chapter2 fluid me.ppt

7/29/2019 chapter2 fluid me.ppt

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CHAPTER 2:

FLUIDS IN RELATIVE EQUILIBRIUM:

HYDROSTATIC PRESSURE ANDBUOYANCY


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2.1 PRESSURE

Pressure is defined as a normal forceexerted by a fluid per unit area.Pressure is defined as force perunitarea, it has the unit of newtons per

square meter (N/m2), which is called apascal (Pa). That is,

1 Pa = 1 N/m2

ALIZA@SESI 2 SEM 2012/2013


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2.1 PRESSURE

Three other pressure units are bar, standardatmosphere, and kilogram-force per square

centimeter:

1 bar = 105 Pa = 0.1 Mpa = 100kPa1 atm = 101, 325 Pa = 101.325 kPa = 1.01325 bars

1 kgf/cm2 = 9.807 N/cm2

= 9.807 x 104 N/m2

= 9.807 x 104 Pa= 0.9807 bar= 0.9679 atm



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21 PRESSURE

The normal stress (orpressure) on thefeet of a chubbyperson is muchgreater than on the

feet of a slim person.



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21 PRESSURE

The actual pressure at a given position iscalled the absolute pressure, and it is

measured relative to absolute vacuum(i.e., absolute zero pressure)



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21 PRESSURE

Most pressure-measuringdevices, however, arecalibrated to read zero inthe atmosphere and so theyindicate the differencebetween the absolutepressure and the localatmospheric pressure.

This difference is called thegage pressure.



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21 PRESSURE

Pressures below atmosphericpressure are called vacuum

pressures and are measured byvacuum gages that indicate thedifferencebetween the atmosphericpressure and the absolute pressure.



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21 PRESSURE

Pgage = Pabs- Patm

Pvac = Patm- Pabs


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21 PRESSURE

EXAMPLE 21Absolute Pressure of a Vacuum Chamber

A vacuum gage connected to a chamber reads 5.8 psi at alocation where the atmospheric pressure is 14.5 psi. Determine

the absolute pressure in the chamber.AnalysisThe absolute pressure is easily determined.DiscussionNote that the localvalue of the atmospheric pressure

is used when determining the absolute pressure.

Pabs = Patm - Pvac= 14.5 - 5.8= 8.7 psi



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21-1 Variation of Pressure with Depth

Pressure in a fluidincreases with depth

because more fluidrests on deeperlayers, and the effectof this extra weight

on a deeper layer isbalanced by anincrease in pressure

The pressure of a fluid atrest increases with depth(as a result of addedweight).



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To obtain a relationfor the variation of

pressure withdepth, consider arectangular fluidelement in

equilibrium,Assuming the density of the

fluid r to be constant, a force balance in theverticalz-direction gives



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If we take point 1 to be at

the free surface of a liquid

open to the atmosphere

where the pressure is theatmospheric pressure Patm,

then the pressure at a depth

h from the free surface

becomes P2

P = Patm + gh or Pgage = gh



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Pressure in a fluid at rest is independentof the shape or cross section of thecontainer. It changes with the verticaldistance, but remains constant in otherdirections. Therefore, the pressure is thesame at all points on a horizontal plane in

a given fluid.



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Since there can be noshearing forces for a fluid atrest, and there will be noaccelerating forces, the sumof the forces in any directionmust therefore, be zero. The

forces acting are due to thepressures on thesurrounding and the gravityforce.



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Force due to Px = Px x Area ABEF =Pxdydz

Horizontal component of force due to Ps= - (Ps x Area ABCD) sin(q) = - Psdsdzdy/ds = -Psdydz

As Py has no component in the x

direction, the element will be inequilibrium, if

Pxdydz + (-Psdydz) = 0,i.e. Px = Ps



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Similarly in the y direction, force dueto Py = Pydxdz Since dx, dy, and dzare very small quantities, dxdydz isnegligible in comparison with othertwo vertical force terms, and theequation reduces to,

Component of force due to Ps = - (Psx Area ABCD) cos() = - Psdsdz dx/ds= - Psdxdz



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Force due to weight of element= - mg= - rVg= - r (dxdydz/2) g

Since dx, dy, and dz are very small quantities,dxdydz is negligible in comparison with other twovertical force terms, and the equation reduces

to,Py = Ps. Therefore, Px = Py = Ps i.e. pressure at a point is same in all directions.

