the scale

Surface tension causes a capillary rise in small tubes If a U tube is used with a meniscus in each leg the surface-tension effects cancel

25 FORCES ON PLANE AREAS

In the preceding sections variations oF pressure throughout a fluid have been considered

The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force insofar as external reactions to the force system are concerned

In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration by formula and by use of the concept of the pressure prism

Horizontal Surfaces

A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure

The magnitude of the force acting on one side of the surface is

The elemental forces pdA acting on A are all parallel and in the same sense a scalar summation of all such elements yields the magnitude of the resultant force Its direction is normal to the surface and toward the surface if p is positive

Fig 210 arbitrary xy axes - to find the line of action of the resultant ie the point in the area where the moment of the distributed force about any axis through the point is zero

Then since the moment of the resultant must equal the moment of the distributed force system about any axis say the y axis

xrsquo ndash the distance from the y axis to the resultant

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Figure 21 Free-body diagram of wedge-shaped particle

There can be no shear forces the only forces are the normal surface forces and gravity the equations of motion in the x and y directions

px py ps are the average pressures on the three faces γ is the unit gravity force of the fluid ρ is its density and ax ay are the accelerations

When the limit is taken as the free body is reduced to zero size by allowing the inclined face to approach (x y) while maintaining the same angle θ and using

the equations simplify to

Last term of the second equation ndash infinitestimal of higher of smallness may be neglected

When divided by δy and δx respectively the equations can be combined

θ is any arbitrary angle this equation proves that the pressure is the same in all directions at a point in a static fluid

Although the proof was carried out for a two-dimensional case it may be demonstrated for the three-dimensional case with the equilibrium equations for a small tetrahedron of fluid with three faces in the coordinate planes and the fourth face inclined arbitrarily

If the fluid is in motion (one layer moves relative lo an adjacent layer) shear stresses occur and the normal stresses are no longer the same in all directions at a point the pressure is defined as the average of any three mutually perpendicular normal compressive stresses at a point

Fictitious fluid of zero viscosity (frictionless fluid) no shear stresses can occur at a point the pressure is the same in all directions

22 BASIC EQUATION OF FLUID STATICS

Pressure Variation in a Static Fluid

Force balance The forces acting on an element of fluid at rest (Fig 22) surface force

s and body forces With gravity the only body force acting and by taking the y axis vertic

ally upward it is -γ δx δy δz in the y direction With pressure p at its center (x y z) the approximate force exerted on

the side normal to the y axis closest to the origin and the opposite e side are approximately

δy2 ndash the distance from center to a face normal to y

Figure 22

Rectangular parallelepiped element of fluid at rest

Summing the forces acting on the element in the y direction

For the x and z directions since no body forces act

The elemental force vector δF

If the element is reduced to zero size alter dividing through by δx δy δz = δV the expression becomes exact

This is the resultant force per unit volume at a point which must be equated to zero for a fluid at rest

The gradient isnabla

- p is the vector field f or the surface pressure force per unit volume nabla

The fluid static law or variation of pressure is then

For an inviscid fluid in motion or a fluid so moving that the shear stress is everywhere zero Newtons second law takes the form

a is the acceleration of the fluid element f - jγ is the resultant fluid force when gravity is the only body force acting

In component form Eq (224) becomes

The partials for variation in horizontal directions are one form of Pascals law they state that two points at the same elevation in the same continuous mass or fluid at rest have the same pressure

Since p is a function of y only

relates the change of pressure to unit gravity force and change of elevation and holds for both compressible and incompressible fluids

For fluids that may be considered homogeneous and incompressible γ is constant and the above equation when integrated becomes

in which c is the constant of integration The hydrostatic law of variation of pressure is frequently written in the form

h = -y p is the increase in pressure from that at the free surface

Example 21 An oceanographer is to design a sea lab 5 m high to withstand submersion to 100 m measured from sea level to the top of the sea lab Find the pressure variation on a side of the container and the pressure on the top if the relative density of salt water is 1020

At the top h = 100 m and

If y is measured from the top of the sea lab downward the pressure variation is

Pressure Variation in a Compressible Fluid

When the fluid is a perfect gas at rest at constant temperature

When the value of γ in Eq (227) is replaced by ρg and ρ is eliminated between Eqs (227) and (229)

If P = P0 when ρ = ρ0 integration between limits

- the equation for variation of pressure with elevation in an isothermal gas

- constant temperature gradient of atmosphere

Example 22 Assuming isothermal conditions to prevail in the atmosphere compute the pressure and density at 2000 m elevation if P = 105Pa ρ = 124 kgm3 at sea level

From Eq (2212)

Then from Eq (229)

23 UNITS AND SCALES OF PRESSURE MEASUREMENT

Pressure may be expressed with reference to any arbitrary datum absolute zero local atmospheric pressure

Absolute pressure difference between its value and a complete vacuum

Gage pressure difference between its value and the local atmospheric pressure

Figure 23 Bourdon gage

The bourdon gage (Fig 23) typical of the devices used for measuring gage pressures pressure element is a hollow curved flat metallic tube closed at one e

nd the other end is connected to the pressure to be measured when the internal pressure is increased the tube tends to straighten

pulling on a linkage to which is attached a pointer and causing the pointer to move

the dial reads zero when the inside and outside of the tube are at the same pressure regardless of its particular value

the gage measures pressure relative to the pressure of the medium surrounding the tube which is the local atmosphere

Figure 24 Units and scales for pressure measurement

Figure 24 the data and the relations of the common units of pressure measurement

Standard atmospheric pressure is the mean pressure at sea level 760 mm Hg

A pressure expressed in terms of the length of a column of liquid is equivalent to the force per unit area at the base of the column The relation for variation of pressure with altitude in a liquid p = γh [Eq (228)] (p is in pascals γ in newtons per cubic metre and h in metres)

With the unit gravity force of any liquid expressed as its relative density S times the unit gravity force of water

Water γ may be taken as 9806 Nm3

Figure 25 Mercury barometer

Local atmospheric pressure is measured by mercury barometer aneroid barometer (measures the difference in

pressure between the atmosphere and an evacuated box or tube in a manner analogous to the bourdon gage except that the tube is evacuated and sealed)

Mercury barometer glass tube closed at one end filled with mercury and inverted so that the open end is submerged in mercury

It has a scale the height of column R can be determined

The space above the mercury contains mercury vapor If the pressure of the mercury vapor hv is given in millimetres of mercury and R is measured in the same units the pressure at A may be expressed as (mm Hg)

Figure 24 a pressure may be located vertically on the chart which indicates its relation to absolute zero and to local atmospheric pressure

If the point is below the local-atmospheric-pressure line and is referred to gage datum it is called negative suction or vacuum

Example the pressure 460 mm Hg abs as at 1 with barometer reading 720 mm may be expressed as -260 mm Hg 260 mm Hg suction or 260 mm Hg vacuum

Note

Pabs = pbar + pgage

Absolute pressures P gage pressures p

Example 23 The rate of temperature change in the atmosphere with change in elevation is called its lapse rate The motion of a parcel of air depends on the density of the parcel relative to the density of the surrounding (ambient) air However as the parcel ascends through the atmosphere the air pressure decreases the parcel expands and its temperature decreases at a rate known as the dry adiabatic lapse rate A firm wants lo burn a large quantity of refuse It is estimated that the temperature of the smoke plume at 10 m above the ground will be 11oC greater than that of the ambient air For the following conditions determine what will happen to the smoke

(a) At standard atmospheric lapse rate β = -000651oC per meter and t0 =20oC

(b) At an inverted lapse rate β = 000365oC per meter

By combining Eqs (227) and (2214)

The relation between pressure and temperature for a mass of gas expanding without heat transfer (isentropic relation Sec 61) is

in which T1 is the initial smoke absolute temperature and P0 the initial absolute pressure k is the specific heat ratio 14 for air and other diatomic gases

Eliminating PP0 in the last two equations

Since the gas will rise until its temperature is equal to the ambient temperature

the last two equations may be solved for y Let

Then

For β = -000651oC per metre R = 287 mN(kgK) a = 2002 and y = 3201 m For the atmospheric temperature inversion β = -000365oC per metre a = -02721 and y = 8092 m

24 MANOMETERS

Manometers are devices that employ liquid columns for determining differences in pressure

Figure 26a the most elementary manometer ndash piezometer It measures the pressure in a liquid when it is above zero gage Glass tube is mounted vertically so that it is connected to the space

within the container Liquid rises in the tube until equilibrium is reached The pressure is then given by the vertical distance h from the menis

cus (liquid surface) to the point where the pressure is to be measured expressed in units of length of the liquid in the container

Piezometer would not work for negative gage pressures because air would flow into the container through the tube

Figure 26 Simple manometers

Figure 26b for small negative or positive gage pressures in a liquid With this arrangement the meniscus may come to rest below A as sho

wn Since the pressure at the meniscus is zero gage and since pressure decreases with elevation

units of length H2O Figure 26c for greater negative or positive gage pressures (a second li

quid of greater relative density employed) It must be immiscible in the first fluid which may now be a gas If the relative density of the fluid at A is S1 (based on water) and the re

lative density of the manometer liquid is S2 the equation for pressure at A

hA - the unknown pressure expressed in length units of water

h1 h2 - in length units

A general procedure in working all manometer problems

1 Start at one end (or any meniscus if the circuit is continuous) and write the pressure there in an appropriate unit (say pascals) or in an appropriate symbol if it is unknown

2 Add to this the change in pressure in the same unit from one meniscus to the next (plus if the next meniscus is lower minus if higher) (For pascals this is the product of the difference in elevation in metres and the unit gravity force of the fluid in newtons per cubic metre)

3 Continue until the other end of the gage (or the starting meniscus) is reached and equate the expression to the pressure at that point known or unknown

The expression will contain one unknown for a simple manometer or will give a difference in pressures for the differential manometer In equation form

Figure 27 Differential manometers

A differential manometer (Fig 27) determines the difference in pressures at two points A and B when the actual pressure at any point in the system cannot be determined

Application of the procedure outlined above to Fig 27a produces

For Fig 27b

If the pressures at A and B are expressed in length of the water column the above results can be written for Fig 27a

For Fig 27b

Example 24 In Fig 27a the liquids at A and B are water and the manometer liquid is oil S = 080 h1 = 300 mm h2 = 200 mm and h3 = 600 mm (a) Determine pA - pB in pacals (b) If pB = 50kPa and the barometer reading is 730 mm Hg find the pressure at A in meters of water absolute (a)

(b)

(a)

Micromanometers

For determining very small differences in pressure or determining large pressure differences precisely ndash several types of manometers

One type very accurately measures the differences in elevation of two menisci of a manometer

By means of small telescopes with horizontal cross hairs mounted along the tubes on a rack which is raised and lowered by a pinion and slow motion screw so that the cross hairs can be set accurately the difference in elevation of menisci (the gage difference) can be read with verniers

Figure 28 Micromanometer using two gage liquids

Fig 28 two gage liquids immiscible in each other and in the fluid to be measured a large gage difference R can be produced for a small pressure difference

The heavier gage liquid fills the lower U tube up to 0-0 then the lighter gage liquid is added to both sides filling the larger reservoirs up to 1-1

The gas or liquid in the system fills the space above 1-1 When the pressure at C is slightly greater than at D the menisci move as indicated in Fig 28

The volume of liquid displaced in each reservoir equals the displacement in the U tube

Manometer equation

γ1 γ2 and γ3 are the unit gravity force

Example 25 In the micromanometer of Fig 28 the pressure difference is wanted in pascals when air is in the system S2 = 10 S3 = 110 aA = 001 R = 5 mm t = 20oC and the barometer reads 760 mm Hg

The term γ1(aA) may be neglected Substituting into Eq (241) gives

Figure 29

Inclined manometer

The inclined manometer frequently used for measuring small differences in gas pressures

Adjusted to read zero by moving the inclined scale when A and B are open Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube it affords greater accuracy in reading the scale

Surface tension causes a capillary rise in small tubes If a U tube is used with a meniscus in each leg the surface-tension effects cancel

25 FORCES ON PLANE AREAS

In the preceding sections variations oF pressure throughout a fluid have been considered

The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force insofar as external reactions to the force system are concerned

In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration by formula and by use of the concept of the pressure prism

Horizontal Surfaces

A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure

The magnitude of the force acting on one side of the surface is

The elemental forces pdA acting on A are all parallel and in the same sense a scalar summation of all such elements yields the magnitude of the resultant force Its direction is normal to the surface and toward the surface if p is positive

Fig 210 arbitrary xy axes - to find the line of action of the resultant ie the point in the area where the moment of the distributed force about any axis through the point is zero

Then since the moment of the resultant must equal the moment of the distributed force system about any axis say the y axis

xrsquo ndash the distance from the y axis to the resultant

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

There can be no shear forces the only forces are the normal surface forces and gravity the equations of motion in the x and y directions

px py ps are the average pressures on the three faces γ is the unit gravity force of the fluid ρ is its density and ax ay are the accelerations

When the limit is taken as the free body is reduced to zero size by allowing the inclined face to approach (x y) while maintaining the same angle θ and using

the equations simplify to

Last term of the second equation ndash infinitestimal of higher of smallness may be neglected

When divided by δy and δx respectively the equations can be combined

θ is any arbitrary angle this equation proves that the pressure is the same in all directions at a point in a static fluid

Although the proof was carried out for a two-dimensional case it may be demonstrated for the three-dimensional case with the equilibrium equations for a small tetrahedron of fluid with three faces in the coordinate planes and the fourth face inclined arbitrarily

If the fluid is in motion (one layer moves relative lo an adjacent layer) shear stresses occur and the normal stresses are no longer the same in all directions at a point the pressure is defined as the average of any three mutually perpendicular normal compressive stresses at a point

Fictitious fluid of zero viscosity (frictionless fluid) no shear stresses can occur at a point the pressure is the same in all directions

22 BASIC EQUATION OF FLUID STATICS

Pressure Variation in a Static Fluid

Force balance The forces acting on an element of fluid at rest (Fig 22) surface force

s and body forces With gravity the only body force acting and by taking the y axis vertic

ally upward it is -γ δx δy δz in the y direction With pressure p at its center (x y z) the approximate force exerted on

the side normal to the y axis closest to the origin and the opposite e side are approximately

δy2 ndash the distance from center to a face normal to y

Figure 22

Rectangular parallelepiped element of fluid at rest

Summing the forces acting on the element in the y direction

For the x and z directions since no body forces act

The elemental force vector δF

If the element is reduced to zero size alter dividing through by δx δy δz = δV the expression becomes exact

This is the resultant force per unit volume at a point which must be equated to zero for a fluid at rest

The gradient isnabla

- p is the vector field f or the surface pressure force per unit volume nabla

The fluid static law or variation of pressure is then

For an inviscid fluid in motion or a fluid so moving that the shear stress is everywhere zero Newtons second law takes the form

a is the acceleration of the fluid element f - jγ is the resultant fluid force when gravity is the only body force acting

In component form Eq (224) becomes

The partials for variation in horizontal directions are one form of Pascals law they state that two points at the same elevation in the same continuous mass or fluid at rest have the same pressure

Since p is a function of y only

relates the change of pressure to unit gravity force and change of elevation and holds for both compressible and incompressible fluids

For fluids that may be considered homogeneous and incompressible γ is constant and the above equation when integrated becomes

in which c is the constant of integration The hydrostatic law of variation of pressure is frequently written in the form

h = -y p is the increase in pressure from that at the free surface

Example 21 An oceanographer is to design a sea lab 5 m high to withstand submersion to 100 m measured from sea level to the top of the sea lab Find the pressure variation on a side of the container and the pressure on the top if the relative density of salt water is 1020

At the top h = 100 m and

If y is measured from the top of the sea lab downward the pressure variation is

Pressure Variation in a Compressible Fluid

When the fluid is a perfect gas at rest at constant temperature

When the value of γ in Eq (227) is replaced by ρg and ρ is eliminated between Eqs (227) and (229)

If P = P0 when ρ = ρ0 integration between limits

- the equation for variation of pressure with elevation in an isothermal gas

- constant temperature gradient of atmosphere

Example 22 Assuming isothermal conditions to prevail in the atmosphere compute the pressure and density at 2000 m elevation if P = 105Pa ρ = 124 kgm3 at sea level

From Eq (2212)

Then from Eq (229)

23 UNITS AND SCALES OF PRESSURE MEASUREMENT

Pressure may be expressed with reference to any arbitrary datum absolute zero local atmospheric pressure

Absolute pressure difference between its value and a complete vacuum

Gage pressure difference between its value and the local atmospheric pressure

Figure 23 Bourdon gage

The bourdon gage (Fig 23) typical of the devices used for measuring gage pressures pressure element is a hollow curved flat metallic tube closed at one e

nd the other end is connected to the pressure to be measured when the internal pressure is increased the tube tends to straighten

pulling on a linkage to which is attached a pointer and causing the pointer to move

the dial reads zero when the inside and outside of the tube are at the same pressure regardless of its particular value

the gage measures pressure relative to the pressure of the medium surrounding the tube which is the local atmosphere

Figure 24 Units and scales for pressure measurement

Figure 24 the data and the relations of the common units of pressure measurement

Standard atmospheric pressure is the mean pressure at sea level 760 mm Hg

A pressure expressed in terms of the length of a column of liquid is equivalent to the force per unit area at the base of the column The relation for variation of pressure with altitude in a liquid p = γh [Eq (228)] (p is in pascals γ in newtons per cubic metre and h in metres)

With the unit gravity force of any liquid expressed as its relative density S times the unit gravity force of water

Water γ may be taken as 9806 Nm3

Figure 25 Mercury barometer

Local atmospheric pressure is measured by mercury barometer aneroid barometer (measures the difference in

pressure between the atmosphere and an evacuated box or tube in a manner analogous to the bourdon gage except that the tube is evacuated and sealed)

Mercury barometer glass tube closed at one end filled with mercury and inverted so that the open end is submerged in mercury

It has a scale the height of column R can be determined

The space above the mercury contains mercury vapor If the pressure of the mercury vapor hv is given in millimetres of mercury and R is measured in the same units the pressure at A may be expressed as (mm Hg)

Figure 24 a pressure may be located vertically on the chart which indicates its relation to absolute zero and to local atmospheric pressure

If the point is below the local-atmospheric-pressure line and is referred to gage datum it is called negative suction or vacuum

Example the pressure 460 mm Hg abs as at 1 with barometer reading 720 mm may be expressed as -260 mm Hg 260 mm Hg suction or 260 mm Hg vacuum

Note

Pabs = pbar + pgage

Absolute pressures P gage pressures p

Example 23 The rate of temperature change in the atmosphere with change in elevation is called its lapse rate The motion of a parcel of air depends on the density of the parcel relative to the density of the surrounding (ambient) air However as the parcel ascends through the atmosphere the air pressure decreases the parcel expands and its temperature decreases at a rate known as the dry adiabatic lapse rate A firm wants lo burn a large quantity of refuse It is estimated that the temperature of the smoke plume at 10 m above the ground will be 11oC greater than that of the ambient air For the following conditions determine what will happen to the smoke

(a) At standard atmospheric lapse rate β = -000651oC per meter and t0 =20oC

(b) At an inverted lapse rate β = 000365oC per meter

By combining Eqs (227) and (2214)

The relation between pressure and temperature for a mass of gas expanding without heat transfer (isentropic relation Sec 61) is

in which T1 is the initial smoke absolute temperature and P0 the initial absolute pressure k is the specific heat ratio 14 for air and other diatomic gases

Eliminating PP0 in the last two equations

Since the gas will rise until its temperature is equal to the ambient temperature

the last two equations may be solved for y Let

Then

For β = -000651oC per metre R = 287 mN(kgK) a = 2002 and y = 3201 m For the atmospheric temperature inversion β = -000365oC per metre a = -02721 and y = 8092 m

24 MANOMETERS

Manometers are devices that employ liquid columns for determining differences in pressure

Figure 26a the most elementary manometer ndash piezometer It measures the pressure in a liquid when it is above zero gage Glass tube is mounted vertically so that it is connected to the space

within the container Liquid rises in the tube until equilibrium is reached The pressure is then given by the vertical distance h from the menis

cus (liquid surface) to the point where the pressure is to be measured expressed in units of length of the liquid in the container

Piezometer would not work for negative gage pressures because air would flow into the container through the tube

Figure 26 Simple manometers

Figure 26b for small negative or positive gage pressures in a liquid With this arrangement the meniscus may come to rest below A as sho

wn Since the pressure at the meniscus is zero gage and since pressure decreases with elevation

units of length H2O Figure 26c for greater negative or positive gage pressures (a second li

quid of greater relative density employed) It must be immiscible in the first fluid which may now be a gas If the relative density of the fluid at A is S1 (based on water) and the re

lative density of the manometer liquid is S2 the equation for pressure at A

hA - the unknown pressure expressed in length units of water

h1 h2 - in length units

A general procedure in working all manometer problems

1 Start at one end (or any meniscus if the circuit is continuous) and write the pressure there in an appropriate unit (say pascals) or in an appropriate symbol if it is unknown

2 Add to this the change in pressure in the same unit from one meniscus to the next (plus if the next meniscus is lower minus if higher) (For pascals this is the product of the difference in elevation in metres and the unit gravity force of the fluid in newtons per cubic metre)

3 Continue until the other end of the gage (or the starting meniscus) is reached and equate the expression to the pressure at that point known or unknown

The expression will contain one unknown for a simple manometer or will give a difference in pressures for the differential manometer In equation form

Figure 27 Differential manometers

A differential manometer (Fig 27) determines the difference in pressures at two points A and B when the actual pressure at any point in the system cannot be determined

Application of the procedure outlined above to Fig 27a produces

For Fig 27b

If the pressures at A and B are expressed in length of the water column the above results can be written for Fig 27a

For Fig 27b

Example 24 In Fig 27a the liquids at A and B are water and the manometer liquid is oil S = 080 h1 = 300 mm h2 = 200 mm and h3 = 600 mm (a) Determine pA - pB in pacals (b) If pB = 50kPa and the barometer reading is 730 mm Hg find the pressure at A in meters of water absolute (a)

(b)

(a)

Micromanometers

For determining very small differences in pressure or determining large pressure differences precisely ndash several types of manometers

One type very accurately measures the differences in elevation of two menisci of a manometer

By means of small telescopes with horizontal cross hairs mounted along the tubes on a rack which is raised and lowered by a pinion and slow motion screw so that the cross hairs can be set accurately the difference in elevation of menisci (the gage difference) can be read with verniers

Figure 28 Micromanometer using two gage liquids

Fig 28 two gage liquids immiscible in each other and in the fluid to be measured a large gage difference R can be produced for a small pressure difference

The heavier gage liquid fills the lower U tube up to 0-0 then the lighter gage liquid is added to both sides filling the larger reservoirs up to 1-1

The gas or liquid in the system fills the space above 1-1 When the pressure at C is slightly greater than at D the menisci move as indicated in Fig 28

The volume of liquid displaced in each reservoir equals the displacement in the U tube

Manometer equation

γ1 γ2 and γ3 are the unit gravity force

Example 25 In the micromanometer of Fig 28 the pressure difference is wanted in pascals when air is in the system S2 = 10 S3 = 110 aA = 001 R = 5 mm t = 20oC and the barometer reads 760 mm Hg

The term γ1(aA) may be neglected Substituting into Eq (241) gives

Figure 29

Inclined manometer

The inclined manometer frequently used for measuring small differences in gas pressures

Adjusted to read zero by moving the inclined scale when A and B are open Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube it affords greater accuracy in reading the scale

Surface tension causes a capillary rise in small tubes If a U tube is used with a meniscus in each leg the surface-tension effects cancel

25 FORCES ON PLANE AREAS

In the preceding sections variations oF pressure throughout a fluid have been considered

The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force insofar as external reactions to the force system are concerned

In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration by formula and by use of the concept of the pressure prism

Horizontal Surfaces

A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure

The magnitude of the force acting on one side of the surface is

The elemental forces pdA acting on A are all parallel and in the same sense a scalar summation of all such elements yields the magnitude of the resultant force Its direction is normal to the surface and toward the surface if p is positive

Fig 210 arbitrary xy axes - to find the line of action of the resultant ie the point in the area where the moment of the distributed force about any axis through the point is zero

Then since the moment of the resultant must equal the moment of the distributed force system about any axis say the y axis

xrsquo ndash the distance from the y axis to the resultant

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

When divided by δy and δx respectively the equations can be combined

θ is any arbitrary angle this equation proves that the pressure is the same in all directions at a point in a static fluid

Although the proof was carried out for a two-dimensional case it may be demonstrated for the three-dimensional case with the equilibrium equations for a small tetrahedron of fluid with three faces in the coordinate planes and the fourth face inclined arbitrarily

If the fluid is in motion (one layer moves relative lo an adjacent layer) shear stresses occur and the normal stresses are no longer the same in all directions at a point the pressure is defined as the average of any three mutually perpendicular normal compressive stresses at a point

Fictitious fluid of zero viscosity (frictionless fluid) no shear stresses can occur at a point the pressure is the same in all directions

22 BASIC EQUATION OF FLUID STATICS

Pressure Variation in a Static Fluid

Force balance The forces acting on an element of fluid at rest (Fig 22) surface force

s and body forces With gravity the only body force acting and by taking the y axis vertic

ally upward it is -γ δx δy δz in the y direction With pressure p at its center (x y z) the approximate force exerted on

the side normal to the y axis closest to the origin and the opposite e side are approximately

δy2 ndash the distance from center to a face normal to y

Figure 22

Rectangular parallelepiped element of fluid at rest

Summing the forces acting on the element in the y direction

For the x and z directions since no body forces act

The elemental force vector δF

If the element is reduced to zero size alter dividing through by δx δy δz = δV the expression becomes exact

This is the resultant force per unit volume at a point which must be equated to zero for a fluid at rest

The gradient isnabla

- p is the vector field f or the surface pressure force per unit volume nabla

The fluid static law or variation of pressure is then

For an inviscid fluid in motion or a fluid so moving that the shear stress is everywhere zero Newtons second law takes the form

a is the acceleration of the fluid element f - jγ is the resultant fluid force when gravity is the only body force acting

In component form Eq (224) becomes

The partials for variation in horizontal directions are one form of Pascals law they state that two points at the same elevation in the same continuous mass or fluid at rest have the same pressure

Since p is a function of y only

relates the change of pressure to unit gravity force and change of elevation and holds for both compressible and incompressible fluids

For fluids that may be considered homogeneous and incompressible γ is constant and the above equation when integrated becomes

in which c is the constant of integration The hydrostatic law of variation of pressure is frequently written in the form

h = -y p is the increase in pressure from that at the free surface

Example 21 An oceanographer is to design a sea lab 5 m high to withstand submersion to 100 m measured from sea level to the top of the sea lab Find the pressure variation on a side of the container and the pressure on the top if the relative density of salt water is 1020

At the top h = 100 m and

If y is measured from the top of the sea lab downward the pressure variation is

Pressure Variation in a Compressible Fluid

When the fluid is a perfect gas at rest at constant temperature

When the value of γ in Eq (227) is replaced by ρg and ρ is eliminated between Eqs (227) and (229)

If P = P0 when ρ = ρ0 integration between limits

- the equation for variation of pressure with elevation in an isothermal gas

- constant temperature gradient of atmosphere

Example 22 Assuming isothermal conditions to prevail in the atmosphere compute the pressure and density at 2000 m elevation if P = 105Pa ρ = 124 kgm3 at sea level

From Eq (2212)

Then from Eq (229)

23 UNITS AND SCALES OF PRESSURE MEASUREMENT

Pressure may be expressed with reference to any arbitrary datum absolute zero local atmospheric pressure

Absolute pressure difference between its value and a complete vacuum

Gage pressure difference between its value and the local atmospheric pressure

Figure 23 Bourdon gage

The bourdon gage (Fig 23) typical of the devices used for measuring gage pressures pressure element is a hollow curved flat metallic tube closed at one e

nd the other end is connected to the pressure to be measured when the internal pressure is increased the tube tends to straighten

pulling on a linkage to which is attached a pointer and causing the pointer to move

the dial reads zero when the inside and outside of the tube are at the same pressure regardless of its particular value

the gage measures pressure relative to the pressure of the medium surrounding the tube which is the local atmosphere

Figure 24 Units and scales for pressure measurement

Figure 24 the data and the relations of the common units of pressure measurement

Standard atmospheric pressure is the mean pressure at sea level 760 mm Hg

A pressure expressed in terms of the length of a column of liquid is equivalent to the force per unit area at the base of the column The relation for variation of pressure with altitude in a liquid p = γh [Eq (228)] (p is in pascals γ in newtons per cubic metre and h in metres)

With the unit gravity force of any liquid expressed as its relative density S times the unit gravity force of water

Water γ may be taken as 9806 Nm3

Figure 25 Mercury barometer

Local atmospheric pressure is measured by mercury barometer aneroid barometer (measures the difference in

pressure between the atmosphere and an evacuated box or tube in a manner analogous to the bourdon gage except that the tube is evacuated and sealed)

