CHAP02 Munson

1

FUNDAMENTALS OFFUNDAMENTALS OF FLUID MECHANICS FLUID MECHANICS

Chapter 2 Fluids at Rest Chapter 2 Fluids at Rest

- Pressure and its Effect- Pressure and its Effect

2

MAIN TOPICSMAIN TOPICS

Pressure at a PointPressure at a Point Basic Equation for Pressure FieldBasic Equation for Pressure Field Pressure variation in a Fluid at RestPressure variation in a Fluid at Rest Standard AtmosphereStandard Atmosphere Measurement of PressureMeasurement of Pressure ManometryManometry Mechanical and Electronic Pressure Measuring DevicesMechanical and Electronic Pressure Measuring Devices Hydrostatic Force on a Plane SurfaceHydrostatic Force on a Plane Surface Pressure PrismPressure Prism Hydrostatic Force on a Curved SurfaceHydrostatic Force on a Curved Surface Buoyancy, Floating, and StabilityBuoyancy, Floating, and Stability Rigid-Body MotionRigid-Body Motion

3

Pressure at a Point Pressure at a Point 1/41/4

Pressure ? Pressure ? Indicating the normal force per unit area at a given point Indicating the normal force per unit area at a given point

acting on a given plane within the fluid mass of interest.acting on a given plane within the fluid mass of interest. How the pressure at a point varies with the orientation of How the pressure at a point varies with the orientation of

the plane passing through the point ?the plane passing through the point ?

4


Consider the free-body diagram within a fluid mass. Consider the free-body diagram within a fluid mass. In which there are no shearing stress, the only external In which there are no shearing stress, the only external

forces acting on the wedge are due to the pressure and the forces acting on the wedge are due to the pressure and the weight.weight.

5


The equation of motion (Newton’s second law, F=ma) in The equation of motion (Newton’s second law, F=ma) in the y and z direction are,the y and z direction are,

zSzZ

ySyy

a2

δxδyδzρ

2

δxδyδzγ-δxδscosθPδxδyPF

a2

δxδyδzρδxδssinθPδxδzPF

2

δyρaPP

2

δzγρaPP ySyZSZ

sinszcossy

δx=0δx=0 、、 δy=0δy=0 、、 δz=0δz=0 SyZ PPP

6


The The pressure at a pointpressure at a point in a fluid at rest, or in motion, is in a fluid at rest, or in motion, is independent of the directionindependent of the direction as long as there are no as long as there are no shearing stresses present.shearing stresses present.

The result is known asThe result is known as Pascal’s lawPascal’s law named in honor ofnamed in honor of Blaise PascalBlaise Pascal (1623-1662). (1623-1662).

7

Basic Equation for Pressure FieldBasic Equation for Pressure Field

To obtain an basic equation for pressure field in a static To obtain an basic equation for pressure field in a static fluid.fluid.

Apply Newton’s second law to a differential fluid massApply Newton’s second law to a differential fluid mass

aδmFδ There are two types of There are two types of

forces acting on the mass forces acting on the mass of fluid: of fluid: surface force and surface force and body force.body force.

Vm

8

Body Force on ElementBody Force on Element

VgmgFB

kzyxkFB

Vm

Where ρ is the density.Where ρ is the density.

gg is the local is the local

gravitational acceleration.gravitational acceleration.

9

Surface Forces Surface Forces 1/41/4

No shear stresses, the only No shear stresses, the only surface force is the surface force is the pressure force.pressure force.

10


The pressure at the left faceThe pressure at the left face

The pressure at the right faceThe pressure at the right face

The pressure force in y The pressure force in y directiondirection

δxδyδzy

pδxδz

2

dy

y

ppδxδz

2

dy

y

ppδFy

2

dy

y

pp

2

dy

y

ppyy

y

ppp LL

2

dy

y

pp

2

dy

y

ppyy

y

ppp RR

11


The pressure force in x directionThe pressure force in x direction

The pressure force in z directionThe pressure force in z direction

δxδyδzx

pδyδz

2

dx

x

ppδyδz

2

dx

x

ppδFx

δxδyδzz

pδxδy

2

dz

z

ppδxδy

2

dz

z

ppδFz

12


The net surface forces acting on the elementThe net surface forces acting on the element

δxδyδzkz

pj

y

pi

x

pkδFjδFiδFFδ ZYXs

kz

pj

y

pi

x

ppgradp

δxδyδzpδxδyδz)(gradpFδ s

13

General Equation of MotionGeneral Equation of Motion

zyxakzyxzyxp

Vd)gp(

zyx)gp(FFF BS

akp

The general equation of motion for a fluid in which there are no shearing stresses

14

Pressure Variation in a Fluid at RestPressure Variation in a Fluid at Rest

For a fluid at rest For a fluid at rest aa=0=0 0akp

directionz...0gz

p

directiony...0gy

p

directionx...0gx

p

z

y

x

z

P

0y

P

0x

P

gg

,0g,0g

z

yx

gdz

dp

15

Pressure-Height RelationPressure-Height Relation

The basic pressure-height relation of static fluid :The basic pressure-height relation of static fluid :

Restriction:Restriction:Static fluid.Static fluid.Gravity is the only body force.Gravity is the only body force.The z axis is vertical and upward.The z axis is vertical and upward.

Integrated to determine the Integrated to determine the pressure distribution in a pressure distribution in a static fluidstatic fluid with appropriatewith appropriateboundary conditions.boundary conditions.

How the specific weight varies with z?How the specific weight varies with z?

gdz

dp

16

Pressure in Incompressible FluidPressure in Incompressible Fluid

A fluid with constant density is called an incompressible A fluid with constant density is called an incompressible fluid.fluid.

