1

CHAPTER IIFLUID STATICS

Prepared by: Balewgize A. Zeru (MTech-IITM) & Henok M. (MSc)

2 Outline

• Introduction, • Pressure specifications, • Hydrostatic pressure distributions, • Manometry, • Hydrostatic Forces on plane surfaces, • Hydrostatic forces on curved surfaces, • Buoyancy and Stability, • Pressure variation with rigid body motion

3 Introduction

Many fluid problems do not involve motion. They concern the pressuredistribution in a static fluid and its effect on solid surfaces and on floating and submerged bodies.

When the fluid velocity is zero, denoted as the hydrostatic condition, the pressure variation is due only to the weight of the fluid.

Important applications in this chapter are Pressure distribution in the atmosphere and the oceans, The design of manometer pressure instruments, Forces on submerged flat and curved surfaces, Buoyancy on a submerged body, and The behavior of floating bodies.

The last two result in Archimedes’ principles.

4 Pressure and Pressure Gradient

The normal stress on any plane through a fluid element at rest is equal to a unique value called the fluid pressure p, taken positive for compression by common convention.

Figure shows a small wedge of fluid at rest of size Δx by Δz by Δs and depth b into the paper.

There is no shear by definition, but we postulate that the pressures px , py, and pn may be different on each face.

5 …continuedSummation of forces must equal zero (no acceleration) in both the x and z directions.

but the geometry of the wedge is such that

rearrangement give

These relations illustrate two important principles of the hydrostatic, or shear-free, condition: (1) There is no pressure change in the horizontal direction, and (2) There is a vertical change in pressure proportional to the density, gravity, and depth

change.

In the limit as the fluid wedge shrinks to a “point,’’ Δz → 0

Since θ is arbitrary, we conclude that the pressure p at a point in a static fluid is independent of orientation.

6 Pressure Force on a Fluid Element

Pressure (or any other stress, for that matter) causes no net force on a fluid element unless it varies spatially.Let the pressure vary arbitrarily:

Figure. Net x force on an elementdue to pressure variation.

Consider the pressure acting on the two x faces.The net force in the x direction on the element is given by:

The net force vector on the element due to pressure is:

7 …continued

The term in parentheses is the negative vector gradient of p.Denoting f as the net force per unit element volume;

Thus it is not the pressure but the pressure gradient causing a net force which must be balanced by gravity or acceleration or some other effect in the fluid.

The pressure gradient is a surface force which acts on the sides of the element.

8 Body force

In addition to the surface force due to pressure difference, there may also be a body force, due to electromagnetic or gravitational potentials, acting on the entire mass of the element.

Consider only the gravity force, or weight of the element:

Note: g denotes the acceleration of gravity, a vector acting towards the center of the earth.

9 Surface force due to Gradient of Viscous Stress

In general, there may also be a surface force due to the gradient, if any, of the viscous stresses. For completeness, we write this term here without derivation.For an incompressible fluid with constant viscosity, the net viscous force is:

where VS stands for viscous stresses and μ is the coefficient of viscosity.

10 Resultant force

The total vector resultant of these three forces—pressure, gravity, and viscous stress—must either keep the element in equilibrium or cause it to move with acceleration a. From Newton’s law:

Re-writing:where B is a short notation for the vector sum on the right-hand side.

Equating the Components of the gradient and integrating gives solution for pressure distribution for known values of B.

11 Special cases of fluid analysis

There are at least four special cases:1. Flow at rest or at constant velocity: The acceleration and viscous

terms vanish identically, and p depends only upon gravity and density. This is the hydrostatic condition.

2. Rigid-body translation and rotation: The viscous term vanishes identically, and p depends only upon the term (g-a).

3. Irrotational motion ( V =0):The viscous term vanishes identically, and an exact integral called Bernoulli’s equation can be found for the pressure distribution.

4. Arbitrary viscous motion: Nothing helpful happens, no general rules apply, but still the integration is quite straightforward

12 Absolute Pressure, Gage Pressure and Vacuum Pressure

Pressure can be specified as 1. the absolute or total magnitude or 2. the value relative to the local ambient atmosphere.– Gage Pressure – Vacuum Pressure

The second case occurs because many pressure instruments are ofDifferential type and record, not an absolute magnitude, but the difference between the fluid pressure and the atmosphere.

i.e. when

13 …continued

Figure. Illustration of absolute, gage, and vacuum pressure readings.