Mercury barometer glass tube closed at one end filled with mercury and inverted so that the open end is submerged in mercury

It has a scale the height of column R can be determined

The space above the mercury contains mercury vapor If the pressure of the mercury vapor hv is given in millimetres of mercury and R is measured in the same units the pressure at A may be expressed as (mm Hg)

Figure 24 a pressure may be located vertically on the chart which indicates its relation to absolute zero and to local atmospheric pressure

If the point is below the local-atmospheric-pressure line and is referred to gage datum it is called negative suction or vacuum

Example the pressure 460 mm Hg abs as at 1 with barometer reading 720 mm may be expressed as -260 mm Hg 260 mm Hg suction or 260 mm Hg vacuum

Note

Pabs = pbar + pgage

Absolute pressures P gage pressures p

Example 23 The rate of temperature change in the atmosphere with change in elevation is called its lapse rate The motion of a parcel of air depends on the density of the parcel relative to the density of the surrounding (ambient) air However as the parcel ascends through the atmosphere the air pressure decreases the parcel expands and its temperature decreases at a rate known as the dry adiabatic lapse rate A firm wants lo burn a large quantity of refuse It is estimated that the temperature of the smoke plume at 10 m above the ground will be 11oC greater than that of the ambient air For the following conditions determine what will happen to the smoke

(a) At standard atmospheric lapse rate β = -000651oC per meter and t0 =20oC

(b) At an inverted lapse rate β = 000365oC per meter

By combining Eqs (227) and (2214)

The relation between pressure and temperature for a mass of gas expanding without heat transfer (isentropic relation Sec 61) is

in which T1 is the initial smoke absolute temperature and P0 the initial absolute pressure k is the specific heat ratio 14 for air and other diatomic gases

Eliminating PP0 in the last two equations

Since the gas will rise until its temperature is equal to the ambient temperature

the last two equations may be solved for y Let

Then

For β = -000651oC per metre R = 287 mN(kgK) a = 2002 and y = 3201 m For the atmospheric temperature inversion β = -000365oC per metre a = -02721 and y = 8092 m

24 MANOMETERS

Manometers are devices that employ liquid columns for determining differences in pressure

Figure 26a the most elementary manometer ndash piezometer It measures the pressure in a liquid when it is above zero gage Glass tube is mounted vertically so that it is connected to the space

within the container Liquid rises in the tube until equilibrium is reached The pressure is then given by the vertical distance h from the menis

cus (liquid surface) to the point where the pressure is to be measured expressed in units of length of the liquid in the container

Piezometer would not work for negative gage pressures because air would flow into the container through the tube

Figure 26 Simple manometers

Figure 26b for small negative or positive gage pressures in a liquid With this arrangement the meniscus may come to rest below A as sho

wn Since the pressure at the meniscus is zero gage and since pressure decreases with elevation

units of length H2O Figure 26c for greater negative or positive gage pressures (a second li

quid of greater relative density employed) It must be immiscible in the first fluid which may now be a gas If the relative density of the fluid at A is S1 (based on water) and the re

lative density of the manometer liquid is S2 the equation for pressure at A

hA - the unknown pressure expressed in length units of water

h1 h2 - in length units

A general procedure in working all manometer problems

1 Start at one end (or any meniscus if the circuit is continuous) and write the pressure there in an appropriate unit (say pascals) or in an appropriate symbol if it is unknown

2 Add to this the change in pressure in the same unit from one meniscus to the next (plus if the next meniscus is lower minus if higher) (For pascals this is the product of the difference in elevation in metres and the unit gravity force of the fluid in newtons per cubic metre)

3 Continue until the other end of the gage (or the starting meniscus) is reached and equate the expression to the pressure at that point known or unknown

The expression will contain one unknown for a simple manometer or will give a difference in pressures for the differential manometer In equation form

Figure 27 Differential manometers

A differential manometer (Fig 27) determines the difference in pressures at two points A and B when the actual pressure at any point in the system cannot be determined

Application of the procedure outlined above to Fig 27a produces

For Fig 27b

If the pressures at A and B are expressed in length of the water column the above results can be written for Fig 27a

For Fig 27b

Example 24 In Fig 27a the liquids at A and B are water and the manometer liquid is oil S = 080 h1 = 300 mm h2 = 200 mm and h3 = 600 mm (a) Determine pA - pB in pacals (b) If pB = 50kPa and the barometer reading is 730 mm Hg find the pressure at A in meters of water absolute (a)

(b)

(a)

Micromanometers

For determining very small differences in pressure or determining large pressure differences precisely ndash several types of manometers

One type very accurately measures the differences in elevation of two menisci of a manometer

By means of small telescopes with horizontal cross hairs mounted along the tubes on a rack which is raised and lowered by a pinion and slow motion screw so that the cross hairs can be set accurately the difference in elevation of menisci (the gage difference) can be read with verniers

Figure 28 Micromanometer using two gage liquids

Fig 28 two gage liquids immiscible in each other and in the fluid to be measured a large gage difference R can be produced for a small pressure difference

The heavier gage liquid fills the lower U tube up to 0-0 then the lighter gage liquid is added to both sides filling the larger reservoirs up to 1-1

The gas or liquid in the system fills the space above 1-1 When the pressure at C is slightly greater than at D the menisci move as indicated in Fig 28

The volume of liquid displaced in each reservoir equals the displacement in the U tube

Manometer equation

γ1 γ2 and γ3 are the unit gravity force

Example 25 In the micromanometer of Fig 28 the pressure difference is wanted in pascals when air is in the system S2 = 10 S3 = 110 aA = 001 R = 5 mm t = 20oC and the barometer reads 760 mm Hg

The term γ1(aA) may be neglected Substituting into Eq (241) gives

Figure 29

Inclined manometer

The inclined manometer frequently used for measuring small differences in gas pressures

Adjusted to read zero by moving the inclined scale when A and B are open Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube it affords greater accuracy in reading the scale

Surface tension causes a capillary rise in small tubes If a U tube is used with a meniscus in each leg the surface-tension effects cancel

25 FORCES ON PLANE AREAS

In the preceding sections variations oF pressure throughout a fluid have been considered

The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force insofar as external reactions to the force system are concerned

In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration by formula and by use of the concept of the pressure prism

Horizontal Surfaces

A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure

The magnitude of the force acting on one side of the surface is

The elemental forces pdA acting on A are all parallel and in the same sense a scalar summation of all such elements yields the magnitude of the resultant force Its direction is normal to the surface and toward the surface if p is positive

Fig 210 arbitrary xy axes - to find the line of action of the resultant ie the point in the area where the moment of the distributed force about any axis through the point is zero

Then since the moment of the resultant must equal the moment of the distributed force system about any axis say the y axis

xrsquo ndash the distance from the y axis to the resultant

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

22 BASIC EQUATION OF FLUID STATICS

Pressure Variation in a Static Fluid

Force balance The forces acting on an element of fluid at rest (Fig 22) surface force

s and body forces With gravity the only body force acting and by taking the y axis vertic

ally upward it is -γ δx δy δz in the y direction With pressure p at its center (x y z) the approximate force exerted on

the side normal to the y axis closest to the origin and the opposite e side are approximately

δy2 ndash the distance from center to a face normal to y

Figure 22

Rectangular parallelepiped element of fluid at rest

Summing the forces acting on the element in the y direction

For the x and z directions since no body forces act

The elemental force vector δF

If the element is reduced to zero size alter dividing through by δx δy δz = δV the expression becomes exact

This is the resultant force per unit volume at a point which must be equated to zero for a fluid at rest

The gradient isnabla

- p is the vector field f or the surface pressure force per unit volume nabla

The fluid static law or variation of pressure is then

For an inviscid fluid in motion or a fluid so moving that the shear stress is everywhere zero Newtons second law takes the form

a is the acceleration of the fluid element f - jγ is the resultant fluid force when gravity is the only body force acting

In component form Eq (224) becomes

The partials for variation in horizontal directions are one form of Pascals law they state that two points at the same elevation in the same continuous mass or fluid at rest have the same pressure

Since p is a function of y only

relates the change of pressure to unit gravity force and change of elevation and holds for both compressible and incompressible fluids

For fluids that may be considered homogeneous and incompressible γ is constant and the above equation when integrated becomes

in which c is the constant of integration The hydrostatic law of variation of pressure is frequently written in the form

h = -y p is the increase in pressure from that at the free surface

Example 21 An oceanographer is to design a sea lab 5 m high to withstand submersion to 100 m measured from sea level to the top of the sea lab Find the pressure variation on a side of the container and the pressure on the top if the relative density of salt water is 1020

At the top h = 100 m and

If y is measured from the top of the sea lab downward the pressure variation is

Pressure Variation in a Compressible Fluid

When the fluid is a perfect gas at rest at constant temperature

When the value of γ in Eq (227) is replaced by ρg and ρ is eliminated between Eqs (227) and (229)

If P = P0 when ρ = ρ0 integration between limits

- the equation for variation of pressure with elevation in an isothermal gas

- constant temperature gradient of atmosphere

Example 22 Assuming isothermal conditions to prevail in the atmosphere compute the pressure and density at 2000 m elevation if P = 105Pa ρ = 124 kgm3 at sea level

From Eq (2212)

Then from Eq (229)

23 UNITS AND SCALES OF PRESSURE MEASUREMENT

Pressure may be expressed with reference to any arbitrary datum absolute zero local atmospheric pressure

Absolute pressure difference between its value and a complete vacuum

Gage pressure difference between its value and the local atmospheric pressure

Figure 23 Bourdon gage

The bourdon gage (Fig 23) typical of the devices used for measuring gage pressures pressure element is a hollow curved flat metallic tube closed at one e

nd the other end is connected to the pressure to be measured when the internal pressure is increased the tube tends to straighten

pulling on a linkage to which is attached a pointer and causing the pointer to move

the dial reads zero when the inside and outside of the tube are at the same pressure regardless of its particular value

the gage measures pressure relative to the pressure of the medium surrounding the tube which is the local atmosphere

Figure 24 Units and scales for pressure measurement

Figure 24 the data and the relations of the common units of pressure measurement

Standard atmospheric pressure is the mean pressure at sea level 760 mm Hg

A pressure expressed in terms of the length of a column of liquid is equivalent to the force per unit area at the base of the column The relation for variation of pressure with altitude in a liquid p = γh [Eq (228)] (p is in pascals γ in newtons per cubic metre and h in metres)

With the unit gravity force of any liquid expressed as its relative density S times the unit gravity force of water

Water γ may be taken as 9806 Nm3

Figure 25 Mercury barometer

Local atmospheric pressure is measured by mercury barometer aneroid barometer (measures the difference in

pressure between the atmosphere and an evacuated box or tube in a manner analogous to the bourdon gage except that the tube is evacuated and sealed)

Mercury barometer glass tube closed at one end filled with mercury and inverted so that the open end is submerged in mercury

It has a scale the height of column R can be determined

The space above the mercury contains mercury vapor If the pressure of the mercury vapor hv is given in millimetres of mercury and R is measured in the same units the pressure at A may be expressed as (mm Hg)

Figure 24 a pressure may be located vertically on the chart which indicates its relation to absolute zero and to local atmospheric pressure

If the point is below the local-atmospheric-pressure line and is referred to gage datum it is called negative suction or vacuum

Example the pressure 460 mm Hg abs as at 1 with barometer reading 720 mm may be expressed as -260 mm Hg 260 mm Hg suction or 260 mm Hg vacuum

Note

Pabs = pbar + pgage

Absolute pressures P gage pressures p

Example 23 The rate of temperature change in the atmosphere with change in elevation is called its lapse rate The motion of a parcel of air depends on the density of the parcel relative to the density of the surrounding (ambient) air However as the parcel ascends through the atmosphere the air pressure decreases the parcel expands and its temperature decreases at a rate known as the dry adiabatic lapse rate A firm wants lo burn a large quantity of refuse It is estimated that the temperature of the smoke plume at 10 m above the ground will be 11oC greater than that of the ambient air For the following conditions determine what will happen to the smoke

(a) At standard atmospheric lapse rate β = -000651oC per meter and t0 =20oC

(b) At an inverted lapse rate β = 000365oC per meter

By combining Eqs (227) and (2214)

The relation between pressure and temperature for a mass of gas expanding without heat transfer (isentropic relation Sec 61) is

in which T1 is the initial smoke absolute temperature and P0 the initial absolute pressure k is the specific heat ratio 14 for air and other diatomic gases

Eliminating PP0 in the last two equations

Since the gas will rise until its temperature is equal to the ambient temperature

the last two equations may be solved for y Let

Then

For β = -000651oC per metre R = 287 mN(kgK) a = 2002 and y = 3201 m For the atmospheric temperature inversion β = -000365oC per metre a = -02721 and y = 8092 m

24 MANOMETERS

Manometers are devices that employ liquid columns for determining differences in pressure

Figure 26a the most elementary manometer ndash piezometer It measures the pressure in a liquid when it is above zero gage Glass tube is mounted vertically so that it is connected to the space

within the container Liquid rises in the tube until equilibrium is reached The pressure is then given by the vertical distance h from the menis

cus (liquid surface) to the point where the pressure is to be measured expressed in units of length of the liquid in the container

Piezometer would not work for negative gage pressures because air would flow into the container through the tube

Figure 26 Simple manometers

Figure 26b for small negative or positive gage pressures in a liquid With this arrangement the meniscus may come to rest below A as sho

wn Since the pressure at the meniscus is zero gage and since pressure decreases with elevation

units of length H2O Figure 26c for greater negative or positive gage pressures (a second li

quid of greater relative density employed) It must be immiscible in the first fluid which may now be a gas If the relative density of the fluid at A is S1 (based on water) and the re

lative density of the manometer liquid is S2 the equation for pressure at A

hA - the unknown pressure expressed in length units of water

h1 h2 - in length units

A general procedure in working all manometer problems

1 Start at one end (or any meniscus if the circuit is continuous) and write the pressure there in an appropriate unit (say pascals) or in an appropriate symbol if it is unknown

2 Add to this the change in pressure in the same unit from one meniscus to the next (plus if the next meniscus is lower minus if higher) (For pascals this is the product of the difference in elevation in metres and the unit gravity force of the fluid in newtons per cubic metre)

3 Continue until the other end of the gage (or the starting meniscus) is reached and equate the expression to the pressure at that point known or unknown

The expression will contain one unknown for a simple manometer or will give a difference in pressures for the differential manometer In equation form

Figure 27 Differential manometers

A differential manometer (Fig 27) determines the difference in pressures at two points A and B when the actual pressure at any point in the system cannot be determined

Application of the procedure outlined above to Fig 27a produces

For Fig 27b

If the pressures at A and B are expressed in length of the water column the above results can be written for Fig 27a

For Fig 27b

Example 24 In Fig 27a the liquids at A and B are water and the manometer liquid is oil S = 080 h1 = 300 mm h2 = 200 mm and h3 = 600 mm (a) Determine pA - pB in pacals (b) If pB = 50kPa and the barometer reading is 730 mm Hg find the pressure at A in meters of water absolute (a)

(b)

(a)

Micromanometers

For determining very small differences in pressure or determining large pressure differences precisely ndash several types of manometers

One type very accurately measures the differences in elevation of two menisci of a manometer

By means of small telescopes with horizontal cross hairs mounted along the tubes on a rack which is raised and lowered by a pinion and slow motion screw so that the cross hairs can be set accurately the difference in elevation of menisci (the gage difference) can be read with verniers

Figure 28 Micromanometer using two gage liquids

Fig 28 two gage liquids immiscible in each other and in the fluid to be measured a large gage difference R can be produced for a small pressure difference

The heavier gage liquid fills the lower U tube up to 0-0 then the lighter gage liquid is added to both sides filling the larger reservoirs up to 1-1

The gas or liquid in the system fills the space above 1-1 When the pressure at C is slightly greater than at D the menisci move as indicated in Fig 28

The volume of liquid displaced in each reservoir equals the displacement in the U tube

Manometer equation

γ1 γ2 and γ3 are the unit gravity force

Example 25 In the micromanometer of Fig 28 the pressure difference is wanted in pascals when air is in the system S2 = 10 S3 = 110 aA = 001 R = 5 mm t = 20oC and the barometer reads 760 mm Hg

The term γ1(aA) may be neglected Substituting into Eq (241) gives

Figure 29

Inclined manometer

The inclined manometer frequently used for measuring small differences in gas pressures

Adjusted to read zero by moving the inclined scale when A and B are open Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube it affords greater accuracy in reading the scale

Surface tension causes a capillary rise in small tubes If a U tube is used with a meniscus in each leg the surface-tension effects cancel

25 FORCES ON PLANE AREAS

In the preceding sections variations oF pressure throughout a fluid have been considered

The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force insofar as external reactions to the force system are concerned

In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration by formula and by use of the concept of the pressure prism

Horizontal Surfaces

A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure

The magnitude of the force acting on one side of the surface is

The elemental forces pdA acting on A are all parallel and in the same sense a scalar summation of all such elements yields the magnitude of the resultant force Its direction is normal to the surface and toward the surface if p is positive

Fig 210 arbitrary xy axes - to find the line of action of the resultant ie the point in the area where the moment of the distributed force about any axis through the point is zero

Then since the moment of the resultant must equal the moment of the distributed force system about any axis say the y axis

xrsquo ndash the distance from the y axis to the resultant

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Figure 22

Rectangular parallelepiped element of fluid at rest

Summing the forces acting on the element in the y direction

For the x and z directions since no body forces act

The elemental force vector δF

If the element is reduced to zero size alter dividing through by δx δy δz = δV the expression becomes exact

This is the resultant force per unit volume at a point which must be equated to zero for a fluid at rest

The gradient isnabla

- p is the vector field f or the surface pressure force per unit volume nabla

The fluid static law or variation of pressure is then

For an inviscid fluid in motion or a fluid so moving that the shear stress is everywhere zero Newtons second law takes the form

a is the acceleration of the fluid element f - jγ is the resultant fluid force when gravity is the only body force acting

In component form Eq (224) becomes

The partials for variation in horizontal directions are one form of Pascals law they state that two points at the same elevation in the same continuous mass or fluid at rest have the same pressure

Since p is a function of y only

relates the change of pressure to unit gravity force and change of elevation and holds for both compressible and incompressible fluids

For fluids that may be considered homogeneous and incompressible γ is constant and the above equation when integrated becomes

in which c is the constant of integration The hydrostatic law of variation of pressure is frequently written in the form

h = -y p is the increase in pressure from that at the free surface

Example 21 An oceanographer is to design a sea lab 5 m high to withstand submersion to 100 m measured from sea level to the top of the sea lab Find the pressure variation on a side of the container and the pressure on the top if the relative density of salt water is 1020

At the top h = 100 m and

If y is measured from the top of the sea lab downward the pressure variation is

Pressure Variation in a Compressible Fluid

When the fluid is a perfect gas at rest at constant temperature

When the value of γ in Eq (227) is replaced by ρg and ρ is eliminated between Eqs (227) and (229)

If P = P0 when ρ = ρ0 integration between limits

- the equation for variation of pressure with elevation in an isothermal gas

- constant temperature gradient of atmosphere

Example 22 Assuming isothermal conditions to prevail in the atmosphere compute the pressure and density at 2000 m elevation if P = 105Pa ρ = 124 kgm3 at sea level

From Eq (2212)

Then from Eq (229)

23 UNITS AND SCALES OF PRESSURE MEASUREMENT

Pressure may be expressed with reference to any arbitrary datum absolute zero local atmospheric pressure

Absolute pressure difference between its value and a complete vacuum

Gage pressure difference between its value and the local atmospheric pressure

Figure 23 Bourdon gage

The bourdon gage (Fig 23) typical of the devices used for measuring gage pressures pressure element is a hollow curved flat metallic tube closed at one e

nd the other end is connected to the pressure to be measured when the internal pressure is increased the tube tends to straighten

pulling on a linkage to which is attached a pointer and causing the pointer to move

the dial reads zero when the inside and outside of the tube are at the same pressure regardless of its particular value

the gage measures pressure relative to the pressure of the medium surrounding the tube which is the local atmosphere

Figure 24 Units and scales for pressure measurement

Figure 24 the data and the relations of the common units of pressure measurement

Standard atmospheric pressure is the mean pressure at sea level 760 mm Hg

A pressure expressed in terms of the length of a column of liquid is equivalent to the force per unit area at the base of the column The relation for variation of pressure with altitude in a liquid p = γh [Eq (228)] (p is in pascals γ in newtons per cubic metre and h in metres)

With the unit gravity force of any liquid expressed as its relative density S times the unit gravity force of water

Water γ may be taken as 9806 Nm3

Figure 25 Mercury barometer

Local atmospheric pressure is measured by mercury barometer aneroid barometer (measures the difference in

pressure between the atmosphere and an evacuated box or tube in a manner analogous to the bourdon gage except that the tube is evacuated and sealed)

Mercury barometer glass tube closed at one end filled with mercury and inverted so that the open end is submerged in mercury

It has a scale the height of column R can be determined

The space above the mercury contains mercury vapor If the pressure of the mercury vapor hv is given in millimetres of mercury and R is measured in the same units the pressure at A may be expressed as (mm Hg)

Figure 24 a pressure may be located vertically on the chart which indicates its relation to absolute zero and to local atmospheric pressure

If the point is below the local-atmospheric-pressure line and is referred to gage datum it is called negative suction or vacuum

Example the pressure 460 mm Hg abs as at 1 with barometer reading 720 mm may be expressed as -260 mm Hg 260 mm Hg suction or 260 mm Hg vacuum

Note

Pabs = pbar + pgage

Absolute pressures P gage pressures p

Example 23 The rate of temperature change in the atmosphere with change in elevation is called its lapse rate The motion of a parcel of air depends on the density of the parcel relative to the density of the surrounding (ambient) air However as the parcel ascends through the atmosphere the air pressure decreases the parcel expands and its temperature decreases at a rate known as the dry adiabatic lapse rate A firm wants lo burn a large quantity of refuse It is estimated that the temperature of the smoke plume at 10 m above the ground will be 11oC greater than that of the ambient air For the following conditions determine what will happen to the smoke

(a) At standard atmospheric lapse rate β = -000651oC per meter and t0 =20oC

(b) At an inverted lapse rate β = 000365oC per meter

By combining Eqs (227) and (2214)

The relation between pressure and temperature for a mass of gas expanding without heat transfer (isentropic relation Sec 61) is

in which T1 is the initial smoke absolute temperature and P0 the initial absolute pressure k is the specific heat ratio 14 for air and other diatomic gases

Eliminating PP0 in the last two equations

Since the gas will rise until its temperature is equal to the ambient temperature

the last two equations may be solved for y Let

Then

For β = -000651oC per metre R = 287 mN(kgK) a = 2002 and y = 3201 m For the atmospheric temperature inversion β = -000365oC per metre a = -02721 and y = 8092 m

24 MANOMETERS

Manometers are devices that employ liquid columns for determining differences in pressure

Figure 26a the most elementary manometer ndash piezometer It measures the pressure in a liquid when it is above zero gage Glass tube is mounted vertically so that it is connected to the space

within the container Liquid rises in the tube until equilibrium is reached The pressure is then given by the vertical distance h from the menis

cus (liquid surface) to the point where the pressure is to be measured expressed in units of length of the liquid in the container

Piezometer would not work for negative gage pressures because air would flow into the container through the tube

Figure 26 Simple manometers

Figure 26b for small negative or positive gage pressures in a liquid With this arrangement the meniscus may come to rest below A as sho

wn Since the pressure at the meniscus is zero gage and since pressure decreases with elevation

units of length H2O Figure 26c for greater negative or positive gage pressures (a second li

quid of greater relative density employed) It must be immiscible in the first fluid which may now be a gas If the relative density of the fluid at A is S1 (based on water) and the re

lative density of the manometer liquid is S2 the equation for pressure at A

hA - the unknown pressure expressed in length units of water

h1 h2 - in length units

A general procedure in working all manometer problems

1 Start at one end (or any meniscus if the circuit is continuous) and write the pressure there in an appropriate unit (say pascals) or in an appropriate symbol if it is unknown

2 Add to this the change in pressure in the same unit from one meniscus to the next (plus if the next meniscus is lower minus if higher) (For pascals this is the product of the difference in elevation in metres and the unit gravity force of the fluid in newtons per cubic metre)

3 Continue until the other end of the gage (or the starting meniscus) is reached and equate the expression to the pressure at that point known or unknown

The expression will contain one unknown for a simple manometer or will give a difference in pressures for the differential manometer In equation form

Figure 27 Differential manometers

A differential manometer (Fig 27) determines the difference in pressures at two points A and B when the actual pressure at any point in the system cannot be determined

Application of the procedure outlined above to Fig 27a produces

For Fig 27b

If the pressures at A and B are expressed in length of the water column the above results can be written for Fig 27a

For Fig 27b

Example 24 In Fig 27a the liquids at A and B are water and the manometer liquid is oil S = 080 h1 = 300 mm h2 = 200 mm and h3 = 600 mm (a) Determine pA - pB in pacals (b) If pB = 50kPa and the barometer reading is 730 mm Hg find the pressure at A in meters of water absolute (a)

(b)

(a)

Micromanometers

For determining very small differences in pressure or determining large pressure differences precisely ndash several types of manometers

One type very accurately measures the differences in elevation of two menisci of a manometer

By means of small telescopes with horizontal cross hairs mounted along the tubes on a rack which is raised and lowered by a pinion and slow motion screw so that the cross hairs can be set accurately the difference in elevation of menisci (the gage difference) can be read with verniers

Figure 28 Micromanometer using two gage liquids

Fig 28 two gage liquids immiscible in each other and in the fluid to be measured a large gage difference R can be produced for a small pressure difference

The heavier gage liquid fills the lower U tube up to 0-0 then the lighter gage liquid is added to both sides filling the larger reservoirs up to 1-1

The gas or liquid in the system fills the space above 1-1 When the pressure at C is slightly greater than at D the menisci move as indicated in Fig 28

The volume of liquid displaced in each reservoir equals the displacement in the U tube

Manometer equation

γ1 γ2 and γ3 are the unit gravity force

Example 25 In the micromanometer of Fig 28 the pressure difference is wanted in pascals when air is in the system S2 = 10 S3 = 110 aA = 001 R = 5 mm t = 20oC and the barometer reads 760 mm Hg

The term γ1(aA) may be neglected Substituting into Eq (241) gives

Figure 29

Inclined manometer

The inclined manometer frequently used for measuring small differences in gas pressures

Adjusted to read zero by moving the inclined scale when A and B are open Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube it affords greater accuracy in reading the scale

Surface tension causes a capillary rise in small tubes If a U tube is used with a meniscus in each leg the surface-tension effects cancel

25 FORCES ON PLANE AREAS

In the preceding sections variations oF pressure throughout a fluid have been considered

The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force insofar as external reactions to the force system are concerned

In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration by formula and by use of the concept of the pressure prism

Horizontal Surfaces

A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure

The magnitude of the force acting on one side of the surface is

The elemental forces pdA acting on A are all parallel and in the same sense a scalar summation of all such elements yields the magnitude of the resultant force Its direction is normal to the surface and toward the surface if p is positive

Fig 210 arbitrary xy axes - to find the line of action of the resultant ie the point in the area where the moment of the distributed force about any axis through the point is zero

Then since the moment of the resultant must equal the moment of the distributed force system about any axis say the y axis

xrsquo ndash the distance from the y axis to the resultant

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Summing the forces acting on the element in the y direction

For the x and z directions since no body forces act

The elemental force vector δF

If the element is reduced to zero size alter dividing through by δx δy δz = δV the expression becomes exact

This is the resultant force per unit volume at a point which must be equated to zero for a fluid at rest

The gradient isnabla

- p is the vector field f or the surface pressure force per unit volume nabla

The fluid static law or variation of pressure is then

For an inviscid fluid in motion or a fluid so moving that the shear stress is everywhere zero Newtons second law takes the form

a is the acceleration of the fluid element f - jγ is the resultant fluid force when gravity is the only body force acting

In component form Eq (224) becomes

The partials for variation in horizontal directions are one form of Pascals law they state that two points at the same elevation in the same continuous mass or fluid at rest have the same pressure

Since p is a function of y only

relates the change of pressure to unit gravity force and change of elevation and holds for both compressible and incompressible fluids

For fluids that may be considered homogeneous and incompressible γ is constant and the above equation when integrated becomes

in which c is the constant of integration The hydrostatic law of variation of pressure is frequently written in the form

h = -y p is the increase in pressure from that at the free surface

Example 21 An oceanographer is to design a sea lab 5 m high to withstand submersion to 100 m measured from sea level to the top of the sea lab Find the pressure variation on a side of the container and the pressure on the top if the relative density of salt water is 1020

At the top h = 100 m and

If y is measured from the top of the sea lab downward the pressure variation is

Pressure Variation in a Compressible Fluid

When the fluid is a perfect gas at rest at constant temperature

When the value of γ in Eq (227) is replaced by ρg and ρ is eliminated between Eqs (227) and (229)

