FUNDAMENTALSFUNDAMENTALS OFOF FLUIDFLUID MECHANICSMECHANICS ChapterChapter 55 FlowFlow AnalysisAnalysis UsingUsing ControlControl VolumeVolume JyhJyh--CherngCherng ShiehShieh Department of Bio-Industrial Mechatronics Engineering National Taiwan University 10/19/2009 1 MAINMAIN TOPICSTOPICS

ConservationConservation ofof MassMass NewtonNewton’’ss SecondSecond LawLaw –– TheThe LinearLinear MomentumMomentum EquationsEquations TheThe MomentMoment--ofof--MomentumMomentum EquationsEquations FirstFirst LawLaw ofof ThermodynamicsThermodynamics –– TheThe EnergyEnergy EquationEquation SecondSecond LawLaw ofof ThermodynamicsThermodynamics –– IrreversibleIrreversible FlowFlow

2 LearningLearning ObjectsObjects

SelectSelect anan appropriateappropriate finitefinite CVCV toto solvesolve aa fluidfluid mechanicsmechanics problem.problem. ApplyApply basicbasic lawslaws toto thethe contentscontents ofof aa finitefinite CVCV toto getget importantimportant answers.answers.

How to apply these basic laws? How to express these basic laws based on CV method?

3 ReviewReview ofof ReynoldsReynolds TransportTransport TheoremTheorem

ThisThis isis thethe fundamentalfundamental relationrelation betweenbetween thethe raterate ofof changechange ofof anyany arbitraryarbitrary extensiveextensive property,property, B,B, ofof aa systemsystem andand thethe variationsvariations ofof thisthis propertyproperty associatedassociated withwith aa controlcontrol volume.volume.

dBsys B   CV  bV  ndA dt t  CS    bdV  bV  ndA t  CV  CS 4 ConservationConservation ofof MassMass –– TheThe ContinuityContinuity EquationEquation 1/4 BasicBasic LawLaw forfor ConservationConservation ofof MassMass System method下，描述質量守恒的方程式 dM   0  Msystem  dm  dV dt system M(system) V(system) ForFor aa systemsystem andand aa fixed,fixed, nondeformingnondeforming controlcontrol volumevolume thatthat areare coincidentcoincident atat anan instantinstant ofof time,time, thethe ReynoldsReynolds TransportTransport TheoremTheorem leadsleads toto 第一時間，兩者一致 d   B=M and b =1 dV  dV  V  ndA dt sys t CV CS Time rate of change Time rate of change of the Net rate of flow of of the mass of the = mass of the content of the + mass through the coincident system coincident control volume control surface 依據Chapter 4推導Reynolds transport theorem的步驟 5 ConservationConservation ofof MassMass –– TheThe ContinuityContinuity EquationEquation 2/4 Chapter 4：system與CV在不同時間下的關係

The instant time considered

System and control volume at three different instances of time. (a) System and control volume at time t – δt. (b) System and control volume at time t, coincident condition. (c) System and control volume at time t + δt.

6 ConservationConservation ofof MassMass –– TheThe ContinuityContinuity EquationEquation 3/4 當CV是固定且不變形 ForFor aa fixed,fixed, nondeformingnondeforming controlcontrol volume,volume, thethe controlcontrol volumevolume formulationformulation ofof thethe conservationconservation ofof mass:mass: TheThe continuitycontinuity equationequation  m out  m in dM      dV  V  ndA  0 dt t CV CS 解讀結果 system Rate of increase Net influx of Of mass in CV mass   dV   V  ndA t CV CS

7 ConservationConservation ofof MassMass –– TheThe ContinuityContinuity EquationEquation 4/4 Special case IncompressibleIncompressible FluidsFluids      dV   V  ndA  0  dV  V  ndA  0 t CV CS t CV CS

ForFor SteadySteady flowflow  m out  m in  V  ndA  0 The mass flow rate into a control volume CS must be equal to the mass flow rate out of the control volume.

8 OtherOther DefinitionDefinition 衍生定義 MassMass flowrateflowrate throughthrough aa sectionsection ofof controlcontrol surfacesurface   m  Q  V  ndA  m out  m in  A     TheThe averageaverage velocityvelocity    V  ndA V  A A

9 Fixed,Fixed, NondeformingNondeforming ControlControl VolumeVolume 1/2

Special case WhenWhen thethe flowflow isis steadysteady  dV  0 m out  m in t CV

WhenWhen thethe flowflow isis steadysteady andand incompressibleincompressible

Q out  Q in WhenWhen thethe flowflow isis notnot steadysteady  dV  0 “+” : the mass of the contents of the control volume is increasing t CV “-” : the mass of the contents of the control volume is decreasing.

10 Fixed,Fixed, NondeformingNondeforming ControlControl VolumeVolume 2/2

Special case WhenWhen thethe flowflow isis uniformlyuniformly distributeddistributed overover thethe openingopening inin thethe controlcontrol surfacesurface (one(one dimensionaldimensional flow)flow) m  AV WhenWhen thethe flowflow isis nonuniformlynonuniformly distributeddistributed overover thethe openingopening inin thethe controlcontrol surfacesurface m  AV

11 ExampleExample 5.15.1 ConservationConservation ofof MassMass –– Steady,Steady, IncompressibleIncompressible FlowFlow  Seawater flows steadily through a simple conical-shaped nozzle at the end of a fire hose as illustrated in Figure E5.1. If the nozzle exit velocity must be at least 20 m/s, determine the minimum pumping capacity required in m3/s.

Figure E5.1

12 ExampleExample 5.15.1 SolutionSolution

The continuity equation Steady flow   dV  V  ndA  0 t CV CS   V  ndA  m 2  m1  0 or m 2  m1 CS    

1Q1   2Q 2 With incompressible condition

3 1  2  Q1  Q2  V2A2  ...  0.0251m /s 求minimum pumping capacity

13 ExampleExample 5.25.2 ConservationConservation ofof MassMass –– Steady,Steady, CompressibleCompressible FlowFlow  Air flows steadily between two sections in a long, straight portion of 4-in. inside diameter as indicated in Figure E5.2. The uniformly distributed temperature and pressure at each section are given. If the average air velocity (Nonuniform velocity distribution) at section (2) is 1000ft/s, calculate the average air velocity at section (1).

