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1 FUNDAMENTALS OF FUNDAMENTALS OF FLUID MECHANICS FLUID MECHANICS Chapter 3 Fluids in Motion Chapter 3 Fluids in Motion - - The Bernoulli Equation The Bernoulli Equation
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FUNDAMENTALS OFFUNDAMENTALS OFFLUID MECHANICSFLUID MECHANICS

Chapter 3 Fluids in Motion Chapter 3 Fluids in Motion -- The Bernoulli EquationThe Bernoulli Equation

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MAIN TOPICSMAIN TOPICS

NewtonNewton’’s Second Laws Second LawF=ma Along a StreamlineF=ma Along a StreamlineF=ma Normal to a StreamlineF=ma Normal to a StreamlinePhysical Interpretation Physical Interpretation of of Bernoulli Equation

Static, Stagnation, Dynamic, and Total PressureStatic, Stagnation, Dynamic, and Total PressureApplication of the Bernoulli EquationApplication of the Bernoulli EquationThe Energy Line and the Hydraulic Grade LineThe Energy Line and the Hydraulic Grade LineRestrictions on Use of the Bernoulli Equation

Bernoulli Equation

Restrictions on Use of the Bernoulli Equation

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NewtonNewton’’s Second Law s Second Law 1/51/5

As a fluid particle moves from one location to another, it As a fluid particle moves from one location to another, it experiences an acceleration or deceleration.experiences an acceleration or deceleration.According to According to NewtonNewton’’s second law of motions second law of motion, the net force acting , the net force acting on the fluid particle under consideration must equal its mass tion the fluid particle under consideration must equal its mass times mes its acceleration.its acceleration.

F=maF=maIn this chapter, we consider the motion of In this chapter, we consider the motion of inviscidinviscid fluidsfluids. That is, . That is, the fluid is assumed to have the fluid is assumed to have zero viscosityzero viscosity. For such case, . For such case, it is it is possible to ignore viscous effects.possible to ignore viscous effects.The forces acting on the particle ? Coordinates used ?The forces acting on the particle ? Coordinates used ?

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NewtonNewton’’s Second Law s Second Law 2/52/5

The fluid motion is governed byThe fluid motion is governed byF= Net pressure force + Net gravity forceF= Net pressure force + Net gravity force

To apply NewtonTo apply Newton’’s second law to a fluid, s second law to a fluid, an appropriate an appropriate coordinate system must be chosen to describe the coordinate system must be chosen to describe the motionmotion. In general, the motion will be three. In general, the motion will be three--dimensional dimensional and unsteady so that and unsteady so that three space coordinates and timethree space coordinates and timeare needed to describe it.are needed to describe it.The most often used coordinate systems are The most often used coordinate systems are rectangular rectangular (x,y,z) and cylindrical (r,(x,y,z) and cylindrical (r,θθ,z) system.,z) system.

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NewtonNewton’’s Second Law s Second Law 3/53/5

In this chapter, the flow is confined to be In this chapter, the flow is confined to be twotwo--dimensional dimensional motion.motion.As is done in the study of dynamics, the motion of each As is done in the study of dynamics, the motion of each fluid particle is described in terms of its velocity vector fluid particle is described in terms of its velocity vector VV..As the particle moves, As the particle moves, it it follows a particular path.follows a particular path.The location of the particle The location of the particle along the path along the path is a function is a function of its initial position and of its initial position and velocity.velocity.

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NewtonNewton’’s Second Law s Second Law 4/54/5

For steady flowsFor steady flows, each particle slides along its path, and , each particle slides along its path, and its velocity vector is everywhere tangent to the path. The its velocity vector is everywhere tangent to the path. The lines that are tangent to the velocity vectors throughout the lines that are tangent to the velocity vectors throughout the flow field are called flow field are called streamlinesstreamlines..For such situation, the particle motion is described in For such situation, the particle motion is described in terms of its distance, s=s(t), along the streamline from terms of its distance, s=s(t), along the streamline from some convenient origin and the local radius of curvature some convenient origin and the local radius of curvature of the streamline, R=R(s).of the streamline, R=R(s).

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NewtonNewton’’s Second Law s Second Law 5/55/5

The The distance along the streamline is related to the distance along the streamline is related to the particleparticle’’s speeds speed by by VV==dsds//dtdt, and the radius of curvature is , and the radius of curvature is related to shape of the streamline.related to shape of the streamline.The acceleration is the time rate of change of the velocity The acceleration is the time rate of change of the velocity of the particleof the particle

The The components of accelerationcomponents of acceleration in the in the s and ns and n directiondirection

nRVs

dsdVVnasa

dtVda

2

nsrrrr

rr

+=+==

RVa

dsdVVa

2

ns == CHAPTER 04CHAPTER 04

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F=ma along a Streamline F=ma along a Streamline 1/31/3

Consider the small fluid particle of fluid particle of size of size of δδs by s by δδnn in the plane of the figure and δy normal to the figure. For steady flow, the component of Newton’s second law along the streamline direction s

sVVV

sVmVmaF SS ∂

∂ρδ=

∂∂

δ=δ=δ∑

Where represents the sum of the s components of all the force acting on the particle.

∑δ SF

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F=ma along a Streamline F=ma along a Streamline 2/32/3

The The gravity force (weight)gravity force (weight) on the particle in the on the particle in the streamline directionstreamline direction

The The net pressure forcenet pressure force on the particle in the streamline on the particle in the streamline directiondirection

θγδ−=θδ−=δ sinVsinWWs

( ) ( ) Vspynp2ynppynppF SSSps δ∂∂

−=δδδ−=δδδ+−δδδ−=δ

VspsinFWF psss δ⎟⎠⎞

⎜⎝⎛

∂∂

−θγ−=δ+δ=δ

sasVV

spsin ρ=

∂∂

ρ=∂∂

−θγ−Equation of motion Equation of motion along the streamline along the streamline directiondirection

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F=ma along a Streamline F=ma along a Streamline 3/33/3

A change in fluid particle speed is accomplished by the A change in fluid particle speed is accomplished by the appropriate combination of pressure gradient and particle appropriate combination of pressure gradient and particle weight along the streamline.weight along the streamline.For fluid For fluid static situationstatic situation, the balance between pressure , the balance between pressure and gravity force is such that no change in particle speed and gravity force is such that no change in particle speed is produced.is produced.

sasVV

spsin ρ=

∂∂

ρ=∂∂

−θγ−

0spsin =∂∂

−θγ−

IntegrationIntegration…………

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IntegrationIntegration……....

sasVV

spsin ρ=

∂∂

ρ=∂∂

−θγ− Rearranged and IntegratedRearranged and Integrated…………

( )

CgzV21dp

0dzVd21dp

dsdV

21

dsdp

dsdz

2

22

=++ρ

>>>>

=γ+ρ+>>ρ=−γ−

∫ along a streamlinealong a streamline

Where Where C is a constant of integration C is a constant of integration to be to be determined by the conditions at some point on determined by the conditions at some point on the streamline.the streamline.