1

2

2

1

z

z

p

pdzdp

pp11 －－ pp22 = γ(z = γ(z22 －－ zz11)=γh)=γhpp11=γh +p=γh +p22 　　

h= zh= z22 －－ zz11 ，， h is the depth of fluid h is the depth of fluid measured downward from the measured downward from the location of plocation of p22..

This type of pressure distribution is This type of pressure distribution is called a hydrostatic distribution.called a hydrostatic distribution.

gdz

dp

17

Pressure Head in Static Fluid Pressure Head in Static Fluid 1/21/2

The pressure difference between two points in a fluid at The pressure difference between two points in a fluid at rest: prest: p11 －－ pp22 = γ(z = γ(z22 －－ zz11)=γh)=γh

21 pph h is called the pressure head h is called the pressure head

and is interpreted as the and is interpreted as the height of a column of fluid of height of a column of fluid of specific weight γrequired to specific weight γrequired to give a pressure difference pgive a pressure difference p11--pp22..

18

Pressure Head in Static Fluid Pressure Head in Static Fluid 2/22/2

The pressure p at any depth h below the free surface is The pressure p at any depth h below the free surface is given by given by p = γh + pp = γh + poo

The pressure in a homogeneous, The pressure in a homogeneous, incompressible fluid at rest incompressible fluid at rest depends on the depth of the fluid depends on the depth of the fluid relative to some reference plane, relative to some reference plane, and and it is not influenced by the it is not influenced by the size or shape of the tanksize or shape of the tank or or container in which the fluid is container in which the fluid is held.held.

19

Example 2.1 Pressure-Depth RelationshipExample 2.1 Pressure-Depth Relationship

Because of a leak in a buried gasoline storage tank, water has Because of a leak in a buried gasoline storage tank, water has seeped in to the depth shown in Figure E2.1. If the specific gravity seeped in to the depth shown in Figure E2.1. If the specific gravity of the gasoline is SG=0.68. Determine the pressure at the gasoline-of the gasoline is SG=0.68. Determine the pressure at the gasoline-water interface and at the bottom of the tank. Express the pressure in water interface and at the bottom of the tank. Express the pressure in units of lb/ftunits of lb/ft22, lb/in, lb/in22, and as pressure head in feet of water., and as pressure head in feet of water.

Determine the pressure at the gasoline-Determine the pressure at the gasoline-water interface and at the bottom of the water interface and at the bottom of the tanktank

20

Example 2.1 SolutionSolution1/2

20

03

0OH1

ft/lbp721

p)ft17)(ft/lb4.62)(68.0(phSGp2

The pressure at the interface isThe pressure at the interface is

ppo o =0=0

ft6.11ft/lb4.62

ft/lb721p

in/lb01.5ft/in144

ft/lb721ft/lb721p

3

2

OH

1

222

22

1

2

ppoo is the pressure at the free surface of the gasoline.is the pressure at the free surface of the gasoline.

21


ft6.14ft/lb4.62

ft/lb908p

in/lb31.6ft/in144

ft/lb908p

ft/lb908ft/lb721)ft3)(ft/lb4.62(php

3

2

OH

2

222

2

2

2231OHOH2

2

22

The pressure at the tank The pressure at the tank bottombottom

22

Transmission of Fluid PressureTransmission of Fluid Pressure

The The required equality of pressure at equal elevations required equality of pressure at equal elevations throughout a systemthroughout a system is important for the operation of is important for the operation of hydraulic jacks, lifts, and presses, as well as hydraulic hydraulic jacks, lifts, and presses, as well as hydraulic controls on aircraft and other type of heavy machinery.controls on aircraft and other type of heavy machinery.

11

222211 F

A

AFpAFpAF

The transmission of fluid pressure throughout a stationary fluid is the principle upon which many hydraulic deviceshydraulic devices are based.

23

Pressure In Compressible Fluid Pressure In Compressible Fluid 1/21/2

For compressible fluid, For compressible fluid, ρ=ρ(P,T)ρ=ρ(P,T) , how to determine the , how to determine the pressure variation? pressure variation?

The density must be expressed as a function of one of the The density must be expressed as a function of one of the other variable in the equation.other variable in the equation.

For example: Determine the pressure variation in the For example: Determine the pressure variation in the ideal ideal gas.gas.

RT

gp

dz

dpRTp

1

2

2

1

z

z1

2p

p T

dz

R

g

p

pln

p

dp

gdz

dp

24

Pressure In Compressible Fluid Pressure In Compressible Fluid 2/22/2

T=TT=T00=constant=constant

o

1212 RT

)zz(gexppp

T=Ta-T=Ta-ββzz

R/g

aa

R/g

aa

aa

z

0a

p

p

a

)T

T(p)

T

z1(pp

)T

z1ln(

R

g

p

pln

)zT(R

gdz

p

dp

dz)mzT(R

pgdz

RT

pggdzdp

a

Pa is the absolute pressure at z=0

25

Example 2.2 Incompressible and Example 2.2 Incompressible and Isothermal Pressure-Depth VariationsIsothermal Pressure-Depth Variations

The Empire State Building in New York City, one of the tallest The Empire State Building in New York City, one of the tallest building in the world, rises to a height of approximately 1250ft. building in the world, rises to a height of approximately 1250ft. Estimate the ratio of the pressure at the top of the building to the Estimate the ratio of the pressure at the top of the building to the pressure at its base, assuming the air to be at a common temperature pressure at its base, assuming the air to be at a common temperature of 59of 59°F°F. Compare this result with that obtained by assuming the air . Compare this result with that obtained by assuming the air to be incompressible with to be incompressible with =0.0765fb/ft=0.0765fb/ft33 at 14.7psi (abs). at 14.7psi (abs).