This is Pascal's law. This applies to fluid at rest.



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Example 2.2Variation of Pressure with Depth

What is the pressure difference in 1m of sea water

compared with 100m of sea water ( = 1 x 103 kgm-3)

P1 = atmospheric pressure = 1.013 x 105Pa

At 1m

P2 = P1 + gh= 1.013 x 105Pa + 1 x 103 kgm-3(9.81 ms-2) (1m)

= 1.11 x 105Pa



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At 100m

P2 = P1 + gh

= 1.013 x 105

Pa + 1 x 103

kgm-3(9.81 ms-2) (100m)= 1.08 x 106Pa

Pressure difference = 1.08 x 106Pa - 1.11 x 105Pa =

9.75 105Pa



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2.2 PRESSURE MEASUREMENT

2.2.1 Manometer

Plastic / glass u-tube

Water, oil, mercury,

alcohol

P2= P1

P2 = Patm + gh



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EXAMPLE 23 MeasuringPressure with a Manometer

A manometer is used to measure thepressure in a tank. The fluid usedhas a specific gravity of 0.85, and themanometer column height is 55 cm,as shown. If the local atmosphericpressure is 96 kPa, determine theabsolute pressure within the tank.

A B



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AssumptionsThe fluid in the tank is a gas whosedensity is much lower than the density ofmanometer fluid.

PropertiesThe specific gravity of the manometerfluid is given to be 0.85.We take the standarddensity of water to be 1000 kg/m3.

AnalysisThe density of the fluid is obtained bymultiplying its specific gravity by the density ofwater, which is taken to be 1000 kg/m3:



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fluid =SG (H2O) = (0.85)(1000 kgm3) = 850 kgm3

then using

ghP

PP

atm

BA

....

22

23

1000

1

1.1

155.081.985096

m

N

kPa

s

mkg

NmmskgmkPa

PA= 100.6 kPa



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In stacked-up fluid layers, thepressure change across a fluidlayer of density, and height, his gh.

Patm + 1gh1 + 2gh2 + 3gh3 = P1


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A relation for the pressure

difference P1 - P2 can beobtained by starting

at point 1 with P1, moving along

the tube by adding or

subtracting the ghterms until we reach point 2,

and setting the result equal to

P2:

Assume that , PA=PB, so

P1 +1g(a+h) = P2

P1 P2 = (2- 1) gh

+ 2gh+ 1ga



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EXAMPLE 24Measuring Pressure with aMultifluid Manometer

The water in a tank is pressurized by air, and thepressure is measured by a multifluid manometer asshown in. The tank is located on a mountain at analtitude of 1400 m where the atmospheric pressureis 85.6 kPa.

Determine the air pressure in the tank ifh1 0.1 m,h2 0.2 m, and h3 0.35 m. Take the densities ofwater, oil, and mercury to be 1000 kg/m3,850kg/m3, and 13,600 kg/m3, respectively.



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a b



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AssumptionThe air pressure in the tank is uniform (i.e.,itsvariation with elevation is negligible due to its lowdensity), and thus we can determine the pressure at the

airwater interface.AnalysisStarting with the pressure at point 1 at the air

water interface,moving along the tube by adding orsubtracting the ghterms until we reach point 2, andsetting the result equal to Patm since the tube is open to

the atmosphere gives



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3121

3121

ghghghPP

PghghghP

PP

mercuryoilwateratm

atmmercuryoilwater

BA

223332

213

/10001

/.112.0/8501.0(/100035.0/13600/81.96.85

mNkPa

smkgNmmkgmmkgmmkgsmkPa

hhhgP oilwatermercuryatm

So,

= 130 kPa



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DiscussionNote that jumping horizontallyfrom one tube to the next and realizing thatpressure remains the same in the same

fluid simplifies the analysis considerably.Also note that mercury is a toxic fluid, andmercury manometers and thermometersare being replaced by ones with safer fluidsbecause of the risk of exposure to mercury

vapor during an accident.