In general it is not possible to integrate the pressure term becIn general it is not possible to integrate the pressure term because ause the density may not be constant and, therefore, cannot be removethe density may not be constant and, therefore, cannot be removed d from under the integral sign.from under the integral sign.

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Example 3.1 Pressure Variation along A Example 3.1 Pressure Variation along A StreamlineStreamline

Consider the Consider the inviscidinviscid, incompressible, steady flow, incompressible, steady flow along the along the horizontal streamline Ahorizontal streamline A--B in front of the sphere of radius a, as B in front of the sphere of radius a, as shown in Figure E3.1(a). From a more advanced theory of flow passhown in Figure E3.1(a). From a more advanced theory of flow past t a sphere, the fluid velocity along this streamline isa sphere, the fluid velocity along this streamline is

Determine the pressure variation along the streamline from pDetermine the pressure variation along the streamline from point A oint A far in front of the sphere (far in front of the sphere (xxAA==--∞∞ and Vand VAA= V= V00) to point B on the ) to point B on the sphere (sphere (xxAA==--aa and Vand VBB=0)=0)

⎟⎟⎠

⎞⎜⎜⎝

⎛+= 3

3

0 xa1VV

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Example 3.1 Example 3.1 SolutionSolution1/21/2

The equation of motion along the streamline (The equation of motion along the streamline (sinsinθθ=0)=0)

The acceleration term

sVV

sp

∂∂

ρ−=∂∂

(1)(1) sasVV

spsin ρ=

∂∂

ρ=∂∂

−θγ−

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛+−=⎟⎟

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛+=

∂∂

=∂∂

4

3

3

32

04

30

3

3

0 xa

xa1V3

xaV3

xa1V

xVV

sVV

The pressure gradient along the streamline is

( )4

3320

3

xx/a1Va3

sp +ρ=

∂∂

(2)(2)

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Example 3.1 Example 3.1 SolutionSolution2/22/2

The pressure gradient along the streamline

( )4

3320

3

xx/a1Va3

sp +ρ=

∂∂

(2)(2)

The pressure distribution along the streamline

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎠⎞

⎜⎝⎛ρ−=

2)x/a(

xaVp

622

0

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Bernoulli Equation Along a StreamlineBernoulli Equation Along a Streamline

For the special case of For the special case of incompressible flowincompressible flow

Restrictions : Steady flow.Restrictions : Steady flow.Incompressible flow.Incompressible flow.Frictionless flow. Frictionless flow. NO NO ����

Flow along a streamline.Flow along a streamline.

ttanconsz2

Vp2

=γ+ρ+ BERNOULLI EQUATIONBERNOULLI EQUATION

CgzV21dp 2 =++

ρ∫

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Example 3.2 The Bernoulli EquationExample 3.2 The Bernoulli Equation

Consider the flow of air around a bicyclist moving through stillConsider the flow of air around a bicyclist moving through still air air with with velocity Vvelocity V00, as is shown in Figure E3.2. Determine the , as is shown in Figure E3.2. Determine the difference in the pressure between points (1) and (2).difference in the pressure between points (1) and (2).

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Example 3.2 Example 3.2 SolutionSolution

The Bernoulli’s equation applied along the streamline that passes along the streamline that passes through (1) and (2)through (1) and (2)

2

22

21

21

1 z2

Vpz2

Vp γ+ρ+=γ+ρ+

z1=z2(1) is in the free stream VV11=V=V00(2) is at the tip of the bicyclist’s nose V2=0

2V

2Vpp

20

21

12 ρ=ρ=−

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F=ma Normal to a StreamlineF=ma Normal to a Streamline1/21/2

For steady flow, the For steady flow, the component of Newtoncomponent of Newton’’s s second law in the normal second law in the normal direction n direction n

RVV

RmVF

22

nρδ

=δ∑Where represents the Where represents the sum of the n components of all sum of the n components of all the force acting on the particle.the force acting on the particle.

∑δ nF

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F=ma Normal to a StreamlineF=ma Normal to a Streamline2/22/2

The The gravity force (weight)gravity force (weight) on the particle in the on the particle in the normal normal directiondirection

The The net pressure forcenet pressure force on the particle in the on the particle in the normal normal directiondirection

θγδ−=θδ−=δ cosVcosWWn

( ) Vnpysp2ys)pp(ysppF nnnpn δ∂∂

−=δδδ−=δδδ+−δδδ−=δ

VRVV

npcosFWF

2

pnnn δρδγδδδ =⎟⎠⎞

⎜⎝⎛

∂∂

−−=+=

RV

npcos

2ρθγ =∂∂

−− Equation of motion Equation of motion normal to the streamlinenormal to the streamlineNormal directionNormal direction

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IntegrationIntegration……....

RV

dndp

dndz 2ρ

=−γ−

RV

npcos

2ρ=

∂∂

−θγ−RearrangedRearranged

across the streamlineacross the streamline

In general it is not possible to In general it is not possible to integrate the pressure term because integrate the pressure term because the density may not be constant and, the density may not be constant and, therefore, cannot be removed from therefore, cannot be removed from under the integral sign.under the integral sign.

IntegratedIntegrated……

A change in the direction of flow of a fluid particle is A change in the direction of flow of a fluid particle is accomplished by the appropriate combination of pressure accomplished by the appropriate combination of pressure gradient and particle weight normal to the streamlinegradient and particle weight normal to the streamline

CgzdnRVdp 2

=++ρ∫ ∫

Without knowing the n dependent Without knowing the n dependent in V=V(s,n) and R=R(s,n) this in V=V(s,n) and R=R(s,n) this integration cannot be completed.integration cannot be completed.

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Bernoulli Equation Normal to a StreamlineBernoulli Equation Normal to a Streamline

For the special case of For the special case of incompressible flowincompressible flow

Restrictions : Steady flow.Restrictions : Steady flow.Incompressible flow.Incompressible flow.Frictionless flow. Frictionless flow. NO NO ����Flow normal to a streamline.Flow normal to a streamline.

CzdnRVp

2

=γ+ρ+ ∫ BERNOULLI EQUATIONBERNOULLI EQUATION

CgzdnR

Vdp 2

=+∫ ∫+ρ

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Example 3.3 Pressure Variation Normal to Example 3.3 Pressure Variation Normal to a Streamlinea Streamline

Shown in Figure E3.3 (a) and (b) are two flow fields with circulShown in Figure E3.3 (a) and (b) are two flow fields with circular ar streamlines. The streamlines. The velocity distributionsvelocity distributions are are

)b(r

C)r(V)a(rC)r(V 21 ==

Assuming the flows are steady, inviscid, and incompressible with streamlines in the horizontal plane (dz/dn=0).