26


955.0)ft/in144)(in/lb7.14(

)ft1250)(ft/lb765.0(1

p

zz1

p

p

orzzpp

222

3

1

12

1

2

1212

For For isothermal conditionsisothermal conditions

For For incompressible conditionsincompressible conditions

956.0]R)46059)[(Rslug/lbft1716(

)ft1250)(s/ft2.32(exp

RT

)zz(gexp

p

p

2

o

12

1

2

27


Note that there is little difference between the two results. Since the Note that there is little difference between the two results. Since the pressure difference between the bottom and top of the building is pressure difference between the bottom and top of the building is small, it follows that the variation in fluid density is small and, small, it follows that the variation in fluid density is small and, therefore, the compressible fluid and incompressible fluid analyses therefore, the compressible fluid and incompressible fluid analyses yield essentially the same result.yield essentially the same result.

By repeating the calculation, for various values of height, h, the By repeating the calculation, for various values of height, h, the results shown in figure are obtained.results shown in figure are obtained.

28

Standard AtmosphereStandard Atmosphere1/21/2

Standard atmosphere was first developed in the Standard atmosphere was first developed in the 1920s1920s. . The currently accepted Standard atmosphere is based on a The currently accepted Standard atmosphere is based on a report report published in 1962 and updated in 1976published in 1962 and updated in 1976..

The so-called U.S. standard atmosphere is an The so-called U.S. standard atmosphere is an idealized idealized representationrepresentation of of middle-latitude, year-around mean middle-latitude, year-around mean conditions of the earth’s atmosphereconditions of the earth’s atmosphere. .

29

Standard AtmosphereStandard Atmosphere2/22/2

For example, in the For example, in the troposphere, the temperature troposphere, the temperature variation is of the form variation is of the form

T = TT = Taa– βz– βz

where Ta is the temperature at where Ta is the temperature at sea level (z=0) and β is the sea level (z=0) and β is the lapse rate (the rate of change of lapse rate (the rate of change of temperature with elevation).temperature with elevation).

R/g

aa T

z1pp Pa is the absolute pressure at z=0.Pa is the absolute pressure at z=0.

30

Measurement of Pressure: Measurement of Pressure: Absolute and GageAbsolute and Gage

Absolute pressure: measured with respect to vacuum.Absolute pressure: measured with respect to vacuum. Gage pressure: measured with respect to atmospheric Gage pressure: measured with respect to atmospheric

pressure. pressure.

atmosphereabsolutegage ppp

31

BarometersBarometers

Mercury Barometer is used to measure atmosphere Mercury Barometer is used to measure atmosphere pressure:pressure:

PPatmatm ＝＝ γh +Pγh +Pvaporvapor ＝＝ γh γh

PPvaporvapor ＝＝ 0.000023 lb / in0.000023 lb / in22@68@68ooFF

γγ ：： specific weight of mercuryspecific weight of mercury

ghpp vaporatm

The height of a mercury column is The height of a mercury column is converted to atmosphere pressure by converted to atmosphere pressure by usingusing

32

Example 2.3 Barometric Pressure Example 2.3 Barometric Pressure

A mountain lake has an average temperature of 10 A mountain lake has an average temperature of 10 ℃℃ and and a maximum depth of 40 m. For a barometric pressure of a maximum depth of 40 m. For a barometric pressure of 598 mm Hg, determine the absolute pressure (in pascals) 598 mm Hg, determine the absolute pressure (in pascals) at the deepest part of the lake .at the deepest part of the lake .

33

Example 2.3 Example 2.3 SolutionSolution1/21/2

m598.0mm598P

Hg

barometric

0php

3

Hgm/kN133

The pressure in the lake at any depth, h

p0 is the local barometric expressed in a consistent of units .

230 m/kN5.79)m/kN133)(m598.0(p

34

30 /804.9

2mkNH

kPa472m/kN5.79m/kN392

m/kN5.79)m40)(m/kN804.9(p22

23

From Table B.2,From Table B.2, at 10 at 10 00CC


35

Manometry

A standard technique for A standard technique for measuring pressuremeasuring pressure involves the involves the use of use of liquid column in vertical or inclined tubesliquid column in vertical or inclined tubes..

Pressure measuring devices based on this technique are Pressure measuring devices based on this technique are

called called manometersmanometers. . The mercury barometer is an The mercury barometer is an example of one type of manometerexample of one type of manometer, but there are many , but there are many other configuration possible, depending on the other configuration possible, depending on the particular particular application.application. Piezometer Tube.Piezometer Tube. U-Tube manometer.U-Tube manometer. Inclined-Tube manometer.Inclined-Tube manometer.

36

Piezometer Tube

The fundamental equation isThe fundamental equation isP = P0 + γh >> PA = γ1 h1 PA : gage pressure ( P0=0)

γ1 :the specific weight of the liquid in the

container h1: measured from the meniscus at the upper

surface to point(1)Only suitable if Only suitable if the pressure in the container is the pressure in the container is greater than atmospheric pressuregreater than atmospheric pressure, and the , and the pressure to be measured must be relatively small so pressure to be measured must be relatively small so the required height of the column is reasonable. the required height of the column is reasonable. The The fluid in the container must be a liquid rather than fluid in the container must be a liquid rather than a gasa gas..

37

Simple U-Tube Manometer

The fluid in the manometer is called the gage fluid.The fluid in the manometer is called the gage fluid.

A(1)(2)(3)Open

PA + γ1 h 1 – γ2h 2 = 0

>> PA =γ2h 2 –γ1 h 1

If pipe A contains a gas

then γ1h 1 0≒

>> PA =γ2h 2

38

Example 2.4 Simple U-Tube ManometerExample 2.4 Simple U-Tube Manometer

A closed tank contains compressed A closed tank contains compressed air and oil (SGair and oil (SGoiloil = 0.90) as is = 0.90) as is

shown in Figure E2.4. A U-tube shown in Figure E2.4. A U-tube manometer using mercury (SGmanometer using mercury (SGoiloil = =

13.6) is connected to the tank as 13.6) is connected to the tank as shown. For column heights hshown. For column heights h11 = 36 = 36

in., hin., h22 = 6 in., and h = 6 in., and h33 = 9 in., = 9 in.,

determine the pressure reading (in determine the pressure reading (in psi) of the gage.psi) of the gage.