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2.2.2 Barometer Atmospheric pressure ismeasured by a device called a

barometer; thus, the

atmospheric pressure is oftenreferred to as the barometric

pressure.

Writing a force balance

in the vertical direction gives

Patm = gh



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EXAMPLE 25 Measuring AtmosphericPressure with a Barometer

Determine the atmospheric pressure at a locationwhere the barometric reading is 740 mm Hg and thegravitational acceleration is g 9.81 m/s2. Assumethe temperature of mercury to be 10C, at which itsdensity is 13,570 kg/m3.



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AnalysisFrom equation , the atmosphericpressure is determined to be

Patm = gh= (13,570 kgm3)(9.81 ms2)(0.74 m)

( 1 N/ kg. ms2)( 1kPa/1000 Nm= 98.5 kPa

DiscussionNote that density changes withtemperature, and thus this effect should beconsidered in calculations.



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2.3 HYDROSTATIC FORCES

2.3.1 ON SUBMERGED PLANE SURFACES

A plate exposed to a liquid, such as a gate

valve in a dam, the wall of a liquid storagetank, or the hull of a ship at rest, is subjectedto fluid pressure distributed over its surface

On a planesurface, the hydrostatic forces

form a system of parallel forces, and we oftenneed to determine the magnitudeof the forceand its point of application, which is calledthe center of pressure.


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When analyzinghydrostatic forces onsubmerged surfaces, theatmospheric pressure canbe subtracted forsimplicity when it acts onboth sides of thestructure.




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Hydrostatic force on an inclined plane surface completelysubmerged in a liquid.




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Based on the diagram above,

P = P0 + gh = P0 gy sin

The pressure at the centroid of asurface is equivalent to the averagepressure on the surface.

FR = (P0 gyCsin )A

= (P0ghC)A

= PCA

= PaveA



2 3 HYDROSTATIC FORCES


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The magnitude of theresultant force actingon a plane surface of a

completely submergedplate in homogeneous(constant density) fluidis equal to the productof the pressure P

Cat

the centroid of thesurface and the areaAof the surface




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Pressure acts normal to the surface, and thehydrostatic forces acting on a flat plate of any shapeform a volume whose base is the plate area andwhose height is the linearly varying pressure, asshown below.This virtual pressure prismhas aninteresting physical interpretation: its volumeis equalto the magnitudeof the resultant hydrostatic force

acting on the plate since V P dA, and the line ofaction of this force passes through the centroidofthis homogeneous prism.



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The projection of the centroid

on the plate is thepressure

center. Therefore, with theconcept of pressure prism, the

problem of describing the

resultant hydrostatic force on a

plane surface reduces to

finding the volume and the two

coordinates of the centroid

of this pressure prism.



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The resultant hydrostaticforce on the upper

surface is equal to theaverage pressure, whichis the pressure at themidpoint of the surface,times the surface areaA.

That is,


Special Case: Submerged Rectangular Plate

FR =PC A= [P0g(s + b/2) sin ]ab


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When the upper edge of the plate is at the free surface

and thus s = 0,

Tilted rectangular plate (s = 0),FR= [P0 g(b sin )/2]ab



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EXAMPLE 26Hydrostatic Force Acting onthe Door of a Submerged Car

A heavy car plunges into a lake during an accidentand lands at the Bottom of the lake on its wheelsas shown. The door is 1.2 m high and 1 m wide,and the top edge of the door is 8 m below the freesurface of the water. Determine the hydrostaticforce on the door and the location of the pressurecenter, and discuss if the driver can open the door.