If P = P0 when ρ = ρ0 integration between limits

- the equation for variation of pressure with elevation in an isothermal gas

- constant temperature gradient of atmosphere

Example 22 Assuming isothermal conditions to prevail in the atmosphere compute the pressure and density at 2000 m elevation if P = 105Pa ρ = 124 kgm3 at sea level

From Eq (2212)

Then from Eq (229)

23 UNITS AND SCALES OF PRESSURE MEASUREMENT

Pressure may be expressed with reference to any arbitrary datum absolute zero local atmospheric pressure

Absolute pressure difference between its value and a complete vacuum

Gage pressure difference between its value and the local atmospheric pressure

Figure 23 Bourdon gage

The bourdon gage (Fig 23) typical of the devices used for measuring gage pressures pressure element is a hollow curved flat metallic tube closed at one e

nd the other end is connected to the pressure to be measured when the internal pressure is increased the tube tends to straighten

pulling on a linkage to which is attached a pointer and causing the pointer to move

the dial reads zero when the inside and outside of the tube are at the same pressure regardless of its particular value

the gage measures pressure relative to the pressure of the medium surrounding the tube which is the local atmosphere

Figure 24 Units and scales for pressure measurement

Figure 24 the data and the relations of the common units of pressure measurement

Standard atmospheric pressure is the mean pressure at sea level 760 mm Hg

A pressure expressed in terms of the length of a column of liquid is equivalent to the force per unit area at the base of the column The relation for variation of pressure with altitude in a liquid p = γh [Eq (228)] (p is in pascals γ in newtons per cubic metre and h in metres)

With the unit gravity force of any liquid expressed as its relative density S times the unit gravity force of water

Water γ may be taken as 9806 Nm3

Figure 25 Mercury barometer

Local atmospheric pressure is measured by mercury barometer aneroid barometer (measures the difference in

pressure between the atmosphere and an evacuated box or tube in a manner analogous to the bourdon gage except that the tube is evacuated and sealed)

Mercury barometer glass tube closed at one end filled with mercury and inverted so that the open end is submerged in mercury

It has a scale the height of column R can be determined

The space above the mercury contains mercury vapor If the pressure of the mercury vapor hv is given in millimetres of mercury and R is measured in the same units the pressure at A may be expressed as (mm Hg)

Figure 24 a pressure may be located vertically on the chart which indicates its relation to absolute zero and to local atmospheric pressure

If the point is below the local-atmospheric-pressure line and is referred to gage datum it is called negative suction or vacuum

Example the pressure 460 mm Hg abs as at 1 with barometer reading 720 mm may be expressed as -260 mm Hg 260 mm Hg suction or 260 mm Hg vacuum

Note

Pabs = pbar + pgage

Absolute pressures P gage pressures p

Example 23 The rate of temperature change in the atmosphere with change in elevation is called its lapse rate The motion of a parcel of air depends on the density of the parcel relative to the density of the surrounding (ambient) air However as the parcel ascends through the atmosphere the air pressure decreases the parcel expands and its temperature decreases at a rate known as the dry adiabatic lapse rate A firm wants lo burn a large quantity of refuse It is estimated that the temperature of the smoke plume at 10 m above the ground will be 11oC greater than that of the ambient air For the following conditions determine what will happen to the smoke

(a) At standard atmospheric lapse rate β = -000651oC per meter and t0 =20oC

(b) At an inverted lapse rate β = 000365oC per meter

By combining Eqs (227) and (2214)

The relation between pressure and temperature for a mass of gas expanding without heat transfer (isentropic relation Sec 61) is

in which T1 is the initial smoke absolute temperature and P0 the initial absolute pressure k is the specific heat ratio 14 for air and other diatomic gases

Eliminating PP0 in the last two equations

Since the gas will rise until its temperature is equal to the ambient temperature

the last two equations may be solved for y Let

Then

For β = -000651oC per metre R = 287 mN(kgK) a = 2002 and y = 3201 m For the atmospheric temperature inversion β = -000365oC per metre a = -02721 and y = 8092 m

24 MANOMETERS

Manometers are devices that employ liquid columns for determining differences in pressure

Figure 26a the most elementary manometer ndash piezometer It measures the pressure in a liquid when it is above zero gage Glass tube is mounted vertically so that it is connected to the space

within the container Liquid rises in the tube until equilibrium is reached The pressure is then given by the vertical distance h from the menis

cus (liquid surface) to the point where the pressure is to be measured expressed in units of length of the liquid in the container

Piezometer would not work for negative gage pressures because air would flow into the container through the tube

Figure 26 Simple manometers

Figure 26b for small negative or positive gage pressures in a liquid With this arrangement the meniscus may come to rest below A as sho

wn Since the pressure at the meniscus is zero gage and since pressure decreases with elevation

units of length H2O Figure 26c for greater negative or positive gage pressures (a second li

quid of greater relative density employed) It must be immiscible in the first fluid which may now be a gas If the relative density of the fluid at A is S1 (based on water) and the re

lative density of the manometer liquid is S2 the equation for pressure at A

hA - the unknown pressure expressed in length units of water

h1 h2 - in length units

A general procedure in working all manometer problems

1 Start at one end (or any meniscus if the circuit is continuous) and write the pressure there in an appropriate unit (say pascals) or in an appropriate symbol if it is unknown

2 Add to this the change in pressure in the same unit from one meniscus to the next (plus if the next meniscus is lower minus if higher) (For pascals this is the product of the difference in elevation in metres and the unit gravity force of the fluid in newtons per cubic metre)

3 Continue until the other end of the gage (or the starting meniscus) is reached and equate the expression to the pressure at that point known or unknown

The expression will contain one unknown for a simple manometer or will give a difference in pressures for the differential manometer In equation form

Figure 27 Differential manometers

A differential manometer (Fig 27) determines the difference in pressures at two points A and B when the actual pressure at any point in the system cannot be determined

Application of the procedure outlined above to Fig 27a produces

For Fig 27b

If the pressures at A and B are expressed in length of the water column the above results can be written for Fig 27a

For Fig 27b

Example 24 In Fig 27a the liquids at A and B are water and the manometer liquid is oil S = 080 h1 = 300 mm h2 = 200 mm and h3 = 600 mm (a) Determine pA - pB in pacals (b) If pB = 50kPa and the barometer reading is 730 mm Hg find the pressure at A in meters of water absolute (a)

(b)

(a)

Micromanometers

For determining very small differences in pressure or determining large pressure differences precisely ndash several types of manometers

One type very accurately measures the differences in elevation of two menisci of a manometer

By means of small telescopes with horizontal cross hairs mounted along the tubes on a rack which is raised and lowered by a pinion and slow motion screw so that the cross hairs can be set accurately the difference in elevation of menisci (the gage difference) can be read with verniers

Figure 28 Micromanometer using two gage liquids

Fig 28 two gage liquids immiscible in each other and in the fluid to be measured a large gage difference R can be produced for a small pressure difference

The heavier gage liquid fills the lower U tube up to 0-0 then the lighter gage liquid is added to both sides filling the larger reservoirs up to 1-1

The gas or liquid in the system fills the space above 1-1 When the pressure at C is slightly greater than at D the menisci move as indicated in Fig 28

The volume of liquid displaced in each reservoir equals the displacement in the U tube

Manometer equation

γ1 γ2 and γ3 are the unit gravity force

Example 25 In the micromanometer of Fig 28 the pressure difference is wanted in pascals when air is in the system S2 = 10 S3 = 110 aA = 001 R = 5 mm t = 20oC and the barometer reads 760 mm Hg

The term γ1(aA) may be neglected Substituting into Eq (241) gives

Figure 29

Inclined manometer

The inclined manometer frequently used for measuring small differences in gas pressures

Adjusted to read zero by moving the inclined scale when A and B are open Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube it affords greater accuracy in reading the scale

Surface tension causes a capillary rise in small tubes If a U tube is used with a meniscus in each leg the surface-tension effects cancel

25 FORCES ON PLANE AREAS

In the preceding sections variations oF pressure throughout a fluid have been considered

The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force insofar as external reactions to the force system are concerned

In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration by formula and by use of the concept of the pressure prism

Horizontal Surfaces

A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure

The magnitude of the force acting on one side of the surface is

The elemental forces pdA acting on A are all parallel and in the same sense a scalar summation of all such elements yields the magnitude of the resultant force Its direction is normal to the surface and toward the surface if p is positive

Fig 210 arbitrary xy axes - to find the line of action of the resultant ie the point in the area where the moment of the distributed force about any axis through the point is zero

Then since the moment of the resultant must equal the moment of the distributed force system about any axis say the y axis

xrsquo ndash the distance from the y axis to the resultant

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

- p is the vector field f or the surface pressure force per unit volume nabla

The fluid static law or variation of pressure is then

For an inviscid fluid in motion or a fluid so moving that the shear stress is everywhere zero Newtons second law takes the form

a is the acceleration of the fluid element f - jγ is the resultant fluid force when gravity is the only body force acting

In component form Eq (224) becomes

The partials for variation in horizontal directions are one form of Pascals law they state that two points at the same elevation in the same continuous mass or fluid at rest have the same pressure

Since p is a function of y only

relates the change of pressure to unit gravity force and change of elevation and holds for both compressible and incompressible fluids

For fluids that may be considered homogeneous and incompressible γ is constant and the above equation when integrated becomes

in which c is the constant of integration The hydrostatic law of variation of pressure is frequently written in the form

h = -y p is the increase in pressure from that at the free surface

Example 21 An oceanographer is to design a sea lab 5 m high to withstand submersion to 100 m measured from sea level to the top of the sea lab Find the pressure variation on a side of the container and the pressure on the top if the relative density of salt water is 1020

At the top h = 100 m and

If y is measured from the top of the sea lab downward the pressure variation is

Pressure Variation in a Compressible Fluid

When the fluid is a perfect gas at rest at constant temperature

When the value of γ in Eq (227) is replaced by ρg and ρ is eliminated between Eqs (227) and (229)

If P = P0 when ρ = ρ0 integration between limits

- the equation for variation of pressure with elevation in an isothermal gas

- constant temperature gradient of atmosphere

Example 22 Assuming isothermal conditions to prevail in the atmosphere compute the pressure and density at 2000 m elevation if P = 105Pa ρ = 124 kgm3 at sea level

From Eq (2212)

Then from Eq (229)

23 UNITS AND SCALES OF PRESSURE MEASUREMENT

Pressure may be expressed with reference to any arbitrary datum absolute zero local atmospheric pressure

Absolute pressure difference between its value and a complete vacuum

Gage pressure difference between its value and the local atmospheric pressure

Figure 23 Bourdon gage

The bourdon gage (Fig 23) typical of the devices used for measuring gage pressures pressure element is a hollow curved flat metallic tube closed at one e

nd the other end is connected to the pressure to be measured when the internal pressure is increased the tube tends to straighten

pulling on a linkage to which is attached a pointer and causing the pointer to move

the dial reads zero when the inside and outside of the tube are at the same pressure regardless of its particular value

the gage measures pressure relative to the pressure of the medium surrounding the tube which is the local atmosphere

Figure 24 Units and scales for pressure measurement

Figure 24 the data and the relations of the common units of pressure measurement

Standard atmospheric pressure is the mean pressure at sea level 760 mm Hg

A pressure expressed in terms of the length of a column of liquid is equivalent to the force per unit area at the base of the column The relation for variation of pressure with altitude in a liquid p = γh [Eq (228)] (p is in pascals γ in newtons per cubic metre and h in metres)

With the unit gravity force of any liquid expressed as its relative density S times the unit gravity force of water

Water γ may be taken as 9806 Nm3

Figure 25 Mercury barometer

Local atmospheric pressure is measured by mercury barometer aneroid barometer (measures the difference in

pressure between the atmosphere and an evacuated box or tube in a manner analogous to the bourdon gage except that the tube is evacuated and sealed)

Mercury barometer glass tube closed at one end filled with mercury and inverted so that the open end is submerged in mercury

It has a scale the height of column R can be determined

The space above the mercury contains mercury vapor If the pressure of the mercury vapor hv is given in millimetres of mercury and R is measured in the same units the pressure at A may be expressed as (mm Hg)

Figure 24 a pressure may be located vertically on the chart which indicates its relation to absolute zero and to local atmospheric pressure

If the point is below the local-atmospheric-pressure line and is referred to gage datum it is called negative suction or vacuum

Example the pressure 460 mm Hg abs as at 1 with barometer reading 720 mm may be expressed as -260 mm Hg 260 mm Hg suction or 260 mm Hg vacuum

Note

Pabs = pbar + pgage

Absolute pressures P gage pressures p

Example 23 The rate of temperature change in the atmosphere with change in elevation is called its lapse rate The motion of a parcel of air depends on the density of the parcel relative to the density of the surrounding (ambient) air However as the parcel ascends through the atmosphere the air pressure decreases the parcel expands and its temperature decreases at a rate known as the dry adiabatic lapse rate A firm wants lo burn a large quantity of refuse It is estimated that the temperature of the smoke plume at 10 m above the ground will be 11oC greater than that of the ambient air For the following conditions determine what will happen to the smoke

(a) At standard atmospheric lapse rate β = -000651oC per meter and t0 =20oC

(b) At an inverted lapse rate β = 000365oC per meter

By combining Eqs (227) and (2214)

The relation between pressure and temperature for a mass of gas expanding without heat transfer (isentropic relation Sec 61) is

in which T1 is the initial smoke absolute temperature and P0 the initial absolute pressure k is the specific heat ratio 14 for air and other diatomic gases

Eliminating PP0 in the last two equations

Since the gas will rise until its temperature is equal to the ambient temperature

the last two equations may be solved for y Let

Then

For β = -000651oC per metre R = 287 mN(kgK) a = 2002 and y = 3201 m For the atmospheric temperature inversion β = -000365oC per metre a = -02721 and y = 8092 m

24 MANOMETERS

Manometers are devices that employ liquid columns for determining differences in pressure

Figure 26a the most elementary manometer ndash piezometer It measures the pressure in a liquid when it is above zero gage Glass tube is mounted vertically so that it is connected to the space

within the container Liquid rises in the tube until equilibrium is reached The pressure is then given by the vertical distance h from the menis

cus (liquid surface) to the point where the pressure is to be measured expressed in units of length of the liquid in the container

Piezometer would not work for negative gage pressures because air would flow into the container through the tube

Figure 26 Simple manometers

Figure 26b for small negative or positive gage pressures in a liquid With this arrangement the meniscus may come to rest below A as sho

wn Since the pressure at the meniscus is zero gage and since pressure decreases with elevation

units of length H2O Figure 26c for greater negative or positive gage pressures (a second li

quid of greater relative density employed) It must be immiscible in the first fluid which may now be a gas If the relative density of the fluid at A is S1 (based on water) and the re

lative density of the manometer liquid is S2 the equation for pressure at A

hA - the unknown pressure expressed in length units of water

h1 h2 - in length units

A general procedure in working all manometer problems

1 Start at one end (or any meniscus if the circuit is continuous) and write the pressure there in an appropriate unit (say pascals) or in an appropriate symbol if it is unknown

2 Add to this the change in pressure in the same unit from one meniscus to the next (plus if the next meniscus is lower minus if higher) (For pascals this is the product of the difference in elevation in metres and the unit gravity force of the fluid in newtons per cubic metre)

3 Continue until the other end of the gage (or the starting meniscus) is reached and equate the expression to the pressure at that point known or unknown

The expression will contain one unknown for a simple manometer or will give a difference in pressures for the differential manometer In equation form

Figure 27 Differential manometers

A differential manometer (Fig 27) determines the difference in pressures at two points A and B when the actual pressure at any point in the system cannot be determined

Application of the procedure outlined above to Fig 27a produces

For Fig 27b

If the pressures at A and B are expressed in length of the water column the above results can be written for Fig 27a

For Fig 27b

Example 24 In Fig 27a the liquids at A and B are water and the manometer liquid is oil S = 080 h1 = 300 mm h2 = 200 mm and h3 = 600 mm (a) Determine pA - pB in pacals (b) If pB = 50kPa and the barometer reading is 730 mm Hg find the pressure at A in meters of water absolute (a)

(b)

(a)

Micromanometers

For determining very small differences in pressure or determining large pressure differences precisely ndash several types of manometers

One type very accurately measures the differences in elevation of two menisci of a manometer

By means of small telescopes with horizontal cross hairs mounted along the tubes on a rack which is raised and lowered by a pinion and slow motion screw so that the cross hairs can be set accurately the difference in elevation of menisci (the gage difference) can be read with verniers

Figure 28 Micromanometer using two gage liquids

Fig 28 two gage liquids immiscible in each other and in the fluid to be measured a large gage difference R can be produced for a small pressure difference

The heavier gage liquid fills the lower U tube up to 0-0 then the lighter gage liquid is added to both sides filling the larger reservoirs up to 1-1

The gas or liquid in the system fills the space above 1-1 When the pressure at C is slightly greater than at D the menisci move as indicated in Fig 28

The volume of liquid displaced in each reservoir equals the displacement in the U tube

Manometer equation

γ1 γ2 and γ3 are the unit gravity force

Example 25 In the micromanometer of Fig 28 the pressure difference is wanted in pascals when air is in the system S2 = 10 S3 = 110 aA = 001 R = 5 mm t = 20oC and the barometer reads 760 mm Hg

The term γ1(aA) may be neglected Substituting into Eq (241) gives

Figure 29

Inclined manometer

The inclined manometer frequently used for measuring small differences in gas pressures

Adjusted to read zero by moving the inclined scale when A and B are open Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube it affords greater accuracy in reading the scale

Surface tension causes a capillary rise in small tubes If a U tube is used with a meniscus in each leg the surface-tension effects cancel

25 FORCES ON PLANE AREAS

In the preceding sections variations oF pressure throughout a fluid have been considered

The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force insofar as external reactions to the force system are concerned

In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration by formula and by use of the concept of the pressure prism

Horizontal Surfaces

A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure

The magnitude of the force acting on one side of the surface is

The elemental forces pdA acting on A are all parallel and in the same sense a scalar summation of all such elements yields the magnitude of the resultant force Its direction is normal to the surface and toward the surface if p is positive

Fig 210 arbitrary xy axes - to find the line of action of the resultant ie the point in the area where the moment of the distributed force about any axis through the point is zero

Then since the moment of the resultant must equal the moment of the distributed force system about any axis say the y axis

xrsquo ndash the distance from the y axis to the resultant

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

In component form Eq (224) becomes

The partials for variation in horizontal directions are one form of Pascals law they state that two points at the same elevation in the same continuous mass or fluid at rest have the same pressure

Since p is a function of y only

relates the change of pressure to unit gravity force and change of elevation and holds for both compressible and incompressible fluids

For fluids that may be considered homogeneous and incompressible γ is constant and the above equation when integrated becomes

in which c is the constant of integration The hydrostatic law of variation of pressure is frequently written in the form

h = -y p is the increase in pressure from that at the free surface

Example 21 An oceanographer is to design a sea lab 5 m high to withstand submersion to 100 m measured from sea level to the top of the sea lab Find the pressure variation on a side of the container and the pressure on the top if the relative density of salt water is 1020

At the top h = 100 m and

If y is measured from the top of the sea lab downward the pressure variation is

Pressure Variation in a Compressible Fluid

When the fluid is a perfect gas at rest at constant temperature

When the value of γ in Eq (227) is replaced by ρg and ρ is eliminated between Eqs (227) and (229)

If P = P0 when ρ = ρ0 integration between limits

- the equation for variation of pressure with elevation in an isothermal gas

- constant temperature gradient of atmosphere

Example 22 Assuming isothermal conditions to prevail in the atmosphere compute the pressure and density at 2000 m elevation if P = 105Pa ρ = 124 kgm3 at sea level

From Eq (2212)

Then from Eq (229)

23 UNITS AND SCALES OF PRESSURE MEASUREMENT

Pressure may be expressed with reference to any arbitrary datum absolute zero local atmospheric pressure

Absolute pressure difference between its value and a complete vacuum

Gage pressure difference between its value and the local atmospheric pressure

Figure 23 Bourdon gage

The bourdon gage (Fig 23) typical of the devices used for measuring gage pressures pressure element is a hollow curved flat metallic tube closed at one e

nd the other end is connected to the pressure to be measured when the internal pressure is increased the tube tends to straighten

pulling on a linkage to which is attached a pointer and causing the pointer to move

the dial reads zero when the inside and outside of the tube are at the same pressure regardless of its particular value

the gage measures pressure relative to the pressure of the medium surrounding the tube which is the local atmosphere

Figure 24 Units and scales for pressure measurement

Figure 24 the data and the relations of the common units of pressure measurement

Standard atmospheric pressure is the mean pressure at sea level 760 mm Hg

A pressure expressed in terms of the length of a column of liquid is equivalent to the force per unit area at the base of the column The relation for variation of pressure with altitude in a liquid p = γh [Eq (228)] (p is in pascals γ in newtons per cubic metre and h in metres)

With the unit gravity force of any liquid expressed as its relative density S times the unit gravity force of water

Water γ may be taken as 9806 Nm3

Figure 25 Mercury barometer

Local atmospheric pressure is measured by mercury barometer aneroid barometer (measures the difference in

pressure between the atmosphere and an evacuated box or tube in a manner analogous to the bourdon gage except that the tube is evacuated and sealed)

Mercury barometer glass tube closed at one end filled with mercury and inverted so that the open end is submerged in mercury

It has a scale the height of column R can be determined

The space above the mercury contains mercury vapor If the pressure of the mercury vapor hv is given in millimetres of mercury and R is measured in the same units the pressure at A may be expressed as (mm Hg)

Figure 24 a pressure may be located vertically on the chart which indicates its relation to absolute zero and to local atmospheric pressure

If the point is below the local-atmospheric-pressure line and is referred to gage datum it is called negative suction or vacuum

Example the pressure 460 mm Hg abs as at 1 with barometer reading 720 mm may be expressed as -260 mm Hg 260 mm Hg suction or 260 mm Hg vacuum

Note

Pabs = pbar + pgage

Absolute pressures P gage pressures p

Example 23 The rate of temperature change in the atmosphere with change in elevation is called its lapse rate The motion of a parcel of air depends on the density of the parcel relative to the density of the surrounding (ambient) air However as the parcel ascends through the atmosphere the air pressure decreases the parcel expands and its temperature decreases at a rate known as the dry adiabatic lapse rate A firm wants lo burn a large quantity of refuse It is estimated that the temperature of the smoke plume at 10 m above the ground will be 11oC greater than that of the ambient air For the following conditions determine what will happen to the smoke

(a) At standard atmospheric lapse rate β = -000651oC per meter and t0 =20oC

(b) At an inverted lapse rate β = 000365oC per meter

By combining Eqs (227) and (2214)

The relation between pressure and temperature for a mass of gas expanding without heat transfer (isentropic relation Sec 61) is

in which T1 is the initial smoke absolute temperature and P0 the initial absolute pressure k is the specific heat ratio 14 for air and other diatomic gases

Eliminating PP0 in the last two equations

Since the gas will rise until its temperature is equal to the ambient temperature

the last two equations may be solved for y Let

Then

For β = -000651oC per metre R = 287 mN(kgK) a = 2002 and y = 3201 m For the atmospheric temperature inversion β = -000365oC per metre a = -02721 and y = 8092 m

24 MANOMETERS

Manometers are devices that employ liquid columns for determining differences in pressure

Figure 26a the most elementary manometer ndash piezometer It measures the pressure in a liquid when it is above zero gage Glass tube is mounted vertically so that it is connected to the space

within the container Liquid rises in the tube until equilibrium is reached The pressure is then given by the vertical distance h from the menis

cus (liquid surface) to the point where the pressure is to be measured expressed in units of length of the liquid in the container

Piezometer would not work for negative gage pressures because air would flow into the container through the tube

Figure 26 Simple manometers

Figure 26b for small negative or positive gage pressures in a liquid With this arrangement the meniscus may come to rest below A as sho

wn Since the pressure at the meniscus is zero gage and since pressure decreases with elevation

units of length H2O Figure 26c for greater negative or positive gage pressures (a second li

quid of greater relative density employed) It must be immiscible in the first fluid which may now be a gas If the relative density of the fluid at A is S1 (based on water) and the re

lative density of the manometer liquid is S2 the equation for pressure at A

hA - the unknown pressure expressed in length units of water

h1 h2 - in length units

A general procedure in working all manometer problems

1 Start at one end (or any meniscus if the circuit is continuous) and write the pressure there in an appropriate unit (say pascals) or in an appropriate symbol if it is unknown

2 Add to this the change in pressure in the same unit from one meniscus to the next (plus if the next meniscus is lower minus if higher) (For pascals this is the product of the difference in elevation in metres and the unit gravity force of the fluid in newtons per cubic metre)

3 Continue until the other end of the gage (or the starting meniscus) is reached and equate the expression to the pressure at that point known or unknown

The expression will contain one unknown for a simple manometer or will give a difference in pressures for the differential manometer In equation form

Figure 27 Differential manometers

A differential manometer (Fig 27) determines the difference in pressures at two points A and B when the actual pressure at any point in the system cannot be determined

Application of the procedure outlined above to Fig 27a produces

For Fig 27b

If the pressures at A and B are expressed in length of the water column the above results can be written for Fig 27a

For Fig 27b

Example 24 In Fig 27a the liquids at A and B are water and the manometer liquid is oil S = 080 h1 = 300 mm h2 = 200 mm and h3 = 600 mm (a) Determine pA - pB in pacals (b) If pB = 50kPa and the barometer reading is 730 mm Hg find the pressure at A in meters of water absolute (a)

(b)

(a)

Micromanometers

For determining very small differences in pressure or determining large pressure differences precisely ndash several types of manometers

One type very accurately measures the differences in elevation of two menisci of a manometer

By means of small telescopes with horizontal cross hairs mounted along the tubes on a rack which is raised and lowered by a pinion and slow motion screw so that the cross hairs can be set accurately the difference in elevation of menisci (the gage difference) can be read with verniers

Figure 28 Micromanometer using two gage liquids

Fig 28 two gage liquids immiscible in each other and in the fluid to be measured a large gage difference R can be produced for a small pressure difference

The heavier gage liquid fills the lower U tube up to 0-0 then the lighter gage liquid is added to both sides filling the larger reservoirs up to 1-1

The gas or liquid in the system fills the space above 1-1 When the pressure at C is slightly greater than at D the menisci move as indicated in Fig 28

The volume of liquid displaced in each reservoir equals the displacement in the U tube

Manometer equation

γ1 γ2 and γ3 are the unit gravity force

Example 25 In the micromanometer of Fig 28 the pressure difference is wanted in pascals when air is in the system S2 = 10 S3 = 110 aA = 001 R = 5 mm t = 20oC and the barometer reads 760 mm Hg

The term γ1(aA) may be neglected Substituting into Eq (241) gives

Figure 29

Inclined manometer

The inclined manometer frequently used for measuring small differences in gas pressures

Adjusted to read zero by moving the inclined scale when A and B are open Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube it affords greater accuracy in reading the scale

Surface tension causes a capillary rise in small tubes If a U tube is used with a meniscus in each leg the surface-tension effects cancel

25 FORCES ON PLANE AREAS

In the preceding sections variations oF pressure throughout a fluid have been considered

The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force insofar as external reactions to the force system are concerned

In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration by formula and by use of the concept of the pressure prism

Horizontal Surfaces

A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure

The magnitude of the force acting on one side of the surface is

The elemental forces pdA acting on A are all parallel and in the same sense a scalar summation of all such elements yields the magnitude of the resultant force Its direction is normal to the surface and toward the surface if p is positive

Fig 210 arbitrary xy axes - to find the line of action of the resultant ie the point in the area where the moment of the distributed force about any axis through the point is zero

Then since the moment of the resultant must equal the moment of the distributed force system about any axis say the y axis

xrsquo ndash the distance from the y axis to the resultant

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Example 21 An oceanographer is to design a sea lab 5 m high to withstand submersion to 100 m measured from sea level to the top of the sea lab Find the pressure variation on a side of the container and the pressure on the top if the relative density of salt water is 1020

At the top h = 100 m and

If y is measured from the top of the sea lab downward the pressure variation is

Pressure Variation in a Compressible Fluid

When the fluid is a perfect gas at rest at constant temperature

When the value of γ in Eq (227) is replaced by ρg and ρ is eliminated between Eqs (227) and (229)

If P = P0 when ρ = ρ0 integration between limits

- the equation for variation of pressure with elevation in an isothermal gas

- constant temperature gradient of atmosphere

Example 22 Assuming isothermal conditions to prevail in the atmosphere compute the pressure and density at 2000 m elevation if P = 105Pa ρ = 124 kgm3 at sea level

From Eq (2212)

Then from Eq (229)

23 UNITS AND SCALES OF PRESSURE MEASUREMENT

Pressure may be expressed with reference to any arbitrary datum absolute zero local atmospheric pressure