14 Hydrostatic PressureDistributions

The hydrostatic pressure distribution for a fluid at rest or at constant velocity reduces to;

This is a hydrostatic distribution and is correct for all fluids at rest, regardless of their viscosity, because the viscous term vanishes identically.From vector analysis, the gradient of pressure ∇p: expresses the magnitude and direction of the maximum spatial rate

of increase of the scalar property p. is perpendicular everywhere to surfaces of constant p.

Therefore, states that a fluid in hydrostatic equilibrium will align its constant-pressure surfaces everywhere normal to the local-gravity vector.

15 …continued

In our customary coordinate system, axis z is “up.’’ Thus the local-gravity vector becomes; , where g = 9.807m/s.Equating the components of =

i.e. p is independent of x and yThis is the solution to the hydrostatic problem. The integration requires an assumption about the density and gravity distribution. Gases and liquids are usually treated differently.

From this solution, the pressure is the same at all points on a given horizontal plane in the fluid. The pressure increases with depth in the fluid

16 …continuedPressure in a continuously distributed uniform static fluid varies only with vertical distance and is independent of the shape of the container. .

Figure. Hydrostatic pressure distribution. Points a, b, c, and d are at equal depths in water and therefore have identical pressures. Points A, B, and Care also at equal depths in water and have identical pressures higher than a, b, c, and d. However, point D, although at the same depthas A, B, and C, has a different pressure because it lies beneath a different fluid, mercury.

17 Hydrostatic Pressure in Liquids

Liquids are so nearly incompressible that we can neglect their density variation in hydrostatics.

Where, is called the specific weight of the

fluid, with dimensions of weight per unit volume.

The quantity p/is a length called the pressure head of the fluid.

18 Hydrostatic-pressure distribution in oceans and atmospheres.

Eg. Lake Shala , an alkaline lake in Ethiopia, has a maximum depth of 266m, and the mean atmospheric pressure is 82 kPa. Estimate the absolute pressure in kPa at this maximum depth.Take specific weight of the lake water = 9790N/m3Solution:

m) x2686kPa = 2604kPa (gauge)

Taking the reference at (P=Pa, z=0)For any negative depth, the hydrostatic pressure is given by a general formula as:

19 The Mercury BarometerThe simplest practical application of the hydrostatic formula is the barometer which measures atmospheric pressure.

A tube is filled with mercury and inverted while submerged in a reservoir. This causes a near vacuum in the closed upper end because mercury has an extremely small vapor pressure at room temperatures.

Since atmospheric pressure forces a mercury column to rise a distance h into the tube, the upper mercury surface is at zero pressure.

Mercury is used because it is the heaviest common liquid.

20 …continuedDetermine the height of mercury column required to measure the atmospheric pressure.

Apply equation between to point 1 and point 2.

i.e. to measure atmospheric pressure, 761mm of mercury column is required while a water barometer would be 34 ft high.

21 Hydrostatic Pressure in Gases

Gases are compressible, with density nearly proportional to pressure. Thus density must be considered as a variable if the integration carries over large pressure changes.Apply the perfect gas law as; i.e.

Separate the variables and integrating with an assumption of constant temperature with respect to z (isothermal atmosphere with T=T0) gives;

22 …continuedActually the earth’s mean atmospheric temperature drops off nearly linearly with z up to an altitude of about 36,000 ft (11,000m) called the troposphere.This variation is approximated as:

23 …continued

Example: If sea-level pressure is 101,350 Pa, compute the standard pressure at an altitude of 5000 m, using (a) the exact formula and (b) an isothermal assumption at a standard sea-level temperature of 15°C. Is the isothermal approximation adequate?

Case B. ISOTHERMAL THEMPERATURE

This is 11% higher than the exact value. Therefore the isothermal assumption is not accurate.

Case A. EXACT FORMULA

24 Application toManometry

A manometer is device used to measure pressure differences between two points using a static column of one or more liquids or gases.

The figure shows a simple open manometer for measuring relative to atmospheric pressure.

Pascal’s law:Any two points at the same elevation in a continuous mass of the same static fluid will be at the same pressure.

Simply,

Or, apply thebasic hydrostatic formula

Neglecting the height for the gases; i.e.