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Assumptions1 The bottom surface of the lakeis horizontal. 2 The passenger cabin is well-sealed so that no water leaks inside. 3 The door

can be approximated as a vertical rectangularplate. 4 The pressure in the passenger cabinremains at atmospheric value since there is nowater leaking in, andthus no compression of theair inside. Therefore, atmospheric pressurecancels out in the calculations since it acts on

both sides of the door. 5 The weight of the car islarger than the buoyant force acting on it.



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Analysis The average pressure on the door is thepressure value at the centroid (midpoint) of the door

and is determined to be. Then the resultant hydrostaticforce on the door becomes

Pave = PC = ghC =g(s +b/2)= (1000 kg/m3)(9.81 ms2)(8+ 1.2/2 m)

(1 kN/1000 kg.ms2)= 84.4 kNm2



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Then the resultant hydrostatic force on the doorbecomes

FR = PaveA =(84.4 kN/m2) (1 m x1.2 m)= 101.3 kN

The pressure center is directly under the midpointof the door, assume P

0

= 0

yp = s+ b/2 +b2/(12( s+ b))

= 8 +1.2/2 + 1.22/ (12(8 + 1.2))

= 8.61 m



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EXAMPLE 2-7

A rectangular gate that leansagainst the floor with an angle

of 45

with the horizontal is tobe opened from its lower edgeby applying a normal force atits center. The minimum force Frequired to open the water gateis to be determined.




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Assumptions1 The atmospheric pressure acts onboth sides of the gate, and thus it can be ignored incalculations for convenience. 2 Friction at the hingeis negligible.

AnalysisThe length of the gate and the distance ofthe upper edge of the gate (point B) from the freesurface in the plane of the gate are




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The average pressure on a surface is the

pressure at the centroid (midpoint) of the surface,

and multiplying it by the plate area gives

the resultant hydrostatic on the surface,



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The distance of the pressure center from the free

surface of water along the plane of the gate is



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The distance of the pressure center from the hinge

at point B is




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Taking the moment about point B and setting it equal to

zero gives

Solving forFand substituting, the required force is

determined to be


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Discussion The applied force is inverselyproportional to the distance of the point of

application from the

hinge, and the required force can be reduced by

applying the force at a lower point on the gate.



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EXAMPLE 2.8

The density of a wood log is to

be measured by tying leadweights to it until both the logand the weights arecompletely submerged, andthen weighing them separately

in air. The average density ofa given log is to bedetermined by this approach.



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Analysis The weight of a body is equal to the buoyantforce when the body is floating in a fluid while being

completely submerged in it (a consequence of verticalforce balance from static equilibrium). In this case

the average density of the body must be equal to the

density of the fluid since



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Substituting, the volume and density of the log are

determined to be



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Discussion Note that the log must be completelysubmerged for this analysis to be valid. Ideally, the

lead weights must also be completely submerged,but this is not very critical because of the small

volume of the lead weights.



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2.3.2 ON SUBMERGED CURVED SURFACES

For a submerged curved surface, the determination of theresultant hydrostatic force is more involved since ittypically requires the integration of the pressure forcesthat change direction along the curved surface. Theconcept of the pressure prism in this case is not much

help either because of the complicated shapes involved.



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The easiest way to determine the resultanthydrostatic force FRacting on a two-dimensionalcurved surface is to determine the horizontal and

vertical components FHand FVseparately. This isdone by considering the free-body diagram of theliquid block enclosed by the curved surface and thetwo plane surfaces (one horizontal and one vertical)passing through the two ends of the curved surface,

as shown



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Fluid block is in static equilibrium, the forcebalances in the horizontal and vertical directionsgive

Horizontal force componenton curved surface, FH = Fx

Vertical force componenton curved surface, FV = Fy + W



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Thus, we conclude that1. The horizontal component of the hydrostatic force

acting on a curved surface is equal (in both

magnitude and the line of action) to the hydrostaticforce acting on the vertical projection of the curvedsurface.