求section (1)平均速度

Figure E5.2 14 ExampleExample 5.25.2 SolutionSolution

The continuity equation

Steady flow   dV  V  ndA  0 t CV CS   V  ndA  m 2  m1  0  m 2  m1 CS    

1A1V1   2 A 2 V2 假設氣體為理想氣體 p T  2 2 1 Since A =A V  V V1  V2  ...219ft / s 1 2 1 2 p T 1 1 2 The ideal gas equation p  

RT 15 ExampleExample 5.35.3 ConservationConservation ofof MassMass –– TwoTwo FluidsFluids  Moist air (a mixture of dry air and water vapor) enters a dehumidifier at the rate of 22 slugs/hr. Liquid water drains out of the dehumidifier at a rate of 0.5 slugs/hr. Determine the mass flowrate of the dry air and the water vapor leaving the dehumidifier. 求出口處的mass flowrate

Figure E5.3 16 ExampleExample 5.35.3 SolutionSolution

The continuity equation Steady flow   dV  V  ndA  0 t CV CS

   V  ndA  m1  m 2  m 3  0 CS   

m 2  m 1  m 3  22slugs / hr  0.5slugs / hr  21.5slugs / hr

17 ExampleExample 5.45.4 ConservationConservation ofof MassMass –– NonuniformNonuniform VelocityVelocity ProfilesProfiles  Incompressible, laminar water flow develops in a straight pipe having radius R as indicated in Figure E5.4. At section (1), the velocity profile is uniform; the velocity is equal to a constant value U and is parallel to the pipe axis everywhere. At section (2), the velocity profile is axisymmetric and parabolic, with zero velocity at

the pipe wall and a maximum value of umax at the centerline. How are U and umax related? How are the average velocity at section

(2), V 2 , and umax related?

18 ExampleExample 5.45.4 SolutionSolution

The continuity equation Steady flow   dV  V  ndA  0 t CV CS  R   A U  V  ndA  0   A U   u 2rdr  0 1 1  1 1 2  2 A 2 0 2 With incompressible condition 1  2   r   u  u max 1     2   R   R   r      A1U  2u max 1    rdr  0 0   R  

u max  2U V2  u max / 2

19 ExampleExample 5.55.5 ConservationConservation ofof MassMass –– UnsteadyUnsteady FlowFlow  A bathtub is being filled with water from a faucet. The rate of flow from the faucet is steady at 9 gal/min. The tub volume is approximated by a rectangular space as indicate Figure E5.5a. Estimate the time rate of change of the depth of water in the tub,  h /  t , in in./min at any instant. 求水深的時間改變率

Figure E5.5 20 ExampleExample 5.55.5 SolutionSolution1/21/2

The continuity equation

  dV  V  ndA  0 t CV CS    air dVair   water dVwater  m water  m air t air volume t water volume  For air air dVair  m air  0 t air volume

21 ExampleExample 5.55.5 SolutionSolution2/22/2

 For water water volume  water dVwater  m water t

CV內水的體積 water volume  water dVwater   water [h(2ft)(5ft)  (1.5ft  h)A j ]

2 h   water (10ft  A j )  m water t

h Q water (9gal / min)(12in. / ft)  2  3 2 t (10ft  A j ) (7.48gal / ft )(10ft )

2 A j 10ft

22 FilmsFilms

Flow through a Sink flow Vacuum filter contraction

23 Moving,Moving, NondeformingNondeforming ControlControl VolumeVolume1/2 CV是移動，不變形 WhenWhen aa movingmoving controlcontrol volumevolume isis used,used, thethe fluidfluid velocityvelocity relativerelative toto thethe movingmoving controlcontrol isis anan importantimportant variable.variable. WW isis thethe relativerelative fluidfluid velocityvelocity seenseen byby anan observerobserver movingmoving withwith thethe controlcontrol volume.volume. 觀察者站在CV上看到的流體速度 VVcv isis thethe controlcontrol volumevolume velocivelocityty asas seenseen fromfrom aa fixedfixed coordinatecoordinate system.system. 站在固定座標看到的CV移動速度 VV isis thethe absoluteabsolute fluidfluid velocityvelocity seenseen byby aa stationarystationary observerobserver inin aa fixedfixed coordinatecoordinate system.system. 站在固定 座標看到 的流體速 度（絕對）

24 Moving,Moving, NondeformingNondeforming ControlControl VolumeVolume2/2

   dMsys   V  W  V  dV  W  ndA CV dt t  CV  C.S.   dV  W  ndA  0  CV  C.S. t 在移動的CV上架設一參考座標系

Velocities seen from the control volume reference frame (relative velocities)

25 ExampleExample 5.65.6 ConservationConservation ofof MassMass -- CompressibleCompressible FlowFlow withwith aa MovingMoving ControlControl VolumeVolume

 An airplane moves forward at speed of 971 km/hr as shown in Figure E5.6a. The frontal intake area of the jet engine is 0.80m2 and the entering air density is 0.736 kg/m3. A stationary observer determines that relative to the earth, the jet engine exhaust gases move away from the engine with a speed of 1050 km/hr. The engine exhaust area is 0.558 m2, and the exhaust gas density is 0.515 kg/m3. Estimate the mass flowrate of fuel into the engine in kg/hr.

Determine the mass flowrate of fuel into the engine in kg/hr

CV選在移動的引擎

Figure E5.6 26 ExampleExample 5.65.6 SolutionSolution

The intake velocity, W , relative to the moving The continuity equation The intake velocity, W1, relative to the moving control volume. The exhaust velocity, W2, also =0   needs to be measured relative to the moving dV  W  ndA  0 control volume.  CV  C.S. t W1是從CV觀察到的進氣速度 W2是從CV觀察到的排氣速度 Assuming one-dimensional flow 在地面觀察者看到的排氣離開引擎的速度  m   A W   A W  0  fuel in 1 1 1 2 2 2 飛機飛行速度

 m fuel in  2A2W2  1A1W1

W2  V2  Vplane  1050km / hr  971km / hr  201km / hr 3 2  m fuel in  (0.515kg / m )(0.558m )(2021km / hr)(1000m / km)  ...  9100kg / hr

27 W2是從CV觀察到的排氣速度 ExampleExample 5.75.7 ConservationConservation ofof MassMass -- RelativeRelative VelocityVelocity CV選在轉動的撒水器部分  Water enters a rotating lawn sprinkler through its base at the steady rate of 1000 ml/s as sketched in Figure E5.7. If the exit area of each of the two nozzle is 30 mm2 , determine the average speed of the water leaving each nozzle, relative to the nozzle, if (a) the rotary sprinkler head is (a) the rotary sprinkler head is Figure E5.7 stationary, (b) the sprinkler head rotates at 60 rpm, and (c) the Determine the average speed sprinkler head accelerates from 0 to of the water leaving each 600 rpm. nozzle, relative to the nozzle…