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RV

dndp

dndz 2ργ =−−Example 3.3 Example 3.3 SolutionSolution

For flow in the horizontal plane (dz/dn=0). The streamlines are circles ∂/∂n=-∂/∂rThe radius of curvature R=r

rV

rp 2ρ=

∂∂

For case (a) this gives

( ) 02

022

1 prrC21p +−ρ=rC

rp 2

1ρ=∂∂

For case (b) this gives

3

22

rC

rp ρ=

∂∂

0220

22 p

r1

r1C

21p +⎟⎟

⎞⎜⎜⎝

⎛−ρ=

24

Physical InterpreterPhysical Interpreter1/21/2

Under the basic assumptions: Under the basic assumptions: the flow is steady and the fluid is the flow is steady and the fluid is inviscidinviscid and incompressible.and incompressible.Application of F=ma and integration of equation of motion along Application of F=ma and integration of equation of motion along and normal to the streamline result inand normal to the streamline result in

To To produce an acceleration, there must be an unbalance of produce an acceleration, there must be an unbalance of the resultant forcethe resultant force, , of which only pressure and gravity were of which only pressure and gravity were considered to be importantconsidered to be important. Thus, there are three process . Thus, there are three process involved in the flow involved in the flow –– mass times acceleration (the mass times acceleration (the ρρVV22/2 term), /2 term), pressure (the p term), and weight (the pressure (the p term), and weight (the γγz term).z term).

CzdnR

Vp2

=+∫+ γρCz2

Vp2

=++ γρ

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Physical InterpreterPhysical Interpreter2/22/2

The Bernoulli equation is a mathematical statement of The Bernoulli equation is a mathematical statement of ““The The work done on a particle of all force acting on the particle is ework done on a particle of all force acting on the particle is equal qual to the change of the kinetic energy of the particleto the change of the kinetic energy of the particle””..

Work done by force : FWork done by force : F××d.d.Work done by weight: Work done by weight: γγz z Work done by pressure force: pWork done by pressure force: p

Kinetic energy: Kinetic energy: ρρVV22/2/2

26

The Bernoulli Equation can be written in The Bernoulli Equation can be written in terms of heights called headsHeadHead terms of heights called heads

An alternative but equivalent form of the Bernoulli An alternative but equivalent form of the Bernoulli equation is obtained by dividing each term by equation is obtained by dividing each term by γγ

czg2

VP 2

=++γ

27

Example 3.4 Kinetic, Potential, and Example 3.4 Kinetic, Potential, and Pressure EnergyPressure Energy

Consider the flow of water from the syringe Consider the flow of water from the syringe shown in Figure E3.4. A force applied to the shown in Figure E3.4. A force applied to the plunger will produce a pressure greater than plunger will produce a pressure greater than atmospheric at point (1) within the syringe. atmospheric at point (1) within the syringe. The water flows from the needle, point (2), The water flows from the needle, point (2), with relatively high velocity and coasts up to with relatively high velocity and coasts up to point (3) at the top of its trajectory. Discuss point (3) at the top of its trajectory. Discuss the energy of the fluid at point (1), (2), and (3) the energy of the fluid at point (1), (2), and (3) by using the Bernoulli equation. by using the Bernoulli equation.

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Example 3.4 Solution

The sum of the three types of energy (kinetic, potential, and prThe sum of the three types of energy (kinetic, potential, and pressure) essure) or heads (velocity, elevation, and pressure) must remain constanor heads (velocity, elevation, and pressure) must remain constant. t.

streamlinethealongttanconszV21p 2 =γ+ρ+

The motion results in a change in the magnitude of each type of energy as the fluid flows from one location to another.

The pressure gradient between (1) and (2) The pressure gradient between (1) and (2) produces an acceleration to eject the water produces an acceleration to eject the water form the needle. form the needle. Gravity acting on the particle between (2) and Gravity acting on the particle between (2) and (3) produces a deceleration to cause the water (3) produces a deceleration to cause the water to come to a momentary stop at the top of its to come to a momentary stop at the top of its flight.flight.

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Example 3.5 Pressure Variation in a Example 3.5 Pressure Variation in a Flowing StreamFlowing Stream

Consider the Consider the inviscidinviscid, incompressible, steady flow shown in Figure , incompressible, steady flow shown in Figure E3.5. From section A to B the streamlines are straight, while frE3.5. From section A to B the streamlines are straight, while from C om C to D they follow circular paths. to D they follow circular paths. Describe the pressure variation Describe the pressure variation between points (1) and (2)and points(3) and (4)between points (1) and (2)and points(3) and (4)

30

Example 3.5 Example 3.5 SolutionSolution1/21/2

R= ∞∞ , , for the portion from for the portion from A to BA to B

ttanconsrzp =+ Point (1)~(2)

Using p2=0,z1=0,and z2=h2-1

1221221 rhp)zz(rpp −+=−+=

Since the radius of curvature of the streamline is infinite, the pressure variation in the vertical direction is the same as if the fluids were stationary.

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Example 3.5 Example 3.5 SolutionSolution2/22/2

For the portion from For the portion from C to DC to D

∫ρ−= −4

3

z

z

2

343 dzRVrhp

With pp44=0 and z=0 and z44--zz33=h=h44--33 ,this becomes

334z

z

2

4 rzprz)dz(R

Vp4

3

+=+−ρ+ ∫ Point (3)~(4)

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Static, Stagnation, Dynamic, and Static, Stagnation, Dynamic, and Total PressureTotal Pressure1/51/5

Each term in the Bernoulli equation can be interpreted as a Each term in the Bernoulli equation can be interpreted as a form of pressure.form of pressure.

pp is the actual thermodynamic pressure of the fluid as it is the actual thermodynamic pressure of the fluid as it flows. To measure this pressure, one must move along flows. To measure this pressure, one must move along with the fluid, thus being with the fluid, thus being ““staticstatic”” relative to the moving relative to the moving fluid. Hence, it is termed the fluid. Hence, it is termed the static pressurestatic pressure …… seen by the seen by the fluid particle as it movesfluid particle as it moves. .

Cz2

Vp2

=γ+ρ+ Each term can be interpreted as a form of pressure

33

Static, Stagnation, Dynamic, and Static, Stagnation, Dynamic, and Total PressureTotal Pressure2/52/5

The static pressureThe static pressure is measured in a flowing fluid using a is measured in a flowing fluid using a wall pressure wall pressure ““taptap””, or a static pressure probe., or a static pressure probe.

hhhphp 34133131 γ=γ+γ=+γ= −−−The static pressureThe static pressureγγzz is termed the is termed the hydrostatic hydrostatic pressurepressure. It is not actually a . It is not actually a pressure but does represent the pressure but does represent the change in pressure possible due change in pressure possible due to potential energy variations of to potential energy variations of the fluid as a result of elevation the fluid as a result of elevation changes.changes.

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Static, Stagnation, Dynamic, and Static, Stagnation, Dynamic, and Total PressureTotal Pressure3/53/5

ρρVV22/2/2 is termed the is termed the dynamic pressuredynamic pressure. It can be interpreted . It can be interpreted as the pressure at the end of a small tube inserted into the as the pressure at the end of a small tube inserted into the flow and pointing upstream. flow and pointing upstream. After the initial transient After the initial transient motion has died out, the liquid will fill the tube to a height motion has died out, the liquid will fill the tube to a height of H.of H.The fluid in the tube, including that at its tip (2), will be The fluid in the tube, including that at its tip (2), will be stationary. That is, Vstationary. That is, V22=0, or point (2) is a stagnation point.=0, or point (2) is a stagnation point.