= + (

25 Home work onManometery

For the multiple-fluid manometer shown bellow for find the difference in pressure between two chambers A and B.

-

26 Example onManometry

Pressure gage B is to measure the pressure at point A in a water flow. If the pressure at B is 87 kPa, estimate the pressure at A, in kPa. Assume all fluids are at 20°C.

From Table:

Now proceed from A to B, calculating the pressure change in each fluid and adding:

since i.e.

Solve for the unknown.

→

27 Hydrostatic Forces on Plane Surfaces

• The pressure on any submerged surface varies linearly with depth if the change in density is neglected.

• For a plane surface, the linear stress distribution is exactly analogous to combined bending and compression of a beam in strength-of-materials theory.

• The hydrostatic problem thus reduces to simple formulas involving the centroid and moments of inertia of the plate cross-sectional area.

Figure shows a plane panel of arbitrary shape completely submerged in a liquid. θ is an arbitrary angle b/n the panel plane and the horizontal free surface, so that the depth varies over the panel surface.h is the depth to any element area dAof the plate.Therefore, the pressure is:

28 …continuedThen the total hydrostatic force on one side of the plate is given by

by definition, the centroidal slant distance from the surface to the plate is

but

If the depth straight down fromthe surface to the plate centroid

THEREFORE,The force on one side of any plane submerged surface in a uniform fluid equals the pressure at the plate centroid times the plate area, independent of the shape of the plate or the angle θ at which it is slanted.

29 …continuedCenter of Pressure, CP

i.e.

The negative sign shows that is below the centroid at a deeper leveland, unlike F, depends upon angle θ.

The resultant force F does not act through the centroid but below it toward the high-pressure side through the center of pressure CP of the plate.To find the coordinates (), take the sum moments of the elemental force pdA about the centroid and equate to the moment of the resultant F.

30 …continuedCenter of Pressure, CP

If the plate moves deeper, reaches centroid because every term remains constant except which increases.

Determination of is similar to i.e,

Where is the product of inertia of the plate.

For positive , is negative because the dominant pressure force acts in the third, or lower left, quadrant of the panel. If = 0, usually implying symmetry, = 0 and the center of pressure lies directly below the centroid on the y axis.

31…continued

Centroidal Moments of Inertia: (a) rectangle, (b) circle, (c) triangle, and (d) semicircle

32 …continuedGage-Pressure Formulas

In most cases the ambient pressure is neglected because it acts on both sides of the plate; e.g., the other side of the plate is inside a ship or on the dry side of a gate or dam.Therefore,

the center of pressure becomes independent of specific weight, i.e.

33 Example 1

The gate in figure shown is 5 ft wide, is hinged at point B, and rests against a smooth wall at point A. Compute (a) the force on the gate due to seawater

pressure, (b) the horizontal force P exerted by the

wall at point A, and (c) the reactions at the hinge B.

34 Solution toexample 1

a)By geometry the gate is 10 ft long from A to B, and its centroid is halfway between, or at elevation 3 ft above point B.i.e the depth hCG is thus 15 - 3 = 12 ft.

Neglect pa as acting on both sides of the gate.

b)First find the center of pressure of F for the rectangular gate as shown in the free body diagram.

Find the distance l from CG to CP.

i.e, the distance from B to for force F( CP) is 10 – 5 – 0.417 = 4.583 ft Summing moments counterclockwise about B gives;

…continued

c)With F and P known, the reactions Bxand Bz are found by summing forces on the gate:

Similarly;

35

36 Example 2

A tank of oil has a right-triangular panel near the bottom, as in the figure. Omitting , find the (a) hydrostatic force and (b) Centre of Pressure CP on the panel

37 Solution toexample 2

a) The centroid is one-third up (4 m) and one-third over (2 m) from the lower left corner

38 …continued

The CP position is

The resultant force F=2.54 MN acts through this point, which is down and to the right of the centroid.

39 Hydrostatic Forces onCurved Surfaces

The resultant pressure force on a curved surface is most easily computed by separating it into horizontal and vertical components.

The incremental pressure forces, being normal to the local area element, vary in direction along the surface and thus cannot be added numerically.

The laborious calculations procedure can be avoided by using a simple free body diagram and apply static equilibrium condition.