CV跟著轉動 28 ExampleExample 5.75.7 SolutionSolution

W是相對於轉動的Nozzle的速度 The continuity equation =0   dV  W  ndA  0 t  CV  C.S.  W  ndA  m  m  0  C.S.  in  out

m out  2A2W2  m in  Q Q (1000ml /s)(0.001m3 / liter)(106 mm2 / m2 ) W2   2  16.7m /s  W2 2A2 (1000ml / liter)(2)(30mm )

The value of W2 is independent of the speed of rotation of the sprinkler head and represents the average speed of the water exiting from each nozzle with

respect to the nozzle for case (a), (b), (c). 29 DeformingDeforming ControlControl VolumeVolume 1/21/2

CV變形，CS當然就出現移動 AA deformingdeforming controlcontrol volumevolume involvesinvolves changingchanging volumevolume sizesize andand controlcontrol surfacesurface movement.movement. TheThe ReynoldsReynolds transporttransport theoremtheorem forfor aa deformingdeforming controlcontrol volumevolume cancan bebe usedused forfor thisthis case.case.

dMsys    dV  W  ndA dt t  CV  C.S.    V  W  VCS

Vcs is the velocity of the control surface as seen by a fixed observer. W is the relative velocity referenced to the control surface.

30 DeformingDeforming ControlControl VolumeVolume 2/22/2

dMsys    dV  W  ndA dt t  CV  C.S.

相對於變形的control volume表面 通常不是零。因CV範圍 的速度。CV表面速度並非每一點都 與時俱變，必須妥善處 理。 相同。因此，相對速度自然較為複 雜。

31 ExampleExample 5.85.8 ConservationConservation ofof MassMass –– DeformingDeforming ControlControl VolumeVolume 1/21/2  A syringe is used to inoculate a cow. The plunger has a face area of 500 mm2. If the liquid in the syringe is to be injected steadily at a rate of 300 cm3/min, at what speed should the plunger be advanced? The leakage rate pastpast thethe plplunger is 0.01 times the volume flowrate out of the needle. Leakage rate Determine the speed of the plunger be advanced

Figure E5.8 32 ExampleExample 5.85.8 SolutionSolution

外漏加上從針頭注出者=貫穿CS進出CV的量 A1  Ap    The continuity equation CVdV  C.S.W  ndA  0 t      CVdV  m 2  Qleak  0  CVdV  (A1  Vneedle ) t     dV  A t CV 1 t m 2  Q2  Let   Vp  A1Vp  m 2  Qleak  0 t

 A1Vp  Q2  Qleak  0

Q2  Qleak Vp   ...  660mm / min A1 33 TheThe LinearLinear MomentumMomentum EquationsEquations 1/41/4

 Newton’s second law for a system moving relative to an inertial coordinate system. 利用System method描述Newton’s second law Time rate of change of Time rate of change of Sum of external forces the linear momentum of = actingacting onon the systemsystem the systemsystem     d  dP  F  F  F  VdV    sys  S  B dt sys dt  system    Psystem  Vdm  VdV M(system) V(system)

34 TheThe LinearLinear MomentumMomentum EquationsEquations 2/42/4

 When a control volume is coincident with a system at an instant of time, the force acting on the system and the force acting on the contents of the coincident control volume are instantaneously identical.   Fsys  Fcontents of the coincident control volume

在時間 t 的瞬間，作用在CV上的external force等於作用在system者。透過這層連 結，才能得到…… 想一想，為何Continuity equation不需要這 種連結？（原因在於：兩邊都等於0）

External forces acting on system and

coincident control volume 35 ReviewReview ofof ReynoldsReynolds TransportTransport TheoremTheorem

ThisThis isis thethe fundamentalfundamental relationrelation betweenbetween thethe raterate ofof changechange ofof anyany arbitraryarbitrary extensiveextensive property,property, B,B, ofof aa systemsystem andand thethe variationsvariations ofof thisthis propertyproperty associatedassociated withwith aa controlcontrol volume.volume.

dBsys B   CV  bV  ndA dt t  CS    bdV  bV  ndA t  CV  CS 36 TheThe LinearLinear MomentumMomentum EquationsEquations 3/43/4

 For the system and a fixed, nondeforming control volume that are coincident at an instant of time, the Reynolds Transport Theorem leads to 依據Chapter 4推導Reynolds transport theorem的步驟 mass flow rate  d       B=P and b  V VdV  VdV  VV  ndA B=P and sys CV CS dt t    F  F contents of the coincident control volume sys d       VdV  VdV  VV  ndA 因為質量進出所 解讀 dt sys t CV CS 引發的動量進出 Time rate of change Time rate of change of the Net rate of flow of of the linear = linear momentum of the + linear momentum momentum of the content of the coincident through the control coincident system control volume surface

37 TheThe LinearLinear MomentumMomentum EquationsEquations 4/44/4 對固定、不變形的CV而言 ForFor aa fixedfixed andand nondeformingnondeforming controlcontrol volume,volume, the control volume formulation of Newton’s second law

Linear momentum equation

     VdV  VV  ndA  F CV CS  Contents of the coincident t mass flow rate control volume

這是基於CV method，描述牛頓第二定律的方程式

38 Linear momentum equation written for a moving control volume

39 Moving,Moving, NondeformingNondeforming ControlControl VolumeVolume1/3 對移動、不變形的CV而言 Chapter 4: Reynolds transport dBsys    equation for a control volume   bdV   bW  ndA moving with constant velocity is dt t CV CS mass flow rate d      VdV  VdV  VW  ndA dt sys t CV CS        VdV   VW  ndA  F Contents of the coincident t CV CS control volume    V  W  VCV         (W  VCV )dV  (W  VCV )W  ndA  F Contents of the CV CS  t coincident control volume 40 Moving,Moving, NondeformingNondeforming ControlControl VolumeVolume2/3 特例 For a constant control volume velocity, Vcv, and steady flow in the control volume reference frame    W  VCV dV  0 STEADY FLOW t CV =0

          W  VCV W  ndA  WW  ndA  VCV W  ndA CS CS CS

ForFor steadysteady flowflow, continuitycontinuity equationequation  W  ndA  0  C.S. dMsys    dV  W  ndA  0 dt t  CV  C.S. STEADY FLOW

41 Moving,Moving, NondeformingNondeforming ControlControl VolumeVolume3/3

For aa movingmoving, nondeforming control volume, the linear momentum equation of steadysteady flowflow    WW  ndA  F CS  Contents of the coincident control volume mass flow rate