2112 V

21pp ρ+=Stagnation pressure

Static pressure Dynamic pressureDynamic pressure

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Static, Stagnation, Dynamic, and Static, Stagnation, Dynamic, and Total PressureTotal Pressure4/54/5

There is a stagnation point on any stationary body that is There is a stagnation point on any stationary body that is placed into a flowing fluid. Some of the fluid flows placed into a flowing fluid. Some of the fluid flows ““overover””and some and some ““underunder”” the object.the object.The dividing line is termed the The dividing line is termed the stagnation streamlinestagnation streamline and and terminates at the stagnation point on the body.terminates at the stagnation point on the body.Neglecting the elevation Neglecting the elevation effects, effects, the stagnation the stagnation pressure is the largest pressure is the largest pressure obtainable along a pressure obtainable along a given streamlinegiven streamline..

stagnation pointstagnation point

36

Static, Stagnation, Dynamic, and Static, Stagnation, Dynamic, and Total PressureTotal Pressure5/55/5

The The sum of the static pressure, dynamic pressure, and sum of the static pressure, dynamic pressure, and hydrostatic pressure is termed the total pressurehydrostatic pressure is termed the total pressure..The Bernoulli equation is a statement that the total The Bernoulli equation is a statement that the total pressure remains constant along a streamline.pressure remains constant along a streamline.

ttanconspz2

Vp T

2

==γ+ρ+ Constant along a streamline

37

The The PitotPitot--static Tube static Tube 1/51/5

ρ−=>>

ρ=−

==≈

ρ+==

/)pp(2V

2/Vpp

pppzz

2/Vppp

43

243

14

41

232

Knowledge of the values of the static and Knowledge of the values of the static and stagnation pressure in a fluid implies that the stagnation pressure in a fluid implies that the fluid speed can be calculated.fluid speed can be calculated.This is This is the principle on which the the principle on which the PitotPitot--static tube is based.static tube is based.

Static pressureStatic pressure

Stagnation pressureStagnation pressure

PitotPitot--static static stubesstubes measure measure fluid velocity by converting fluid velocity by converting velocity into pressure.velocity into pressure.

38

The The PitotPitot--static Tube static Tube 2/52/5

39

The The PitotPitot--static Tube static Tube 3/53/5

The use of The use of pitotpitot--static tube depends on the ability to static tube depends on the ability to measure the static and stagnation pressure.measure the static and stagnation pressure.An accurate measurement of static pressure requires that An accurate measurement of static pressure requires that none of the fluidnone of the fluid’’s kinetic energy be converted into a s kinetic energy be converted into a pressure rise at the point of measurement.pressure rise at the point of measurement.This requires a smooth hole with no burrs or imperfections.This requires a smooth hole with no burrs or imperfections.

Incorrect and correct design of static pressure taps.Incorrect and correct design of static pressure taps.

40

The The PitotPitot--static Tube static Tube 4/54/5

Typical pressure distribution along a Typical pressure distribution along a PitotPitot--static tube.static tube.

The pressure along the surface of an object varies from the The pressure along the surface of an object varies from the stagnation pressure at its stagnation point to value that stagnation pressure at its stagnation point to value that may be less than free stream static pressure.may be less than free stream static pressure.It is important that the pressure tapes be properly located It is important that the pressure tapes be properly located to ensure that the pressure measured is actually the static to ensure that the pressure measured is actually the static pressure.pressure.

41

The The PitotPitot--static Tube static Tube 5/55/5

Three pressure taps are drilled into a small circular Three pressure taps are drilled into a small circular cylinder, fitted with small tubes, and connected to three cylinder, fitted with small tubes, and connected to three pressure transducers. pressure transducers. The cylinder is rotated until the The cylinder is rotated until the pressures in the two side holes are equalpressures in the two side holes are equal, thus indicating , thus indicating that the center hole points directly upstream.that the center hole points directly upstream.

If θ=0

( ) 21

12

31

PP2V

PP

⎥⎦

⎤⎢⎣

⎡ρ−

=

=

Directional-finding Pitot-static tube.

42

Example 3.6 Example 3.6 PitotPitot--Static TubeStatic Tube

An airplane flies 100mi/hr at an elevation of 10,000 ft in a An airplane flies 100mi/hr at an elevation of 10,000 ft in a standard atmosphere as shown in Figure E3.6. Determine standard atmosphere as shown in Figure E3.6. Determine the pressure at point (1) far ahead of the airplane, point (2), the pressure at point (1) far ahead of the airplane, point (2), and the pressure difference indicated by a and the pressure difference indicated by a PitotPitot--static static probe attached to the fuselage. probe attached to the fuselage.

43

Example 3.6 Example 3.6 Solution Solution 1/21/2

2Vpp

21

12ρ

+=

psia11.10)abs(ft/lb1456p 21 ==

The static pressure and density at the altitude

If the flow is steady, inviscid, and incompressible and elevation changes are neglected. The Bernoulli equation

3ft/slug001756.0=ρ

With V1=100mi/hr=146.6ft/s and V2=0

)abs(ft/lb)9.181456(

2/)s/ft7.146)(ft/slugs001756.0(ft/lb1456p2

222322

+=

+=

44

Example 3.6 Example 3.6 Solution Solution 2/22/2

In terms of gage pressure

psi1313.0ft/lb9.18p 22 ==

The pressure difference indicated by the Pitot-static tube

psi1313.02Vpp

21

12 =ρ

=−

45

Application of Bernoulli Equation Application of Bernoulli Equation 1/21/2

The Bernoulli equation can be appliedThe Bernoulli equation can be applied between any two between any two points on a streamline providedpoints on a streamline provided that the other that the other three three restrictionsrestrictions are satisfied. The result isare satisfied. The result is

2

22

21

21

1 z2Vpz

2Vp γ+

ρ+=γ+

ρ+

Restrictions : Steady flow.Restrictions : Steady flow.Incompressible flow.Incompressible flow.Frictionless flow.Frictionless flow.Flow along a streamline.Flow along a streamline.

46

Application of Bernoulli Equation Application of Bernoulli Equation 2/22/2

Free jet.Free jet.Confined flow.Confined flow.FlowrateFlowrate measurementmeasurement

47

2

22

21

21

1 z2Vpz

2Vp γ+

ρ+=γ+

ρ+Free Jets Free Jets 1/31/3

Application of the Bernoulli equation between points (1) Application of the Bernoulli equation between points (1) and (2) on the streamlineand (2) on the streamline

gh2h2V

2Vh

2

=ργ

=

ρ=γ

At point (5)

)Hh(g2V +=

48

2

22

21

21

1 z2Vpz

2Vp γ+

ρ+=γ+

ρ+Free Jets Free Jets 2/32/3

For a sharpFor a sharp--edged orifice, edged orifice, a a vena vena contractacontracta effecteffect occurs.occurs.The effect is the result of the The effect is the result of the inability of the fluid to turn the inability of the fluid to turn the sharp 90sharp 90°° corner.corner.

For the horizontal nozzle, the For the horizontal nozzle, the velocity at the centerline, Vvelocity at the centerline, V22, , will be greater than that at the will be greater than that at the top Vtop V11..In general, d<<h and use the VIn general, d<<h and use the V22

as average velocity.as average velocity.