40 …continued

The desired forces and are exerted by the surface on the fluid column.

Forces are shown due to fluid weight. Horizontal force is due to horizontal

pressure on the vertical sides of this column.

The column of fluid must be in static equilibrium. On the upper part of the column bcde, the

horizontal components exactly balance. On the curved portion abc; the desired

force due to the curved surface is exactly equal to the force on the vertical left side of the fluid column.

41 …continued

i.e. the left-side force can be computed by the plane surface formula based on a vertical projection of the area of the curved surface:

• The horizontal component of force on a curved surface equals the force on the plane area formed by the projection of the curved surface onto a vertical plane normal to the component.

Summation of vertical forces on the fluid free body gives:

• The vertical component of pressure force on a curved surface equals in magnitude and direction the weight of the entire column of fluid, both liquid and atmosphere, above the curved surface.

42 Example 1

A dam has a parabolic shape = as shown in figure, with = 10 ft and = 24ft. The fluid is water, = 62.4 lbf/, and atmospheric pressure may be omitted. Compute the forces and on the dam and the position CP where they act. The width of the dam is 50 ft.

Parabolic dam: = = 10 ft= 24ftWidth = 50 ft= 62.4 lbf/and

43 Solution toexample 1

The vertical projection of this curved surface is a rectangle 24ft high and 50ft wide, with its centroid halfway down, or = 12 ft.The force becomes;

CP is located for at 12 + 4 = 16ft i.e. two-thirds, down from the free surface. evident by inspection of the triangular pressure distribution.

44 …continued

The vertical component FV equals the weight of the parabolic portion of fluid above the curved surface.

Fv acts down wards at a distance of

The total pressure acting on the dam is;

The force acts downwards at

The location of the resultant force is at CG;

Or, at a center of pressure, on the dam, of

45 Pressure diagram

For the vertical wall of the tank containing a liquid as in the figure, ABC is the pressure diagram, pressure being plotted horizontally against depth vertically.The center of pressure, CP, is the centroid of the pressure diagram.

The area of this triangle will be the product of depth and pressure, and will represent, to scale, the resultant force R on unit width of the immersed surface perpendicular to the plane of the diagram .

46 Pressure diagram foran inclined submerged surface

the resultant force Ris represented by the shaded area, instead of the whole triangle, and acts through the centroid P of this area.

If the plane surface is inclined and submerged below the surface, the pressure diagram is drawn perpendicular to the immersed surface and will be a straight line extending from p = 0 at the free surface to p = ρgH at depth H.

47 Example 2

A closed tank (Fig. 3.10), rectangular in plan with vertical sides, is 1.8 m deep and contains water to a depth of 1.2 m. Air is pumped into the space above the water until the air pressure is 35 kNm−2. If the length of one wall of the tank is 3 m, determine the resultant force on this wall and the height of the centre of pressure above the base.

48 Solution toexample 2

49 ExerciseForce on a curved surface

A sluice gate is in the form of a circular arc of radius 6 m as shown in Figure. Calculate the magnitude and direction of the resultant force on the gate, and the location with respect to O of a point on its line of action.

HINT : The vertical force is equal to the weight of the fluid displacedANSWER: Resultant force=179.46 kN at an angle of 10.27 degrees from the horizontal

50 Buoyancy and Stability

The same principles used to compute hydrostatic forces on surfaces can be applied to the net pressure force on a completely submerged or floating body. The results are the two laws of buoyancy discovered by Archimedes in the third century B.C.:• A body immersed in a fluid experiences a vertical buoyant force equal to the

weight of the fluid it displaces.• A floating body displaces its own weight in the fluid in which it floats.

(a) forces on upper and lower curved surfaces

Since the body lies between an upper curved surface 1 and a lower curved surface 2, the Archimedes law can be derived as:

51 …continuedAlternatively, as in the figure, we can sum the vertical forces on elemental vertical slices through the immersed body:

The same as in the previous derivation for a constant specific weight.

The line of action of the buoyant force passes through the center of volume of the displaced body; i.e., its center of mass is computed as if it had uniform density. This point through which acts is called the center of buoyancy.

(b) summation of elemental vertical pressure forces.

52 ..continuedStatic equilibrium of a floating body

Floating bodies are a special case; only a portion of the body is submerged, with the remainder poking up out of the free surface. the shaded portion is the displaced volume.