42 VectorVector FormForm ofof MomentumMomentum EquationEquation

TheThe sumsum ofof allall forcesforces (surface(surface andand bodybody forces)forces) actingacting onon aa NonNon--acceleratingaccelerating controlcontrol volumevolume isis equalequal toto thethe sumsum ofof thethe rate of change of momentum inside the control volume rate of change of momentum inside the control volume 解讀 andand thethe netnet raterate ofof fluxflux ofof momentummomentum outout throughthrough thethe controlcontrol surfacesurface..    F  F  F  contents of the coincident control volume  S  B      VdV  VV  ndA t CV CS    FB  Bdm  BdV  CV   Where the velocities are measured Relative to the control volume. FS  - pdA Relative to the control volume. 43 A FILMSFILMS

煙囪雲煙 Smokestack Force due to plume momentum a water jet Fire hose

船舶推力 Running on Jelly fish Marine propulsion water 水母

44 LinearLinear MomentumMomentum EquationsEquations 應用及注意事項應用及注意事項 線動量、力具方向性線動量、力具方向性，，其其正負要正負要 與選用的座標系統相符。與選用的座標系統相符。 流體進出流體進出CVCV，要注意流體速度與，要注意流體速度與  表面法向向量表面法向向量，，V  n（＋（＋ forfor flowflow outout ofof thethe CVCV，－，－ forfor flowflow intointo thethe CVCV））。。

45 ApplicationApplication forfor FIXINGFIXING CVCV

5.10~5.165.10~5.16

46 ExampleExample 5.105.10 LinearLinear MomentumMomentum –– ChangeChange inin FlowFlow DirectionDirection

 As shown in Figure E5.10a, a horizontal jet of water exits a nozzle

with a uniform speed of V1=10 ft/s, strike a vane, and is turned through an angleθ. Determine the anchoring force needed to hold the vane stationary. Neglect gravity and viscous effects.

Determine the anchoring force needed to hold the vane stationary.

47 ExampleExample 5.105.10 SolutionSolution

The x and z direction components of linear momentum equation    udV  uV  ndA  Fx t CV CS     V  ui  wk    wdV  wV  ndA  Fz t CV CS 

V1(V1)A1  V1 cos(V1)A2  FAx

(0)(V1)A1  V1 sin (V1)A2  FAz 2  FAx  V 1A1(1 cos)  ..  11.64(1 cos) lb 2  FAz  V 1A1 sin   ...  11.64sin  lb

48 ExampleExample 5.115.11 LinearLinear MomentumMomentum –– Weight,Weight, pressure,pressure, andand ChangeChange inin SpeedSpeed

 Determine the anchoring force required to hold in place a conical nozzle attached to the end of a laboratory sin faucet when the water flowrate is 0.6 liter/s. The nozzle mass is 0.1kg. The nozzle inlet and exit diameters are 16mm and 5mm, respectively. The nozzle axis is vertical and the axial distance between section (1) and (2) is 30mm. The pressure at section (1) is 464 kPa. to hold the vane stationary. Neglect gravity and viscous effects.

49 ExampleExample 5.115.11 SolutionSolution1/31/3

50 ExampleExample 5.115.11 SolutionSolution2/32/3

The z direction component of linear moment equation    wdV  wV  ndA  FA  Wn  p1A1  Ww  p2A2 t CV CS  V  ndA   w dA With the “+” used for flow out of the control volume and “-” used for flow in.

(m 1)(w1)  m 2 (w2 )  Wn  p1A1  Ww  p2A2

FA  m (w1  w2 )  Wn  p1A1  Ww  p2A2 m 1  m 2  m m 1  m 2  m  w1A1  Q  ...  0.599kg / s

51 ExampleExample 5.115.11 SolutionSolution3/33/3

Q Q w1   2  ...  2.98m /s A1 D1 / 4 Q Q w2   2  ...  30.6m A2 D2 / 4 2 Wn  mng  (0.1kg)(9.81m /s )  0.981N

 1 2 2  Ww  Vwg   h(D 1  D 2  D1D2 ) Vwg  ...  0.0278N 12 

FA  m (w1  w2 )  Wn  p1A1  Ww  p2A2  (0.599kg / s)(...)  ...  77.8N

52 ExampleExample 5.125.12 LinearLinear MomentumMomentum –– PressurePressure ,, ChangeChange inin Speed,Speed, andand FrictionFriction

 Water flows through a horizontal, 180° pipe bend. The flow cross- section area is constant at a value of 0.1ft2 through the bend. The magnitude of the flow velocity everywhere in the bend is axial and 50ft/s. The absolute pressure at the entrance and exit of the bend are 30 psia and 24 psia, respectively. Calculate the horizontal (x and y) components of the anchoring force required to hold the bend in place.

53 ExampleExample 5.125.12 SolutionSolution1/21/2

The x direction component of linear moment equation    udV  uV  ndA  FAx t CV CS At section (1) and (2), the flow is in the y direction and therefore u=0 at both sections.

FAx  0 The y direction component of linear moment equation

   vdV  vV  ndA  FAy  p1A1  p2A2 t CV CS

54 ExampleExample 5.125.12 SolutionSolution2/22/2

For one-dimensional flow

(v1)(m 1)  (v2 )(m 2 )  FAy  p1A1  p2A2

 m (v1  v2 )  FAy  p1A1  p2A2

FAy  m (v1  v2 )  p1A1  p2A2  ...  1324lb m 1  m 2  m  v1A1  ...  9.70slugs / s

55 ExampleExample 5.135.13 LinearLinear MomentumMomentum –– Weight,Weight, pressure,pressure, andand ChangeChange inin SpeedSpeed

 Air flows steadily between two cross sections in a long, straight portion of 4-in. inside diameter pipe as indicated in Figure E5.13, where the uniformly distributed temperature and pressure at each cross section are given, If the average air velocity at section (2) is 1000 ft/s, we found in Example 5.2 that the average air velocity at section (1) must be 219 ft/s. Assuming uniform velocity distributions at sections (1) and (2), determine the frictional force exerted by the pipe wall on the air flow between sections (1) and (2).

56 ExampleExample 5.135.13 SolutionSolution1/21/2

The axial component of linear moment equation    udV  uV  ndA  R x  p1A1  p2A2 t CV CS

(u1)(m 1)  (u2 )(m 2 )  R x  p1A1  p2A2

m (u2  u1)  R x  A2 (p1  p2 )

R x  A2 (p1  p2 )  m (u2  u1)

2  p  D 2   2  m 1  m 2  m    u2  ...  0.297slugs / s  RT2  4  57 ExampleExample 5.135.13 SolutionSolution2/22/2

R x  A2 (p1  p2 )  m (u2  u1)

R x  A2 (p1  p2 )  m (u2  u1)  ...  793lb

p2 2  RT2 2 D 2 A  2 4

58 ExampleExample 5.145.14 LinearLinear MomentumMomentum –– Weight,Weight, Pressure,Pressure,……  If the flow of Example 5.4 is vertically upward, develop an expression for the fluid pressure drop that occurs between sections (1) and (2).