39


221oilair1 p)hh(pp

As we move from level (2) to the open end, the pressure must As we move from level (2) to the open end, the pressure must decrease by decrease by γγHgHghh33, and at the open end the pressure is zero. Thus, the manometer equation can be expressed as

0h)hh(phpp 3Hg21oilair3Hg21

or 0h))(SG()hh)()(SG(p 3OHHg21OHoilair 22

The pressure at level (1) is

40

Exmaple 2.4 Exmaple 2.4 2/22/2

tf12

9)lb/ft 4.62)(6.13( ft

12

636)lb/ft 4.62)(9.0(p 33

air

2air lb/ft 440p

The value for pThe value for pairair

So thatSo that

The pressure reading (in psi) of the gageThe pressure reading (in psi) of the gage

psi 06.3/ftin. 144

lb/ft 440p

22

2

gage

41

Differential U-Tube Manometer

AA(1)(1)(2)(2)(3)(3)(4)(4)(5)(5)BB

PPAA ＋＋ γγ11hh11 －－ γγ22hh22 －－ γγ33hh33 ＝＝ PPBB

The pressure difference isThe pressure difference is

PPAA －－ PPBB ＝＝ γγ22hh22 ＋＋ γγ33hh33 －－ γγ11hh11

42

Example 2.5 U-Tube ManometerExample 2.5 U-Tube Manometer

As As will be discussed in Chapter 3will be discussed in Chapter 3, the volume rate of flow, Q, , the volume rate of flow, Q, through a pipe can be determined by means of a flow nozzle located through a pipe can be determined by means of a flow nozzle located in the pipes as illustrated in Figure. the nozzle creates a pressure in the pipes as illustrated in Figure. the nozzle creates a pressure drop, pdrop, pAA －－ ppBB, along the pipe which is related to the flow through , along the pipe which is related to the flow through the equation , where K is a constant depending on the the equation , where K is a constant depending on the pipe and nozzle size. The pressure drop is frequently measured with pipe and nozzle size. The pressure drop is frequently measured with a differential U-tube manometer of the type illustrated. a differential U-tube manometer of the type illustrated.

(a) (a) Determine an equation for pDetermine an equation for pAA －－ ppBB in terms of the specific in terms of the specific weight of the flowing fluidweight of the flowing fluid, , γγ11, the specific weight of the gage fluid, , the specific weight of the gage fluid, γγ22, and the various heights indicated. (b) For γ, and the various heights indicated. (b) For γ11= 9.80kN/m= 9.80kN/m33 , γ , γ22 = = 15.6 kN/m15.6 kN/m33 , h , h11 = 1.0m, and h = 1.0m, and h22 = 0.5m, what is the value of the = 0.5m, what is the value of the pressure drop, ppressure drop, pAA －－ ppBB??

BAppKQ

43

Example 2.5 Example 2.5 SolutionSolution

(Ans)

we start at point A and move vertically upward to level (1), the we start at point A and move vertically upward to level (1), the pressure will decrease by γpressure will decrease by γ11hh11 and will be equal to pressure at (2) and will be equal to pressure at (2)

and (3). We can now move from (3) to (4) where the pressure has and (3). We can now move from (3) to (4) where the pressure has been further reduced by γbeen further reduced by γ22hh22 . The pressure at levels (4) and (5) are . The pressure at levels (4) and (5) are

equal, and as we move from (5) to equal, and as we move from (5) to B B the pressure will increase the pressure will increase byγbyγ11(h(h11 + h + h22))

)(hpp

p)hh(hhp

212BA

B2112211A

44

Inclined-Tube Manometer

To measure small pressure change, an inclined-tube To measure small pressure change, an inclined-tube manometer is frequently used:manometer is frequently used:PA +γ1h1 –γ2l2sinθ –γ3h3 = PB

PA – PB =γ2l2sinθ +γ3h3 –γ1h1

If pipe A and B contain a gas thenγ3h3 γ≒ 1h1 0≒

>> l2 = ( PA – PB ) /γ2 sinθ

45

Mechanical and Electronic Devices

Manometers are Manometers are not well suitednot well suited for measuring very high for measuring very high pressures, or pressures that are pressures, or pressures that are changing rapidlychanging rapidly with with time.time.

To overcome some of these problems numerous other To overcome some of these problems numerous other types of pressure-measuring instruments have been types of pressure-measuring instruments have been developed. Most of these developed. Most of these make use of the idea that when make use of the idea that when a pressure acts on an elastic structure, the structure a pressure acts on an elastic structure, the structure will deform, and this deformation can be related to the will deform, and this deformation can be related to the magnitude of the pressure.magnitude of the pressure.

46

Bourdon Pressure Gage

Bourdon tube pressure gage uses a hollow, elastic, and Bourdon tube pressure gage uses a hollow, elastic, and curved tube to measure pressure.curved tube to measure pressure.

As the pressure within the tube increases the tube tends to As the pressure within the tube increases the tube tends to straighten, and although the deformation is small, it can be straighten, and although the deformation is small, it can be translated into the motion of a pointer on dial.translated into the motion of a pointer on dial.

Connected to the pressure source

47

Aneroid Barometer

The Aneroid barometer is used for The Aneroid barometer is used for measuring atmospheric measuring atmospheric pressure.pressure.

The Aneroid barometer contains a hallow, closed, elastic The Aneroid barometer contains a hallow, closed, elastic elements which is evacuated so that the pressure inside the elements which is evacuated so that the pressure inside the element is near absolute zero.element is near absolute zero.