2. The vertical component of the hydrostatic force

acting on a curved surface is equal to thehydrostatic force acting on the horizontal projectionof the curved surface, plus (minus, if acting in theopposite direction) the weight of the fluid block.



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When the curved surface is a circular arc(fullcircle or any part of it), the resultant hydrostaticforce acting on the surface always passes through

the center of the circle. This is because thepressure forces are normal to the surface, and alllines normal to the surface of a circle passthrough the center

of the circle.



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Thus, the pressure forces form a

concurrent force system

at the center, which can be

reduced to a single equivalent

force at that point



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hydrostatic forces acting on a plane or curved surface

submerged in a multilayered fluid of differentdensities can be determined by considering different

parts of surfaces in different fluids as differentsurfaces, finding the force on each part, and thenadding them using vector addition. For a planesurface, it can be expressed as



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is the pressure at thecentroid of the portion of thesurface in fluid iandAiis the area of the plate inthat fluid. The line of action of this equivalent force

can be determined from the requirement thatthe

moment of the equivalent force about any point is

equal to the sum ofthe moments of the individualforces about the same point.

Pc,i= Po+ighc,i



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EXAMPLE 2.9The height of a waterreservoir is controlled by

a cylindrical gate hingedto the reservoir. Thehydrostatic force on thecylinder and the weightof the cylinder per ftlength are to bedetermined.



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Assumptions

1. The hinge is frictionless. 2 The atmosphericpressure acts on both sides of the gate, and thus it

can be ignored in calculations for convenience.

Properties

We take the density of water to be 62.4 lbm/ft3throughout.



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Then the magnitude and direction of the

hydrostatic force acting on the cylindrical

surface become



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Therefore, the magnitude of the hydrostatic

force acting on the cylinder is 2521 lbf per ftlength of the cylinder, and its line of action

passes through the center of the cylindermaking an angle 46.6 upwards from thehorizontal.



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(b) When the water level is 15-ft high, the gateopens and the reaction force at the bottom ofthe cylinder becomes zero. Then the forces other

than those at the hinge acting on the cylinderare its weight, acting through the center, and thehydrostatic force exerted by water. Taking amoment about the pointAwhere the hinge isand equating it to zero gives



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DiscussionThe weight of the cylinder per ftlength is determined to be 1832 lbf, whichcorresponds to a mass of 1832 lbm, and to a

density of 296 lbm/ft3 for the material of thecylinder.



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EXAMPLE 210A Gravity-Controlled

Cylindrical Gate

A long solid cylinder of radius 0.8 m hinged at point

Ais used as an automatic gate, as shown below.When the water level reaches 5 m, the gate opens byturning about the hinge at pointA. Determine

(a) the hydrostatic force acting on the cylinder and

its line of action when the gate opens and(b) the weight of the cylinder per m length of thecylinder.



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Assumptions 1. Friction at the hinge is negligible.

2.Atmospheric pressure acts on both sides of thegate, and thus it cancels out.

Analysis(a) We consider the free-body diagram ofthe liquid block enclosed by the circular surface ofthe cylinder and its vertical and horizontalprojections. The hydrostatic forces acting on the

vertical and horizontal plane surfaces as well as theweight of the liquid block are determined asHorizontal force on vertical surface:


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Then the magnitude and direction of the hydrostaticforce acting on the cylindrical surface becomes

Therefore, the magnitude of the hydrostatics force

acting on the cylinder is 52.3 kN per m length of thecylinder, and its line of action passes through thecentre of the cylinder making an angle 46.40 with thehorizontal



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b) When the water level is 15 m high, the gate isabout to open and thus the reaction force at thebottom of the cylinder is zero. Then the forces other

than those at the hinge acting on the cylinder areits weight, acting through the centre, and thehydrostatic force exerted by water. Taking amoment about point A at the location of the hingeand equating it to zero gives



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Discussion The weight of the cylinder per m length isdetermined to be 37.9 kN. It can be shown that thiscorresponds to a mass of 3863 kg per m length and toa density of 1921 kg/m3 for the material of the cylinder.