59 ExampleExample 5.145.14 SolutionSolution

The axial component of linear moment equation    wdV  wV  ndA  p1A1  R z  W  p2A2 t CV CS

 (w1)(m1)  (w2 )(w2dA2 )  p1A1  R z  W  p2A2  CS 2   r   2 w  2w 1  w2  2w11 (r / R)  2 1       R   2 R 2 2 R (w2 )(w2dA2 )   w2 2rdr  4w1 CS 0 3 4  w 2R 2  w 2R  p A  R  W  p A 1 3 1 1 1 z 2 2 w 2 R W  p  p  1  z  1 2 60 3 A1 A1 ExampleExample 5.155.15 LinearLinear MomentumMomentum -- TrustTrust

 A static thrust as sketched in Figure E5.15 is to be designed for testing a jet engine. The following conditions are known for a typical test: Intake air velocity = 200 m/s; exhaust gas velocity= = 500 m/s; intake cross-section area = 1m2; intake static pressure = - 22.5 kPa=78.5 kPa (abs); intake static temperature = 268K; exhaust static pressure =0 kPa=101 kPa (abs). Estimate the normal trust for which to design.

61 ExampleExample 5.155.15 SolutionSolution

The x direction component of linear moment equation

   udV  uV  ndA  p1A1  Fth  p2A2  patm (A1  A2 ) t CV CS

 (u1)(m 1)  (u2 )(m 2 )  (p1  patm )A1  (p2  patm )A2  Fth

m  m 1  1A1u1  m 2  2A2u2

 m (u2  u1)  p1A1  p2A2  Fth

Fth  1A1  2A2  m (u2  u1)  ...  83700N

p1 1  m  1A1u1  ...  204kg / s RT1

62 ExampleExample 5.165.16 LinearLinear MomentumMomentum –– NomuniformNomuniform PressurePressure

 A sluice gate across a channel of width b is shown in the closed and open position in Figure 5.16a and b. Is the anchoring force required to hold the gate in place larger when the gate is closed or when it is open?

63 ExampleExample 5.165.16 SolutionSolution

When the gate is closed, the horizontal forces acting on the contents of the control volume are identified in Figure E5.16c.

  1 2 1 2 uV  ndA  H b  R x  R x  H b CS 2 2 When the gate is open, the horizontal forces acting on the contents of the control volume are identified in Figure E5.16d.

  1 2 1 2 uV  ndA  H b  R x  h b  Ff CS 2 2 1 1  u 2Hb  u 2hb  H2b  R  h2b  F 1 2 2 x 2 f

1 2 1 2 2 For H  h and u1  u2  R x  H b  h b  Ff  u2 hb 2 2 64 ApplicationApplication forfor MOVINGMOVING CVCV

5.175.17

65 ExampleExample 5.175.17 LinearLinear MomentumMomentum -- MovingMoving ControlControl VolumeVolume 1/2

 A vane on wheels move with a constant velocity V when a stream .. 0 of water having a nozzle exit velocity of V1 is turned 45° by the vane as indicated in Figure E5.17a. Note that this is the same moving vane considered in Section 4.4.6 earlier . Determine the magnitude and direction of the force, F, exerted by the stream of water on the vane surface. The speed of the water jet leaving the nozzle is 100ft/s, and the vane is moving to the right with a constant speed of 20 ft/s.

CV移動速度Vo V1是流體離開Nozzle的絕對速度

66 ExampleExample 5.175.17 LinearLinear MomentumMomentum -- MovingMoving ControlControl VolumeVolume 2/2

..

移動的CV

67 ExampleExample 5.175.17 SolutionSolution1/21/2

The x direction component of linear moment equation   .. WxW  ndA  R x CS

 (W1)(m 1)  (W2 cos45)(m 2 )  R x

m 1  1W1A1 m 2  2W2A2 The z direction component of linear moment equation   WzW  ndA  R z  WW CS

 (W2 sin 45)(m 2 )  R z  Ww

m 1  1W1A1  m 2  2W2A2 W  W  V  V  ... 1 2 1 0 68 ExampleExample 5.175.17 SolutionSolution2/22/2

2 R x  W1 A1(1 cos45)  ...  21.8..lb 2 R z  W1 A1 sin 45  Ww  ...  53lb 2 2 Ww  gA1 R  R x  R z  ...  57.3lb  R   tan1 z R x

69 FromFrom thethe ProceedingProceeding ExamplesExamples

AA flowingflowing fluidfluid cancan bebe forcedforced toto changechange direction,direction, SpeedSpeed upup oror slowslow down,down, havehave aa velocityvelocity profileprofile change,change, dodo onlyonly somesome oror allall ofof thethe above,above, dodo nobenobe ofof thethe above.above. AA netnet forceforce onon thethe fluidisfluidis requiredrequired forfor achievingachieving anyany oror allall ofof thethe firstfirst fourfour above.above. TheThe forcesforces onon aa flowingflowing fluidfluid balancebalance outout withwith nono netnet forceforce forfor thethe fifth.fifth. TypicalTypical forceforce consideredconsidered includeinclude pressure,pressure, friction,friction, weight.weight.

70 MomentMoment--ofof--MomentumMomentum EquationEquation1/41/4

 Applying Newton’s second law of motion to a particle of fluid 牛頓第二定律 針對一個Particle 對 d   座 The velocity measured in an inertial reference system (VV)  Fparticle 標 dt Acting on the particle 原 點  Taking moment of each side with respect to the origin of an inertial 的 coordinate system r是質點與座標原點的距離 力  d   dr    矩 r  (VV)  r  F  V V  V  0 dt particle dt d  dr  d  (r  V)V   VV  r  (VV) dt dt dt d    (r  V)V r  Fparticle dt 71 MomentMoment--ofof--MomentumMomentum EquationEquation2/42/4

d     (r  V)V r  F 適用到所有particles dt particle d  d  積分 (r  V)dV  (r  V)V  sys dt dt  sys d    (r  V)dV  (r  F) dt  sys  sys 適用所有質點，擴及整個System The time rate of change of the Sum of external torques Moment-of-momentum of the system Acting on system d   (r  V)V  (r  F) Based on system method dt   sys sys 72 MomentMoment--ofof--MomentumMomentum EquationEquation3/43/4