As the external atmospheric pressure changes, the element As the external atmospheric pressure changes, the element deflects, and this motion can be translated into the deflects, and this motion can be translated into the movement of an attached dialmovement of an attached dial..

48

Bourdon Gage + LVDT Bourdon Gage + LVDT

Combining a linear variable differential transformer (LVDT) with a Combining a linear variable differential transformer (LVDT) with a Bourdon pressure gage, converts the pressure into an electric output.Bourdon pressure gage, converts the pressure into an electric output.

The core of the LVDT is connected The core of the LVDT is connected to the free end of the Bourdon so to the free end of the Bourdon so that as a pressure is applied, the that as a pressure is applied, the resulting motion of the end of the resulting motion of the end of the tube moves the core through the coil tube moves the core through the coil and an output voltage develops.and an output voltage develops.

This voltage is a linear function of This voltage is a linear function of the pressure and could be recorded the pressure and could be recorded on an oscillograph or digitized for on an oscillograph or digitized for storage or processing on a computer.storage or processing on a computer.

49

Hydrostatic Forces on a Plane 1/2

When a surface is submerged in a fluid, forces develop on When a surface is submerged in a fluid, forces develop on the surface due to the hydrostatic pressure distribution of the surface due to the hydrostatic pressure distribution of the fluid.the fluid.

The determination of these forces is important in the The determination of these forces is important in the design of storage tanks, ships, dams, and other hydraulic design of storage tanks, ships, dams, and other hydraulic structures.structures.

Pressure distribution and resultant Pressure distribution and resultant hydrostatic force on the bottom of an open hydrostatic force on the bottom of an open tank.tank.

Pressure distribution on the ends of an Pressure distribution on the ends of an open tank.open tank.

50

Hydrostatic Forces on a Plane 2/2

Specifying the magnitude of the force.Specifying the magnitude of the force. Specifying the direction of the force.Specifying the direction of the force. Specifying the line of action of the force.Specifying the line of action of the force. To determine completely the resultant force acting on a To determine completely the resultant force acting on a

submerged force.submerged force.

51

On a Submerged SurfacesOn a Submerged Surfaces

The hydrostatic force on The hydrostatic force on any any elementelement of the surface acts of the surface acts normal to the surface dF = normal to the surface dF = pdA.pdA.

The resultant forceThe resultant force

For constant For constant and and

AAR dAsinyhdAF

Where h=ysin

AR ydAsinF

AyydA CAFirst moment of the area wrt the x-axis >>>First moment of the area wrt the x-axis >>>

x

y

52

Resultant ForceResultant Force

The magnitude of the resultant force is equal to pressure The magnitude of the resultant force is equal to pressure acting at the centroid of the area multiplied by the total acting at the centroid of the area multiplied by the total area.area.

AhsinAyF ccR yycc is the y coordinate of the is the y coordinate of the centroid centroid of the area A. of the area A.hhcc is the vertical distance from the fluid surface to the is the vertical distance from the fluid surface to the centroidcentroid of the area. of the area.

53

Location of Resultant ForceLocation of Resultant Force1/21/2

How to determine the location (xHow to determine the location (xRR,y,yRR) of the resultant force ) of the resultant force ??

A

2

ARR dAysinydFyF

The moment of the resultant force must equal the moment of the distributed pressure force.

AhsinAyF ccR

cc

xc

c

x

c

A

2

R yAy

I

Ay

I

Ay

dAyy

IIxx is the second moment of the area (moment of inertia). is the second moment of the area (moment of inertia).By parallel axis theorem…By parallel axis theorem… 2

cxcx AyII

The resultant force does The resultant force does not pass through the not pass through the centroid but is always centroid but is always below it.below it.

54

Location of Resultant ForceLocation of Resultant Force2/22/2

AARR xydAsinxdFxF AhsinAyF ccR

cc

xyc

c

xy

c

AR x

Ay

I

Ay

I

Ay

xydAx

IIxyxy is the product of inertia wrt the x and y. is the product of inertia wrt the x and y.By parallel axis theorem…By parallel axis theorem…

ccxycxy yAxII

IIxycxyc is the product of inertia with respect to an orthogonal is the product of inertia with respect to an orthogonal coordinate system passing through the centroid of the area and coordinate system passing through the centroid of the area and formed by a translation of the x-y coordinate system.formed by a translation of the x-y coordinate system.

If the submerged area is symmetrical with respect to an axis If the submerged area is symmetrical with respect to an axis passing through the centroid and parallel to either the x or y passing through the centroid and parallel to either the x or y axes, Iaxes, Ixycxyc=0.=0.

55

Geometric Properties of Common Geometric Properties of Common ShapesShapes

Figure 2.18

56

Example 2.6 Hydrostatic Pressure Example 2.6 Hydrostatic Pressure Force on a Plane Circular SurfaceForce on a Plane Circular Surface The The 4-m-diameter circular gate4-m-diameter circular gate of Figure E2.6 (a) is located in the of Figure E2.6 (a) is located in the

inclined wall of a large reservoir containing water (inclined wall of a large reservoir containing water (=9.80kN/m=9.80kN/m33). ). The gate is mounted on a shaft along its horizontal diameter. For a The gate is mounted on a shaft along its horizontal diameter. For a water depth hwater depth hcc=10m above the shaft determine : =10m above the shaft determine : (a) the magnitude (a) the magnitude and location of the resultant force exerted on the gate by the water, and location of the resultant force exerted on the gate by the water, and (b) the moment that would have to be applied to the shaft to and (b) the moment that would have to be applied to the shaft to open the gateopen the gate

57


The vertical distance from the fluid surface to the centroid of The vertical distance from the fluid surface to the centroid of the area is 10mthe area is 10m

c

c

xyc

Rx

Ay

Ix

(a) The magnitude of the force of the water(a) The magnitude of the force of the water