Absolute pressure difference between its value and a complete vacuum

Gage pressure difference between its value and the local atmospheric pressure

Figure 23 Bourdon gage

The bourdon gage (Fig 23) typical of the devices used for measuring gage pressures pressure element is a hollow curved flat metallic tube closed at one e

nd the other end is connected to the pressure to be measured when the internal pressure is increased the tube tends to straighten

pulling on a linkage to which is attached a pointer and causing the pointer to move

the dial reads zero when the inside and outside of the tube are at the same pressure regardless of its particular value

the gage measures pressure relative to the pressure of the medium surrounding the tube which is the local atmosphere

Figure 24 Units and scales for pressure measurement

Figure 24 the data and the relations of the common units of pressure measurement

Standard atmospheric pressure is the mean pressure at sea level 760 mm Hg

A pressure expressed in terms of the length of a column of liquid is equivalent to the force per unit area at the base of the column The relation for variation of pressure with altitude in a liquid p = γh [Eq (228)] (p is in pascals γ in newtons per cubic metre and h in metres)

With the unit gravity force of any liquid expressed as its relative density S times the unit gravity force of water

Water γ may be taken as 9806 Nm3

Figure 25 Mercury barometer

Local atmospheric pressure is measured by mercury barometer aneroid barometer (measures the difference in

pressure between the atmosphere and an evacuated box or tube in a manner analogous to the bourdon gage except that the tube is evacuated and sealed)

Mercury barometer glass tube closed at one end filled with mercury and inverted so that the open end is submerged in mercury

It has a scale the height of column R can be determined

The space above the mercury contains mercury vapor If the pressure of the mercury vapor hv is given in millimetres of mercury and R is measured in the same units the pressure at A may be expressed as (mm Hg)

Figure 24 a pressure may be located vertically on the chart which indicates its relation to absolute zero and to local atmospheric pressure

If the point is below the local-atmospheric-pressure line and is referred to gage datum it is called negative suction or vacuum

Example the pressure 460 mm Hg abs as at 1 with barometer reading 720 mm may be expressed as -260 mm Hg 260 mm Hg suction or 260 mm Hg vacuum

Note

Pabs = pbar + pgage

Absolute pressures P gage pressures p

Example 23 The rate of temperature change in the atmosphere with change in elevation is called its lapse rate The motion of a parcel of air depends on the density of the parcel relative to the density of the surrounding (ambient) air However as the parcel ascends through the atmosphere the air pressure decreases the parcel expands and its temperature decreases at a rate known as the dry adiabatic lapse rate A firm wants lo burn a large quantity of refuse It is estimated that the temperature of the smoke plume at 10 m above the ground will be 11oC greater than that of the ambient air For the following conditions determine what will happen to the smoke

(a) At standard atmospheric lapse rate β = -000651oC per meter and t0 =20oC

(b) At an inverted lapse rate β = 000365oC per meter

By combining Eqs (227) and (2214)

The relation between pressure and temperature for a mass of gas expanding without heat transfer (isentropic relation Sec 61) is

in which T1 is the initial smoke absolute temperature and P0 the initial absolute pressure k is the specific heat ratio 14 for air and other diatomic gases

Eliminating PP0 in the last two equations

Since the gas will rise until its temperature is equal to the ambient temperature

the last two equations may be solved for y Let

Then

For β = -000651oC per metre R = 287 mN(kgK) a = 2002 and y = 3201 m For the atmospheric temperature inversion β = -000365oC per metre a = -02721 and y = 8092 m

24 MANOMETERS

Manometers are devices that employ liquid columns for determining differences in pressure

Figure 26a the most elementary manometer ndash piezometer It measures the pressure in a liquid when it is above zero gage Glass tube is mounted vertically so that it is connected to the space

within the container Liquid rises in the tube until equilibrium is reached The pressure is then given by the vertical distance h from the menis

cus (liquid surface) to the point where the pressure is to be measured expressed in units of length of the liquid in the container

Piezometer would not work for negative gage pressures because air would flow into the container through the tube

Figure 26 Simple manometers

Figure 26b for small negative or positive gage pressures in a liquid With this arrangement the meniscus may come to rest below A as sho

wn Since the pressure at the meniscus is zero gage and since pressure decreases with elevation

units of length H2O Figure 26c for greater negative or positive gage pressures (a second li

quid of greater relative density employed) It must be immiscible in the first fluid which may now be a gas If the relative density of the fluid at A is S1 (based on water) and the re

lative density of the manometer liquid is S2 the equation for pressure at A

hA - the unknown pressure expressed in length units of water

h1 h2 - in length units

A general procedure in working all manometer problems

1 Start at one end (or any meniscus if the circuit is continuous) and write the pressure there in an appropriate unit (say pascals) or in an appropriate symbol if it is unknown

2 Add to this the change in pressure in the same unit from one meniscus to the next (plus if the next meniscus is lower minus if higher) (For pascals this is the product of the difference in elevation in metres and the unit gravity force of the fluid in newtons per cubic metre)

3 Continue until the other end of the gage (or the starting meniscus) is reached and equate the expression to the pressure at that point known or unknown

The expression will contain one unknown for a simple manometer or will give a difference in pressures for the differential manometer In equation form

Figure 27 Differential manometers

A differential manometer (Fig 27) determines the difference in pressures at two points A and B when the actual pressure at any point in the system cannot be determined

Application of the procedure outlined above to Fig 27a produces

For Fig 27b

If the pressures at A and B are expressed in length of the water column the above results can be written for Fig 27a

For Fig 27b

Example 24 In Fig 27a the liquids at A and B are water and the manometer liquid is oil S = 080 h1 = 300 mm h2 = 200 mm and h3 = 600 mm (a) Determine pA - pB in pacals (b) If pB = 50kPa and the barometer reading is 730 mm Hg find the pressure at A in meters of water absolute (a)

(b)

(a)

Micromanometers

For determining very small differences in pressure or determining large pressure differences precisely ndash several types of manometers

One type very accurately measures the differences in elevation of two menisci of a manometer

By means of small telescopes with horizontal cross hairs mounted along the tubes on a rack which is raised and lowered by a pinion and slow motion screw so that the cross hairs can be set accurately the difference in elevation of menisci (the gage difference) can be read with verniers

Figure 28 Micromanometer using two gage liquids

Fig 28 two gage liquids immiscible in each other and in the fluid to be measured a large gage difference R can be produced for a small pressure difference

The heavier gage liquid fills the lower U tube up to 0-0 then the lighter gage liquid is added to both sides filling the larger reservoirs up to 1-1

The gas or liquid in the system fills the space above 1-1 When the pressure at C is slightly greater than at D the menisci move as indicated in Fig 28

The volume of liquid displaced in each reservoir equals the displacement in the U tube

Manometer equation

γ1 γ2 and γ3 are the unit gravity force

Example 25 In the micromanometer of Fig 28 the pressure difference is wanted in pascals when air is in the system S2 = 10 S3 = 110 aA = 001 R = 5 mm t = 20oC and the barometer reads 760 mm Hg

The term γ1(aA) may be neglected Substituting into Eq (241) gives

Figure 29

Inclined manometer

The inclined manometer frequently used for measuring small differences in gas pressures

Adjusted to read zero by moving the inclined scale when A and B are open Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube it affords greater accuracy in reading the scale

Surface tension causes a capillary rise in small tubes If a U tube is used with a meniscus in each leg the surface-tension effects cancel

25 FORCES ON PLANE AREAS

In the preceding sections variations oF pressure throughout a fluid have been considered

The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force insofar as external reactions to the force system are concerned

In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration by formula and by use of the concept of the pressure prism

Horizontal Surfaces

A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure

The magnitude of the force acting on one side of the surface is

The elemental forces pdA acting on A are all parallel and in the same sense a scalar summation of all such elements yields the magnitude of the resultant force Its direction is normal to the surface and toward the surface if p is positive

Fig 210 arbitrary xy axes - to find the line of action of the resultant ie the point in the area where the moment of the distributed force about any axis through the point is zero

Then since the moment of the resultant must equal the moment of the distributed force system about any axis say the y axis

xrsquo ndash the distance from the y axis to the resultant

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Pressure Variation in a Compressible Fluid

When the fluid is a perfect gas at rest at constant temperature

When the value of γ in Eq (227) is replaced by ρg and ρ is eliminated between Eqs (227) and (229)

If P = P0 when ρ = ρ0 integration between limits

- the equation for variation of pressure with elevation in an isothermal gas

- constant temperature gradient of atmosphere

Example 22 Assuming isothermal conditions to prevail in the atmosphere compute the pressure and density at 2000 m elevation if P = 105Pa ρ = 124 kgm3 at sea level

From Eq (2212)

Then from Eq (229)

23 UNITS AND SCALES OF PRESSURE MEASUREMENT

Pressure may be expressed with reference to any arbitrary datum absolute zero local atmospheric pressure

Absolute pressure difference between its value and a complete vacuum

Gage pressure difference between its value and the local atmospheric pressure

Figure 23 Bourdon gage

The bourdon gage (Fig 23) typical of the devices used for measuring gage pressures pressure element is a hollow curved flat metallic tube closed at one e

nd the other end is connected to the pressure to be measured when the internal pressure is increased the tube tends to straighten

pulling on a linkage to which is attached a pointer and causing the pointer to move

the dial reads zero when the inside and outside of the tube are at the same pressure regardless of its particular value

the gage measures pressure relative to the pressure of the medium surrounding the tube which is the local atmosphere

Figure 24 Units and scales for pressure measurement

Figure 24 the data and the relations of the common units of pressure measurement

Standard atmospheric pressure is the mean pressure at sea level 760 mm Hg

A pressure expressed in terms of the length of a column of liquid is equivalent to the force per unit area at the base of the column The relation for variation of pressure with altitude in a liquid p = γh [Eq (228)] (p is in pascals γ in newtons per cubic metre and h in metres)

With the unit gravity force of any liquid expressed as its relative density S times the unit gravity force of water

Water γ may be taken as 9806 Nm3

Figure 25 Mercury barometer

Local atmospheric pressure is measured by mercury barometer aneroid barometer (measures the difference in

pressure between the atmosphere and an evacuated box or tube in a manner analogous to the bourdon gage except that the tube is evacuated and sealed)

Mercury barometer glass tube closed at one end filled with mercury and inverted so that the open end is submerged in mercury

It has a scale the height of column R can be determined

The space above the mercury contains mercury vapor If the pressure of the mercury vapor hv is given in millimetres of mercury and R is measured in the same units the pressure at A may be expressed as (mm Hg)

Figure 24 a pressure may be located vertically on the chart which indicates its relation to absolute zero and to local atmospheric pressure

If the point is below the local-atmospheric-pressure line and is referred to gage datum it is called negative suction or vacuum

Example the pressure 460 mm Hg abs as at 1 with barometer reading 720 mm may be expressed as -260 mm Hg 260 mm Hg suction or 260 mm Hg vacuum

Note

Pabs = pbar + pgage

Absolute pressures P gage pressures p

Example 23 The rate of temperature change in the atmosphere with change in elevation is called its lapse rate The motion of a parcel of air depends on the density of the parcel relative to the density of the surrounding (ambient) air However as the parcel ascends through the atmosphere the air pressure decreases the parcel expands and its temperature decreases at a rate known as the dry adiabatic lapse rate A firm wants lo burn a large quantity of refuse It is estimated that the temperature of the smoke plume at 10 m above the ground will be 11oC greater than that of the ambient air For the following conditions determine what will happen to the smoke

(a) At standard atmospheric lapse rate β = -000651oC per meter and t0 =20oC

(b) At an inverted lapse rate β = 000365oC per meter

By combining Eqs (227) and (2214)

The relation between pressure and temperature for a mass of gas expanding without heat transfer (isentropic relation Sec 61) is

in which T1 is the initial smoke absolute temperature and P0 the initial absolute pressure k is the specific heat ratio 14 for air and other diatomic gases

Eliminating PP0 in the last two equations

Since the gas will rise until its temperature is equal to the ambient temperature

the last two equations may be solved for y Let

Then

For β = -000651oC per metre R = 287 mN(kgK) a = 2002 and y = 3201 m For the atmospheric temperature inversion β = -000365oC per metre a = -02721 and y = 8092 m

24 MANOMETERS

Manometers are devices that employ liquid columns for determining differences in pressure

Figure 26a the most elementary manometer ndash piezometer It measures the pressure in a liquid when it is above zero gage Glass tube is mounted vertically so that it is connected to the space

within the container Liquid rises in the tube until equilibrium is reached The pressure is then given by the vertical distance h from the menis

cus (liquid surface) to the point where the pressure is to be measured expressed in units of length of the liquid in the container

Piezometer would not work for negative gage pressures because air would flow into the container through the tube

Figure 26 Simple manometers

Figure 26b for small negative or positive gage pressures in a liquid With this arrangement the meniscus may come to rest below A as sho

wn Since the pressure at the meniscus is zero gage and since pressure decreases with elevation

units of length H2O Figure 26c for greater negative or positive gage pressures (a second li

quid of greater relative density employed) It must be immiscible in the first fluid which may now be a gas If the relative density of the fluid at A is S1 (based on water) and the re

lative density of the manometer liquid is S2 the equation for pressure at A

hA - the unknown pressure expressed in length units of water

h1 h2 - in length units

A general procedure in working all manometer problems

1 Start at one end (or any meniscus if the circuit is continuous) and write the pressure there in an appropriate unit (say pascals) or in an appropriate symbol if it is unknown

2 Add to this the change in pressure in the same unit from one meniscus to the next (plus if the next meniscus is lower minus if higher) (For pascals this is the product of the difference in elevation in metres and the unit gravity force of the fluid in newtons per cubic metre)

3 Continue until the other end of the gage (or the starting meniscus) is reached and equate the expression to the pressure at that point known or unknown

The expression will contain one unknown for a simple manometer or will give a difference in pressures for the differential manometer In equation form

Figure 27 Differential manometers

A differential manometer (Fig 27) determines the difference in pressures at two points A and B when the actual pressure at any point in the system cannot be determined

Application of the procedure outlined above to Fig 27a produces

For Fig 27b

If the pressures at A and B are expressed in length of the water column the above results can be written for Fig 27a

For Fig 27b

Example 24 In Fig 27a the liquids at A and B are water and the manometer liquid is oil S = 080 h1 = 300 mm h2 = 200 mm and h3 = 600 mm (a) Determine pA - pB in pacals (b) If pB = 50kPa and the barometer reading is 730 mm Hg find the pressure at A in meters of water absolute (a)

(b)

(a)

Micromanometers

For determining very small differences in pressure or determining large pressure differences precisely ndash several types of manometers

One type very accurately measures the differences in elevation of two menisci of a manometer

By means of small telescopes with horizontal cross hairs mounted along the tubes on a rack which is raised and lowered by a pinion and slow motion screw so that the cross hairs can be set accurately the difference in elevation of menisci (the gage difference) can be read with verniers

Figure 28 Micromanometer using two gage liquids

Fig 28 two gage liquids immiscible in each other and in the fluid to be measured a large gage difference R can be produced for a small pressure difference

The heavier gage liquid fills the lower U tube up to 0-0 then the lighter gage liquid is added to both sides filling the larger reservoirs up to 1-1

The gas or liquid in the system fills the space above 1-1 When the pressure at C is slightly greater than at D the menisci move as indicated in Fig 28

The volume of liquid displaced in each reservoir equals the displacement in the U tube

Manometer equation

γ1 γ2 and γ3 are the unit gravity force

Example 25 In the micromanometer of Fig 28 the pressure difference is wanted in pascals when air is in the system S2 = 10 S3 = 110 aA = 001 R = 5 mm t = 20oC and the barometer reads 760 mm Hg

The term γ1(aA) may be neglected Substituting into Eq (241) gives

Figure 29

Inclined manometer

The inclined manometer frequently used for measuring small differences in gas pressures

Adjusted to read zero by moving the inclined scale when A and B are open Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube it affords greater accuracy in reading the scale

Surface tension causes a capillary rise in small tubes If a U tube is used with a meniscus in each leg the surface-tension effects cancel

25 FORCES ON PLANE AREAS

In the preceding sections variations oF pressure throughout a fluid have been considered

The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force insofar as external reactions to the force system are concerned

In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration by formula and by use of the concept of the pressure prism

Horizontal Surfaces

A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure

The magnitude of the force acting on one side of the surface is

The elemental forces pdA acting on A are all parallel and in the same sense a scalar summation of all such elements yields the magnitude of the resultant force Its direction is normal to the surface and toward the surface if p is positive

Fig 210 arbitrary xy axes - to find the line of action of the resultant ie the point in the area where the moment of the distributed force about any axis through the point is zero

Then since the moment of the resultant must equal the moment of the distributed force system about any axis say the y axis

xrsquo ndash the distance from the y axis to the resultant

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Example 22 Assuming isothermal conditions to prevail in the atmosphere compute the pressure and density at 2000 m elevation if P = 105Pa ρ = 124 kgm3 at sea level

From Eq (2212)

Then from Eq (229)

23 UNITS AND SCALES OF PRESSURE MEASUREMENT

Pressure may be expressed with reference to any arbitrary datum absolute zero local atmospheric pressure

Absolute pressure difference between its value and a complete vacuum

Gage pressure difference between its value and the local atmospheric pressure

Figure 23 Bourdon gage

The bourdon gage (Fig 23) typical of the devices used for measuring gage pressures pressure element is a hollow curved flat metallic tube closed at one e

nd the other end is connected to the pressure to be measured when the internal pressure is increased the tube tends to straighten

pulling on a linkage to which is attached a pointer and causing the pointer to move

the dial reads zero when the inside and outside of the tube are at the same pressure regardless of its particular value

the gage measures pressure relative to the pressure of the medium surrounding the tube which is the local atmosphere

Figure 24 Units and scales for pressure measurement

Figure 24 the data and the relations of the common units of pressure measurement

Standard atmospheric pressure is the mean pressure at sea level 760 mm Hg

A pressure expressed in terms of the length of a column of liquid is equivalent to the force per unit area at the base of the column The relation for variation of pressure with altitude in a liquid p = γh [Eq (228)] (p is in pascals γ in newtons per cubic metre and h in metres)

With the unit gravity force of any liquid expressed as its relative density S times the unit gravity force of water

Water γ may be taken as 9806 Nm3

Figure 25 Mercury barometer

Local atmospheric pressure is measured by mercury barometer aneroid barometer (measures the difference in

pressure between the atmosphere and an evacuated box or tube in a manner analogous to the bourdon gage except that the tube is evacuated and sealed)

Mercury barometer glass tube closed at one end filled with mercury and inverted so that the open end is submerged in mercury

It has a scale the height of column R can be determined

The space above the mercury contains mercury vapor If the pressure of the mercury vapor hv is given in millimetres of mercury and R is measured in the same units the pressure at A may be expressed as (mm Hg)

Figure 24 a pressure may be located vertically on the chart which indicates its relation to absolute zero and to local atmospheric pressure

If the point is below the local-atmospheric-pressure line and is referred to gage datum it is called negative suction or vacuum

Example the pressure 460 mm Hg abs as at 1 with barometer reading 720 mm may be expressed as -260 mm Hg 260 mm Hg suction or 260 mm Hg vacuum

Note

Pabs = pbar + pgage

Absolute pressures P gage pressures p

Example 23 The rate of temperature change in the atmosphere with change in elevation is called its lapse rate The motion of a parcel of air depends on the density of the parcel relative to the density of the surrounding (ambient) air However as the parcel ascends through the atmosphere the air pressure decreases the parcel expands and its temperature decreases at a rate known as the dry adiabatic lapse rate A firm wants lo burn a large quantity of refuse It is estimated that the temperature of the smoke plume at 10 m above the ground will be 11oC greater than that of the ambient air For the following conditions determine what will happen to the smoke

(a) At standard atmospheric lapse rate β = -000651oC per meter and t0 =20oC

(b) At an inverted lapse rate β = 000365oC per meter

By combining Eqs (227) and (2214)

The relation between pressure and temperature for a mass of gas expanding without heat transfer (isentropic relation Sec 61) is

in which T1 is the initial smoke absolute temperature and P0 the initial absolute pressure k is the specific heat ratio 14 for air and other diatomic gases

Eliminating PP0 in the last two equations

Since the gas will rise until its temperature is equal to the ambient temperature

the last two equations may be solved for y Let

Then

For β = -000651oC per metre R = 287 mN(kgK) a = 2002 and y = 3201 m For the atmospheric temperature inversion β = -000365oC per metre a = -02721 and y = 8092 m

24 MANOMETERS

Manometers are devices that employ liquid columns for determining differences in pressure

Figure 26a the most elementary manometer ndash piezometer It measures the pressure in a liquid when it is above zero gage Glass tube is mounted vertically so that it is connected to the space

within the container Liquid rises in the tube until equilibrium is reached The pressure is then given by the vertical distance h from the menis

cus (liquid surface) to the point where the pressure is to be measured expressed in units of length of the liquid in the container

Piezometer would not work for negative gage pressures because air would flow into the container through the tube

Figure 26 Simple manometers

Figure 26b for small negative or positive gage pressures in a liquid With this arrangement the meniscus may come to rest below A as sho

wn Since the pressure at the meniscus is zero gage and since pressure decreases with elevation

units of length H2O Figure 26c for greater negative or positive gage pressures (a second li

quid of greater relative density employed) It must be immiscible in the first fluid which may now be a gas If the relative density of the fluid at A is S1 (based on water) and the re

lative density of the manometer liquid is S2 the equation for pressure at A

hA - the unknown pressure expressed in length units of water

h1 h2 - in length units

A general procedure in working all manometer problems

1 Start at one end (or any meniscus if the circuit is continuous) and write the pressure there in an appropriate unit (say pascals) or in an appropriate symbol if it is unknown

2 Add to this the change in pressure in the same unit from one meniscus to the next (plus if the next meniscus is lower minus if higher) (For pascals this is the product of the difference in elevation in metres and the unit gravity force of the fluid in newtons per cubic metre)

3 Continue until the other end of the gage (or the starting meniscus) is reached and equate the expression to the pressure at that point known or unknown

The expression will contain one unknown for a simple manometer or will give a difference in pressures for the differential manometer In equation form

Figure 27 Differential manometers

A differential manometer (Fig 27) determines the difference in pressures at two points A and B when the actual pressure at any point in the system cannot be determined

Application of the procedure outlined above to Fig 27a produces

For Fig 27b

If the pressures at A and B are expressed in length of the water column the above results can be written for Fig 27a

For Fig 27b

Example 24 In Fig 27a the liquids at A and B are water and the manometer liquid is oil S = 080 h1 = 300 mm h2 = 200 mm and h3 = 600 mm (a) Determine pA - pB in pacals (b) If pB = 50kPa and the barometer reading is 730 mm Hg find the pressure at A in meters of water absolute (a)

(b)

(a)

Micromanometers

For determining very small differences in pressure or determining large pressure differences precisely ndash several types of manometers

One type very accurately measures the differences in elevation of two menisci of a manometer

By means of small telescopes with horizontal cross hairs mounted along the tubes on a rack which is raised and lowered by a pinion and slow motion screw so that the cross hairs can be set accurately the difference in elevation of menisci (the gage difference) can be read with verniers

Figure 28 Micromanometer using two gage liquids

Fig 28 two gage liquids immiscible in each other and in the fluid to be measured a large gage difference R can be produced for a small pressure difference

The heavier gage liquid fills the lower U tube up to 0-0 then the lighter gage liquid is added to both sides filling the larger reservoirs up to 1-1

The gas or liquid in the system fills the space above 1-1 When the pressure at C is slightly greater than at D the menisci move as indicated in Fig 28

The volume of liquid displaced in each reservoir equals the displacement in the U tube

Manometer equation

γ1 γ2 and γ3 are the unit gravity force

Example 25 In the micromanometer of Fig 28 the pressure difference is wanted in pascals when air is in the system S2 = 10 S3 = 110 aA = 001 R = 5 mm t = 20oC and the barometer reads 760 mm Hg

The term γ1(aA) may be neglected Substituting into Eq (241) gives

Figure 29

Inclined manometer

The inclined manometer frequently used for measuring small differences in gas pressures

Adjusted to read zero by moving the inclined scale when A and B are open Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube it affords greater accuracy in reading the scale

Surface tension causes a capillary rise in small tubes If a U tube is used with a meniscus in each leg the surface-tension effects cancel

25 FORCES ON PLANE AREAS

In the preceding sections variations oF pressure throughout a fluid have been considered

The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force insofar as external reactions to the force system are concerned

In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration by formula and by use of the concept of the pressure prism

Horizontal Surfaces

A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure

The magnitude of the force acting on one side of the surface is

The elemental forces pdA acting on A are all parallel and in the same sense a scalar summation of all such elements yields the magnitude of the resultant force Its direction is normal to the surface and toward the surface if p is positive

Fig 210 arbitrary xy axes - to find the line of action of the resultant ie the point in the area where the moment of the distributed force about any axis through the point is zero

Then since the moment of the resultant must equal the moment of the distributed force system about any axis say the y axis

xrsquo ndash the distance from the y axis to the resultant

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

23 UNITS AND SCALES OF PRESSURE MEASUREMENT

Pressure may be expressed with reference to any arbitrary datum absolute zero local atmospheric pressure

Absolute pressure difference between its value and a complete vacuum

Gage pressure difference between its value and the local atmospheric pressure

Figure 23 Bourdon gage

The bourdon gage (Fig 23) typical of the devices used for measuring gage pressures pressure element is a hollow curved flat metallic tube closed at one e

nd the other end is connected to the pressure to be measured when the internal pressure is increased the tube tends to straighten

pulling on a linkage to which is attached a pointer and causing the pointer to move

the dial reads zero when the inside and outside of the tube are at the same pressure regardless of its particular value

the gage measures pressure relative to the pressure of the medium surrounding the tube which is the local atmosphere

Figure 24 Units and scales for pressure measurement

Figure 24 the data and the relations of the common units of pressure measurement

Standard atmospheric pressure is the mean pressure at sea level 760 mm Hg

A pressure expressed in terms of the length of a column of liquid is equivalent to the force per unit area at the base of the column The relation for variation of pressure with altitude in a liquid p = γh [Eq (228)] (p is in pascals γ in newtons per cubic metre and h in metres)

With the unit gravity force of any liquid expressed as its relative density S times the unit gravity force of water

Water γ may be taken as 9806 Nm3

Figure 25 Mercury barometer

Local atmospheric pressure is measured by mercury barometer aneroid barometer (measures the difference in

pressure between the atmosphere and an evacuated box or tube in a manner analogous to the bourdon gage except that the tube is evacuated and sealed)

Mercury barometer glass tube closed at one end filled with mercury and inverted so that the open end is submerged in mercury

It has a scale the height of column R can be determined

The space above the mercury contains mercury vapor If the pressure of the mercury vapor hv is given in millimetres of mercury and R is measured in the same units the pressure at A may be expressed as (mm Hg)

Figure 24 a pressure may be located vertically on the chart which indicates its relation to absolute zero and to local atmospheric pressure

If the point is below the local-atmospheric-pressure line and is referred to gage datum it is called negative suction or vacuum

Example the pressure 460 mm Hg abs as at 1 with barometer reading 720 mm may be expressed as -260 mm Hg 260 mm Hg suction or 260 mm Hg vacuum

Note

Pabs = pbar + pgage

Absolute pressures P gage pressures p

Example 23 The rate of temperature change in the atmosphere with change in elevation is called its lapse rate The motion of a parcel of air depends on the density of the parcel relative to the density of the surrounding (ambient) air However as the parcel ascends through the atmosphere the air pressure decreases the parcel expands and its temperature decreases at a rate known as the dry adiabatic lapse rate A firm wants lo burn a large quantity of refuse It is estimated that the temperature of the smoke plume at 10 m above the ground will be 11oC greater than that of the ambient air For the following conditions determine what will happen to the smoke

(a) At standard atmospheric lapse rate β = -000651oC per meter and t0 =20oC

(b) At an inverted lapse rate β = 000365oC per meter

By combining Eqs (227) and (2214)

The relation between pressure and temperature for a mass of gas expanding without heat transfer (isentropic relation Sec 61) is

in which T1 is the initial smoke absolute temperature and P0 the initial absolute pressure k is the specific heat ratio 14 for air and other diatomic gases

Eliminating PP0 in the last two equations

Since the gas will rise until its temperature is equal to the ambient temperature

the last two equations may be solved for y Let

Then

For β = -000651oC per metre R = 287 mN(kgK) a = 2002 and y = 3201 m For the atmospheric temperature inversion β = -000365oC per metre a = -02721 and y = 8092 m

24 MANOMETERS

Manometers are devices that employ liquid columns for determining differences in pressure

Figure 26a the most elementary manometer ndash piezometer It measures the pressure in a liquid when it is above zero gage Glass tube is mounted vertically so that it is connected to the space

within the container Liquid rises in the tube until equilibrium is reached The pressure is then given by the vertical distance h from the menis

cus (liquid surface) to the point where the pressure is to be measured expressed in units of length of the liquid in the container