What if the fluid is layered?For a layered fluid (LF) by summing the weights of each layer of density displaced by the immersed body.

53 Example 1

A rectangular pontoon has a width B of 6m, a length l of 12 m, and a draught D of 1.5m in fresh water (density 1000 kgm-3). Calculate (a) the weight of the pontoon, (b) its draught in sea water (density 1025

kg m−3) and (c) the load (in kN) that can be supported

by the pontoon in fresh water if the maximum draught permissible is 2m.

Pontoon

B

D

54 Solution toExample 1

When the pontoon is floating in an unloaded condition,Upthrust on immersed volume = Weight of pantoon

Weight of pontoon = Weight of fluid displaced W = ρgBlD(a). In fresh water, ρ =1000 kg m−3and D =1.5 m;

(b). In sea water, ρ =1025 kg m−3;

(c) For the maximum draught of 2 m in fresh water,

55 Stability

Steps in the basic principle of the static-stability calculation.1. The basic floating position is calculated from the buoyance force computation

based on the weight of fluid displaced. The body’s center of mass G and center of buoyancy B are computed.

2. The body is tilted a small angle Δθ , and a new waterline is established for the body to float at this angle. The new position B’ of the center of buoyancy is calculated. A vertical line drawn upward from B’ intersects the line of symmetry at a point M, called the metacenter, which is independent of Δθ for small angles.a. If point M is above G, that is, if the metacentric height MG is positive, a

restoring moment is present and the original position is stable. b. If M is below G, (said to be negative MG), an overturning moment is

present and the body is unstable and will overturn if disturbed. Stability increases with increasing MG.

c. If M coincides with G, the body is in neutral equilibrium.The following figures explains the above cases.

56 …continued

57 Determination of the Metacenteric HeightTilting the body a small angle Δθ then submerges small wedge Obd and uncovers an equal wedge cOa,The new position B’ of the center of buoyancy is calculated as the centroid of the submerged portion aObde of the body.

If is the area moment of inertia of the waterline footprint of the body about its tilt axis O;

58 …continued

The first integral vanishes because of the symmetry of the original submerged portion cOdea. The remaining two “wedge” integrals combine into when we notice that L dx equals an element of waterline area.

Thus we determine the desired distance from Mto B.

i.e.

tan θ

The distance from G to B is obtained from the basic shape and design of the floating body

Calculate and is the area moment of inertia of

the waterline footprint of the body about its tilt axis O

59 Example 1

A barge has a uniform rectangular cross section of width 2L and vertical draft of height H, as in Figure. Determine (a) the metacentric height for a small

tilt angle and (b) the range of ratio L/H for which the

barge is statically stable if G is exactly at the waterline as shown.

Solution:If the barge has length b into the paper, the waterline area, relative to tilt axis O, has a base b and a height 2L; therefore, and

The barge can thus be stable only if,

The wider the barge relative to its draft, the more stable it is.

60 Pressure variation with rigid body motion

In rigid-body motion, all particles are in combined translation and rotation, and there is no relative motion between particles. With no relative motion, there are no strains or strain rates, so that the viscous term , in the net force balance equation, vanishes and it becomes;

The pressure gradient acts in the direction ( − ), and lines of constant pressure 𝒈 𝒂(including the free surface, if any) are perpendicular to this direction.

In this case there are two simple special cases; Uniform Linear Acceleration

Acceleration a have the same magnitude and direction for all particles.

Rigid-Body RotationThe fluid as a rigid body is rotating at an angular rotation of Ω about an axis with out any translation.

61 Rigid-body withUniform Linear Acceleration

The parallelogram sum of g and -a gives the direction of the pressure gradient or greatest rate of increase of p.

The surfaces of constant pressure must be perpendicular to this and are thus tilted at a downward angle θ such that

One of these tilted lines is the free surface, which is found by the requirement that the fluid retain its volume unless it spills out. The rate of increase of pressure in the direction g -a is greater than in ordinary hydrostatics and is

These results are independent of the size or shape of the container as long as the fluid is continuously connected throughout the container.

62 Example 1

A drag racer rests her coffee mug on a horizontal tray while she accelerates at 7m/s2. The mug is 10 cm deep and 6 cm in diameter and contains coffee 7 cm deep at rest. (a) Assuming rigid body acceleration of the

coffee, determine whether it will spill out of the mug.