The point (center of pressure) through which FThe point (center of pressure) through which FRR acts acts

AhsinAyF ccR

MN23.1)m4)(m10)(m/N1080.9(F 233R

cc

xc

c

x

c

A

2

R yAy

I

Ay

I

Ay

dAyy

58


The distance below the shaft to the center of pressure isThe distance below the shaft to the center of pressure is

m6.11m55.11m0866.0

60sin

m10

)m4)(60sin/m10(

)m2(4/y

2

4

R

4

RI

4

xc

The area is symmetrical and the center of pressure must lie along the The area is symmetrical and the center of pressure must lie along the diameter A-A. diameter A-A. xxRR=0=0

m0866.0yy cR

The force The force acts through a point along its diameter A-A at a distance acts through a point along its diameter A-A at a distance of 0.0866m below the shaft.of 0.0866m below the shaft.

59


mN1007.1

)m0866.0)(MN23.1(yyFM5

CRR

0Mc (b) Sum moments about the shaft(b) Sum moments about the shaft

The moment required to open the gateThe moment required to open the gate

60

Example 2.7 Hydrostatic Pressure Force Example 2.7 Hydrostatic Pressure Force on a Plane Triangular Surfaceon a Plane Triangular Surface A large A large fish-holding tankfish-holding tank contains seawater (γ=64.0lb/ft contains seawater (γ=64.0lb/ft33) to a ) to a

depth of 10 ft as shown in Figure E2.7. To repair some damage to depth of 10 ft as shown in Figure E2.7. To repair some damage to one corner of the tank, a triangular section is replaced with a new one corner of the tank, a triangular section is replaced with a new section as illustrated. section as illustrated. Determine the magnitude and location of Determine the magnitude and location of the force of the seawater on this triangular areathe force of the seawater on this triangular area..

61


yyc c = h= hc c = 9 ft, and the magnitude of the force= 9 ft, and the magnitude of the force

FFR R =γh=γhc c A = (64.0 lb/ ftA = (64.0 lb/ ft33)(9 ft)(9/2 ft)(9 ft)(9/2 ft22) = 2590 lb) = 2590 lb

The y coordinate of the center of pressure (CP)The y coordinate of the center of pressure (CP)

c

c

xc

Ry

Ay

Iy

43

xc ft36

81

36

)ft3)(ft3(I

62


Similarly, Similarly,

ft9)ft2/9)(ft9(

ft36/81y

2

4

R = 0.0556 = 0.0556 ft + 9 ft = 9.06 ftft + 9 ft = 9.06 ft

c

c

xyc

Rx

Ay

Ix

4

2

xycft

72

81)ft3(

72

)ft3)(ft3(I

ft0278.00)ft2/9)(ft9(

ft72/81x

2

4

R

The center of pressure is 0.0278 ft to the right of and The center of pressure is 0.0278 ft to the right of and 0.0556 ft below the centroid of the area0.0556 ft below the centroid of the area

63

Pressure Prism Pressure Prism for vertical rectangular areafor vertical rectangular area

A2

h)bh)(h(

2

1volume

A2

hApF avR

2211AR

21R

yFyFyF

FFF

2

hhy 21

1

1

122 h

3

)hh(2y

64

Pressure Prism Pressure Prism for inclined plane areafor inclined plane area

The pressure developed depend The pressure developed depend on the vertical distances.on the vertical distances.

65

Pressure Prism Pressure Prism effect of atmospheric pressureeffect of atmospheric pressure

The resultant fluid force on the surface is that due The resultant fluid force on the surface is that due only to the gage pressure contribution of the liquid in only to the gage pressure contribution of the liquid in contact with the surface – contact with the surface – the atmospheric pressure the atmospheric pressure does not contribute to this resultantdoes not contribute to this resultant..

66

Example 2.8 Use of the Pressure Prism Example 2.8 Use of the Pressure Prism ConceptConcept

A pressurized contains oil (SG = 0.90) and has a square, 0.6-m by A pressurized contains oil (SG = 0.90) and has a square, 0.6-m by 0.6-m plate bolted to its side, as is illustrated in Figure E2.8 (a). 0.6-m plate bolted to its side, as is illustrated in Figure E2.8 (a). When the pressure gage on the top of the tank reads 50kPa, what is When the pressure gage on the top of the tank reads 50kPa, what is the magnitude and location of the resultant force on the attached the magnitude and location of the resultant force on the attached plate? The outside of the tank is atmospheric pressure.plate? The outside of the tank is atmospheric pressure.

67


N 10 24.4

)m)](0.36m )(2N/m10 1(0.90)(9.8 N/m10 [50

)Ah (pF

3

23323

1s1

The resultant force on the plate (having an area A) is due to the components, F1 and F2 , where F1 and F2 are due to the rectangular and triangular portions of the pressure distribution, respectively.

N10 0.954

))(0.36m2

0.6m)(N/m10 1(0.90)(9.8 A )

2

h - h( F

3

233212

68


25.4kN N10 25.4 F F F 321R

The magnitude of the resultant force, The magnitude of the resultant force, FFRR, is therefore, is therefore

The vertical location of FR can be obtained by summing moments around an axis through point O

(0.2m) F (0.3m) F yF 210R

0.296m N10 25.4

N)(0.2m)10 (0.954 N)(0.3m)10 (24.4 y

3

33

O

69

On a Curved SurfacesOn a Curved Surfaces

Consider the curved section Consider the curved section BC of the open tank.BC of the open tank.

FF11 and F and F22 can be determined can be determined from the relationships for from the relationships for planar surfaces.planar surfaces.The weight W is simply the The weight W is simply the specific weight of the fluid specific weight of the fluid times the enclosed volume and times the enclosed volume and acts through the center of acts through the center of gravity (CG) of the mass of gravity (CG) of the mass of fluid contained within the fluid contained within the volume.volume.