WhenWhen aa controlcontrol volumevolume isis coincidentcoincident withwith aa systemsystem atat anan instantinstant ofof time,time, thethe torquetorque acactingting onon thethe systemsystem andand thethe torquetorque actingacting onon thethe contentscontents ofof thethe coincidentcoincident controlcontrol volumevolume areare instantaneouslyinstantaneously identicalidentical     在時間 t 的瞬間，作用在CV上的 (r  F)sys  (r  F)cv external force所產生的力矩等於   作用在system者。透過這層連結， 才能得到…… ForFor fixedfixed andand nondeformingnondeforming controlcontrol volume,volume, thethe momentmoment-- ofof--momentummomentum equation:equation:          CV (r  V)dV  CS (r  V)V  ndA   (r  F) t Contents of the coincident control volume Based on Control Volume 73 ReviewReview ofof ReynoldsReynolds TransportTransport TheoremTheorem

ThisThis isis thethe fundamentalfundamental relationrelation betweenbetween thethe raterate ofof changechange ofof anyany arbitraryarbitrary extensiveextensive property,property, B,B, ofof aa systemsystem andand thethe variationsvariations ofof thisthis propertyproperty associatedassociated withwith aa controlcontrol volume.volume.

dBsys B   CV  bV  ndA dt t  CS    bdV  bV  ndA t  CV  CS 74 MomentMoment--ofof--MomentumMomentum EquationEquation4/44/4 依據 Chapter 4 推導 Reynolds transport theorem 的步驟 ForFor thethe systemsystem andand thethe contentscontents ofof thethe coincidentcoincident controlcontrol volumevolume thatthat isis fixedfixed andand nondeformingnondeforming,, TheThe ReynoldsReynolds transporttransport theoremtheorem leadsleads toto 選擇一固定且不變形的CV = SYSTEM d      (r  V)dV  (r  V)dV  (r  V)V  ndA dt sys t CV CS 解 Time rate of change Time rate of change of the Net rate of flow of 讀 Time rate of change Time rate of change of the Net rate of flow of of the moment-of- moment-of-momentum of moment-of-momentum momentum of the = the content of the + through the control system coincident control volume surface   b  r  V B  bm  r  Vm 75 ApplicationApplication1/81/8 CV是固定的，注意所涵蓋範圍    Consider the rotating sprinkler. (r  V)dV  0 t CV 假 The flows are one-dimensional. 設 The flows are steady or steady-in-the-mean. 條 件 Using the axial component of the moment-of-momentum equation to analyze this flow CV Using the fixed and non-deforming control volume which contains within its boundaries the spinning or stationary sprinkler head and the portion of the water flowing CV through the sprinkler contained in the control volume.

76 ApplicationApplication2/82/8

          (r  V)dV   (r  V)V  ndA   (r  F) t CV CS   This term can be nonzero only where fluid is   crossing the control surface. Everywhere else on CS(r  V)V  ndA crossing the control surface. Everywhere else on the control surface this term will be zero because 解  構 At section (1) r  V  0 There is no axial moment-of- 此 momentum flow in section (1) 項  的  轉軸到噴嘴中心的距離 At section (2) r  V  r2V2 內 流體流出噴嘴的絕對切線速度 容 r2 is the radius from the axis of rotation to the nozzle centerline and V2 is the tangential component of the velocity of the flow exiting each nozzle as observed from a frame of reference attached to the fixed and nondeforming

control volume. 77 ApplicationApplication3/83/8

W：流體相對Moving nozzle的相對速度

V：流體流出噴嘴的絕對速度

U：Moving nozzle相對於固定座標的速度

U is the velocity of the moving nozzle as measured relative to    the fixed control surface. V  W  U W is relative velocity of exit flow as viewed from the nozzle V is the absolute velocity of exit flow relative to a fixed control

surface. 78 ApplicationApplication4/84/8

      “-” for flow into CS(r  V)ρV  ndA V  n “+” for flow out

     (r  V)ρV  ndA  (r2Vθ2 ) m  CS axial Where m is the total mass flowrate through both nozzles. The mass flowrate is the same whether the sprinkler rotates or not. 如何判斷正負     “+” or “-”ascertained by r  V r  V using the right-hand rule

右手定則 79 ApplicationApplication5/85/8 Turbine（渦輪機）的Torque 為『負』   The torque term  (r  F)content of the control volume  (r  F)  T  r V m  content of the CV shaft 2 2  axial  Acting on the shaft r  V 『負』表示Torque與旋轉方向相反  The correct algebraic sign of the axial component of  r  V can be easily remembered in the following way:

IfIf VVθ andand UU areare inin thethe samesame direction,direction, useuse ++ IfIf VVθ andand UU areare inin oppositeopposite direction,direction, useuse --

80 ApplicationApplication6/86/8

Acting on the CV ShaftShaft power?power? w W m  U V W shaft Tshaft   r2V2m  shaft shaft  2 2

Sprinkler speed U  r2 Negative shaft work is work out of the control volume, that is, work done by the fluid on the rotor and thus its shaft. Torque與旋轉方向相反『負』的力矩導致『負』的軸 功：表示shaft work is out of the CV，是水『做』功在 渦輪機的轉子上！

81 ApplicationApplication7/87/8

         CV (r  V)dV  CS(r  V)V  ndA  (r  F) t Contents of the Control volume GeneralGeneral casecase

Tshaft   m in  rinVθin  m out  routVθout 

The “-” is used with mass flowrate into the control

volume, min, and the “+” is used with mass flowrate out of the control volume, m , to acount for the sign of the   out dot product V  n .

The “+” or “-” is used with the rV product depends   on the direction of r  Vaxial If V and U are in the same direction, use + θ 82 If Vθ and U are in opposite direction, use - ApplicationApplication8/88/8

U  r2 The shaft power

W shaft Tshaft   (m in )(rinVθin )  (m out )(routVθout )

W shaft   m in  UinVθin  m out  UoutVθout

質量守衡 m  m in  m out

wshaft   UinVθin   UoutVθout  When shaft torque and shaft rotation are in the same (opposite) direction, power is into (out of ) the fluid. 83 判斷 rVrV 正負

A simple way to determine the signsign ofof thethe rVrV productproduct is to compare the direction of V and the blade speed U.

If V and U are in the same direction, the product rV is positive. If V and U are in opposite direction, the product rV is negative.