Vena Vena contractacontracta effects effects for sharpfor sharp--edged edged orifice.orifice.

49

2

22

21

21

1 z2Vpz

2Vp γ+

ρ+=γ+

ρ+Free Jets Free Jets 3/33/3

Typical Typical flow patterns and flow patterns and contraction coefficientscontraction coefficients for for various round exit various round exit configuration.configuration.The diameter of a fluid jet is The diameter of a fluid jet is often smaller than that of the often smaller than that of the hole from which it flows.hole from which it flows.

Define Cc=contraction coefficient Define Cc=contraction coefficient

h

jc A

AC =

AjAj=area of the jet at the vena =area of the jet at the vena contractacontractaAh=area of the holeAh=area of the hole

50

Example 3.7 Flow From a TankExample 3.7 Flow From a Tank�� GravityGravity

A stream of water of diameter d = 0.1m flows steadily from a tanA stream of water of diameter d = 0.1m flows steadily from a tank k of Diameter D = 1.0m as shown in Figure E3.7 (a). of Diameter D = 1.0m as shown in Figure E3.7 (a). Determine the Determine the flowrateflowrate, Q,, Q, needed from the inflow pipe if needed from the inflow pipe if the water depth remains the water depth remains constantconstant, h = 2.0m., h = 2.0m.

51

Example 3.7 Example 3.7 SolutionSolution1/21/2

22

221

2

11 zV21pzV

21p γ+ρ+=γ+ρ+

The Bernoulli equation applied between points The Bernoulli equation applied between points (1) and (2)(1) and (2) isis2

22

21

21

1 z2Vpz

2Vp γ+

ρ+=γ+

ρ+

(1)(1)

With With pp11 = p= p22 = 0, z= 0, z11 = h, and z= h, and z22 = 0= 0

22

21 V

21ghV

21

=+ (2)(2)

For steady and incompressible flow, For steady and incompressible flow, conservation of mass requiresconservation of mass requiresQQ11 = Q= Q22, where Q = AV. Thus, A, where Q = AV. Thus, A11VV11 =A=A22VV2 2 , or , or

22

12 Vd

4VD

22

1 V)Dd(V = (3)

52

Example 3.7 Example 3.7 SolutionSolution2/22/2

Combining Equation 1 and 3

s/m26.6)m1/m1.0(1

)m0.2)(s/m81.9(2)D/d(1

gh2V4

2

42 =−

=−

=

Thus,

s/m0492.0)s/m26.6()m1.0(4

VAVAQ 322211 =

π===

VV11≠≠0 (Q) vs. V0 (Q) vs. V11��0 (Q0 (Q00))

4

4

2

2

0 )/(11

2])/(1/[2

DdghDdgh

VV

QQ

D −=

−==

∞=

53

Example 3.8 Flow from a TankExample 3.8 Flow from a Tank--PressurePressure

Air flows steadily from a tank, through a hose of diameter Air flows steadily from a tank, through a hose of diameter D=0.03m and exits to the atmosphere from a nozzle of D=0.03m and exits to the atmosphere from a nozzle of diameter d=0.01m as shown in Figure E3.8. diameter d=0.01m as shown in Figure E3.8. The pressure The pressure in the tank remains constant at 3.0kPa (gage)in the tank remains constant at 3.0kPa (gage) and the and the atmospheric conditions are standard temperature and atmospheric conditions are standard temperature and pressure. pressure. Determine the Determine the flowrateflowrate and the pressure in and the pressure in the hose.the hose.

54

Example 3.8 Example 3.8 SolutionSolution1/21/2

For steady, For steady, inviscidinviscid, and incompressible flow, the Bernoulli equation , and incompressible flow, the Bernoulli equation along the streamlinealong the streamline

32332

2221

211 zV

21pzV

21pzV

21p γ+ρ+=γ+ρ+=γ+ρ+

With zWith z11 =z=z22 = z= z33 , V, V11 = 0, and p= 0, and p33=0=0

2212

13 V

21ppandp2V ρ−=

ρ= (1)(1)

The density of the air in the tank is obtained from the perfect The density of the air in the tank is obtained from the perfect gas law gas law

33

2

1

m/kg26.1K)27315)(Kkg/mN9.286(

kN/N10]m/kN)1010.3[(RTp

=+⋅⋅

×+==ρ

55

Example 3.8 Example 3.8 SolutionSolution2/22/2

Thus, Thus,

s/m0.69m/kg26.1

)m/N100.3(2p2V 3

231

3 =×

= s/m00542.0Vd4

VAQ 33

233 =

π==oror

The pressure within the hose can be obtained from The pressure within the hose can be obtained from EqEq. 1. 1and the continuity equationand the continuity equation

s/m67.7A/VAV,HenceVAVA 23323322 ===

22

23232212

m/N2963m/N)1.373000(

)s/m67.7)(m/kg26.1(21m/N100.3V

21pp

=−=

−×=ρ−=

56

Example 3.9 Flow in a Variable Area PipeExample 3.9 Flow in a Variable Area Pipe

Water flows through a pipe reducer as is shown in Figure E3.9. TWater flows through a pipe reducer as is shown in Figure E3.9. The he static pressures at (1) and (2) are measured by the inverted Ustatic pressures at (1) and (2) are measured by the inverted U--tube tube manometer containing oil of specific gravity, SG, less than one.manometer containing oil of specific gravity, SG, less than one.Determine the manometer reading, h.Determine the manometer reading, h.

57

Example 3.9 Example 3.9 SolutionSolution1/21/2

For steady, For steady, inviscidinviscid, incompressible flow, the Bernoulli equation along , incompressible flow, the Bernoulli equation along the streamlinethe streamline

22221

211 zpV

21pzpV

21p γ++=γ++

The continuity equationThe continuity equation

2211 VAVAQ ==

Combining these two equations Combining these two equations

])A/A(1[pV21)zz(pp 2

122

21221 −+−γ=− (1)(1)

58

Example 3.9 Example 3.9 SolutionSolution2/22/2

This pressure difference is measured by the manometer and determThis pressure difference is measured by the manometer and determine ine by using the pressureby using the pressure--depth ideas developed in depth ideas developed in Chapter 2Chapter 2..

↑- ↓+

h)SG1()zz(pp 1221 γ−+−γ=−

2121 phSGh)zz(p =γ+γ+γ−γ−−γ− ll

oror(2)(2)

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−=γ−

2

1

222 A

A1pV21h)SG1(

( ) ( )SG1g2)A/A(1A/Qh

2122

2 −−

=

Since VSince V22=Q/A=Q/A22

be independent of be independent of θθ

59

Confined Flows Confined Flows 1/41/4

When the fluid is physically constrained within a device, When the fluid is physically constrained within a device, its pressure cannot be prescribed a priori as was done for its pressure cannot be prescribed a priori as was done for the free jet.the free jet.Such casesSuch cases include include nozzle and pipesnozzle and pipes of of various diametervarious diameterfor which the fluid velocity changes because the flow area for which the fluid velocity changes because the flow area is different from one section to another.is different from one section to another.For such situations, it is necessary to use the concept of For such situations, it is necessary to use the concept of conservation of mass (the continuity equation) along with conservation of mass (the continuity equation) along with the Bernoulli equation.the Bernoulli equation.