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2.4 BUOYANCY & STABILITY

The force that tendsto lift the body iscalled the buoyantforce and is denoted

by FB. The buoyantforce is caused bythe increase ofpressure in a fluid

with depth.



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The difference between these two forces is a netupward force, which is the buoyant force,

FB = FbottomFtop

=fg(s +h)A -fgsA=fghA

=f

gV



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Where V = hA is the volume of the plate. Butthe relation fgVis simply the weight of theliquid whose volume is equal to the volume of

the plate. Thus, we conclude that the buoyantforce acting on the plate is equal to the weightof the liquid displaced by the plate.Note thatthe buoyant force is independent of thedistance of the body from the free surface. It isalso independent of the density of the solidbody.



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Consider an arbitrarily shaped solid bodysubmerged in a fluid at rest and compare it to abody of fluid of the same shape indicated bydotted lines at the same distance from the free

surface as shown below.

The buoyant forces acting on these two bodies arethe same since the pressure distributions, whichdepend only on depth, are the same at the

boundaries of both. The imaginary fluid body is instatic equilibrium, and thus the net force and netmoment acting on it are zero.



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Therefore,the upward buoyant

force must be equal to the

weight of the imaginary fluid

body whose volume is equal tothe volume of the solid body.

Further, the weight and the

buoyant force must have the

same line of action to have a

zero moment. This is known as

Archimedes principle,



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For floatingbodies, the weight of the entire bodymust be equal to the buoyant force, which is theweight of the fluid whose volume is equal to thevolume of the submerged portion of the floating

body. That is, of uniform density, its weight Wsalso acts through the centroid, but its

FB = W

fgVsub = ave,bodygVtotal

Vsub/Vtotal = ave,body/f


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It follows from these discussions

that a body immersed in a fluid (1)

remains at rest at any point in the

fluid when its density is equal tothe density of the fluid, (2) sinks

to the bottom when its density is

greater than the density of the

fluid, and (3) rises to the surface

of the fluid and floats when thedensity of the body is less than

the density of the fluid



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EXAMPLE 211Weight Loss of an Objectin Seawater

A crane is used to lower weights into the sea

(density 1025 kg/m3) for an underwater constructionproject as shown.Determine the tension in the ropeof the crane due to a rectangular 0.4-m x 0.4-m x3-m concrete block (density = 2300 kg/m3) when it

is (a) suspended in the air and (b) completelyimmersed in water.



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Assumptions1 The buoyancy of air is negligible.2 The weight of the ropes is negligible.

PropertiesThe densities are given to be 1025kg/m3 for seawater and 2300 kg/m3 for concrete.

Analysis(a) Consider the free-body diagram of theconcrete block. The forces acting on the concreteblock in air are its weight and the upward pull action

(tension) by the rope. These two forces must balanceeach other, and thus the tension in the rope must beequal to the weight of the block:

3 4 BUOYANCY & STABILITY


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V =(0.4 m)(0.4 m)(3 m) = 0.48 m3

FT,air= W =concretegV



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(b) When the block is immersed in water, there isthe additional force of buoyancy acting upward. Theforce balance in this case gives

FB=fgV

FT ,water= WFB= 10.8 -4.8 = 6.0 kN



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DiscussionNote that the weight of the concreteblock, and thus the tension of the rope, decreasesby (10.8 6.0)/10.8 55 percent in water.



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EXAMPLE 2.12

The height of theportion of a cubic ice

block that extendsabove the watersurface is measured.The height of the ice

block below the surfaceis to be determined.