84 ExampleExample 5.185.18 MomentMoment ofof MomentumMomentum ––TorqueTorque 1/21/2  Water enters a rotating lawn sprinkler through its base at the steady ..……………………………… rate of 1000 ml/s as sketched in Figure E5.18. The exit area of each nozzle is in the tangential direction. The radius from the axis of rotation to the centerline of each nozzle is 200mm. (a) The resisting torque required to hold the sprinkler head stationary.(b) The resisting torque associated with the sprinkler rotating with a constant speed of 500rev/min. (c) The speed of the sprinkler if no resisting torque is applied.

85 ExampleExample 5.185.18 MomentMoment ofof MomentumMomentum ––TorqueTorque 2/22/2

..………………………………

86 ExampleExample 5.185.18 SolutionSolution1/21/2



(a) Tshaft  r2V 2m V 2  V2  T  r V m where V 16.7m / s from Example 5.7 shaft 2 2  2 ..……………………………… (1000ml /s)(103 m3 / liter)(999kg / m3 ) m  Q   0.999kg /s (1000ml / liter) (200mm)(16.7m /s)(0.999kg /s)[1(N / kg) /(m / s2 )] T   3.34N  m shaft (1000ml / liter) (b) V2  W2  U2

where W2  16.7m / s U2  r2 (200mm)(500rev / min)(2 rad / rev) V2  16.7m /s   6.2m /s (1000mm / m)(60s / min) 87 ExampleExample 5.185.18 SolutionSolution2/22/2

Tshaft  r2V2m

(200mm)(6.2m /s)(0.999kg / s)[1(N / kg) /(m /s2 )] T    1.24N  m shaft (1000ml / liter)

..……………………………… (c)

Tshaft  r2 (W2  r2)m  0 W (16.7m /s)(1000mm / m)   2   83.5 rad /s  797rpm r2 (200mm)

88 ExampleExample 5.195.19 MomentMoment ofof MomentumMomentum –– PowerPower 1/21/2  An air fan has a bladed rotor of 12-in. outside diameter and 10-in. ..……………………………… inside diameter as illustrated in Figure E5.19a. The height of each rotor is constant at 1 in. from blade inlet to outlet. The flowrate is steady, on a time-average basis, at 230 ft3/min, and the absolute

velocity of the air at blade inlet, V1, is radial. The blade discharge angle is 30° from the tangential direction. If the rotor rotates at a constant speed of 1725 rpm, estimate the power required to run the fan.

89 ExampleExample 5.195.19 MomentMoment ofof MomentumMomentum –– PowerPower 2/22/2

..………………………………

90 ExampleExample 5.195.19 SolutionSolution

0 (V1 is radial) ..……………………………… W shaft   m 1  U1Vθ1  m 2  U2Vθ2  m  Q  ...  0.00912slug / s (6in.)(1725rpm)(2rad / rev) U  r    90.3ft / s 2 2 (12in./ ft)(60s / min)

V2  W2  U2 V2  U2  W2 cos30 W2 cos30  Vr2 m  Q  A2Vr2  2r2hVr2

 W2  m /(2r2hsin 30)  ...  29.3ft / s

W shaft  m U2V2  ...  0.972hp

91 FirstFirst LawLaw ofof ThermodynamicsThermodynamics –– TheThe EnergyEnergy EquationEquation1/5 Based on system method  The first law of thermodynamics forfor aa systemsystem is Time rate of increase Net time rate of energy Net time rate of energy of the total stored = addition by heat transfer + addition by work energy of the system into the system transfer into the system d edV  Q  Q   W  W   Q  W  dt  sys in out sys in out sys net / in net / in sys d 注意『正』、『負』 or edV  Q  W dt  sys net in net in sys “+” going into system V2 “-” coming out e  uˆ   gz 2 The net rate of work transfer Total stored energy per unit into the system mass for each particle in the system The net rate of heat transfer into the system92 FirstFirst LawLaw ofof ThermodynamicsThermodynamics –– TheThe EnergyEnergy EquationEquation2/5

ForFor thethe controlcontrol volumevolume thatthat isis coincidentcoincident withwith thethe systemsystem atat anan instantinstant ofof time.time. 在時間 t 的瞬間，進出 CV 的 Q 與 W 等於進 出 system 者。透過這層連結，才能得到……

(Q net in  W net in )sys  (Q net in  W net in )coincident control volume

93 ReviewReview ofof ReynoldsReynolds TransportTransport TheoremTheorem

ThisThis isis thethe fundamentalfundamental relationrelation betweenbetween thethe raterate ofof changechange ofof anyany arbitraryarbitrary extensiveextensive property,property, B,B, ofof aa systemsystem andand thethe variationsvariations ofof thisthis propertyproperty associatedassociated withwith aa controlcontrol volume.volume.

dBsys B   CV  bV  ndA dt t  CS    bdV  bV  ndA t  CV  CS 94 FirstFirst LawLaw ofof ThermodynamicsThermodynamics –– TheThe EnergyEnergy EquationEquation3/5 依據 Chapter 4 推導 Reynolds transport theorem 的步驟  For the system and the contents of the coincident control volume that is fixed and nondeforming -- Reynolds Transport Theorem leads to 選擇一固定且不變形的CV = SYSTEM d   edV  edV  eV  ndA dt  sys t  CV  C.S. 解 讀 Net time rate of increase The net rate of flow of the Time rate of increase of the total stored energy total stored energy out of of the total stored = of the contents of the + the control volume through energy of the system control volume the control surface b  e B  e  m 95 FirstFirst LawLaw ofof ThermodynamicsThermodynamics –– TheThe EnergyEnergy EquationEquation4/5

TheThe controlcontrol volumevolume formulaformula forfor thethe firstfirst lawlaw ofof thermodynamics:thermodynamics:

   cvedV  eV  ndA  (Q net in  W net in ) t  CS CV Based on Control Volume

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96 RateRate ofof WorkWork donedone byby CVCV

W  W Shaft  W normal  W shear  W other

ShaftShaft workwork W  Shaft :: thethe raterate ofof workwork transferredtransferred intointo throthroughugh thethe CSCS byby thethe shaftshaft workwork (( negativenegative forfor workwork transferredtransferred out,out, positivepositive forfor workwork inputinput required)required) 藉由shaft傳遞的功 WorkWork donedone byby normalnormal stressesstresses atat thethe CS:CS:       W normal  Fnormal  V   nn V  ndA   pV  ndA CS CS WorkWork donedone byby shearshear stressesstresses atat thethe CS:CS:   Negligibly small W shear   V  ndA CS OtherOther workwork  +輸入系統者，-輸出系統者

     cvedV  eV  ndA Q net in  W shaft net in  pV  ndA t  CS CS 97 FirstFirst LawLaw ofof ThermodynamicsThermodynamics –– TheThe EnergyEnergy EquationEquation5/5

     edV  eV  ndA  Q net in  W Shaf net in  pV  ndA t CV CS CS EnergyEnergy equationequation

2  p V   edV  (uˆ    gz)V  ndA  Q net / in  W Shaft / in t CV CS  2

V2 e  uˆ   gz 2

98 ApplicationApplication ofof EnergyEnergy EquationEquation1/31/3  When the flow is steady edV  0 t  CV The integral of

 p V2   uˆ    gz V  ndA  CS     2  ???