Tools:Bernoulli equation + Continuity equation

60

Confined Flows Confined Flows 2/42/4

Consider a fluid flowing through a fixed volume that has Consider a fluid flowing through a fixed volume that has one inlet and one outlet.one inlet and one outlet.Conservation of mass requiresConservation of mass requiresFor For incompressible flowincompressible flow, the continuity equation is, the continuity equation is

222111 VAVA ρ=ρ

212211 QQVAVA ==

61

Confined Flows Confined Flows 3/43/4

If the fluid velocity is increased, If the fluid velocity is increased, the pressure will decrease.the pressure will decrease.This pressure decrease can be This pressure decrease can be large enough so that the large enough so that the pressure in the liquid is reduced pressure in the liquid is reduced to its to its vapor pressure.vapor pressure.

Pressure variation and cavitationin a variable area pipe.

62

Confined Flows Confined Flows 4/4 example of 4/4 example of cavitationcavitation

A example of A example of cavitationcavitation can be demonstrated with a can be demonstrated with a garden hose.garden hose. If If the hose is the hose is ““kinked,kinked,”” a restriction in the flow area will result.a restriction in the flow area will result.The water The water velocityvelocity through the restriction will be through the restriction will be relatively large.relatively large.With a sufficient amount of restriction the sound of the flowingWith a sufficient amount of restriction the sound of the flowingwater will change water will change –– a definite a definite ““hissinghissing’’ sound will be produced.sound will be produced.

The sound is a result of The sound is a result of cavitationcavitation..

63

Damage from Damage from CavitationCavitation

Cavitation from propeller

64

Example 3.10 Siphon and Example 3.10 Siphon and CavitationCavitation

Water at 60Water at 60��is siphoned from a large tank through a constant is siphoned from a large tank through a constant diameter hose as shown in Figure E3.10. Determine the maximum diameter hose as shown in Figure E3.10. Determine the maximum height of the hill, H, over which the water can be siphoned withheight of the hill, H, over which the water can be siphoned without out cavitationcavitation occurring. The end of the siphon is 5 ft below the bottom occurring. The end of the siphon is 5 ft below the bottom of the tank. Atmospheric pressure is 14.7 of the tank. Atmospheric pressure is 14.7 psiapsia..

The value of H is a function of both the The value of H is a function of both the specific weight of the fluid, specific weight of the fluid, γγ, and its , and its vapor pressure, vapor pressure, ppvv..

65

Example 3.10 Example 3.10 SolutionSolution1/21/2

For ready, For ready, inviscidinviscid, and incompressible flow, the Bernoulli equation , and incompressible flow, the Bernoulli equation along the streamline from (1) to (2) to (3) along the streamline from (1) to (2) to (3)

32332

2221

211 zV

21pzV

21pzV

21p γ+ρ+=γ+ρ+=γ+ρ+ (1)(1)

With With zz1 1 = 15 ft, z= 15 ft, z22 = H, and z= H, and z33 = = --5 ft. Also, V5 ft. Also, V11 = 0 (large tank), p= 0 (large tank), p11 = 0 = 0 (open tank), p(open tank), p33 = 0 (free jet), and from the continuity equation A= 0 (free jet), and from the continuity equation A22VV22 = = AA33VV33, or because the hose is constant diameter V, or because the hose is constant diameter V22 = V= V33.. The speed of The speed of the fluid in the hose is determined from the fluid in the hose is determined from EqEq. 1 to be. 1 to be

22

313 Vs/ft9.35ft)]5(15)[s/ft2.32(2)zz(g2V ==−−=−= VV22=V=V33

66

Example 3.10 Example 3.10 SolutionSolution2/22/2

22212

221

2112 V

21)zz(zV

21zV

21pp ρ−−γ=γ−ρ−γ+ρ+=

Use of Use of EqEq. 1 . 1 between point (1) and (2) then gives the pressure pbetween point (1) and (2) then gives the pressure p22 at the at the toptop of the hill asof the hill as

(2)(2)

The The vapor pressure of water at 60vapor pressure of water at 60��is 0.256 is 0.256 psiapsia. . Hence, for incipient Hence, for incipient cavitationcavitation the lowest pressure in the system will be p = 0.256 the lowest pressure in the system will be p = 0.256 psiapsia..UUsing gage pressure: sing gage pressure: pp11 = 0,= 0, pp22=0.256 =0.256 –– 14.7 = 14.7 = --14.4 14.4 psipsi

233222 )s/ft9.35)(ft/slugs94.1(21ft)H15)(ft/lb4.62()ft/.in144)(.in/lb4.14( −−=−

ftH 2.28=

67

FlowrateFlowrate Measurement Measurement inin pipes 1/5pipes 1/5

Various flow meters are Various flow meters are governed by the governed by the Bernoulli Bernoulli and continuity equationsand continuity equations..

2211

222

211

VAVAQ

V21pV

21p

==

ρ+=ρ+

The theoretical The theoretical flowrateflowrate

( )[ ]2

12

212 )A/A(1

pp2AQ−

−=

ρ Typical devices for measuring Typical devices for measuring flowrateflowrate in pipesin pipes

68

Example 3.11 Example 3.11 VenturiVenturi MeterMeter

Kerosene (SG = 0.85) flows through the Kerosene (SG = 0.85) flows through the VenturiVenturi meter shown in meter shown in Figure E3.11 with Figure E3.11 with flowratesflowrates between 0.005 and 0.050 mbetween 0.005 and 0.050 m33/s. /s. Determine the range in pressure difference, Determine the range in pressure difference, pp11 –– pp22, needed to , needed to measure these measure these flowratesflowrates..

Known Q, Determine pKnown Q, Determine p11--pp22

69

Example 3.11 Example 3.11 SolutionSolution1/21/2

22

2

A2])A/A(1[Q

pp2

12

21

−ρ=−

For steady, For steady, inviscidinviscid, and incompressible flow, the relationship between , and incompressible flow, the relationship between flowrateflowrate and pressure and pressure

( )[ ]2

12

212 )A/A(1

pp2AQ−ρ

−= EqEq. 3.20. 3.20

The density of the flowing fluidThe density of the flowing fluid

33O2H kg/m850)kg/m1000(85.0SG ==ρ=ρ

The area ratioThe area ratio

36.0)m10.0/m006.0()D/D(/AA 221212 ===

70

Example 3.11 Example 3.11 SolutionSolution2/22/2

The pressure difference for the The pressure difference for the smallest smallest flowrateflowrate isis

kPa16.1N/m1160])m06.0)(4/[(2

)36.01()kg/m850()/sm005.0(pp2

22

2323

21

==π

−=−

The pressure difference for the The pressure difference for the largest largest flowrateflowrate isis

22

22

21 ])m06.0)(4/[(2)36.01()850)(05.0(pp

π−

=−

kPa116N/m1016.1 25 =×=

kPa116-ppkPa16.1 21 ≤≤

71

FlowrateFlowrate Measurement Measurement sluice gate 2/5sluice gate 2/5

The sluice gate is often used to regulate and measure the The sluice gate is often used to regulate and measure the flowrateflowrate in in an open channel.an open channel.The The flowrateflowrate, Q, , Q, is function of the water depth upstream, zis function of the water depth upstream, z11, the , the width of the gate, b, and the gate opening, a.width of the gate, b, and the gate opening, a.