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Assumptions1 The buoyancy force in air isnegligible. 2 The top surface of the ice block isparallel to the surface of the sea.

PropertiesThe specific gravities of ice andseawater are given to be 0.92 and 1.025,respectively, and thus the corresponding densitiesare 920 kg/m3 and 1025 kg/m3.



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AnalysisThe weight of a body floating in a fluid isequal to the buoyant force acting on it (aconsequence of vertical force balance from static

equilibrium). Therefore, in this case the averagedensity of the body must be equal to the density ofthe fluid since



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where his the height of the ice block below thesurface. Solving for hgives

h= 0.876 m = 87.6 cm

DiscussionNote that the 0.92/1.025 = 88% of thevolume of an ice block remains under water. Forsymmetrical ice blocks this also represents the

fraction of height that remains under water.



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EXAMPLE 2.13

An irregularly shapedbody is weighed in air

and then in water witha spring scale. Thevolume and the averagedensity of the body are

to be determined.



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The difference between the weights in air and inwater is due to the buoyancy force in water,



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Discussion The volume of the body can also bemeasured by observing the change in the volume of

the container when the body is dropped in it

(assuming the body is not porous).



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Stability of Immersed and Floating Bodies


For floating bodiessuch as ships,

stability is an

important

consideration for

safety.



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ball on the flooranalogy to explain the fundamentalconcepts of stability and instability.


stable since any smalldisturbance (someone moves

the ball to the right or left)

generates a restoring force

(due to gravity) that returns it

to its initial position.



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neutrally stablebecause ifsomeone moves the ball tothe right or left, it would stayput at its new location. It hasno tendency to move back toits original location, nor doesit continue to move away.




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is a situation in which the ball

may be at rest at the moment,

but any disturbance, even an

infinitesimal one, causes theball to roll off the

hillit does not return to its

original position; rather it

diverges from it. Thissituation is unstable.



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ball is on an inclinedfloor?




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The rotational stability criteria are similar forfloating bodies.Again, if the floating body isbottom-heavy and thus the center of gravity Gis

directly below the center of buoyancy B, the body isalways stable. But unlike immersed bodies, afloating body may still be stable when Gis directlyabove Bas shown before.




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A rotational disturbance of the body in suchcases produces a restoring momentto return thebody to its original stable position. Thus, a stabledesign for a submarine calls for the engines andthe cabins for the crew to be located at the lowerhalf in order to shift the weight to the bottom asmuch as possible.




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An immersed body whose center of gravity Gisdirectly above point Bis unstable, and anydisturbance will cause this body to turn upsidedown. A body for which Gand Bcoincide isneutrally stable. This is the case for bodieswhose density is constant throughout. For suchbodies, there is no tendency to overturn orright themselves.




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What about a case

where the center of

gravity is notvertically aligned

with the center of

buoyancy????



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This is because the centroid of the displacedvolume shifts to the side to a point Bduring arotational disturbance while the center of

gravity Gof the body remains unchanged. Ifpoint Bis sufficiently far, these two forcescreate a restoring moment and return the bodyto the original position





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A measure of stability for floating bodies is themetacentric heightGM, which is the distancebetween the center of gravity Gand the

metacenter Mthe intersection point of thelines of action of the buoyant force through thebody before and after rotation. The metacentermay be considered to be a fixed point for mosthull shapes for small rolling angles up to about20.





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A floating body is stable if point Mis above pointG, and thus GMis positive, and unstable if pointMis below point G, and thus GMis negative. Inthe latter case, the weight and the buoyant force

acting on the tilted body generate an overturningmoment instead of a restoring moment,causingthe body to capsize. The length of themetacentric height GMabove Gis a measure ofthe stability: the larger it is, the more stable is thefloating body.





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A floating body is stable if the

body is bottom-heavy and thus

the center of gravity G is belowthe centroid B of the body, or ifthe metacenter, Mis above pointG. However, the body is unstableif point Mis below point G.


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