2V

2HR

1V2H

FFF

WFFFF

70

Example 2.9Example 2.9 　　 Hydrostatic Pressure Hydrostatic Pressure Force on a Curved SurfaceForce on a Curved Surface The 6-ft-diameter drainage conduit of figure a is half full of water at The 6-ft-diameter drainage conduit of figure a is half full of water at

rest. Determine the magnitude and line of action of the resultant rest. Determine the magnitude and line of action of the resultant force that the water exerts on a 1-ft length of the curved section force that the water exerts on a 1-ft length of the curved section BCBC of the conduit wall.of the conduit wall.

71


lb281)ft3)(ft2

3)(ft/lb4.62(AhF 23

c1

The magnitude of The magnitude of FF11 is found form the equation is found form the equation

The weight, The weight, WW, is, is

lb441)ft1)(ft4/9)(ft/lb4.62(volW 23

lb281FF 1H lb441WFv

The magnitude of the resultant forceThe magnitude of the resultant force

lb523)F()F(F 2V

2HR

72

BUOYANCY BUOYANCY 1/21/2

Buoyancy: The net vertical force acting on any body Buoyancy: The net vertical force acting on any body which which is immersed in a liquid, or floating on its surfaceis immersed in a liquid, or floating on its surface due to liquid pressure. Fdue to liquid pressure. FBB

Consider a body of arbitrary Consider a body of arbitrary shape, having a volume V, that shape, having a volume V, that is immersed in a fluid,is immersed in a fluid,

We enclose the body in a We enclose the body in a parallelepiped and draw a free-parallelepiped and draw a free-body diagram of parallelpiped body diagram of parallelpiped with body removed as shown in with body removed as shown in (b).(b).

73

BUOYANCY BUOYANCY 2/22/2

VA)hh(A)hh(F

A)hh(FF

WFFF

1212B

1212

12B

VgFB

FFBB is the force the body is exerting on the fluid. is the force the body is exerting on the fluid.W is the weight of the shaded fluid volume W is the weight of the shaded fluid volume (parallelepiped minus body).(parallelepiped minus body).A is the horizontal area of the upper (or lower) A is the horizontal area of the upper (or lower) surface of the parallelepiped.surface of the parallelepiped.

For a For a submerged bodysubmerged body, the , the buoyancy force of the fluid is equal buoyancy force of the fluid is equal to the weight of displaced fluidto the weight of displaced fluid

A

74

Archimedes’ PrincipleArchimedes’ Principle

For a submerged body, the buoyancy force of the fluid is For a submerged body, the buoyancy force of the fluid is equal to the weight of displaced fluid and is directly equal to the weight of displaced fluid and is directly vertically upward.vertically upward.

The relation reportedly was used by Archimedes in 220 The relation reportedly was used by Archimedes in 220 B.C. to determine the gold content in the crown of King B.C. to determine the gold content in the crown of King Hiero II.Hiero II.

VgFB

75

Example 2.10 Buoyant Force on a Example 2.10 Buoyant Force on a Submerged ObjectSubmerged Object

A spherical buoy has a diameter of 1.5 m, weighs 8.50kN, and is A spherical buoy has a diameter of 1.5 m, weighs 8.50kN, and is anchored to the seafloor with a cable as is shown in Figure a. anchored to the seafloor with a cable as is shown in Figure a. Although the buoy normally floats on the surface, at certain times Although the buoy normally floats on the surface, at certain times the water depth increases so that the buoy is completely immersed the water depth increases so that the buoy is completely immersed as illustrated. For this condition what is the tension of the cable?as illustrated. For this condition what is the tension of the cable?

76


WFT B

With With γ= 10.1 kN/mγ= 10.1 kN/m3 3 and V = πdand V = πd33/6/6

N 10 1.785 ]m) 1.5 [(k/6)( ) N/m 10 10.1 ( F 4333B

VFB

The tension in the cableThe tension in the cable

kN35.9N10850.0N10785.1T 44

FFBB is the buoyant force acting on the buoy, is the buoyant force acting on the buoy, WW is the weight of the is the weight of the

buoy, and buoy, and TT is the tension in the cable. For Equilibrium is the tension in the cable. For Equilibrium

77

The Line of Action of FThe Line of Action of FBB and C.G. and C.G.

The line of action of buoyancy force, which may be found The line of action of buoyancy force, which may be found using the method of “hydrostatic force on submerged using the method of “hydrostatic force on submerged surfaces,” acts through the centroid of the surfaces,” acts through the centroid of the displaced displaced volumevolume..

The point through which the buoyant force acts is called The point through which the buoyant force acts is called the the center of buoyancycenter of buoyancy..

C.G: The body force due to gravity on an object act C.G: The body force due to gravity on an object act through its center of gravity (C.G.).through its center of gravity (C.G.).

78

StabilityStability

Stability? Stable? Unstable?Stability? Stable? Unstable? A body is said to be in a stable A body is said to be in a stable

equilibrium position if, when equilibrium position if, when displaced, it returns to its displaced, it returns to its equilibrium position. equilibrium position. Conversely, it Conversely, it is an unstable equilibrium position is an unstable equilibrium position if, when displaced (even slightly), it if, when displaced (even slightly), it moves to a new equilibrium moves to a new equilibrium position.position.

79

Stability of Immersed BodyStability of Immersed Body

The location of the line of action of The location of the line of action of the buoyancy force determines the buoyancy force determines stability.stability.

While C.G. is below the center of While C.G. is below the center of buoyancy, a rotation from its buoyancy, a rotation from its equilibrium position will create a equilibrium position will create a restoring couple formed by the restoring couple formed by the weight and the buoyancy force.weight and the buoyancy force.