( )2

12

212 )/(1

2zzzzgbzQ

−−

=

With pWith p11=p=p22=0, the =0, the flowrateflowrate

22221111

22221

211

zbVVAzbVVAQ

zV21pzV

21p

====

γ+ρ+=γ+ρ+

72

FlowrateFlowrate Measurement Measurement sluice gate 3/5sluice gate 3/5

In the limit of In the limit of zz11>>z>>z22, this result simply becomes , this result simply becomes

This limiting result represents the fact that if the depth ratioThis limiting result represents the fact that if the depth ratio, z, z11/z/z22, is , is large, large, the kinetic energy of the fluid upstream of the gate is the kinetic energy of the fluid upstream of the gate is negligiblenegligible and the fluid velocity after it has fallen a distance (zand the fluid velocity after it has fallen a distance (z11--zz22)~z)~z11 is approximately is approximately

ZZ2 2 ?? ?? <a Z<a Z22==CCcca a CCc c

12 gz2bzQ =

12 gz2V =

73

FlowrateFlowrate Measurement Measurement sluice gate 4/5sluice gate 4/5

As we discussed relative to flow from an orifice, As we discussed relative to flow from an orifice, the fluid the fluid cannot turn a sharp 90cannot turn a sharp 90°° corner. A vena corner. A vena contractacontracta results results with a contraction coefficient, Cwith a contraction coefficient, Ccc=z=z22/a, less than 1./a, less than 1.Typically CTypically Ccc~0.61 over the depth ratio range of 0<a/z~0.61 over the depth ratio range of 0<a/z11<0.2.<0.2.For For large value of a/zlarge value of a/z11, the value of C, the value of Ccc increase rapidly. increase rapidly.

74

Example 3.12 Sluice Gate

Water flows under the sluice gate in Figure E3.12 (a). Water flows under the sluice gate in Figure E3.12 (a). DertermineDerterminethe approximate the approximate flowrateflowrate per unit width of the channel.per unit width of the channel.

75

Example 3.12 Example 3.12 SolutionSolution1/21/2

For steady,For steady, inviscidinviscid,, incompreesibleincompreesible flow, theflow, the flowerateflowerate per unit widthper unit width

( )2

12

212 )/(1

2zzzzgbzQ

−−

=( )( )212

212 /1

2zzzzgz

bQ

−−

= EqEq.3.21.3.21

With zWith z11=5.0m and a=0.80m, so the ratio a/z=5.0m and a=0.80m, so the ratio a/z11=0.16<0.20.=0.16<0.20.Assuming contraction coefficient is approximately CAssuming contraction coefficient is approximately Ccc=0.61. =0.61. zz22==CCccaa=0.61(0.80m)=0.488m.=0.61(0.80m)=0.488m.The The flowrateflowrate

( ) ( )( )( ) s/m61.4

m0.5/m488.01m488.0m0.5s/m81.92m488.0

bQ 2

2

2

=−

−=

76

Example 3.12 Example 3.12 SolutionSolution1/21/2

If we consider zIf we consider z11>>z>>z22 and neglect the kinetic energy of the upstream and neglect the kinetic energy of the upstream fluidfluid, we would have, we would have

( )( ) smmsmmgzzbQ /83.40.5/81.92488.02 22

12 ===

77

FlowrateFlowrate Measurement Measurement weirweir 5/55/5

For a typical rectangular, sharpFor a typical rectangular, sharp--crested, the crested, the flowrateflowrate over the top of over the top of the weir plate is dependent on the weir plate is dependent on the weir height,the weir height, PPww, the width of the , the width of the channel, b, and the head, H,channel, b, and the head, H, of the water above the top of the weir.of the water above the top of the weir.

2/3111 Hg2bCgH2HbCAVCQ ===The The flowrateflowrate

Where CWhere C11 is a constant to be determined.is a constant to be determined.

78

Example 3.13 WeirExample 3.13 Weir

Water flows over a triangular weir, as is shown in Figure E3.13.Water flows over a triangular weir, as is shown in Figure E3.13.Based on a simple analysis using the Bernoulli equation, determiBased on a simple analysis using the Bernoulli equation, determine ne the dependence of the dependence of flowrateflowrate on the depth H. on the depth H. If the If the flowrateflowrate is Qis Q00when H=Hwhen H=H00, estimate the , estimate the flowrateflowrate when the depth is increased to when the depth is increased to H=3HH=3H00..

79

Example 3.13 Example 3.13 SolutionSolution

gH2

For steady , For steady , inviscidinviscid , and incompressible flow, the average speed , and incompressible flow, the average speed of the fluid over the triangular notch in the weir plate is of the fluid over the triangular notch in the weir plate is proportional toproportional toThe flow area for a depth of H is H[H tan(The flow area for a depth of H is H[H tan(θθ /2)]/2)]The The flowrateflowrate

where Cwhere C22 is an unknown constant to be determined experimentally.is an unknown constant to be determined experimentally.

( ) 2/522

2 Hg22

tanCgH2C2

tanHVAQ θ=

θ==

An increase in the depth by a factor of the three ( from HAn increase in the depth by a factor of the three ( from H00 to 3Hto 3H00 ) ) results in an increase of the results in an increase of the flowrateflowrate by a factor ofby a factor of

( ) ( )( ) ( )

6.15Hg22/tanCH3g22/tanC

QQ

2/502

2/502

H

H3

0

0 =θθ

=

80

EL & HGL EL & HGL 1/41/4

For For steady, steady, inviscidinviscid, incompressible flow, incompressible flow , the total energy , the total energy remains constant along a streamline.remains constant along a streamline.

Httanconszg2

VP 2==++

γg/p ρ The head due to local static pressure (pressure energy)The head due to local static pressure (pressure energy)

The head due to local dynamic pressure (kinetic energy)The head due to local dynamic pressure (kinetic energy)g2/V 2

z The elevation head ( potential energy )The elevation head ( potential energy )

81

EL & HGL EL & HGL 2/42/4

Energy Line (EL) : represents the total head height.Energy Line (EL) : represents the total head height.

Hydraulic Grade Line (HGL) height: Hydraulic Grade Line (HGL) height: represents the sum of the represents the sum of the elevation and static pressure heads.elevation and static pressure heads.