(b) Calculate the gage pressure in the corner at point A if the density of coffee is 1010 kg/m3.

63 Solution to example 1

The free surface tilts at the angle θ is given by;

With and standard gravity, g

If the mug is symmetric about its central axis, the volume of coffee is conserved if the tilted surface intersects the original rest surface exactly at the centerline.

The deflection at the left side of the mug is:

This is less than the 3-cm clearance available, so the coffee will not spill unless it was sloshed during the start-up of acceleration.

When at rest, the gage pressure at point A is given by;

During acceleration, the pressure at point A is given by;

31% higher than the pressure at rest.

64 Rigid-Body Rotation

Consider rotation of the fluid about the z axis without any translation, as in the figure.

Assume that the container has been rotating long enough at constant Ω for the fluid to have attained rigid-body rotation.

The fluid acceleration will then be the centripetal direction, a.

The force balance becomes;

Equating like components,

The right hand side terms are known functions of r and z.

65 …continuedIntegrate the first equation “partially,’’i.e., holding z constant, with respect to r.

where the “constant’’ of integration is actually a function f(z). Differentiating p with respect to z and compare with the second relation;

Integrating for f(z),

Where C is arbitrary constant.Then substituting this for the relation of p,

This is the pressure distribution in the fluid.

The value of C is found by a boundary condition i.e. specifying the pressure at one point. If at (r, z) = (0, 0), then

→

The pressure is linear in z and parabolic in r. If we wish to plot a constant-pressuresurface, eg at

Thus the surfaces are paraboloids of revolution, concave upward, with their minimum point on the axis of rotation.

66 …continued

As in the previous example of linear acceleration, the position of the free surface is found by conserving the volume of fluid.

As in figure shown, for a cylinder rotating about its central axis, the still-water level is exactly halfway between the high and low points of the free surface.

The center of the fluid drops an amount h/2 and the edges rise an equal amount.

67 Example 1The coffee cup, in the previous example, is removed from the drag racer, placed on a turntable, and rotated about its central axis until a rigid-body mode occurs. Find (a) the angular velocity which will cause the

coffee to just reach the lip of the cup and (b) the gage pressure at point A for this

condition.

Solution:a) The cup contains 7 cm of coffee and the remaining distance of 3 cm up to the lip must equal the distance h/2.

This gives i.e.

b) To compute the pressure, keep the origin of coordinates r and z at the bottom of the free-surface depression

Solve,

68 …continued

This is about 43 percent greater than the still-water pressure 694 Pa.

69 Example 2

A U-tube with a radius of 10 in and containing mercury to a height of 30 in is rotated about its center at 180 r/min until a rigid-body mode is achieved. The diameter of the tubing is negligible. Atmospheric pressure is 2116 lbf/ft2 . Find the pressure at point A in the rotating condition.

Solution:Convert the angular velocity to radians per second,

For mercury,

At this high rotation rate, the free surface will slant upward at a fierce angle [about 84°], but the tubing is so thin that the free surface will remain at approximately the same 30-in height, point B.

70 …solution continuedPlacing our origin of coordinates at this height, we can calculate the constant C in pressure equation from the condition =2116 lbf/ft2 at (r, z) = (10 in, 0)

We then obtain by evaluating the pressure equation at (r, z) = (0, - 30 in):

This is less than atmospheric pressure. The free surface will cross the horizontal portion of the U-tube (where p will be atmospheric) and fall below point A.The actual drop from point B will be;

Thus is about (46-30)=16 inHg

Hence rotation has reduced the pressure at point A by 77 percent. Further rotation can reduce to near-zero pressure, and cavitation can occur.

71 Pressure Measurement

Two types of accurate manometers for precise measurements: (a) tilted tube with eyepiece; (b) micrometer pointer with ammeter detector.

72 Bourdon Tube

A bourdon tube device for mechanical measurement of high pressures.

When pressurized internally, a curved tube with flattened cross section will deflect outward. The deflection can be measured by a linkage attached to a calibrated dial pointer, as shown. Or the deflection can be used to drive electric-output sensors, such as a variable transformer. Similarly, a membrane or diaphragm will deflect under pressure and can either be sensed directly or used to drive another sensor.