If C.G. is above the center of If C.G. is above the center of buoyancy,…..buoyancy,…..

80

Stability of Floating BodyStability of Floating Body

The determination of The determination of stability depends in a stability depends in a complicated fashion on the complicated fashion on the particular geometry and particular geometry and weight distribution of the weight distribution of the body.body.

81

Rigid-Body Motion Rigid-Body Motion Pressure VariationPressure Variation

The entire fluid moves as if it were a rigid body – The entire fluid moves as if it were a rigid body – individual fluid particles, although they may be in motion, individual fluid particles, although they may be in motion, are not deformingare not deforming. This means that . This means that there are no shear there are no shear stresses, as in the case of a static fluidstresses, as in the case of a static fluid..

The general equation of motionThe general equation of motion

akp

z

y

x

az

p

ay

p

ax

p

Based on rectangular Based on rectangular coordinates with the coordinates with the positive z axis being positive z axis being vertically upward.vertically upward.

82

Linear Motion Linear Motion

)ag(z

p

ay

p

0ax

p

z

y

x

The change in pressure The change in pressure between two closely between two closely spaced points located at spaced points located at y, z, and y+dy , z+dzy, z, and y+dy , z+dz

Along a line of constant pressure, dp=0Along a line of constant pressure, dp=0

z

y

zy

ag

a

dy

dz

dz)ag(dyadp

dzz

pdy

y

pdp

Y-Z plane motion

83

Example 2.11 Pressure Variation in an acceleration tank The cross section for the fuel tank of an experimental vehicle is The cross section for the fuel tank of an experimental vehicle is

shown in Figure E2.11. the rectangular tank is vented to the shown in Figure E2.11. the rectangular tank is vented to the atmosphere, and a pressure transducer is located in its side as atmosphere, and a pressure transducer is located in its side as illustrated. During testing of the vehicle, the tank is subjected to be illustrated. During testing of the vehicle, the tank is subjected to be a constant linear acceleration, aa constant linear acceleration, ayy. .

(a) Determine an expression that (a) Determine an expression that relates ay and the pressure (in relates ay and the pressure (in lb/ftlb/ft22) at the transducer for a fuel ) at the transducer for a fuel with a SG = 0.65. (b) What is the with a SG = 0.65. (b) What is the maximum acceleration that can maximum acceleration that can occur before the fuel level drops occur before the fuel level drops below the transducer?below the transducer?

84


g

a

dy

dz y

Since aSince az z = 0. Thus for some arbitrary a= 0. Thus for some arbitrary ayy, the change in depth, z, the change in depth, z11

The slope of the surfaceThe slope of the surface

g

a)ft75.0(z

org

a

ft75.0

z

y

1

y1

85


hp

Where h is the depth of fuel above the transducer.Where h is the depth of fuel above the transducer.

The pressure at the transducer is given by the relationshipThe pressure at the transducer is given by the relationship

g

a4.303.20)]g/a)(ft75.0(ft5.0)[ft/lb4.62)(65.0(p y

y

3

The limiting value for (aThe limiting value for (ayy))maxmax

3

g2)a(or

g

)a()ft75.0(ft5.0

maxy

maxy

=0

86

Angular MotionAngular Motion1/21/2

In terms of cylindrical coordinates, the pressure gradient In terms of cylindrical coordinates, the pressure gradient can be expressed can be expressed

zr ez

pe

p

r

1e

r

pp

0a0aerwa zr2

r

The differential pressure is The differential pressure is dzdrrdzz

pdr

r

pdp 2

87

Angular Motion Angular Motion 2/22/2

ttanconsz2

rpegrationint

dzdrrdzz

pdr

r

pdp

22

2

Along a line of constant pressure, dp=0Along a line of constant pressure, dp=0

ttanconsg2

rz

g

r

dr

dz 222

The equation for surface of constant pressure is The equation for surface of constant pressure is

The pressure distribution in The pressure distribution in a rotating liquida rotating liquid

The equation reveals that the surfaces of constant The equation reveals that the surfaces of constant pressure are parabolicpressure are parabolic

Pressure distribution in a rotating fluidPressure distribution in a rotating fluid

88

Example 2.12 Free Surface Shape of Example 2.12 Free Surface Shape of Liquid in a Rotating TankLiquid in a Rotating Tank

It has been suggested that the angular velocity, It has been suggested that the angular velocity, , of a rotating , of a rotating body or shaft can be measured by attaching an open cylinder of body or shaft can be measured by attaching an open cylinder of liquid, as shown in Figure E2.12, and measuring with some type of liquid, as shown in Figure E2.12, and measuring with some type of depth gage the changes in the fluid level, H-hdepth gage the changes in the fluid level, H-hoo, caused by the , caused by the

rotation of the fluid. Determine the relationship between this change rotation of the fluid. Determine the relationship between this change in fluid level and the angular velocity.in fluid level and the angular velocity.

89


0

22

hg2

rh

The initial volume of fluid in the tankThe initial volume of fluid in the tank HRV 2i

The height, h, of the free surface above the tank bottomThe height, h, of the free surface above the tank bottom

This cylindrical shell is taken at some arbitrary radius, r, and its This cylindrical shell is taken at some arbitrary radius, r, and its volume isvolume is rhdr2Vd The total volumeThe total volume

02

42R

0 0

22

hRg4

Rdrh

g2

rr2V

90


g4

RhHorhR

g4

RHR

22

002

422

Since the volume of the fluid in the tank must remain constantSince the volume of the fluid in the tank must remain constant

The change in depth could indeed be used to determine the rotational The change in depth could indeed be used to determine the rotational speed, although the relationship between the change in depth and speed, although the relationship between the change in depth and speed is not a linear one.speed is not a linear one.

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CHAP02 Munson