Piezometer would not work for negative gage pressures because air would flow into the container through the tube

Figure 26 Simple manometers

Figure 26b for small negative or positive gage pressures in a liquid With this arrangement the meniscus may come to rest below A as sho

wn Since the pressure at the meniscus is zero gage and since pressure decreases with elevation

units of length H2O Figure 26c for greater negative or positive gage pressures (a second li

quid of greater relative density employed) It must be immiscible in the first fluid which may now be a gas If the relative density of the fluid at A is S1 (based on water) and the re

lative density of the manometer liquid is S2 the equation for pressure at A

hA - the unknown pressure expressed in length units of water

h1 h2 - in length units

A general procedure in working all manometer problems

1 Start at one end (or any meniscus if the circuit is continuous) and write the pressure there in an appropriate unit (say pascals) or in an appropriate symbol if it is unknown

2 Add to this the change in pressure in the same unit from one meniscus to the next (plus if the next meniscus is lower minus if higher) (For pascals this is the product of the difference in elevation in metres and the unit gravity force of the fluid in newtons per cubic metre)

3 Continue until the other end of the gage (or the starting meniscus) is reached and equate the expression to the pressure at that point known or unknown

The expression will contain one unknown for a simple manometer or will give a difference in pressures for the differential manometer In equation form

Figure 27 Differential manometers

A differential manometer (Fig 27) determines the difference in pressures at two points A and B when the actual pressure at any point in the system cannot be determined

Application of the procedure outlined above to Fig 27a produces

For Fig 27b

If the pressures at A and B are expressed in length of the water column the above results can be written for Fig 27a

For Fig 27b

Example 24 In Fig 27a the liquids at A and B are water and the manometer liquid is oil S = 080 h1 = 300 mm h2 = 200 mm and h3 = 600 mm (a) Determine pA - pB in pacals (b) If pB = 50kPa and the barometer reading is 730 mm Hg find the pressure at A in meters of water absolute (a)

(b)

(a)

Micromanometers

For determining very small differences in pressure or determining large pressure differences precisely ndash several types of manometers

One type very accurately measures the differences in elevation of two menisci of a manometer

By means of small telescopes with horizontal cross hairs mounted along the tubes on a rack which is raised and lowered by a pinion and slow motion screw so that the cross hairs can be set accurately the difference in elevation of menisci (the gage difference) can be read with verniers

Figure 28 Micromanometer using two gage liquids

Fig 28 two gage liquids immiscible in each other and in the fluid to be measured a large gage difference R can be produced for a small pressure difference

The heavier gage liquid fills the lower U tube up to 0-0 then the lighter gage liquid is added to both sides filling the larger reservoirs up to 1-1

The gas or liquid in the system fills the space above 1-1 When the pressure at C is slightly greater than at D the menisci move as indicated in Fig 28

The volume of liquid displaced in each reservoir equals the displacement in the U tube

Manometer equation

γ1 γ2 and γ3 are the unit gravity force

Example 25 In the micromanometer of Fig 28 the pressure difference is wanted in pascals when air is in the system S2 = 10 S3 = 110 aA = 001 R = 5 mm t = 20oC and the barometer reads 760 mm Hg

The term γ1(aA) may be neglected Substituting into Eq (241) gives

Figure 29

Inclined manometer

The inclined manometer frequently used for measuring small differences in gas pressures

Adjusted to read zero by moving the inclined scale when A and B are open Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube it affords greater accuracy in reading the scale

Surface tension causes a capillary rise in small tubes If a U tube is used with a meniscus in each leg the surface-tension effects cancel

25 FORCES ON PLANE AREAS

In the preceding sections variations oF pressure throughout a fluid have been considered

The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force insofar as external reactions to the force system are concerned

In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration by formula and by use of the concept of the pressure prism

Horizontal Surfaces

A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure

The magnitude of the force acting on one side of the surface is

The elemental forces pdA acting on A are all parallel and in the same sense a scalar summation of all such elements yields the magnitude of the resultant force Its direction is normal to the surface and toward the surface if p is positive

Fig 210 arbitrary xy axes - to find the line of action of the resultant ie the point in the area where the moment of the distributed force about any axis through the point is zero

Then since the moment of the resultant must equal the moment of the distributed force system about any axis say the y axis

xrsquo ndash the distance from the y axis to the resultant

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Figure 23 Bourdon gage

The bourdon gage (Fig 23) typical of the devices used for measuring gage pressures pressure element is a hollow curved flat metallic tube closed at one e

nd the other end is connected to the pressure to be measured when the internal pressure is increased the tube tends to straighten

pulling on a linkage to which is attached a pointer and causing the pointer to move

the dial reads zero when the inside and outside of the tube are at the same pressure regardless of its particular value

the gage measures pressure relative to the pressure of the medium surrounding the tube which is the local atmosphere

Figure 24 Units and scales for pressure measurement

Figure 24 the data and the relations of the common units of pressure measurement

Standard atmospheric pressure is the mean pressure at sea level 760 mm Hg

A pressure expressed in terms of the length of a column of liquid is equivalent to the force per unit area at the base of the column The relation for variation of pressure with altitude in a liquid p = γh [Eq (228)] (p is in pascals γ in newtons per cubic metre and h in metres)

With the unit gravity force of any liquid expressed as its relative density S times the unit gravity force of water

Water γ may be taken as 9806 Nm3

Figure 25 Mercury barometer

Local atmospheric pressure is measured by mercury barometer aneroid barometer (measures the difference in

pressure between the atmosphere and an evacuated box or tube in a manner analogous to the bourdon gage except that the tube is evacuated and sealed)

Mercury barometer glass tube closed at one end filled with mercury and inverted so that the open end is submerged in mercury

It has a scale the height of column R can be determined

The space above the mercury contains mercury vapor If the pressure of the mercury vapor hv is given in millimetres of mercury and R is measured in the same units the pressure at A may be expressed as (mm Hg)

Figure 24 a pressure may be located vertically on the chart which indicates its relation to absolute zero and to local atmospheric pressure

If the point is below the local-atmospheric-pressure line and is referred to gage datum it is called negative suction or vacuum

Example the pressure 460 mm Hg abs as at 1 with barometer reading 720 mm may be expressed as -260 mm Hg 260 mm Hg suction or 260 mm Hg vacuum

Note

Pabs = pbar + pgage

Absolute pressures P gage pressures p

Example 23 The rate of temperature change in the atmosphere with change in elevation is called its lapse rate The motion of a parcel of air depends on the density of the parcel relative to the density of the surrounding (ambient) air However as the parcel ascends through the atmosphere the air pressure decreases the parcel expands and its temperature decreases at a rate known as the dry adiabatic lapse rate A firm wants lo burn a large quantity of refuse It is estimated that the temperature of the smoke plume at 10 m above the ground will be 11oC greater than that of the ambient air For the following conditions determine what will happen to the smoke

(a) At standard atmospheric lapse rate β = -000651oC per meter and t0 =20oC

(b) At an inverted lapse rate β = 000365oC per meter

By combining Eqs (227) and (2214)

The relation between pressure and temperature for a mass of gas expanding without heat transfer (isentropic relation Sec 61) is

in which T1 is the initial smoke absolute temperature and P0 the initial absolute pressure k is the specific heat ratio 14 for air and other diatomic gases

Eliminating PP0 in the last two equations

Since the gas will rise until its temperature is equal to the ambient temperature

the last two equations may be solved for y Let

Then

For β = -000651oC per metre R = 287 mN(kgK) a = 2002 and y = 3201 m For the atmospheric temperature inversion β = -000365oC per metre a = -02721 and y = 8092 m

24 MANOMETERS

Manometers are devices that employ liquid columns for determining differences in pressure

Figure 26a the most elementary manometer ndash piezometer It measures the pressure in a liquid when it is above zero gage Glass tube is mounted vertically so that it is connected to the space

within the container Liquid rises in the tube until equilibrium is reached The pressure is then given by the vertical distance h from the menis

cus (liquid surface) to the point where the pressure is to be measured expressed in units of length of the liquid in the container

Piezometer would not work for negative gage pressures because air would flow into the container through the tube

Figure 26 Simple manometers

Figure 26b for small negative or positive gage pressures in a liquid With this arrangement the meniscus may come to rest below A as sho

wn Since the pressure at the meniscus is zero gage and since pressure decreases with elevation

units of length H2O Figure 26c for greater negative or positive gage pressures (a second li

quid of greater relative density employed) It must be immiscible in the first fluid which may now be a gas If the relative density of the fluid at A is S1 (based on water) and the re

lative density of the manometer liquid is S2 the equation for pressure at A

hA - the unknown pressure expressed in length units of water

h1 h2 - in length units

A general procedure in working all manometer problems

1 Start at one end (or any meniscus if the circuit is continuous) and write the pressure there in an appropriate unit (say pascals) or in an appropriate symbol if it is unknown

2 Add to this the change in pressure in the same unit from one meniscus to the next (plus if the next meniscus is lower minus if higher) (For pascals this is the product of the difference in elevation in metres and the unit gravity force of the fluid in newtons per cubic metre)

3 Continue until the other end of the gage (or the starting meniscus) is reached and equate the expression to the pressure at that point known or unknown

The expression will contain one unknown for a simple manometer or will give a difference in pressures for the differential manometer In equation form

Figure 27 Differential manometers

A differential manometer (Fig 27) determines the difference in pressures at two points A and B when the actual pressure at any point in the system cannot be determined

Application of the procedure outlined above to Fig 27a produces

For Fig 27b

If the pressures at A and B are expressed in length of the water column the above results can be written for Fig 27a

For Fig 27b

Example 24 In Fig 27a the liquids at A and B are water and the manometer liquid is oil S = 080 h1 = 300 mm h2 = 200 mm and h3 = 600 mm (a) Determine pA - pB in pacals (b) If pB = 50kPa and the barometer reading is 730 mm Hg find the pressure at A in meters of water absolute (a)

(b)

(a)

Micromanometers

For determining very small differences in pressure or determining large pressure differences precisely ndash several types of manometers

One type very accurately measures the differences in elevation of two menisci of a manometer

By means of small telescopes with horizontal cross hairs mounted along the tubes on a rack which is raised and lowered by a pinion and slow motion screw so that the cross hairs can be set accurately the difference in elevation of menisci (the gage difference) can be read with verniers

Figure 28 Micromanometer using two gage liquids

Fig 28 two gage liquids immiscible in each other and in the fluid to be measured a large gage difference R can be produced for a small pressure difference

The heavier gage liquid fills the lower U tube up to 0-0 then the lighter gage liquid is added to both sides filling the larger reservoirs up to 1-1

The gas or liquid in the system fills the space above 1-1 When the pressure at C is slightly greater than at D the menisci move as indicated in Fig 28

The volume of liquid displaced in each reservoir equals the displacement in the U tube

Manometer equation

γ1 γ2 and γ3 are the unit gravity force

Example 25 In the micromanometer of Fig 28 the pressure difference is wanted in pascals when air is in the system S2 = 10 S3 = 110 aA = 001 R = 5 mm t = 20oC and the barometer reads 760 mm Hg

The term γ1(aA) may be neglected Substituting into Eq (241) gives

Figure 29

Inclined manometer

The inclined manometer frequently used for measuring small differences in gas pressures

Adjusted to read zero by moving the inclined scale when A and B are open Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube it affords greater accuracy in reading the scale

Surface tension causes a capillary rise in small tubes If a U tube is used with a meniscus in each leg the surface-tension effects cancel

25 FORCES ON PLANE AREAS

In the preceding sections variations oF pressure throughout a fluid have been considered

The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force insofar as external reactions to the force system are concerned

In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration by formula and by use of the concept of the pressure prism

Horizontal Surfaces

A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure

The magnitude of the force acting on one side of the surface is

The elemental forces pdA acting on A are all parallel and in the same sense a scalar summation of all such elements yields the magnitude of the resultant force Its direction is normal to the surface and toward the surface if p is positive

Fig 210 arbitrary xy axes - to find the line of action of the resultant ie the point in the area where the moment of the distributed force about any axis through the point is zero

Then since the moment of the resultant must equal the moment of the distributed force system about any axis say the y axis

xrsquo ndash the distance from the y axis to the resultant

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

The bourdon gage (Fig 23) typical of the devices used for measuring gage pressures pressure element is a hollow curved flat metallic tube closed at one e

nd the other end is connected to the pressure to be measured when the internal pressure is increased the tube tends to straighten

pulling on a linkage to which is attached a pointer and causing the pointer to move

the dial reads zero when the inside and outside of the tube are at the same pressure regardless of its particular value

the gage measures pressure relative to the pressure of the medium surrounding the tube which is the local atmosphere

Figure 24 Units and scales for pressure measurement

Figure 24 the data and the relations of the common units of pressure measurement

Standard atmospheric pressure is the mean pressure at sea level 760 mm Hg

A pressure expressed in terms of the length of a column of liquid is equivalent to the force per unit area at the base of the column The relation for variation of pressure with altitude in a liquid p = γh [Eq (228)] (p is in pascals γ in newtons per cubic metre and h in metres)

With the unit gravity force of any liquid expressed as its relative density S times the unit gravity force of water

Water γ may be taken as 9806 Nm3

Figure 25 Mercury barometer

Local atmospheric pressure is measured by mercury barometer aneroid barometer (measures the difference in

pressure between the atmosphere and an evacuated box or tube in a manner analogous to the bourdon gage except that the tube is evacuated and sealed)

Mercury barometer glass tube closed at one end filled with mercury and inverted so that the open end is submerged in mercury

It has a scale the height of column R can be determined

The space above the mercury contains mercury vapor If the pressure of the mercury vapor hv is given in millimetres of mercury and R is measured in the same units the pressure at A may be expressed as (mm Hg)

Figure 24 a pressure may be located vertically on the chart which indicates its relation to absolute zero and to local atmospheric pressure

If the point is below the local-atmospheric-pressure line and is referred to gage datum it is called negative suction or vacuum

Example the pressure 460 mm Hg abs as at 1 with barometer reading 720 mm may be expressed as -260 mm Hg 260 mm Hg suction or 260 mm Hg vacuum

Note

Pabs = pbar + pgage

Absolute pressures P gage pressures p

Example 23 The rate of temperature change in the atmosphere with change in elevation is called its lapse rate The motion of a parcel of air depends on the density of the parcel relative to the density of the surrounding (ambient) air However as the parcel ascends through the atmosphere the air pressure decreases the parcel expands and its temperature decreases at a rate known as the dry adiabatic lapse rate A firm wants lo burn a large quantity of refuse It is estimated that the temperature of the smoke plume at 10 m above the ground will be 11oC greater than that of the ambient air For the following conditions determine what will happen to the smoke

(a) At standard atmospheric lapse rate β = -000651oC per meter and t0 =20oC

(b) At an inverted lapse rate β = 000365oC per meter

By combining Eqs (227) and (2214)

The relation between pressure and temperature for a mass of gas expanding without heat transfer (isentropic relation Sec 61) is

in which T1 is the initial smoke absolute temperature and P0 the initial absolute pressure k is the specific heat ratio 14 for air and other diatomic gases

Eliminating PP0 in the last two equations

Since the gas will rise until its temperature is equal to the ambient temperature

the last two equations may be solved for y Let

Then

For β = -000651oC per metre R = 287 mN(kgK) a = 2002 and y = 3201 m For the atmospheric temperature inversion β = -000365oC per metre a = -02721 and y = 8092 m

24 MANOMETERS

Manometers are devices that employ liquid columns for determining differences in pressure

Figure 26a the most elementary manometer ndash piezometer It measures the pressure in a liquid when it is above zero gage Glass tube is mounted vertically so that it is connected to the space

within the container Liquid rises in the tube until equilibrium is reached The pressure is then given by the vertical distance h from the menis

cus (liquid surface) to the point where the pressure is to be measured expressed in units of length of the liquid in the container

Piezometer would not work for negative gage pressures because air would flow into the container through the tube

Figure 26 Simple manometers

Figure 26b for small negative or positive gage pressures in a liquid With this arrangement the meniscus may come to rest below A as sho

wn Since the pressure at the meniscus is zero gage and since pressure decreases with elevation

units of length H2O Figure 26c for greater negative or positive gage pressures (a second li

quid of greater relative density employed) It must be immiscible in the first fluid which may now be a gas If the relative density of the fluid at A is S1 (based on water) and the re

lative density of the manometer liquid is S2 the equation for pressure at A

hA - the unknown pressure expressed in length units of water

h1 h2 - in length units

A general procedure in working all manometer problems

1 Start at one end (or any meniscus if the circuit is continuous) and write the pressure there in an appropriate unit (say pascals) or in an appropriate symbol if it is unknown

2 Add to this the change in pressure in the same unit from one meniscus to the next (plus if the next meniscus is lower minus if higher) (For pascals this is the product of the difference in elevation in metres and the unit gravity force of the fluid in newtons per cubic metre)

3 Continue until the other end of the gage (or the starting meniscus) is reached and equate the expression to the pressure at that point known or unknown

The expression will contain one unknown for a simple manometer or will give a difference in pressures for the differential manometer In equation form

Figure 27 Differential manometers

A differential manometer (Fig 27) determines the difference in pressures at two points A and B when the actual pressure at any point in the system cannot be determined

Application of the procedure outlined above to Fig 27a produces

For Fig 27b

If the pressures at A and B are expressed in length of the water column the above results can be written for Fig 27a

For Fig 27b

Example 24 In Fig 27a the liquids at A and B are water and the manometer liquid is oil S = 080 h1 = 300 mm h2 = 200 mm and h3 = 600 mm (a) Determine pA - pB in pacals (b) If pB = 50kPa and the barometer reading is 730 mm Hg find the pressure at A in meters of water absolute (a)

(b)

(a)

Micromanometers

For determining very small differences in pressure or determining large pressure differences precisely ndash several types of manometers

One type very accurately measures the differences in elevation of two menisci of a manometer

By means of small telescopes with horizontal cross hairs mounted along the tubes on a rack which is raised and lowered by a pinion and slow motion screw so that the cross hairs can be set accurately the difference in elevation of menisci (the gage difference) can be read with verniers

Figure 28 Micromanometer using two gage liquids

Fig 28 two gage liquids immiscible in each other and in the fluid to be measured a large gage difference R can be produced for a small pressure difference

The heavier gage liquid fills the lower U tube up to 0-0 then the lighter gage liquid is added to both sides filling the larger reservoirs up to 1-1

The gas or liquid in the system fills the space above 1-1 When the pressure at C is slightly greater than at D the menisci move as indicated in Fig 28

The volume of liquid displaced in each reservoir equals the displacement in the U tube

Manometer equation

γ1 γ2 and γ3 are the unit gravity force

Example 25 In the micromanometer of Fig 28 the pressure difference is wanted in pascals when air is in the system S2 = 10 S3 = 110 aA = 001 R = 5 mm t = 20oC and the barometer reads 760 mm Hg

The term γ1(aA) may be neglected Substituting into Eq (241) gives

Figure 29

Inclined manometer

The inclined manometer frequently used for measuring small differences in gas pressures

Adjusted to read zero by moving the inclined scale when A and B are open Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube it affords greater accuracy in reading the scale

Surface tension causes a capillary rise in small tubes If a U tube is used with a meniscus in each leg the surface-tension effects cancel

25 FORCES ON PLANE AREAS

In the preceding sections variations oF pressure throughout a fluid have been considered

The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force insofar as external reactions to the force system are concerned

In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration by formula and by use of the concept of the pressure prism

Horizontal Surfaces

A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure

The magnitude of the force acting on one side of the surface is

The elemental forces pdA acting on A are all parallel and in the same sense a scalar summation of all such elements yields the magnitude of the resultant force Its direction is normal to the surface and toward the surface if p is positive

Fig 210 arbitrary xy axes - to find the line of action of the resultant ie the point in the area where the moment of the distributed force about any axis through the point is zero

Then since the moment of the resultant must equal the moment of the distributed force system about any axis say the y axis

xrsquo ndash the distance from the y axis to the resultant

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Figure 24 Units and scales for pressure measurement

Figure 24 the data and the relations of the common units of pressure measurement

Standard atmospheric pressure is the mean pressure at sea level 760 mm Hg

A pressure expressed in terms of the length of a column of liquid is equivalent to the force per unit area at the base of the column The relation for variation of pressure with altitude in a liquid p = γh [Eq (228)] (p is in pascals γ in newtons per cubic metre and h in metres)

With the unit gravity force of any liquid expressed as its relative density S times the unit gravity force of water

Water γ may be taken as 9806 Nm3

Figure 25 Mercury barometer

Local atmospheric pressure is measured by mercury barometer aneroid barometer (measures the difference in

pressure between the atmosphere and an evacuated box or tube in a manner analogous to the bourdon gage except that the tube is evacuated and sealed)

Mercury barometer glass tube closed at one end filled with mercury and inverted so that the open end is submerged in mercury

It has a scale the height of column R can be determined

The space above the mercury contains mercury vapor If the pressure of the mercury vapor hv is given in millimetres of mercury and R is measured in the same units the pressure at A may be expressed as (mm Hg)

Figure 24 a pressure may be located vertically on the chart which indicates its relation to absolute zero and to local atmospheric pressure

If the point is below the local-atmospheric-pressure line and is referred to gage datum it is called negative suction or vacuum

Example the pressure 460 mm Hg abs as at 1 with barometer reading 720 mm may be expressed as -260 mm Hg 260 mm Hg suction or 260 mm Hg vacuum

Note

Pabs = pbar + pgage

Absolute pressures P gage pressures p

Example 23 The rate of temperature change in the atmosphere with change in elevation is called its lapse rate The motion of a parcel of air depends on the density of the parcel relative to the density of the surrounding (ambient) air However as the parcel ascends through the atmosphere the air pressure decreases the parcel expands and its temperature decreases at a rate known as the dry adiabatic lapse rate A firm wants lo burn a large quantity of refuse It is estimated that the temperature of the smoke plume at 10 m above the ground will be 11oC greater than that of the ambient air For the following conditions determine what will happen to the smoke

(a) At standard atmospheric lapse rate β = -000651oC per meter and t0 =20oC

(b) At an inverted lapse rate β = 000365oC per meter

By combining Eqs (227) and (2214)

The relation between pressure and temperature for a mass of gas expanding without heat transfer (isentropic relation Sec 61) is

in which T1 is the initial smoke absolute temperature and P0 the initial absolute pressure k is the specific heat ratio 14 for air and other diatomic gases

Eliminating PP0 in the last two equations

Since the gas will rise until its temperature is equal to the ambient temperature

the last two equations may be solved for y Let

Then

For β = -000651oC per metre R = 287 mN(kgK) a = 2002 and y = 3201 m For the atmospheric temperature inversion β = -000365oC per metre a = -02721 and y = 8092 m

24 MANOMETERS

Manometers are devices that employ liquid columns for determining differences in pressure

Figure 26a the most elementary manometer ndash piezometer It measures the pressure in a liquid when it is above zero gage Glass tube is mounted vertically so that it is connected to the space

within the container Liquid rises in the tube until equilibrium is reached The pressure is then given by the vertical distance h from the menis

cus (liquid surface) to the point where the pressure is to be measured expressed in units of length of the liquid in the container

Piezometer would not work for negative gage pressures because air would flow into the container through the tube

Figure 26 Simple manometers

Figure 26b for small negative or positive gage pressures in a liquid With this arrangement the meniscus may come to rest below A as sho

wn Since the pressure at the meniscus is zero gage and since pressure decreases with elevation

units of length H2O Figure 26c for greater negative or positive gage pressures (a second li

quid of greater relative density employed) It must be immiscible in the first fluid which may now be a gas If the relative density of the fluid at A is S1 (based on water) and the re

lative density of the manometer liquid is S2 the equation for pressure at A

hA - the unknown pressure expressed in length units of water

h1 h2 - in length units

A general procedure in working all manometer problems

1 Start at one end (or any meniscus if the circuit is continuous) and write the pressure there in an appropriate unit (say pascals) or in an appropriate symbol if it is unknown

2 Add to this the change in pressure in the same unit from one meniscus to the next (plus if the next meniscus is lower minus if higher) (For pascals this is the product of the difference in elevation in metres and the unit gravity force of the fluid in newtons per cubic metre)

3 Continue until the other end of the gage (or the starting meniscus) is reached and equate the expression to the pressure at that point known or unknown

The expression will contain one unknown for a simple manometer or will give a difference in pressures for the differential manometer In equation form

Figure 27 Differential manometers

A differential manometer (Fig 27) determines the difference in pressures at two points A and B when the actual pressure at any point in the system cannot be determined

Application of the procedure outlined above to Fig 27a produces

For Fig 27b

If the pressures at A and B are expressed in length of the water column the above results can be written for Fig 27a

For Fig 27b

Example 24 In Fig 27a the liquids at A and B are water and the manometer liquid is oil S = 080 h1 = 300 mm h2 = 200 mm and h3 = 600 mm (a) Determine pA - pB in pacals (b) If pB = 50kPa and the barometer reading is 730 mm Hg find the pressure at A in meters of water absolute (a)

(b)

(a)

Micromanometers

For determining very small differences in pressure or determining large pressure differences precisely ndash several types of manometers

One type very accurately measures the differences in elevation of two menisci of a manometer

By means of small telescopes with horizontal cross hairs mounted along the tubes on a rack which is raised and lowered by a pinion and slow motion screw so that the cross hairs can be set accurately the difference in elevation of menisci (the gage difference) can be read with verniers

Figure 28 Micromanometer using two gage liquids

Fig 28 two gage liquids immiscible in each other and in the fluid to be measured a large gage difference R can be produced for a small pressure difference

The heavier gage liquid fills the lower U tube up to 0-0 then the lighter gage liquid is added to both sides filling the larger reservoirs up to 1-1

The gas or liquid in the system fills the space above 1-1 When the pressure at C is slightly greater than at D the menisci move as indicated in Fig 28

The volume of liquid displaced in each reservoir equals the displacement in the U tube

Manometer equation

γ1 γ2 and γ3 are the unit gravity force

Example 25 In the micromanometer of Fig 28 the pressure difference is wanted in pascals when air is in the system S2 = 10 S3 = 110 aA = 001 R = 5 mm t = 20oC and the barometer reads 760 mm Hg

The term γ1(aA) may be neglected Substituting into Eq (241) gives

Figure 29

Inclined manometer

The inclined manometer frequently used for measuring small differences in gas pressures

Adjusted to read zero by moving the inclined scale when A and B are open Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube it affords greater accuracy in reading the scale

Surface tension causes a capillary rise in small tubes If a U tube is used with a meniscus in each leg the surface-tension effects cancel

25 FORCES ON PLANE AREAS

In the preceding sections variations oF pressure throughout a fluid have been considered

The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force insofar as external reactions to the force system are concerned

In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration by formula and by use of the concept of the pressure prism

Horizontal Surfaces

A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure

The magnitude of the force acting on one side of the surface is

The elemental forces pdA acting on A are all parallel and in the same sense a scalar summation of all such elements yields the magnitude of the resultant force Its direction is normal to the surface and toward the surface if p is positive

Fig 210 arbitrary xy axes - to find the line of action of the resultant ie the point in the area where the moment of the distributed force about any axis through the point is zero

Then since the moment of the resultant must equal the moment of the distributed force system about any axis say the y axis

xrsquo ndash the distance from the y axis to the resultant

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Figure 24 the data and the relations of the common units of pressure measurement

Standard atmospheric pressure is the mean pressure at sea level 760 mm Hg

A pressure expressed in terms of the length of a column of liquid is equivalent to the force per unit area at the base of the column The relation for variation of pressure with altitude in a liquid p = γh [Eq (228)] (p is in pascals γ in newtons per cubic metre and h in metres)

With the unit gravity force of any liquid expressed as its relative density S times the unit gravity force of water

Water γ may be taken as 9806 Nm3

Figure 25 Mercury barometer

Local atmospheric pressure is measured by mercury barometer aneroid barometer (measures the difference in

pressure between the atmosphere and an evacuated box or tube in a manner analogous to the bourdon gage except that the tube is evacuated and sealed)

Mercury barometer glass tube closed at one end filled with mercury and inverted so that the open end is submerged in mercury

It has a scale the height of column R can be determined

The space above the mercury contains mercury vapor If the pressure of the mercury vapor hv is given in millimetres of mercury and R is measured in the same units the pressure at A may be expressed as (mm Hg)

Figure 24 a pressure may be located vertically on the chart which indicates its relation to absolute zero and to local atmospheric pressure

If the point is below the local-atmospheric-pressure line and is referred to gage datum it is called negative suction or vacuum

Example the pressure 460 mm Hg abs as at 1 with barometer reading 720 mm may be expressed as -260 mm Hg 260 mm Hg suction or 260 mm Hg vacuum