The difference in heights between the The difference in heights between the EL and the HGLEL and the HGL represents represents the dynamic ( velocity ) headthe dynamic ( velocity ) head

zg2

VP 2++

γ

zP+

γ

g2/V2

82

EL & HGL EL & HGL 3/43/4

Httanconszg2

VP 2==++

γ

84

Example 3.14 Example 3.14 Energy Line and Hydraulic Grade LineEnergy Line and Hydraulic Grade Line

Water is siphoned from the tank shown in Figure E3.14 through a Water is siphoned from the tank shown in Figure E3.14 through a hose of constant diameter. A small hole is found in the hose at hose of constant diameter. A small hole is found in the hose at location (1) as indicate. When the siphon is used, will water lelocation (1) as indicate. When the siphon is used, will water leak out ak out of the hose, or will air leak into the hose, thereby possibly caof the hose, or will air leak into the hose, thereby possibly causing using the siphon to malfunction?the siphon to malfunction?

85

Example 3.14Example 3.14 SolutionSolution1/21/2

Whether air will leak into or water will leak out of the hose dWhether air will leak into or water will leak out of the hose depends epends on on whether the pressure within the hose at (1) is less than or whether the pressure within the hose at (1) is less than or greater than atmosphericgreater than atmospheric. Which happens can be easily determined . Which happens can be easily determined by using the energy line and hydraulic grade line concepts. Withby using the energy line and hydraulic grade line concepts. With the the assumption of steady, incompressible, assumption of steady, incompressible, inviscidinviscid flow it follows that the flow it follows that the total head is constanttotal head is constant--thus, the energy line is horizontal.thus, the energy line is horizontal.

Since the hose diameter is constant, it follows from the continuSince the hose diameter is constant, it follows from the continuity ity equation (AV=constant) that the water velocity in the hose is coequation (AV=constant) that the water velocity in the hose is constant nstant throughout. Thus throughout. Thus the hydraulic grade line is constant distance, Vthe hydraulic grade line is constant distance, V22/2g, /2g, below the energy linebelow the energy line as shown in Figure E3.14. as shown in Figure E3.14.

86

Example 3.14Example 3.14 SolutionSolution2/22/2

Since the pressure at the end of the hose is atmospheric, it folSince the pressure at the end of the hose is atmospheric, it follows that lows that the hydraulic grade line is at the same elevation as the end of the hydraulic grade line is at the same elevation as the end of the hose the hose outlet. outlet. The fluid within the hose at any point above the hydraulic gradeThe fluid within the hose at any point above the hydraulic gradeline will be at less than atmospheric pressure.line will be at less than atmospheric pressure.

Thus, Thus, air will leak into the hose through the hole at point (1). air will leak into the hose through the hole at point (1).

87

Restrictions on Use of the Bernoulli Restrictions on Use of the Bernoulli Equation Equation compressibility effects 1/4compressibility effects 1/4

The assumption of incompressibility is reasonable for The assumption of incompressibility is reasonable for most liquid flows.most liquid flows.In certain instances, the assumption introduce considerable In certain instances, the assumption introduce considerable errors for gases.errors for gases.To account for compressibility effectsTo account for compressibility effects

CgzV21dp 2 =++

ρ∫

88

Restrictions on Use of the Bernoulli Restrictions on Use of the Bernoulli Equation Equation compressibility effects 2/4compressibility effects 2/4

For isothermal flow of perfect gas

For isentropic flow of perfect gas the density and pressure For isentropic flow of perfect gas the density and pressure are related by are related by ��P / P / ρρkk =Ct, where k = Specific heat ratio=Ct, where k = Specific heat ratio

22

2

11

21 z

g2V

PPln

gRTz

g2V

+=⎟⎟⎠

⎞⎜⎜⎝

⎛++

ttanconsgzV21dPPC 2k

1k1

=++∫−

ttanconsgzV21dpRT 2 =++

ρ∫

2

22

2

21

21

1

1 gz2

VP1k

kgz2

VP1k

k++

ρ⎟⎠⎞

⎜⎝⎛

−=++

ρ⎟⎠⎞

⎜⎝⎛

89

Restrictions on Use of the Bernoulli Restrictions on Use of the Bernoulli Equation Equation compressibility effects 3/4compressibility effects 3/4

To find the pressure ratio as a function of Mach number

1111a1 kRT/Vc/VM ==The upstream Mach number

Speed of sound

⎥⎥

⎢⎢

⎡−⎟

⎠⎞

⎜⎝⎛ ++=

− −1M

21k1

ppp 1k

k

21a

1

12Compressible flow

Incompressible flow21a

1

12 M2k

ppp

=−

90

Restrictions on Use of the Bernoulli Restrictions on Use of the Bernoulli Equation Equation compressibility effects 4/4compressibility effects 4/4

21a

1

12 M2k

ppp

=−

⎥⎥

⎢⎢

⎡−⎟

⎠⎞

⎜⎝⎛ ++=

− −1M

21k1

ppp 1k

k

21a

1

12

91

Example 3.15 Compressible Flow Example 3.15 Compressible Flow –– Mach Mach NumberNumber

A Boeing 777 flies at Mach 0.82 at an altitude of 10 km in a A Boeing 777 flies at Mach 0.82 at an altitude of 10 km in a standard atmosphere. Determine the stagnation pressure on the standard atmosphere. Determine the stagnation pressure on the leading edge of its wing if the flow is incompressible; and if tleading edge of its wing if the flow is incompressible; and if the he flow is incompressible isentropic.flow is incompressible isentropic.

For incompressible flow For compressible isentropic flow

kPa7.14....pp

55.0...1M2

1k1p

pp

12

1kk

21a

1

12

==−

==⎥⎥

⎢⎢

⎡−⎟

⎠⎞

⎜⎝⎛ ++=

− −

kPa5.12...pp

471.0...M2k

ppp

12

21a

1

12

=−

===−

92

Restrictions on Use of the Bernoulli Restrictions on Use of the Bernoulli Equation Equation unsteady effectsunsteady effects

For unsteady flow V = V ( s , t )For unsteady flow V = V ( s , t )

sVV

tVaS ∂

∂+

∂∂

=

( ) 0dzVd21dpds

tV 2 =γ+ρ++∂∂

ρ Along a streamline

+ Incompressible condition+ Incompressible condition

2222

S

S12

11 zV21pds

tVzV

21p 2

1

γ+ρ++∂∂

ρ=γ+ρ+ ∫

93

An incompressible, An incompressible, inviscidinviscid liquid liquid is placed in a vertical, constant is placed in a vertical, constant diameter Udiameter U--tube as indicated in tube as indicated in Figure E3.16. When released from Figure E3.16. When released from the the nonequilibriumnonequilibrium position shown, position shown, the liquid column will the liquid column will oscilateoscilate at a at a specific frequency. Determine this specific frequency. Determine this frequencefrequence..

94

Example 3.16Example 3.16 SolutionSolutionLet points (1) and (2) be at the air-water interface of the two columns of the tube and z=0 correspond to the equilibrium position of the interface.Hence z = 0 , p1 =p2 = 0, z1 = 0, z2 = - z , V1 = V2 = V →z = z ( t )

dtdVds

dtdVds

tV 2

1

2

1

S

S

S

Sl==

∂∂

∫∫The total length of the liquid colum

( )

ll

l

/g20zg2dt

zd

gdtdzV

zdtdVz

2

2

=ω⇒=+⇒

ρ=γ=

γ+ρ=−γ

Liquid oscillationLiquid oscillation

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