Note

Pabs = pbar + pgage

Absolute pressures P gage pressures p

Example 23 The rate of temperature change in the atmosphere with change in elevation is called its lapse rate The motion of a parcel of air depends on the density of the parcel relative to the density of the surrounding (ambient) air However as the parcel ascends through the atmosphere the air pressure decreases the parcel expands and its temperature decreases at a rate known as the dry adiabatic lapse rate A firm wants lo burn a large quantity of refuse It is estimated that the temperature of the smoke plume at 10 m above the ground will be 11oC greater than that of the ambient air For the following conditions determine what will happen to the smoke

(a) At standard atmospheric lapse rate β = -000651oC per meter and t0 =20oC

(b) At an inverted lapse rate β = 000365oC per meter

By combining Eqs (227) and (2214)

The relation between pressure and temperature for a mass of gas expanding without heat transfer (isentropic relation Sec 61) is

in which T1 is the initial smoke absolute temperature and P0 the initial absolute pressure k is the specific heat ratio 14 for air and other diatomic gases

Eliminating PP0 in the last two equations

Since the gas will rise until its temperature is equal to the ambient temperature

the last two equations may be solved for y Let

Then

For β = -000651oC per metre R = 287 mN(kgK) a = 2002 and y = 3201 m For the atmospheric temperature inversion β = -000365oC per metre a = -02721 and y = 8092 m

24 MANOMETERS

Manometers are devices that employ liquid columns for determining differences in pressure

Figure 26a the most elementary manometer ndash piezometer It measures the pressure in a liquid when it is above zero gage Glass tube is mounted vertically so that it is connected to the space

within the container Liquid rises in the tube until equilibrium is reached The pressure is then given by the vertical distance h from the menis

cus (liquid surface) to the point where the pressure is to be measured expressed in units of length of the liquid in the container

Piezometer would not work for negative gage pressures because air would flow into the container through the tube

Figure 26 Simple manometers

Figure 26b for small negative or positive gage pressures in a liquid With this arrangement the meniscus may come to rest below A as sho

wn Since the pressure at the meniscus is zero gage and since pressure decreases with elevation

units of length H2O Figure 26c for greater negative or positive gage pressures (a second li

quid of greater relative density employed) It must be immiscible in the first fluid which may now be a gas If the relative density of the fluid at A is S1 (based on water) and the re

lative density of the manometer liquid is S2 the equation for pressure at A

hA - the unknown pressure expressed in length units of water

h1 h2 - in length units

A general procedure in working all manometer problems

1 Start at one end (or any meniscus if the circuit is continuous) and write the pressure there in an appropriate unit (say pascals) or in an appropriate symbol if it is unknown

2 Add to this the change in pressure in the same unit from one meniscus to the next (plus if the next meniscus is lower minus if higher) (For pascals this is the product of the difference in elevation in metres and the unit gravity force of the fluid in newtons per cubic metre)

3 Continue until the other end of the gage (or the starting meniscus) is reached and equate the expression to the pressure at that point known or unknown

The expression will contain one unknown for a simple manometer or will give a difference in pressures for the differential manometer In equation form

Figure 27 Differential manometers

A differential manometer (Fig 27) determines the difference in pressures at two points A and B when the actual pressure at any point in the system cannot be determined

Application of the procedure outlined above to Fig 27a produces

For Fig 27b

If the pressures at A and B are expressed in length of the water column the above results can be written for Fig 27a

For Fig 27b

Example 24 In Fig 27a the liquids at A and B are water and the manometer liquid is oil S = 080 h1 = 300 mm h2 = 200 mm and h3 = 600 mm (a) Determine pA - pB in pacals (b) If pB = 50kPa and the barometer reading is 730 mm Hg find the pressure at A in meters of water absolute (a)

(b)

(a)

Micromanometers

For determining very small differences in pressure or determining large pressure differences precisely ndash several types of manometers

One type very accurately measures the differences in elevation of two menisci of a manometer

By means of small telescopes with horizontal cross hairs mounted along the tubes on a rack which is raised and lowered by a pinion and slow motion screw so that the cross hairs can be set accurately the difference in elevation of menisci (the gage difference) can be read with verniers

Figure 28 Micromanometer using two gage liquids

Fig 28 two gage liquids immiscible in each other and in the fluid to be measured a large gage difference R can be produced for a small pressure difference

The heavier gage liquid fills the lower U tube up to 0-0 then the lighter gage liquid is added to both sides filling the larger reservoirs up to 1-1

The gas or liquid in the system fills the space above 1-1 When the pressure at C is slightly greater than at D the menisci move as indicated in Fig 28

The volume of liquid displaced in each reservoir equals the displacement in the U tube

Manometer equation

γ1 γ2 and γ3 are the unit gravity force

Example 25 In the micromanometer of Fig 28 the pressure difference is wanted in pascals when air is in the system S2 = 10 S3 = 110 aA = 001 R = 5 mm t = 20oC and the barometer reads 760 mm Hg

The term γ1(aA) may be neglected Substituting into Eq (241) gives

Figure 29

Inclined manometer

The inclined manometer frequently used for measuring small differences in gas pressures

Adjusted to read zero by moving the inclined scale when A and B are open Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube it affords greater accuracy in reading the scale

Surface tension causes a capillary rise in small tubes If a U tube is used with a meniscus in each leg the surface-tension effects cancel

25 FORCES ON PLANE AREAS

In the preceding sections variations oF pressure throughout a fluid have been considered

The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force insofar as external reactions to the force system are concerned

In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration by formula and by use of the concept of the pressure prism

Horizontal Surfaces

A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure

The magnitude of the force acting on one side of the surface is

The elemental forces pdA acting on A are all parallel and in the same sense a scalar summation of all such elements yields the magnitude of the resultant force Its direction is normal to the surface and toward the surface if p is positive

Fig 210 arbitrary xy axes - to find the line of action of the resultant ie the point in the area where the moment of the distributed force about any axis through the point is zero

Then since the moment of the resultant must equal the moment of the distributed force system about any axis say the y axis

xrsquo ndash the distance from the y axis to the resultant

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Figure 25 Mercury barometer

Local atmospheric pressure is measured by mercury barometer aneroid barometer (measures the difference in

pressure between the atmosphere and an evacuated box or tube in a manner analogous to the bourdon gage except that the tube is evacuated and sealed)

Mercury barometer glass tube closed at one end filled with mercury and inverted so that the open end is submerged in mercury

It has a scale the height of column R can be determined

The space above the mercury contains mercury vapor If the pressure of the mercury vapor hv is given in millimetres of mercury and R is measured in the same units the pressure at A may be expressed as (mm Hg)

Figure 24 a pressure may be located vertically on the chart which indicates its relation to absolute zero and to local atmospheric pressure

If the point is below the local-atmospheric-pressure line and is referred to gage datum it is called negative suction or vacuum

Example the pressure 460 mm Hg abs as at 1 with barometer reading 720 mm may be expressed as -260 mm Hg 260 mm Hg suction or 260 mm Hg vacuum

Note

Pabs = pbar + pgage

Absolute pressures P gage pressures p

Example 23 The rate of temperature change in the atmosphere with change in elevation is called its lapse rate The motion of a parcel of air depends on the density of the parcel relative to the density of the surrounding (ambient) air However as the parcel ascends through the atmosphere the air pressure decreases the parcel expands and its temperature decreases at a rate known as the dry adiabatic lapse rate A firm wants lo burn a large quantity of refuse It is estimated that the temperature of the smoke plume at 10 m above the ground will be 11oC greater than that of the ambient air For the following conditions determine what will happen to the smoke

(a) At standard atmospheric lapse rate β = -000651oC per meter and t0 =20oC

(b) At an inverted lapse rate β = 000365oC per meter

By combining Eqs (227) and (2214)

The relation between pressure and temperature for a mass of gas expanding without heat transfer (isentropic relation Sec 61) is

in which T1 is the initial smoke absolute temperature and P0 the initial absolute pressure k is the specific heat ratio 14 for air and other diatomic gases

Eliminating PP0 in the last two equations

Since the gas will rise until its temperature is equal to the ambient temperature

the last two equations may be solved for y Let

Then

For β = -000651oC per metre R = 287 mN(kgK) a = 2002 and y = 3201 m For the atmospheric temperature inversion β = -000365oC per metre a = -02721 and y = 8092 m

24 MANOMETERS

Manometers are devices that employ liquid columns for determining differences in pressure

Figure 26a the most elementary manometer ndash piezometer It measures the pressure in a liquid when it is above zero gage Glass tube is mounted vertically so that it is connected to the space

within the container Liquid rises in the tube until equilibrium is reached The pressure is then given by the vertical distance h from the menis

cus (liquid surface) to the point where the pressure is to be measured expressed in units of length of the liquid in the container

Piezometer would not work for negative gage pressures because air would flow into the container through the tube

Figure 26 Simple manometers

Figure 26b for small negative or positive gage pressures in a liquid With this arrangement the meniscus may come to rest below A as sho

wn Since the pressure at the meniscus is zero gage and since pressure decreases with elevation

units of length H2O Figure 26c for greater negative or positive gage pressures (a second li

quid of greater relative density employed) It must be immiscible in the first fluid which may now be a gas If the relative density of the fluid at A is S1 (based on water) and the re

lative density of the manometer liquid is S2 the equation for pressure at A

hA - the unknown pressure expressed in length units of water

h1 h2 - in length units

A general procedure in working all manometer problems

1 Start at one end (or any meniscus if the circuit is continuous) and write the pressure there in an appropriate unit (say pascals) or in an appropriate symbol if it is unknown

2 Add to this the change in pressure in the same unit from one meniscus to the next (plus if the next meniscus is lower minus if higher) (For pascals this is the product of the difference in elevation in metres and the unit gravity force of the fluid in newtons per cubic metre)

3 Continue until the other end of the gage (or the starting meniscus) is reached and equate the expression to the pressure at that point known or unknown

The expression will contain one unknown for a simple manometer or will give a difference in pressures for the differential manometer In equation form

Figure 27 Differential manometers

A differential manometer (Fig 27) determines the difference in pressures at two points A and B when the actual pressure at any point in the system cannot be determined

Application of the procedure outlined above to Fig 27a produces

For Fig 27b

If the pressures at A and B are expressed in length of the water column the above results can be written for Fig 27a

For Fig 27b

Example 24 In Fig 27a the liquids at A and B are water and the manometer liquid is oil S = 080 h1 = 300 mm h2 = 200 mm and h3 = 600 mm (a) Determine pA - pB in pacals (b) If pB = 50kPa and the barometer reading is 730 mm Hg find the pressure at A in meters of water absolute (a)

(b)

(a)

Micromanometers

For determining very small differences in pressure or determining large pressure differences precisely ndash several types of manometers

One type very accurately measures the differences in elevation of two menisci of a manometer

By means of small telescopes with horizontal cross hairs mounted along the tubes on a rack which is raised and lowered by a pinion and slow motion screw so that the cross hairs can be set accurately the difference in elevation of menisci (the gage difference) can be read with verniers

Figure 28 Micromanometer using two gage liquids

Fig 28 two gage liquids immiscible in each other and in the fluid to be measured a large gage difference R can be produced for a small pressure difference

The heavier gage liquid fills the lower U tube up to 0-0 then the lighter gage liquid is added to both sides filling the larger reservoirs up to 1-1

The gas or liquid in the system fills the space above 1-1 When the pressure at C is slightly greater than at D the menisci move as indicated in Fig 28

The volume of liquid displaced in each reservoir equals the displacement in the U tube

Manometer equation

γ1 γ2 and γ3 are the unit gravity force

Example 25 In the micromanometer of Fig 28 the pressure difference is wanted in pascals when air is in the system S2 = 10 S3 = 110 aA = 001 R = 5 mm t = 20oC and the barometer reads 760 mm Hg

The term γ1(aA) may be neglected Substituting into Eq (241) gives

Figure 29

Inclined manometer

The inclined manometer frequently used for measuring small differences in gas pressures

Adjusted to read zero by moving the inclined scale when A and B are open Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube it affords greater accuracy in reading the scale

Surface tension causes a capillary rise in small tubes If a U tube is used with a meniscus in each leg the surface-tension effects cancel

25 FORCES ON PLANE AREAS

In the preceding sections variations oF pressure throughout a fluid have been considered

The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force insofar as external reactions to the force system are concerned

In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration by formula and by use of the concept of the pressure prism

Horizontal Surfaces

A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure

The magnitude of the force acting on one side of the surface is

The elemental forces pdA acting on A are all parallel and in the same sense a scalar summation of all such elements yields the magnitude of the resultant force Its direction is normal to the surface and toward the surface if p is positive

Fig 210 arbitrary xy axes - to find the line of action of the resultant ie the point in the area where the moment of the distributed force about any axis through the point is zero

Then since the moment of the resultant must equal the moment of the distributed force system about any axis say the y axis

xrsquo ndash the distance from the y axis to the resultant

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Figure 24 a pressure may be located vertically on the chart which indicates its relation to absolute zero and to local atmospheric pressure

If the point is below the local-atmospheric-pressure line and is referred to gage datum it is called negative suction or vacuum

Example the pressure 460 mm Hg abs as at 1 with barometer reading 720 mm may be expressed as -260 mm Hg 260 mm Hg suction or 260 mm Hg vacuum

Note

Pabs = pbar + pgage

Absolute pressures P gage pressures p

Example 23 The rate of temperature change in the atmosphere with change in elevation is called its lapse rate The motion of a parcel of air depends on the density of the parcel relative to the density of the surrounding (ambient) air However as the parcel ascends through the atmosphere the air pressure decreases the parcel expands and its temperature decreases at a rate known as the dry adiabatic lapse rate A firm wants lo burn a large quantity of refuse It is estimated that the temperature of the smoke plume at 10 m above the ground will be 11oC greater than that of the ambient air For the following conditions determine what will happen to the smoke

(a) At standard atmospheric lapse rate β = -000651oC per meter and t0 =20oC

(b) At an inverted lapse rate β = 000365oC per meter

By combining Eqs (227) and (2214)

The relation between pressure and temperature for a mass of gas expanding without heat transfer (isentropic relation Sec 61) is

in which T1 is the initial smoke absolute temperature and P0 the initial absolute pressure k is the specific heat ratio 14 for air and other diatomic gases

Eliminating PP0 in the last two equations

Since the gas will rise until its temperature is equal to the ambient temperature

the last two equations may be solved for y Let

Then

For β = -000651oC per metre R = 287 mN(kgK) a = 2002 and y = 3201 m For the atmospheric temperature inversion β = -000365oC per metre a = -02721 and y = 8092 m

24 MANOMETERS

Manometers are devices that employ liquid columns for determining differences in pressure

Figure 26a the most elementary manometer ndash piezometer It measures the pressure in a liquid when it is above zero gage Glass tube is mounted vertically so that it is connected to the space

within the container Liquid rises in the tube until equilibrium is reached The pressure is then given by the vertical distance h from the menis

cus (liquid surface) to the point where the pressure is to be measured expressed in units of length of the liquid in the container

Piezometer would not work for negative gage pressures because air would flow into the container through the tube

Figure 26 Simple manometers

Figure 26b for small negative or positive gage pressures in a liquid With this arrangement the meniscus may come to rest below A as sho

wn Since the pressure at the meniscus is zero gage and since pressure decreases with elevation

units of length H2O Figure 26c for greater negative or positive gage pressures (a second li

quid of greater relative density employed) It must be immiscible in the first fluid which may now be a gas If the relative density of the fluid at A is S1 (based on water) and the re

lative density of the manometer liquid is S2 the equation for pressure at A

hA - the unknown pressure expressed in length units of water

h1 h2 - in length units

A general procedure in working all manometer problems

1 Start at one end (or any meniscus if the circuit is continuous) and write the pressure there in an appropriate unit (say pascals) or in an appropriate symbol if it is unknown

2 Add to this the change in pressure in the same unit from one meniscus to the next (plus if the next meniscus is lower minus if higher) (For pascals this is the product of the difference in elevation in metres and the unit gravity force of the fluid in newtons per cubic metre)

3 Continue until the other end of the gage (or the starting meniscus) is reached and equate the expression to the pressure at that point known or unknown

The expression will contain one unknown for a simple manometer or will give a difference in pressures for the differential manometer In equation form

Figure 27 Differential manometers

A differential manometer (Fig 27) determines the difference in pressures at two points A and B when the actual pressure at any point in the system cannot be determined

Application of the procedure outlined above to Fig 27a produces

For Fig 27b

If the pressures at A and B are expressed in length of the water column the above results can be written for Fig 27a

For Fig 27b

Example 24 In Fig 27a the liquids at A and B are water and the manometer liquid is oil S = 080 h1 = 300 mm h2 = 200 mm and h3 = 600 mm (a) Determine pA - pB in pacals (b) If pB = 50kPa and the barometer reading is 730 mm Hg find the pressure at A in meters of water absolute (a)

(b)

(a)

Micromanometers

For determining very small differences in pressure or determining large pressure differences precisely ndash several types of manometers

One type very accurately measures the differences in elevation of two menisci of a manometer

By means of small telescopes with horizontal cross hairs mounted along the tubes on a rack which is raised and lowered by a pinion and slow motion screw so that the cross hairs can be set accurately the difference in elevation of menisci (the gage difference) can be read with verniers

Figure 28 Micromanometer using two gage liquids

Fig 28 two gage liquids immiscible in each other and in the fluid to be measured a large gage difference R can be produced for a small pressure difference

The heavier gage liquid fills the lower U tube up to 0-0 then the lighter gage liquid is added to both sides filling the larger reservoirs up to 1-1

The gas or liquid in the system fills the space above 1-1 When the pressure at C is slightly greater than at D the menisci move as indicated in Fig 28

The volume of liquid displaced in each reservoir equals the displacement in the U tube

Manometer equation

γ1 γ2 and γ3 are the unit gravity force

Example 25 In the micromanometer of Fig 28 the pressure difference is wanted in pascals when air is in the system S2 = 10 S3 = 110 aA = 001 R = 5 mm t = 20oC and the barometer reads 760 mm Hg

The term γ1(aA) may be neglected Substituting into Eq (241) gives

Figure 29

Inclined manometer

The inclined manometer frequently used for measuring small differences in gas pressures

Adjusted to read zero by moving the inclined scale when A and B are open Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube it affords greater accuracy in reading the scale

Surface tension causes a capillary rise in small tubes If a U tube is used with a meniscus in each leg the surface-tension effects cancel

25 FORCES ON PLANE AREAS

In the preceding sections variations oF pressure throughout a fluid have been considered

The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force insofar as external reactions to the force system are concerned

In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration by formula and by use of the concept of the pressure prism

Horizontal Surfaces

A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure

The magnitude of the force acting on one side of the surface is

The elemental forces pdA acting on A are all parallel and in the same sense a scalar summation of all such elements yields the magnitude of the resultant force Its direction is normal to the surface and toward the surface if p is positive

Fig 210 arbitrary xy axes - to find the line of action of the resultant ie the point in the area where the moment of the distributed force about any axis through the point is zero

Then since the moment of the resultant must equal the moment of the distributed force system about any axis say the y axis

xrsquo ndash the distance from the y axis to the resultant

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Example 23 The rate of temperature change in the atmosphere with change in elevation is called its lapse rate The motion of a parcel of air depends on the density of the parcel relative to the density of the surrounding (ambient) air However as the parcel ascends through the atmosphere the air pressure decreases the parcel expands and its temperature decreases at a rate known as the dry adiabatic lapse rate A firm wants lo burn a large quantity of refuse It is estimated that the temperature of the smoke plume at 10 m above the ground will be 11oC greater than that of the ambient air For the following conditions determine what will happen to the smoke

(a) At standard atmospheric lapse rate β = -000651oC per meter and t0 =20oC

(b) At an inverted lapse rate β = 000365oC per meter

By combining Eqs (227) and (2214)

The relation between pressure and temperature for a mass of gas expanding without heat transfer (isentropic relation Sec 61) is

in which T1 is the initial smoke absolute temperature and P0 the initial absolute pressure k is the specific heat ratio 14 for air and other diatomic gases

Eliminating PP0 in the last two equations

Since the gas will rise until its temperature is equal to the ambient temperature

the last two equations may be solved for y Let

Then

For β = -000651oC per metre R = 287 mN(kgK) a = 2002 and y = 3201 m For the atmospheric temperature inversion β = -000365oC per metre a = -02721 and y = 8092 m

24 MANOMETERS

Manometers are devices that employ liquid columns for determining differences in pressure

Figure 26a the most elementary manometer ndash piezometer It measures the pressure in a liquid when it is above zero gage Glass tube is mounted vertically so that it is connected to the space

within the container Liquid rises in the tube until equilibrium is reached The pressure is then given by the vertical distance h from the menis

cus (liquid surface) to the point where the pressure is to be measured expressed in units of length of the liquid in the container

Piezometer would not work for negative gage pressures because air would flow into the container through the tube

Figure 26 Simple manometers

Figure 26b for small negative or positive gage pressures in a liquid With this arrangement the meniscus may come to rest below A as sho

wn Since the pressure at the meniscus is zero gage and since pressure decreases with elevation

units of length H2O Figure 26c for greater negative or positive gage pressures (a second li

quid of greater relative density employed) It must be immiscible in the first fluid which may now be a gas If the relative density of the fluid at A is S1 (based on water) and the re

lative density of the manometer liquid is S2 the equation for pressure at A

hA - the unknown pressure expressed in length units of water

h1 h2 - in length units

A general procedure in working all manometer problems

1 Start at one end (or any meniscus if the circuit is continuous) and write the pressure there in an appropriate unit (say pascals) or in an appropriate symbol if it is unknown

2 Add to this the change in pressure in the same unit from one meniscus to the next (plus if the next meniscus is lower minus if higher) (For pascals this is the product of the difference in elevation in metres and the unit gravity force of the fluid in newtons per cubic metre)

3 Continue until the other end of the gage (or the starting meniscus) is reached and equate the expression to the pressure at that point known or unknown

The expression will contain one unknown for a simple manometer or will give a difference in pressures for the differential manometer In equation form

Figure 27 Differential manometers

A differential manometer (Fig 27) determines the difference in pressures at two points A and B when the actual pressure at any point in the system cannot be determined

Application of the procedure outlined above to Fig 27a produces

For Fig 27b

If the pressures at A and B are expressed in length of the water column the above results can be written for Fig 27a

For Fig 27b

Example 24 In Fig 27a the liquids at A and B are water and the manometer liquid is oil S = 080 h1 = 300 mm h2 = 200 mm and h3 = 600 mm (a) Determine pA - pB in pacals (b) If pB = 50kPa and the barometer reading is 730 mm Hg find the pressure at A in meters of water absolute (a)

(b)

(a)

Micromanometers

For determining very small differences in pressure or determining large pressure differences precisely ndash several types of manometers

One type very accurately measures the differences in elevation of two menisci of a manometer

By means of small telescopes with horizontal cross hairs mounted along the tubes on a rack which is raised and lowered by a pinion and slow motion screw so that the cross hairs can be set accurately the difference in elevation of menisci (the gage difference) can be read with verniers

Figure 28 Micromanometer using two gage liquids

Fig 28 two gage liquids immiscible in each other and in the fluid to be measured a large gage difference R can be produced for a small pressure difference

The heavier gage liquid fills the lower U tube up to 0-0 then the lighter gage liquid is added to both sides filling the larger reservoirs up to 1-1

The gas or liquid in the system fills the space above 1-1 When the pressure at C is slightly greater than at D the menisci move as indicated in Fig 28

The volume of liquid displaced in each reservoir equals the displacement in the U tube

Manometer equation

γ1 γ2 and γ3 are the unit gravity force

Example 25 In the micromanometer of Fig 28 the pressure difference is wanted in pascals when air is in the system S2 = 10 S3 = 110 aA = 001 R = 5 mm t = 20oC and the barometer reads 760 mm Hg

The term γ1(aA) may be neglected Substituting into Eq (241) gives

Figure 29

Inclined manometer

The inclined manometer frequently used for measuring small differences in gas pressures

Adjusted to read zero by moving the inclined scale when A and B are open Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube it affords greater accuracy in reading the scale

Surface tension causes a capillary rise in small tubes If a U tube is used with a meniscus in each leg the surface-tension effects cancel

25 FORCES ON PLANE AREAS

In the preceding sections variations oF pressure throughout a fluid have been considered

The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force insofar as external reactions to the force system are concerned

In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration by formula and by use of the concept of the pressure prism

Horizontal Surfaces

A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure

The magnitude of the force acting on one side of the surface is

The elemental forces pdA acting on A are all parallel and in the same sense a scalar summation of all such elements yields the magnitude of the resultant force Its direction is normal to the surface and toward the surface if p is positive

Fig 210 arbitrary xy axes - to find the line of action of the resultant ie the point in the area where the moment of the distributed force about any axis through the point is zero

Then since the moment of the resultant must equal the moment of the distributed force system about any axis say the y axis

xrsquo ndash the distance from the y axis to the resultant

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

By combining Eqs (227) and (2214)

The relation between pressure and temperature for a mass of gas expanding without heat transfer (isentropic relation Sec 61) is

in which T1 is the initial smoke absolute temperature and P0 the initial absolute pressure k is the specific heat ratio 14 for air and other diatomic gases

Eliminating PP0 in the last two equations

Since the gas will rise until its temperature is equal to the ambient temperature

the last two equations may be solved for y Let

Then

For β = -000651oC per metre R = 287 mN(kgK) a = 2002 and y = 3201 m For the atmospheric temperature inversion β = -000365oC per metre a = -02721 and y = 8092 m

24 MANOMETERS

Manometers are devices that employ liquid columns for determining differences in pressure

Figure 26a the most elementary manometer ndash piezometer It measures the pressure in a liquid when it is above zero gage Glass tube is mounted vertically so that it is connected to the space

within the container Liquid rises in the tube until equilibrium is reached The pressure is then given by the vertical distance h from the menis

cus (liquid surface) to the point where the pressure is to be measured expressed in units of length of the liquid in the container

Piezometer would not work for negative gage pressures because air would flow into the container through the tube

Figure 26 Simple manometers

Figure 26b for small negative or positive gage pressures in a liquid With this arrangement the meniscus may come to rest below A as sho

wn Since the pressure at the meniscus is zero gage and since pressure decreases with elevation

units of length H2O Figure 26c for greater negative or positive gage pressures (a second li

quid of greater relative density employed) It must be immiscible in the first fluid which may now be a gas If the relative density of the fluid at A is S1 (based on water) and the re

lative density of the manometer liquid is S2 the equation for pressure at A

hA - the unknown pressure expressed in length units of water

h1 h2 - in length units

A general procedure in working all manometer problems

1 Start at one end (or any meniscus if the circuit is continuous) and write the pressure there in an appropriate unit (say pascals) or in an appropriate symbol if it is unknown

2 Add to this the change in pressure in the same unit from one meniscus to the next (plus if the next meniscus is lower minus if higher) (For pascals this is the product of the difference in elevation in metres and the unit gravity force of the fluid in newtons per cubic metre)

3 Continue until the other end of the gage (or the starting meniscus) is reached and equate the expression to the pressure at that point known or unknown

The expression will contain one unknown for a simple manometer or will give a difference in pressures for the differential manometer In equation form

Figure 27 Differential manometers

A differential manometer (Fig 27) determines the difference in pressures at two points A and B when the actual pressure at any point in the system cannot be determined

Application of the procedure outlined above to Fig 27a produces

For Fig 27b

If the pressures at A and B are expressed in length of the water column the above results can be written for Fig 27a

For Fig 27b

Example 24 In Fig 27a the liquids at A and B are water and the manometer liquid is oil S = 080 h1 = 300 mm h2 = 200 mm and h3 = 600 mm (a) Determine pA - pB in pacals (b) If pB = 50kPa and the barometer reading is 730 mm Hg find the pressure at A in meters of water absolute (a)

(b)

(a)

Micromanometers

For determining very small differences in pressure or determining large pressure differences precisely ndash several types of manometers

One type very accurately measures the differences in elevation of two menisci of a manometer

By means of small telescopes with horizontal cross hairs mounted along the tubes on a rack which is raised and lowered by a pinion and slow motion screw so that the cross hairs can be set accurately the difference in elevation of menisci (the gage difference) can be read with verniers

Figure 28 Micromanometer using two gage liquids

Fig 28 two gage liquids immiscible in each other and in the fluid to be measured a large gage difference R can be produced for a small pressure difference

The heavier gage liquid fills the lower U tube up to 0-0 then the lighter gage liquid is added to both sides filling the larger reservoirs up to 1-1

The gas or liquid in the system fills the space above 1-1 When the pressure at C is slightly greater than at D the menisci move as indicated in Fig 28

The volume of liquid displaced in each reservoir equals the displacement in the U tube

Manometer equation

γ1 γ2 and γ3 are the unit gravity force

Example 25 In the micromanometer of Fig 28 the pressure difference is wanted in pascals when air is in the system S2 = 10 S3 = 110 aA = 001 R = 5 mm t = 20oC and the barometer reads 760 mm Hg

The term γ1(aA) may be neglected Substituting into Eq (241) gives

Figure 29

Inclined manometer

The inclined manometer frequently used for measuring small differences in gas pressures

Adjusted to read zero by moving the inclined scale when A and B are open Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube it affords greater accuracy in reading the scale

Surface tension causes a capillary rise in small tubes If a U tube is used with a meniscus in each leg the surface-tension effects cancel

25 FORCES ON PLANE AREAS

In the preceding sections variations oF pressure throughout a fluid have been considered

The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force insofar as external reactions to the force system are concerned

In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration by formula and by use of the concept of the pressure prism

Horizontal Surfaces

A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure

The magnitude of the force acting on one side of the surface is

The elemental forces pdA acting on A are all parallel and in the same sense a scalar summation of all such elements yields the magnitude of the resultant force Its direction is normal to the surface and toward the surface if p is positive

Fig 210 arbitrary xy axes - to find the line of action of the resultant ie the point in the area where the moment of the distributed force about any axis through the point is zero

Then since the moment of the resultant must equal the moment of the distributed force system about any axis say the y axis

xrsquo ndash the distance from the y axis to the resultant

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

24 MANOMETERS

Manometers are devices that employ liquid columns for determining differences in pressure

Figure 26a the most elementary manometer ndash piezometer It measures the pressure in a liquid when it is above zero gage Glass tube is mounted vertically so that it is connected to the space

within the container Liquid rises in the tube until equilibrium is reached The pressure is then given by the vertical distance h from the menis

cus (liquid surface) to the point where the pressure is to be measured expressed in units of length of the liquid in the container

Piezometer would not work for negative gage pressures because air would flow into the container through the tube

Figure 26 Simple manometers

Figure 26b for small negative or positive gage pressures in a liquid With this arrangement the meniscus may come to rest below A as sho

wn Since the pressure at the meniscus is zero gage and since pressure decreases with elevation

units of length H2O Figure 26c for greater negative or positive gage pressures (a second li

quid of greater relative density employed) It must be immiscible in the first fluid which may now be a gas If the relative density of the fluid at A is S1 (based on water) and the re

lative density of the manometer liquid is S2 the equation for pressure at A

hA - the unknown pressure expressed in length units of water

h1 h2 - in length units

A general procedure in working all manometer problems

1 Start at one end (or any meniscus if the circuit is continuous) and write the pressure there in an appropriate unit (say pascals) or in an appropriate symbol if it is unknown

2 Add to this the change in pressure in the same unit from one meniscus to the next (plus if the next meniscus is lower minus if higher) (For pascals this is the product of the difference in elevation in metres and the unit gravity force of the fluid in newtons per cubic metre)

3 Continue until the other end of the gage (or the starting meniscus) is reached and equate the expression to the pressure at that point known or unknown

The expression will contain one unknown for a simple manometer or will give a difference in pressures for the differential manometer In equation form

Figure 27 Differential manometers

A differential manometer (Fig 27) determines the difference in pressures at two points A and B when the actual pressure at any point in the system cannot be determined

Application of the procedure outlined above to Fig 27a produces

For Fig 27b

If the pressures at A and B are expressed in length of the water column the above results can be written for Fig 27a

For Fig 27b

Example 24 In Fig 27a the liquids at A and B are water and the manometer liquid is oil S = 080 h1 = 300 mm h2 = 200 mm and h3 = 600 mm (a) Determine pA - pB in pacals (b) If pB = 50kPa and the barometer reading is 730 mm Hg find the pressure at A in meters of water absolute (a)

(b)

(a)

Micromanometers

For determining very small differences in pressure or determining large pressure differences precisely ndash several types of manometers

One type very accurately measures the differences in elevation of two menisci of a manometer

By means of small telescopes with horizontal cross hairs mounted along the tubes on a rack which is raised and lowered by a pinion and slow motion screw so that the cross hairs can be set accurately the difference in elevation of menisci (the gage difference) can be read with verniers

Figure 28 Micromanometer using two gage liquids

Fig 28 two gage liquids immiscible in each other and in the fluid to be measured a large gage difference R can be produced for a small pressure difference

The heavier gage liquid fills the lower U tube up to 0-0 then the lighter gage liquid is added to both sides filling the larger reservoirs up to 1-1

The gas or liquid in the system fills the space above 1-1 When the pressure at C is slightly greater than at D the menisci move as indicated in Fig 28

The volume of liquid displaced in each reservoir equals the displacement in the U tube

Manometer equation

γ1 γ2 and γ3 are the unit gravity force

Example 25 In the micromanometer of Fig 28 the pressure difference is wanted in pascals when air is in the system S2 = 10 S3 = 110 aA = 001 R = 5 mm t = 20oC and the barometer reads 760 mm Hg

The term γ1(aA) may be neglected Substituting into Eq (241) gives

Figure 29

Inclined manometer

The inclined manometer frequently used for measuring small differences in gas pressures

Adjusted to read zero by moving the inclined scale when A and B are open Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube it affords greater accuracy in reading the scale

Surface tension causes a capillary rise in small tubes If a U tube is used with a meniscus in each leg the surface-tension effects cancel

25 FORCES ON PLANE AREAS

In the preceding sections variations oF pressure throughout a fluid have been considered

The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force insofar as external reactions to the force system are concerned

In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration by formula and by use of the concept of the pressure prism

Horizontal Surfaces

A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure

The magnitude of the force acting on one side of the surface is

The elemental forces pdA acting on A are all parallel and in the same sense a scalar summation of all such elements yields the magnitude of the resultant force Its direction is normal to the surface and toward the surface if p is positive

Fig 210 arbitrary xy axes - to find the line of action of the resultant ie the point in the area where the moment of the distributed force about any axis through the point is zero

Then since the moment of the resultant must equal the moment of the distributed force system about any axis say the y axis

xrsquo ndash the distance from the y axis to the resultant

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Figure 26 Simple manometers

Figure 26b for small negative or positive gage pressures in a liquid With this arrangement the meniscus may come to rest below A as sho

wn Since the pressure at the meniscus is zero gage and since pressure decreases with elevation

units of length H2O Figure 26c for greater negative or positive gage pressures (a second li

quid of greater relative density employed) It must be immiscible in the first fluid which may now be a gas If the relative density of the fluid at A is S1 (based on water) and the re

lative density of the manometer liquid is S2 the equation for pressure at A

hA - the unknown pressure expressed in length units of water

h1 h2 - in length units

A general procedure in working all manometer problems

1 Start at one end (or any meniscus if the circuit is continuous) and write the pressure there in an appropriate unit (say pascals) or in an appropriate symbol if it is unknown

2 Add to this the change in pressure in the same unit from one meniscus to the next (plus if the next meniscus is lower minus if higher) (For pascals this is the product of the difference in elevation in metres and the unit gravity force of the fluid in newtons per cubic metre)

3 Continue until the other end of the gage (or the starting meniscus) is reached and equate the expression to the pressure at that point known or unknown

The expression will contain one unknown for a simple manometer or will give a difference in pressures for the differential manometer In equation form

Figure 27 Differential manometers

A differential manometer (Fig 27) determines the difference in pressures at two points A and B when the actual pressure at any point in the system cannot be determined

Application of the procedure outlined above to Fig 27a produces

For Fig 27b

If the pressures at A and B are expressed in length of the water column the above results can be written for Fig 27a

For Fig 27b

Example 24 In Fig 27a the liquids at A and B are water and the manometer liquid is oil S = 080 h1 = 300 mm h2 = 200 mm and h3 = 600 mm (a) Determine pA - pB in pacals (b) If pB = 50kPa and the barometer reading is 730 mm Hg find the pressure at A in meters of water absolute (a)

(b)

(a)

Micromanometers

For determining very small differences in pressure or determining large pressure differences precisely ndash several types of manometers

One type very accurately measures the differences in elevation of two menisci of a manometer

By means of small telescopes with horizontal cross hairs mounted along the tubes on a rack which is raised and lowered by a pinion and slow motion screw so that the cross hairs can be set accurately the difference in elevation of menisci (the gage difference) can be read with verniers

Figure 28 Micromanometer using two gage liquids

Fig 28 two gage liquids immiscible in each other and in the fluid to be measured a large gage difference R can be produced for a small pressure difference

The heavier gage liquid fills the lower U tube up to 0-0 then the lighter gage liquid is added to both sides filling the larger reservoirs up to 1-1

The gas or liquid in the system fills the space above 1-1 When the pressure at C is slightly greater than at D the menisci move as indicated in Fig 28

The volume of liquid displaced in each reservoir equals the displacement in the U tube

Manometer equation

γ1 γ2 and γ3 are the unit gravity force

Example 25 In the micromanometer of Fig 28 the pressure difference is wanted in pascals when air is in the system S2 = 10 S3 = 110 aA = 001 R = 5 mm t = 20oC and the barometer reads 760 mm Hg

The term γ1(aA) may be neglected Substituting into Eq (241) gives

Figure 29

Inclined manometer

The inclined manometer frequently used for measuring small differences in gas pressures

Adjusted to read zero by moving the inclined scale when A and B are open Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube it affords greater accuracy in reading the scale

Surface tension causes a capillary rise in small tubes If a U tube is used with a meniscus in each leg the surface-tension effects cancel

25 FORCES ON PLANE AREAS

In the preceding sections variations oF pressure throughout a fluid have been considered

The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force insofar as external reactions to the force system are concerned

In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration by formula and by use of the concept of the pressure prism

Horizontal Surfaces

A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure

The magnitude of the force acting on one side of the surface is

The elemental forces pdA acting on A are all parallel and in the same sense a scalar summation of all such elements yields the magnitude of the resultant force Its direction is normal to the surface and toward the surface if p is positive

Fig 210 arbitrary xy axes - to find the line of action of the resultant ie the point in the area where the moment of the distributed force about any axis through the point is zero

Then since the moment of the resultant must equal the moment of the distributed force system about any axis say the y axis

xrsquo ndash the distance from the y axis to the resultant

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Figure 26b for small negative or positive gage pressures in a liquid With this arrangement the meniscus may come to rest below A as sho

wn Since the pressure at the meniscus is zero gage and since pressure decreases with elevation

units of length H2O Figure 26c for greater negative or positive gage pressures (a second li

quid of greater relative density employed) It must be immiscible in the first fluid which may now be a gas If the relative density of the fluid at A is S1 (based on water) and the re

lative density of the manometer liquid is S2 the equation for pressure at A

hA - the unknown pressure expressed in length units of water

h1 h2 - in length units

A general procedure in working all manometer problems

1 Start at one end (or any meniscus if the circuit is continuous) and write the pressure there in an appropriate unit (say pascals) or in an appropriate symbol if it is unknown

2 Add to this the change in pressure in the same unit from one meniscus to the next (plus if the next meniscus is lower minus if higher) (For pascals this is the product of the difference in elevation in metres and the unit gravity force of the fluid in newtons per cubic metre)

3 Continue until the other end of the gage (or the starting meniscus) is reached and equate the expression to the pressure at that point known or unknown

The expression will contain one unknown for a simple manometer or will give a difference in pressures for the differential manometer In equation form

Figure 27 Differential manometers

A differential manometer (Fig 27) determines the difference in pressures at two points A and B when the actual pressure at any point in the system cannot be determined

Application of the procedure outlined above to Fig 27a produces

For Fig 27b

If the pressures at A and B are expressed in length of the water column the above results can be written for Fig 27a

For Fig 27b

Example 24 In Fig 27a the liquids at A and B are water and the manometer liquid is oil S = 080 h1 = 300 mm h2 = 200 mm and h3 = 600 mm (a) Determine pA - pB in pacals (b) If pB = 50kPa and the barometer reading is 730 mm Hg find the pressure at A in meters of water absolute (a)

(b)

(a)

Micromanometers

For determining very small differences in pressure or determining large pressure differences precisely ndash several types of manometers

One type very accurately measures the differences in elevation of two menisci of a manometer

By means of small telescopes with horizontal cross hairs mounted along the tubes on a rack which is raised and lowered by a pinion and slow motion screw so that the cross hairs can be set accurately the difference in elevation of menisci (the gage difference) can be read with verniers

Figure 28 Micromanometer using two gage liquids

Fig 28 two gage liquids immiscible in each other and in the fluid to be measured a large gage difference R can be produced for a small pressure difference

The heavier gage liquid fills the lower U tube up to 0-0 then the lighter gage liquid is added to both sides filling the larger reservoirs up to 1-1

The gas or liquid in the system fills the space above 1-1 When the pressure at C is slightly greater than at D the menisci move as indicated in Fig 28

The volume of liquid displaced in each reservoir equals the displacement in the U tube

Manometer equation

γ1 γ2 and γ3 are the unit gravity force

Example 25 In the micromanometer of Fig 28 the pressure difference is wanted in pascals when air is in the system S2 = 10 S3 = 110 aA = 001 R = 5 mm t = 20oC and the barometer reads 760 mm Hg

The term γ1(aA) may be neglected Substituting into Eq (241) gives

Figure 29

Inclined manometer

The inclined manometer frequently used for measuring small differences in gas pressures

Adjusted to read zero by moving the inclined scale when A and B are open Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube it affords greater accuracy in reading the scale

Surface tension causes a capillary rise in small tubes If a U tube is used with a meniscus in each leg the surface-tension effects cancel

25 FORCES ON PLANE AREAS

In the preceding sections variations oF pressure throughout a fluid have been considered

The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force insofar as external reactions to the force system are concerned

In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration by formula and by use of the concept of the pressure prism

Horizontal Surfaces

A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure

The magnitude of the force acting on one side of the surface is

The elemental forces pdA acting on A are all parallel and in the same sense a scalar summation of all such elements yields the magnitude of the resultant force Its direction is normal to the surface and toward the surface if p is positive

Fig 210 arbitrary xy axes - to find the line of action of the resultant ie the point in the area where the moment of the distributed force about any axis through the point is zero

Then since the moment of the resultant must equal the moment of the distributed force system about any axis say the y axis

xrsquo ndash the distance from the y axis to the resultant

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

A general procedure in working all manometer problems

1 Start at one end (or any meniscus if the circuit is continuous) and write the pressure there in an appropriate unit (say pascals) or in an appropriate symbol if it is unknown

2 Add to this the change in pressure in the same unit from one meniscus to the next (plus if the next meniscus is lower minus if higher) (For pascals this is the product of the difference in elevation in metres and the unit gravity force of the fluid in newtons per cubic metre)

3 Continue until the other end of the gage (or the starting meniscus) is reached and equate the expression to the pressure at that point known or unknown

The expression will contain one unknown for a simple manometer or will give a difference in pressures for the differential manometer In equation form

Figure 27 Differential manometers

A differential manometer (Fig 27) determines the difference in pressures at two points A and B when the actual pressure at any point in the system cannot be determined

Application of the procedure outlined above to Fig 27a produces

For Fig 27b

If the pressures at A and B are expressed in length of the water column the above results can be written for Fig 27a

For Fig 27b

Example 24 In Fig 27a the liquids at A and B are water and the manometer liquid is oil S = 080 h1 = 300 mm h2 = 200 mm and h3 = 600 mm (a) Determine pA - pB in pacals (b) If pB = 50kPa and the barometer reading is 730 mm Hg find the pressure at A in meters of water absolute (a)

(b)

(a)

Micromanometers

For determining very small differences in pressure or determining large pressure differences precisely ndash several types of manometers

One type very accurately measures the differences in elevation of two menisci of a manometer

By means of small telescopes with horizontal cross hairs mounted along the tubes on a rack which is raised and lowered by a pinion and slow motion screw so that the cross hairs can be set accurately the difference in elevation of menisci (the gage difference) can be read with verniers

Figure 28 Micromanometer using two gage liquids

Fig 28 two gage liquids immiscible in each other and in the fluid to be measured a large gage difference R can be produced for a small pressure difference

The heavier gage liquid fills the lower U tube up to 0-0 then the lighter gage liquid is added to both sides filling the larger reservoirs up to 1-1

The gas or liquid in the system fills the space above 1-1 When the pressure at C is slightly greater than at D the menisci move as indicated in Fig 28

The volume of liquid displaced in each reservoir equals the displacement in the U tube

Manometer equation

γ1 γ2 and γ3 are the unit gravity force

Example 25 In the micromanometer of Fig 28 the pressure difference is wanted in pascals when air is in the system S2 = 10 S3 = 110 aA = 001 R = 5 mm t = 20oC and the barometer reads 760 mm Hg

The term γ1(aA) may be neglected Substituting into Eq (241) gives

Figure 29

Inclined manometer

The inclined manometer frequently used for measuring small differences in gas pressures

Adjusted to read zero by moving the inclined scale when A and B are open Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube it affords greater accuracy in reading the scale

Surface tension causes a capillary rise in small tubes If a U tube is used with a meniscus in each leg the surface-tension effects cancel

25 FORCES ON PLANE AREAS

In the preceding sections variations oF pressure throughout a fluid have been considered

The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force insofar as external reactions to the force system are concerned

In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration by formula and by use of the concept of the pressure prism

Horizontal Surfaces

A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure

The magnitude of the force acting on one side of the surface is

The elemental forces pdA acting on A are all parallel and in the same sense a scalar summation of all such elements yields the magnitude of the resultant force Its direction is normal to the surface and toward the surface if p is positive

Fig 210 arbitrary xy axes - to find the line of action of the resultant ie the point in the area where the moment of the distributed force about any axis through the point is zero

Then since the moment of the resultant must equal the moment of the distributed force system about any axis say the y axis

xrsquo ndash the distance from the y axis to the resultant

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Figure 27 Differential manometers

A differential manometer (Fig 27) determines the difference in pressures at two points A and B when the actual pressure at any point in the system cannot be determined

Application of the procedure outlined above to Fig 27a produces

For Fig 27b

If the pressures at A and B are expressed in length of the water column the above results can be written for Fig 27a

For Fig 27b

Example 24 In Fig 27a the liquids at A and B are water and the manometer liquid is oil S = 080 h1 = 300 mm h2 = 200 mm and h3 = 600 mm (a) Determine pA - pB in pacals (b) If pB = 50kPa and the barometer reading is 730 mm Hg find the pressure at A in meters of water absolute (a)

(b)

(a)

Micromanometers

For determining very small differences in pressure or determining large pressure differences precisely ndash several types of manometers

One type very accurately measures the differences in elevation of two menisci of a manometer

By means of small telescopes with horizontal cross hairs mounted along the tubes on a rack which is raised and lowered by a pinion and slow motion screw so that the cross hairs can be set accurately the difference in elevation of menisci (the gage difference) can be read with verniers

Figure 28 Micromanometer using two gage liquids

Fig 28 two gage liquids immiscible in each other and in the fluid to be measured a large gage difference R can be produced for a small pressure difference

The heavier gage liquid fills the lower U tube up to 0-0 then the lighter gage liquid is added to both sides filling the larger reservoirs up to 1-1

The gas or liquid in the system fills the space above 1-1 When the pressure at C is slightly greater than at D the menisci move as indicated in Fig 28

The volume of liquid displaced in each reservoir equals the displacement in the U tube

Manometer equation

γ1 γ2 and γ3 are the unit gravity force

Example 25 In the micromanometer of Fig 28 the pressure difference is wanted in pascals when air is in the system S2 = 10 S3 = 110 aA = 001 R = 5 mm t = 20oC and the barometer reads 760 mm Hg

The term γ1(aA) may be neglected Substituting into Eq (241) gives

Figure 29

Inclined manometer

The inclined manometer frequently used for measuring small differences in gas pressures

Adjusted to read zero by moving the inclined scale when A and B are open Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube it affords greater accuracy in reading the scale

Surface tension causes a capillary rise in small tubes If a U tube is used with a meniscus in each leg the surface-tension effects cancel

25 FORCES ON PLANE AREAS

In the preceding sections variations oF pressure throughout a fluid have been considered

The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force insofar as external reactions to the force system are concerned

In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration by formula and by use of the concept of the pressure prism

Horizontal Surfaces

A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure

The magnitude of the force acting on one side of the surface is

The elemental forces pdA acting on A are all parallel and in the same sense a scalar summation of all such elements yields the magnitude of the resultant force Its direction is normal to the surface and toward the surface if p is positive

Fig 210 arbitrary xy axes - to find the line of action of the resultant ie the point in the area where the moment of the distributed force about any axis through the point is zero

Then since the moment of the resultant must equal the moment of the distributed force system about any axis say the y axis

xrsquo ndash the distance from the y axis to the resultant

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

A differential manometer (Fig 27) determines the difference in pressures at two points A and B when the actual pressure at any point in the system cannot be determined

Application of the procedure outlined above to Fig 27a produces

For Fig 27b

If the pressures at A and B are expressed in length of the water column the above results can be written for Fig 27a

For Fig 27b

Example 24 In Fig 27a the liquids at A and B are water and the manometer liquid is oil S = 080 h1 = 300 mm h2 = 200 mm and h3 = 600 mm (a) Determine pA - pB in pacals (b) If pB = 50kPa and the barometer reading is 730 mm Hg find the pressure at A in meters of water absolute (a)

(b)

(a)

Micromanometers

For determining very small differences in pressure or determining large pressure differences precisely ndash several types of manometers

One type very accurately measures the differences in elevation of two menisci of a manometer

By means of small telescopes with horizontal cross hairs mounted along the tubes on a rack which is raised and lowered by a pinion and slow motion screw so that the cross hairs can be set accurately the difference in elevation of menisci (the gage difference) can be read with verniers

Figure 28 Micromanometer using two gage liquids

Fig 28 two gage liquids immiscible in each other and in the fluid to be measured a large gage difference R can be produced for a small pressure difference

The heavier gage liquid fills the lower U tube up to 0-0 then the lighter gage liquid is added to both sides filling the larger reservoirs up to 1-1

The gas or liquid in the system fills the space above 1-1 When the pressure at C is slightly greater than at D the menisci move as indicated in Fig 28

The volume of liquid displaced in each reservoir equals the displacement in the U tube

Manometer equation

γ1 γ2 and γ3 are the unit gravity force

Example 25 In the micromanometer of Fig 28 the pressure difference is wanted in pascals when air is in the system S2 = 10 S3 = 110 aA = 001 R = 5 mm t = 20oC and the barometer reads 760 mm Hg

The term γ1(aA) may be neglected Substituting into Eq (241) gives

Figure 29

Inclined manometer

The inclined manometer frequently used for measuring small differences in gas pressures

Adjusted to read zero by moving the inclined scale when A and B are open Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube it affords greater accuracy in reading the scale

Surface tension causes a capillary rise in small tubes If a U tube is used with a meniscus in each leg the surface-tension effects cancel

25 FORCES ON PLANE AREAS

In the preceding sections variations oF pressure throughout a fluid have been considered

The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force insofar as external reactions to the force system are concerned

In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration by formula and by use of the concept of the pressure prism

Horizontal Surfaces

A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure

The magnitude of the force acting on one side of the surface is

The elemental forces pdA acting on A are all parallel and in the same sense a scalar summation of all such elements yields the magnitude of the resultant force Its direction is normal to the surface and toward the surface if p is positive

Fig 210 arbitrary xy axes - to find the line of action of the resultant ie the point in the area where the moment of the distributed force about any axis through the point is zero

Then since the moment of the resultant must equal the moment of the distributed force system about any axis say the y axis

xrsquo ndash the distance from the y axis to the resultant

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Example 24 In Fig 27a the liquids at A and B are water and the manometer liquid is oil S = 080 h1 = 300 mm h2 = 200 mm and h3 = 600 mm (a) Determine pA - pB in pacals (b) If pB = 50kPa and the barometer reading is 730 mm Hg find the pressure at A in meters of water absolute (a)

(b)

(a)

Micromanometers

For determining very small differences in pressure or determining large pressure differences precisely ndash several types of manometers

One type very accurately measures the differences in elevation of two menisci of a manometer

By means of small telescopes with horizontal cross hairs mounted along the tubes on a rack which is raised and lowered by a pinion and slow motion screw so that the cross hairs can be set accurately the difference in elevation of menisci (the gage difference) can be read with verniers

Figure 28 Micromanometer using two gage liquids

Fig 28 two gage liquids immiscible in each other and in the fluid to be measured a large gage difference R can be produced for a small pressure difference

The heavier gage liquid fills the lower U tube up to 0-0 then the lighter gage liquid is added to both sides filling the larger reservoirs up to 1-1

The gas or liquid in the system fills the space above 1-1 When the pressure at C is slightly greater than at D the menisci move as indicated in Fig 28

The volume of liquid displaced in each reservoir equals the displacement in the U tube

Manometer equation

γ1 γ2 and γ3 are the unit gravity force

Example 25 In the micromanometer of Fig 28 the pressure difference is wanted in pascals when air is in the system S2 = 10 S3 = 110 aA = 001 R = 5 mm t = 20oC and the barometer reads 760 mm Hg

The term γ1(aA) may be neglected Substituting into Eq (241) gives

Figure 29

Inclined manometer

The inclined manometer frequently used for measuring small differences in gas pressures

Adjusted to read zero by moving the inclined scale when A and B are open Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube it affords greater accuracy in reading the scale

Surface tension causes a capillary rise in small tubes If a U tube is used with a meniscus in each leg the surface-tension effects cancel

25 FORCES ON PLANE AREAS

In the preceding sections variations oF pressure throughout a fluid have been considered

The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force insofar as external reactions to the force system are concerned

In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration by formula and by use of the concept of the pressure prism

Horizontal Surfaces

A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure

The magnitude of the force acting on one side of the surface is

The elemental forces pdA acting on A are all parallel and in the same sense a scalar summation of all such elements yields the magnitude of the resultant force Its direction is normal to the surface and toward the surface if p is positive

Fig 210 arbitrary xy axes - to find the line of action of the resultant ie the point in the area where the moment of the distributed force about any axis through the point is zero

Then since the moment of the resultant must equal the moment of the distributed force system about any axis say the y axis

xrsquo ndash the distance from the y axis to the resultant

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Micromanometers

For determining very small differences in pressure or determining large pressure differences precisely ndash several types of manometers

One type very accurately measures the differences in elevation of two menisci of a manometer

By means of small telescopes with horizontal cross hairs mounted along the tubes on a rack which is raised and lowered by a pinion and slow motion screw so that the cross hairs can be set accurately the difference in elevation of menisci (the gage difference) can be read with verniers

Figure 28 Micromanometer using two gage liquids

Fig 28 two gage liquids immiscible in each other and in the fluid to be measured a large gage difference R can be produced for a small pressure difference

The heavier gage liquid fills the lower U tube up to 0-0 then the lighter gage liquid is added to both sides filling the larger reservoirs up to 1-1

The gas or liquid in the system fills the space above 1-1 When the pressure at C is slightly greater than at D the menisci move as indicated in Fig 28

The volume of liquid displaced in each reservoir equals the displacement in the U tube

Manometer equation

γ1 γ2 and γ3 are the unit gravity force

Example 25 In the micromanometer of Fig 28 the pressure difference is wanted in pascals when air is in the system S2 = 10 S3 = 110 aA = 001 R = 5 mm t = 20oC and the barometer reads 760 mm Hg

The term γ1(aA) may be neglected Substituting into Eq (241) gives

Figure 29

Inclined manometer

The inclined manometer frequently used for measuring small differences in gas pressures

Adjusted to read zero by moving the inclined scale when A and B are open Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube it affords greater accuracy in reading the scale

Surface tension causes a capillary rise in small tubes If a U tube is used with a meniscus in each leg the surface-tension effects cancel

25 FORCES ON PLANE AREAS

In the preceding sections variations oF pressure throughout a fluid have been considered

The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force insofar as external reactions to the force system are concerned

In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration by formula and by use of the concept of the pressure prism

Horizontal Surfaces

A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure

The magnitude of the force acting on one side of the surface is

The elemental forces pdA acting on A are all parallel and in the same sense a scalar summation of all such elements yields the magnitude of the resultant force Its direction is normal to the surface and toward the surface if p is positive

Fig 210 arbitrary xy axes - to find the line of action of the resultant ie the point in the area where the moment of the distributed force about any axis through the point is zero

Then since the moment of the resultant must equal the moment of the distributed force system about any axis say the y axis

xrsquo ndash the distance from the y axis to the resultant

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Figure 28 Micromanometer using two gage liquids

Fig 28 two gage liquids immiscible in each other and in the fluid to be measured a large gage difference R can be produced for a small pressure difference

The heavier gage liquid fills the lower U tube up to 0-0 then the lighter gage liquid is added to both sides filling the larger reservoirs up to 1-1

The gas or liquid in the system fills the space above 1-1 When the pressure at C is slightly greater than at D the menisci move as indicated in Fig 28

The volume of liquid displaced in each reservoir equals the displacement in the U tube

Manometer equation

γ1 γ2 and γ3 are the unit gravity force

Example 25 In the micromanometer of Fig 28 the pressure difference is wanted in pascals when air is in the system S2 = 10 S3 = 110 aA = 001 R = 5 mm t = 20oC and the barometer reads 760 mm Hg

The term γ1(aA) may be neglected Substituting into Eq (241) gives

Figure 29

Inclined manometer

The inclined manometer frequently used for measuring small differences in gas pressures

Adjusted to read zero by moving the inclined scale when A and B are open Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube it affords greater accuracy in reading the scale

Surface tension causes a capillary rise in small tubes If a U tube is used with a meniscus in each leg the surface-tension effects cancel

25 FORCES ON PLANE AREAS

In the preceding sections variations oF pressure throughout a fluid have been considered

The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force insofar as external reactions to the force system are concerned

In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration by formula and by use of the concept of the pressure prism

Horizontal Surfaces

A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure

The magnitude of the force acting on one side of the surface is

The elemental forces pdA acting on A are all parallel and in the same sense a scalar summation of all such elements yields the magnitude of the resultant force Its direction is normal to the surface and toward the surface if p is positive

Fig 210 arbitrary xy axes - to find the line of action of the resultant ie the point in the area where the moment of the distributed force about any axis through the point is zero

Then since the moment of the resultant must equal the moment of the distributed force system about any axis say the y axis

xrsquo ndash the distance from the y axis to the resultant

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Fig 28 two gage liquids immiscible in each other and in the fluid to be measured a large gage difference R can be produced for a small pressure difference

The heavier gage liquid fills the lower U tube up to 0-0 then the lighter gage liquid is added to both sides filling the larger reservoirs up to 1-1

The gas or liquid in the system fills the space above 1-1 When the pressure at C is slightly greater than at D the menisci move as indicated in Fig 28

The volume of liquid displaced in each reservoir equals the displacement in the U tube

Manometer equation

γ1 γ2 and γ3 are the unit gravity force

Example 25 In the micromanometer of Fig 28 the pressure difference is wanted in pascals when air is in the system S2 = 10 S3 = 110 aA = 001 R = 5 mm t = 20oC and the barometer reads 760 mm Hg

The term γ1(aA) may be neglected Substituting into Eq (241) gives

Figure 29

Inclined manometer

The inclined manometer frequently used for measuring small differences in gas pressures

Adjusted to read zero by moving the inclined scale when A and B are open Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube it affords greater accuracy in reading the scale

Surface tension causes a capillary rise in small tubes If a U tube is used with a meniscus in each leg the surface-tension effects cancel

25 FORCES ON PLANE AREAS

In the preceding sections variations oF pressure throughout a fluid have been considered

The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force insofar as external reactions to the force system are concerned

In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration by formula and by use of the concept of the pressure prism

Horizontal Surfaces

A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure

The magnitude of the force acting on one side of the surface is

The elemental forces pdA acting on A are all parallel and in the same sense a scalar summation of all such elements yields the magnitude of the resultant force Its direction is normal to the surface and toward the surface if p is positive

Fig 210 arbitrary xy axes - to find the line of action of the resultant ie the point in the area where the moment of the distributed force about any axis through the point is zero

Then since the moment of the resultant must equal the moment of the distributed force system about any axis say the y axis

xrsquo ndash the distance from the y axis to the resultant

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Example 25 In the micromanometer of Fig 28 the pressure difference is wanted in pascals when air is in the system S2 = 10 S3 = 110 aA = 001 R = 5 mm t = 20oC and the barometer reads 760 mm Hg

The term γ1(aA) may be neglected Substituting into Eq (241) gives

Figure 29

Inclined manometer

The inclined manometer frequently used for measuring small differences in gas pressures

Adjusted to read zero by moving the inclined scale when A and B are open Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube it affords greater accuracy in reading the scale

Surface tension causes a capillary rise in small tubes If a U tube is used with a meniscus in each leg the surface-tension effects cancel

25 FORCES ON PLANE AREAS

In the preceding sections variations oF pressure throughout a fluid have been considered

The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force insofar as external reactions to the force system are concerned

In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration by formula and by use of the concept of the pressure prism

Horizontal Surfaces

A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure

The magnitude of the force acting on one side of the surface is

The elemental forces pdA acting on A are all parallel and in the same sense a scalar summation of all such elements yields the magnitude of the resultant force Its direction is normal to the surface and toward the surface if p is positive

Fig 210 arbitrary xy axes - to find the line of action of the resultant ie the point in the area where the moment of the distributed force about any axis through the point is zero

Then since the moment of the resultant must equal the moment of the distributed force system about any axis say the y axis

xrsquo ndash the distance from the y axis to the resultant

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Figure 29

Inclined manometer

The inclined manometer frequently used for measuring small differences in gas pressures

Adjusted to read zero by moving the inclined scale when A and B are open Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube it affords greater accuracy in reading the scale

Surface tension causes a capillary rise in small tubes If a U tube is used with a meniscus in each leg the surface-tension effects cancel

25 FORCES ON PLANE AREAS

In the preceding sections variations oF pressure throughout a fluid have been considered

The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force insofar as external reactions to the force system are concerned

In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration by formula and by use of the concept of the pressure prism

Horizontal Surfaces

A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure

The magnitude of the force acting on one side of the surface is

The elemental forces pdA acting on A are all parallel and in the same sense a scalar summation of all such elements yields the magnitude of the resultant force Its direction is normal to the surface and toward the surface if p is positive

Fig 210 arbitrary xy axes - to find the line of action of the resultant ie the point in the area where the moment of the distributed force about any axis through the point is zero

Then since the moment of the resultant must equal the moment of the distributed force system about any axis say the y axis

xrsquo ndash the distance from the y axis to the resultant

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

25 FORCES ON PLANE AREAS

In the preceding sections variations oF pressure throughout a fluid have been considered

The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force insofar as external reactions to the force system are concerned

In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration by formula and by use of the concept of the pressure prism

Horizontal Surfaces

A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure

The magnitude of the force acting on one side of the surface is

The elemental forces pdA acting on A are all parallel and in the same sense a scalar summation of all such elements yields the magnitude of the resultant force Its direction is normal to the surface and toward the surface if p is positive

Fig 210 arbitrary xy axes - to find the line of action of the resultant ie the point in the area where the moment of the distributed force about any axis through the point is zero

Then since the moment of the resultant must equal the moment of the distributed force system about any axis say the y axis

xrsquo ndash the distance from the y axis to the resultant

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Horizontal Surfaces

A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure

The magnitude of the force acting on one side of the surface is

The elemental forces pdA acting on A are all parallel and in the same sense a scalar summation of all such elements yields the magnitude of the resultant force Its direction is normal to the surface and toward the surface if p is positive

Fig 210 arbitrary xy axes - to find the line of action of the resultant ie the point in the area where the moment of the distributed force about any axis through the point is zero

Then since the moment of the resultant must equal the moment of the distributed force system about any axis say the y axis

xrsquo ndash the distance from the y axis to the resultant

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Figure 210 Notation for determining the line of action of a force

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Momentum (1) First moment

The moment of an area A about the y axis

The moment about a parallel axis for example x = k the moment

Centroidal axis

Volume center

Mass center center of gravity of a body

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Figure A1 Notation for first and second moments

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

(2) Second moment The second moment of an area A (the moment of inertia of the area)

The moment about a parallel axis for example x = k the moment

Figure A2 Moments of inertia of simple areas about centroidal axes

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

The product of inertia Ixy of an area

- the product of inertia about centroidal axes parallel to the xy axes

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Inclined Surfaces

Fig 211 a plane surface is indicated by its trace ABlsquoit is inclined θo from the horizontal x axis intersection of the plane of the area and the free surface y axis taken in the plane of the area with origin O in the free surface The xy plane portrays the arbitrary inclined area The magnitude direction and line of action of the resultant force due to the liquid acting on one side of the area are sought

For δA Since all such elemental forces are parallel the integral over the area y

ields the magnitude of force F acting on one side of the area

Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid

The presence of a free surface is unnecessary

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Figure 211 Notation for force of liquid on one side of a plane inclined area

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Center of Pressure

Fig 211 the line of action of the resultant force has its piercing point in the surface at a point called the pressure center with coordinates (xp yp) Center of pressure of an inclined surface is not at the centroid To find the pressure center the moments of the resultant xpF ypF are equated to the moment of the distributed forces about the y axis and x axis respectively

- may be evaluated conveniently through graphical integration for simple areas they may be transformed into general formulas

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

When either of the centroidal axes is an axis of symmetry for the surface vanishes and the pressure center lies on x = x- Since may be either positive or negative the pressure center may lie on either side of the line x = x- To determine yp by formula with Eqs (252) and (256)

In the parallel-axis theorem for moments of inertia

in which IG is the second moment or the area about its horizontal centroidal axis If IG is eliminated from Eq (259)

xyIxyI

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Figure 212 Triangular gate

Example 26 The triangular gate CDE (Fig 212) is hinged along CD and is opened by a normal force P applied at E It holds oil relative density 080 above it and is open to the atmosphere on its lower side Neglecting the weight of the gate find (a) the magnitude of force exerted on the gate by integration and by Eq (252) (b) the location of pressure center (c) the force P needed to open the gate

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

(a) By integration with reference to Fig 212

When y = 4 x = 0 and when y = 65 x = 3 with x varying linearly with y thus

in which the coordinates have been substituted to find x in terms of y Solving for a and b gives

Similarly y = 65 x = 3 y = 9 x = 0 and x = 65(9 - y) Hence

Integrating and substituting for γsinθ leads to

By Eq (252)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

(b) With the axes as shown

In Eq (258)

I-xyis zero owing to symmetry about the centroidal axis parallel to the x

axis hence

In Eq (2511)

ie the pressure center is 016 m below the centroid measured in the plane of the area

(c) When moments about CD are taken and the action of the oil is replaced by the resultant

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

The Pressure Prism

Pressure prism another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh h is the vertical distance to the free surface Fig 213 (An imaginary free surface may be used to define h if no real free surface exists) (in the figure γh may be laid off to any convenient scale such that its trace is OM)

The force acting on an elemental area δA is

(2512)

- an element of volume of the pressure prism After integrating F = ϑ From Eqs (255) and (256)

(2513)

xp yp are distances to the centroid of the pressure prism the line of action of the resultant passes through the centroid of the pressure prism

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Figure 213 Pressure prism

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Effects of Atmospheric Pressure on Forces on Plane Areas

In the discussion of pressure forces the pressure datum was not mentioned p = γh the datum taken was gage pressure zero or the local atmospheric pressure

When the opposite side of the surface is open to the atmosphere a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area or P0A based on absolute zero as datum On the liquid side the force is

The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location

So long as the same pressure datum is selected for all sides of a free body the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Example 28 An application of pressure forces on plane areas is given in the design of a gravity dam The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam Figure 215 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 25γ and γ is the unit gravity force of water A 1 m section of dam is considered as a free body the forces are due to the concrete the water the foundation pressure and the hydrostatic uplift Determining amount of hydrostatic uplift is beyond the scope of this treatment but it will be assumed to be one-half the hydrostatic head at the upstream edge decreasing linearly to zero at the downstream edge of the dam Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ + 2625γ - 1750γ = 7625γ N The position of Ry is such that the free body is in equilibrium For moments around O

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Figure 215 Concrete gravity dam

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

It is customary to assume that the foundation pressure varies linearly over the base of the dam ie that the pressure prism is a trapezoid with a volume equal to Ry thus

in which σmax σmin are the maximum and minimum compressive stresses in pascals The centroid of the pressure prism is at the point where x = 448 m By taking moments about 0 to express the position of the centroid in terms of σmax and σmin

Simplifying gives

When the resultant falls within the middle third of the base of the dam σmin will always be a compressive stress Owing to the poor tensile properties of concrete good design requires the resultant to fall within the middle third of the base

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

26 FORCE COMPONENTS ON CURVED SURFACES

When the elemental forces p δA vary in direction as in the case of a curved surface they must be added as vector quantities their components in three mutually perpendicular directions are ad

ded as scalars and then the three components are added vectorially With two horizontal components at right angles and with the vertical c

omponent - all easily computed for a curved surface - the resultant can be determined

The lines of action of the components also are readily determined

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Horizontal Component of Force on a Curved Surface

The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface The vertical plane of projection is normal to the direction of the component

Fig 216 the surface represents any three-dimensional surface and δA an element of its area its normal making the angle θ with the negative x direction Then

Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane

force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection

To find the horizontal component at right angles to the x direction the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Figure 216 Horizontal component of force on a curved surface

Figure 217 Projections of area elements on opposite sides of a body

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

When looking for the horizontal component of pressure force on a closed body the projection of the curved surface on a vertical plane is always zero since on opposite sides of the body the area-element projections have opposite signs (Fig 217)

Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C If the element of area of the body cut by the prism at B is δAB and at C is δAC then

and similarly for all other area elements To find the line of action of a horizontal component of force on a curve

d surface the resultant of the parallel force system composed of the force components from each area element is required This is exactly the resultant of the force on the projected area since the two force systems have an identical distribution of elemental horizontal force components Hence the pressure center is located on the projected area by the methods of Sec 25

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Example 29 The equation of an ellipsoid of revolution submerged in water is x24 + y24 + z29 = 1 The center of the body is located 2 m below the free surface Find the horizontal force components acting on the curved surface that is located in the first octant Consider the xz plane to be horizontal and y to be positive upward

Projection of the surface on the yz plane has an area of

Its centroid is located m below the free surface

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Vertical Component of Force on a Curved Surface

The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface

Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface

In Fig218 an area element is shown with the force p δA acting normal to it Let θ be the angle the normal to the area element makes with the vertical Then the vertical component of force acting on the area element is p cos θ δA and the vertical component of force on the curved surface is given by

(262) p replaced by its equivalent γh cos θ δA is the projection of δA on a horizo

ntal plane Eq (262)

(253-4)

in which δϑ is the volume of the prism of height h and base cos θ δA or the volume of liquid vertically above the area element

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Figure 218 Vertical component of force on a curved surface

Figure 219 Liquid with equivalent free surface

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Fig 219 the liquid is below the curved surface and the pressure magnitude is known at some point (eg O) an imaginary or equivalent free surface s-s can be constructed pγ above O so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point

The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface

In constructing an imaginary free surface the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface otherwise the pressure distribution over the surface will not be correctly represented

With an imaginary liquid above a surface the pressure at a point on the curved surface is equal on both sides but the elemental force components in the vertical direction are opposite in sign the direction of the vertical force component is reversed when an imaginary fluid is above the surface

In some cases a confined liquid may be above the curved surface and an imaginary liquid must be added (or subtracted) to determine the free surface

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force With the axis at O (Fig218)

in which is the distance from 0 to the line of action Since Fv = γϑ

the distance to the centroid of the volume the line of action of the vertical force passes through the centroid of

the volume real or imaginary that extends above the curved surface up to the real or imaginary free surface

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Example 210 A cylindrical barrier (Fig 220) holds water as shown The contact between cylinder and wall is smooth Considering a 1-m length of cylinder determine (a) its gravity force and (b) the force exerted against the wall

(a) For equilibrium the weight of the cylinder must equal the vertical component of force exerted on it by the water (The imaginary free surface for CD is at elevation A) The vertical force on BCD is

The vertical force on AB is

Hence the gravity force per metre of length is

(b) The force exerted against the wall is the horizontal force on ABC minus the horizontal force on CD The horizontal components of force on BC and CD cancel the projection of BCD on a vertical plane is zero

since the projected area is 2 m2 and the pressure at the centroid of the projected area is 9806 Pa

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Figure 220 Semifloating body

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Tensile Stress in a Pipe and Spherical Shell Fig 221 a circular pipe under the action of an internal pressure is in tension

around its periphery assuming that no longitudinal stress occurs the walls are in tension

Consider a section of pipe of unit length (the ring between two planes normal to the axis and unit length apart) If one-half of this ring is taken as a free body the tensions per unit length at top and bottom are respectively T1 and T2

The horizontal component of force acts through the pressure center of the projected area and is 2pr in which p is the pressure at the centerline and r is the internal pipe radius

For high pressures the pressure center may be taken at the pipe center then T1 = T2 and

T is the tensile force per unit length For wall thickness e the tensile stress in the pipe wall is

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Figure 221 Tensile stress in pipe

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Example 211 A 100 mm-ID steel pipe has a 6 mm wall thickness For an allowable tensile stress of 70 MPa what is the maximum pressure

From Eq (266)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

27 BUOYANT FORCE

Buoyant force the resultant force exerted on a body by a static fluid in which it is submerged or floating Always acts vertically upward (there can be no horizontal

component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero)

The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on its upper side

Figure 222

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Figure 222 Buoyant force on floating and submerged bodies

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Fig 222 the upward force on the bottom is equal to the gravity force of liquid real or imaginary which is vertically above the surface ABC indicated by the gravity force of liquid within ABCEFA The downward force on the upper surface equals the gravity force of liquid ADCEFA The difference between the two forces is a force vertically upward due to the gravity force of fluid ABCD that is displaced by the solid In equation form

FB is buoyant force V is the volume of fluid displaced and γ is the unit gravity force of fluid

The same formula holds for floating bodies when V is taken as the volume of liquid displaced

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Fig223 the vertical force exerted on an element of the body in the form of a vertical prism of cross section δA is

δV is the volume of the prism Integrating over the complete body gives

γ is considered constant throughout the volume To find the line of action of the buoyant force moments are taken abo

ut a convenient axis O and are equated to the moment of the resultant thus

i is the distance from the axis to the line of action This equation yields the distance to the centroid of the volume the

buoyant force acts through the centroid of the displaced volume of fluid this holds for both submerged and floating bodies

The centroid of the displaced volume of fluid is called the center of buoyancy

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Figure 223 Vertical force components on element of body

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Determining gravity force on an odd-shaped object suspended in two different fluids yields sufficient data to determine its gravity force volume unit gravity force and relative density

Figure 224 two free-body diagrams for the same object suspended and gravity force determined in two fluids F1 and F2 γ1 and γ2 are the unit gravity forces of the fluids W and V the gravity force and volume of the object are to be found

The equations of equilibrium are written and solved

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Figure 224 Free body diagrams for body suspended in a fluid

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

A hydrometer uses the principle of buoyant force to determine relative densities of liquids

Figure 225 a hydrometer in two liquids with a stem of prismatic cross section a

Considering the liquid on the left to be distilled water (unit relative density S = 100) the hydrometer floats in equilibrium when

V0 is the volume submerged γ is the unit gravity force of water and W is the gravity force of hydrometer

The position of the liquid surface is marked 100 on the stem to indicate unit relative density S When the hydrometer is floated in another 1iquid the equation of equilibrium becomes

where ΔV = aΔh Solving for Δh with Eqs (272) and (273)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Figure 225 Hydrometer in water and in liquid of relative density

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Example 212 A piece of ore having a gravity force of 15 N in air is found to have a gravity force 11 N when submerged in water What is its volume in cubic centimetres and what is its relative density

The buoyant force due to air may be neglected From Fig 224

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

28 STABILITY OF FLOATING AND SUBMERGED BODIES

A body floating in a static liquid has vertical stability A small upward displacement decreases the volume of liquid displaced

an unbalanced downward force which tends to return the body to its original position

Similarly a small downward displacement results in a greater buoyant force which causes an unbalanced upward force

A body has linear stability when a small linear displacement in any direction sets up restoring forces tending to return it to its original position

A body has rotational stability when a restoring couple is set up by any small angular displacement

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Methods for determining rotational stability are developed in the following discussion

A body may float in stable equilibrium unstable equilibrium (any small angular displacement sets up a

couple that tends to increase the angular displacement) neutral equilibrium (any small angular displacement sets up no

couple whatever) Figure 226 three cases of equilibrium

a a light piece of wood with a metal mass at its bottom is stable

b when the metal mass is at the top the body is in equilibrium but any slight angular displacement causes it to assume the position in a

c a homogeneous sphere or right-circular cylinder is in equilibrium for any angular rotation ie no couple results from an angular displacement

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Figure 226 Examples of stable unstable and neutral equilibrium

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

A completely submerged object is rotationally stable only when its center of gravity is below the center of buoyancy (Fig 227a)

When the object is rotated counterclockwise the buoyant force and gravity force produce a couple in the clockwise direction (Fig 227b)

Figure 227 Rotationally stable submerged body

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Normally when a body is too heavy to float it submerges and goes down until it rests on the bottom

Although the unit gravity force of a liquid increases slightly with depth the higher pressure tends to cause the liquid to compress the body or to penetrate into pores of solid substances and thus decrease the buoyancy of the body

Example a ship is sure to go to the bottom once it is completely submerged owing to compression of air trapped in its various parts

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Determination of Rotational Stability of Floating Objects

Any floating object with center of gravity below its center of buoyancy (centroid of displaced volume) floats in stable equilibrium (Fig 226a) Certain floating objects however are in stable equilibrium when their center of gravity is above the center of buoyancy

Figure 228a a cross section of a body with all other parallel cross sections identical The center of buoyancy is always at the centroid of the displaced volume which is at the centroid of the cross-sectional area below liquid surface in this case

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Figure 228 Stability of a prismatic body

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

when the body is tipped (Fig 228b) the center of buoyancy is at the centroid B of the trapezoid ABCD the buoyant force acts upward through B and the gravity force acts downward through G the center of gravity of the body

When the vertical through B intersects the original centerline above C as at M a restoring couple is produced and the body is in stable equilibrium

The intersection of the buoyant force and the centerline is called the metacenter (M)

When M is above G the body is stable when below G it is unstable and when at G it is in neutral equilibrium

The distance MG is called the metacentric height and is a direct measure of the stability of the body The restoring couple is

in which θ is the angular displacement and W the gravity force of the body

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Example 213 In Fig 228 a scow 6 m wide and 20 m long has a gross mass of 200 Mg Its center of gravity is 30 cm above the water surface Find the metacentric height and restoring couple when Δy = 30 cm

The depth of submergence h in the water is

The centroid in the tipped position is located with moments about AB and BC

By similar triangles AEO and BPM

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

G is 197 m from the bottom hence

The scow is stable since is positive the righting moment is

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Nonprismatic Cross Sections

For a floating object of variable cross section (eg a ship) (Fig 239a) a convenient formula can be developed for determination of metacentric height for very small angles of rotation θ

The horizontal shift in center of buoyancy r (Fig 229b) is determined by the change in buoyant forces due to the wedge being submerged which causes an upward force on the left and by the other wedge decreasing the buoyant force by an equal amount ΔFB on the right

The force system consisting of the original buoyant force at B and the couple ΔFB x s due to the wedges must have as resultant the equal buoyant force at B With moments about B to determine the shirt r

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Figure 229 Stability relations in a body of variable cross section

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

The amount of the couple can be determined with moments about O the centerline of the body at the liquid surface

For an element of area δA on the horizontal section through the body at the liquid surface an element of volume of the wedge is xθ δA The buoyant force due to this element is γxθ δA and its moment about O is γx2θ δA in which θ is the small angle of tip in radians

By integrating over the complete original horizontal area at the liquid surface the couple is determined to be

I is the moment of inertia of the area about the axis y-y (Fig229a) Substitution into the above equation produces

V is the total volume of liquid displaced Since θ is very small

and

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Example 214 A barge displacing 1 Gg has the horizontal cross section at the waterline shown in Fig 230 Its center of buoyancy is 20 m below the water surface and its center of gravity is 05 m below the water surface Determine its metacentric height for rolling (about y-y axis) and for pitching (about x-x axis)

GB = 2 ndash 05 = 15 m

For rolling

For pitching

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Figure 230 Horizontal cross section of a ship at

the waterline

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

29 RELATIVE EQUILIBRIUM

Fluid statics no shear stresses the variation of pressure is simple to compute

For fluid motion such that no layer moves relative to an adjacent layer the shear stress is also zero throughout the fluid

A fluid with a translation at uniform velocity still follows the laws of static variation of pressure

When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid) no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body

Two cases are of interest a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus the fluid is said to be in relative equilibrium

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Uniform Linear Acceleration

Fig 231 a liquid in an open vessel is given a uniform linear acceleration a

After some time the liquid adjusts to the acceleration so that it moves as a solid ie the distance between any two fluid particles remains fixed no shear stresses occur

By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig 231a) the z axis is normal to a and there is no acceleration component in that direction

Fig 231b the pressure gradient p is then the vector sum of -ρa and -nablajγ

Since p is in the direction of maximum change in p (the gradient) at nablaright angles to p there is no change in p Surfaces of constant pressurnablae including the free surface must therefore be normal to pnabla

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Figure 231 Acceleration with free surface

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

To obtain a convenient algebraic expression for variation of p with x y and z that is p = p(x y z) Eq (225) is written in component form

Since p is a function of position (x y z) its total differential is

Substituting for the partial differentials gives

which can be integrated for an incompressible fluid

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

To evaluate the constant of integration c let x = 0 y = 0 p = p0 then c = p0 and

When the accelerated incompressible fluid has a free surface its equation is given by setting p = 0 in the above eq Solving it for y gives

The lines of constant pressure p = const have the slope

and are parallel to the free surface The y intercept of the free surface is

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Example 215 The tank in Fig 232 is filled with oil relative density 08 and accelerated as shown There is a small opening in the rank at A Determine the pressure at B and C and the acceleration ax required to make the pressure at B zero

By selecting point A as origin and by applying Eq (292) for ay = 0

At B x = 18 m y = - 12 m and p = 235 kPa Ft C x = -015 m y = -135 m and p = 1118 kPa For zero pressure at B from Eq (292) with origin at A

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Figure 232 Tank completely filled with liquid

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Example 216 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig 233) It is given a constant linear acceleration ax = g2 ay = -g4 Develop an equation for variation of pressure along its base

The free surface has the slope

hence the free surface is located s shown in the figure When the origin is taken at 0 Eq (292) becomes

Then for y = 0 along the bottom

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Figure 233 Uniform linear acceleration of container

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Uniform Rotation about a Vertical Axis

Forced-vortex motion rotation of a fluid moving as a solid about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion in which e

ach particle moves in a circular path with a speed varying inversely as the distance from the center

A liquid in a container when rotated about a vertical axis at constant angular velocity moves like a solid alter some time interval

No shear stresses exist in the liquid and the only acceleration that occurs is directed radially inward toward the axis of rotation

By selecting a coordinate system (Fig 234a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation the following equation may be applied to determine pressure variation throughout the fluid

(225)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Figure 234 Rotation of a fluid about a vertical axis

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

For constant angular velocity w any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r)

Vector addition of -jγ and -ρa (Fig 234b) yields p the pressure gradienablant The pressure does not vary normal to this line at a point if P is taken at the surface the free surface is normal to pnabla

Expanding Eq (225)

k is the unit vector along the z axis (or tangential direction) Then

p is a function of y and r only

For a liquid (γ asymp const) integration yields

c is the constant of integration

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

If the value of pressure at the origin (r = 0 y = 0) is p0 then c = p0 and

When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq is divided by γ

the head or vertical depth varies as the square of the radius The surfaces of equal pressure are paraboloids of revolution

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

When a free surface occurs in a container that is being rotated the fluid volume underneath the paraboloid of revolution is the original fluid volume

The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig 235) The rise of liquid from its vertex to the wall of the cylinder is w2r0

2rg (Eq (296)) for a circular cylinder rotating about its axis

Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder the volume of the liquid above the horizontal plane through the vertex is

When the liquid is at rest this liquid is also above the plane through the vertex to a uniform depth of

Hence the liquid rises along the walls the same amount as the center drops thereby permitting the vertex to be located when w r0 and depth before rotation are given

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Figure 235 Rotation of circular cylinder about its axis

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Example 217 A liquid relative density 12 is rotated at 200 rpm about a vertical axis At one point A in the fluid 1 m from the axis the pressure is 70 kPa What is the pressure at a point B which is 2 m higher than A and 15 m from the axis

When Eq (295) is written for the two points

Then w = 200 x 2π60 = 2095 rads γ = 12 x 9806 = 11767 Nm3 rA = 1 m and rB = 15 m

When the second equation is subtracted from the first and the values are substituted

Hence

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Example 218 A straight tube 2 m long closed at the bottom and filled with water is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 673 rads Draw the paraboloid of zero pressure and determine the pressure at the bottom and midpoint of the tube In Fig 236 the zero-pressure paraboloid passes through point A If the origin is taken at the vertex that is p0 = 0 Eq (296) becomes

which locates the vertex at O 0577 m below A The pressure at the bottom of the tube is or

At the midpoint =289 m and

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Figure 236 Rotation of inclined tube of liquid about a vertical axis

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

Fluid Pressure Forces in Relative Equilibrium

The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface

The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear accel

eration) the magnitude of force is given by the product of pressure at the centroid and area since the volume of the pressure prism is given by pGA

For nonlinear distributions the magnitude and line of action can be found by integration

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

과제물 스캐닝된 파일을 텍스트 파일로 만들 것 (2345 장 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을

유도하라 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라 ( 단순

액주계와 차분 액주계의 예를 포함하여 ) 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을

유도하라 1 차 2 차 모멘트에 대하여 설명하라 부록에 나와있는 도형에 대하여

도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라 부록 B 의 수치해석 기법을 설명하라 프로그램밍 포함 ( 교과서와 다른

책의 내용 참조 ) 예제 27 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여

컴퓨터 프로그램을 작성하여 풀어라

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)

과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 (

프로그램밍 포함 C 및 Fortran)