Special & simple case Uniformly distribution  p V2    p V2   p V2  uˆ    gz V  ndA  uˆ    gzm  uˆ    gzm  CS           2  out   2  in   2 

 p V2   uˆ    gz V  ndA  CS   OnlyOnly oneone streamstream   2  enteringentering andand leavingleaving  p V2   p V2  ˆ ˆ  u    gz m out  u    gz m in  2  2 99 Special & simple case  out  in ApplicationApplication ofof EnergyEnergy EquationEquation2/32/3

If shaft work is involved…. 當 shaft work 包括進來  2 2   p   p  Vout  Vin m uˆ out  uˆ in         gzout  zin        2   out  in  One-dimensional energy equation  Q net in  W shaft net in for steady-in-the-mean flow p Enthalpy hˆ  uˆ  The energy equation is written in terms  of enthalpy. 2 2 ˆ ˆ Vout  Vin  m hout  hin   gzout  zin   Q net / in  W shaft net / in  2 

100 ApplicationApplication ofof EnergyEnergy EquationEquation3/33/3

If shaft work is zero…. 沒有shaft work  2 2   p   p  Vout  Vin m uˆ out  uˆ in         gzout  zin      2   out  in  One-dimensional energy equation  Q net in for steady-in-the-mean flow 2 2 ˆ ˆ Vout  Vin  m hout  hin   gzout  zin   Q net / in  2 

The Pelton wheel is among the most efficient types of water turbines. It was invented by Lester Allan Pelton (1829-1908) in the 1870s 101 ExampleExample 5.205.20 EnergyEnergy –– PumpPump PowerPower 1/2

 A pump delivers water at a steady rate of 300 gal/min as shown in Figure E5.20. Just upstream of the pump [section(1)] where the pipe diameter is 3.5 in., the pressure is 18 psi. Just downstream of the pump [section (2)] where the pipe diameter is 1 in., the pressure is 60 psi. The change in water elevation across the pump is zero. The

rise in internal energy of water, u2-u1, associated with a temperature rise across the pump is 3000 ft·lb/slug. If the pumping process is considered to be adiabatic, determine the power (hp) required by the pump. 絕熱條件：沒有 Q 進出 計算輸入功

102 ExampleExample 5.205.20 EnergyEnergy –– PumpPump PowerPower 2/2

103 ExampleExample 5.205.20 SolutionSolution

One-dimensional energy equation for steady-in-the-mean flow

 2 2   p   p  V2  V1 m uˆ 2  uˆ 1         gz2  z1        2   2  1  =0(Adiabatic flow)  Q net / in  W shaft net / in

(1.94slug / ft3)(300gal / min) m  Q   1.30slugs/ s (7.48gal / ft3 )(60s / min) Q Q Q Q V   2 V1   .....  10.0ft / s V2   ...  123ft / s A D / 4 A1 A2

W shaft net in  (1.30slugs/ s)....  32.3hp

104 ExampleExample 5.215.21 EnergyEnergy –– TurbineTurbine PowerPower perper UnitUnit MassMass ofof FlowFlow

 Steam enters a turbine with a velocity of 30m/s and enthalpy, h1, of 3348 kJ/kg. The steam leaves the turbine as a mixture of vapor and liquid having a velocity of 60 m/s and an enthalpy of 2550 kJ/kg. If the flow through the turbine is adiabatic and changes in elevation are negligible, determine the work output involved per unit mass of steam through-flow. 絕熱條件：沒有 Q 進出 計算輸出功

105 ExampleExample 5.215.21 SolutionSolution

The energy equation in terms of enthalpy. =0(Adiabatic flow) 2 2 ˆ ˆ V2  V1  m h2  h1   gz2  z1   Q net / in  W shaft net / in  2  W 2 2 shaft net in ˆ ˆ V2  V1 w shaft net in   h2  h1  m 2 w shaft net out  w shaft net in 2 2 ˆ ˆ V1  V2 w shaft net out  h1  h2   ...  797kJ / kg 2

106 ExampleExample 5.225.22 EnergyEnergy –– TemperatureTemperature ChangeChange  A 500-ft waterfall involves steady flow from one large body of water to another. Determine the temperature change associated with this flow.

107 ExampleExample 5.225.22 SolutionSolution The temperature change is related to the change of internal energy of the water uˆ  uˆ T  T  2 1 2 1 c where c  1 Btu /(lbm R) is the specific heat of water One-dimensional energy equation for steady-in-the-mean flow without shaft work V =V 2 1 =0(Adiabatic flow)   p   p  V2  V2  ˆ ˆ 2 1 2 1 m u2  u1         gz2  z1   Q net in        2  g(z  z ) (32.2ft / s2 )(500ft) T  T  2 1   0.643R 2 1 c [778ft lb /(lbm R)][32.2(lbm ft) /(lb s2 )]

108 1/4 EnergyEnergy EquationEquation vs.vs. BernoulliBernoulli EquationEquation2  2 2  V  p   p  Vout  Vin m uˆ out  uˆ in         gzout  zin   Q net in  W shaft net in e  uˆ   gz       2   out  in  2 For steady, incompressible flow…One-dimensional energy equation   p   p  V2  V2  沒有 shaft work ˆ ˆ out in out in 有normal stress做的功 m uout  uin         gzout  zin   Q net in        2  2 2 pout Vout pin Vin  m   gzout    gzin  uˆ out  uˆ in  qnet in    2  2 where where qnet in  Q net in / m 參照 chapter 3 For steady, incompressible, frictionless flow… V2 V2 p  out  z  p  in  z Bernoulli equation out 2 out in 2 in Frictionless flow… uˆ out  uˆ in  q net in  0 沒有摩擦損失 109 EnergyEnergy EquationEquation && BernoulliBernoulli EquationEquation 2/4

For steady, incompressible, frictionalfrictional flowflow……

uˆ out  uˆ in  q net in  0 Frictional flow… p V2 Loss發生在in Defining “useful or available energy”…   gz out過程中  2

Defining “loss of useful or available energy”… uˆ out  uˆ in  q net in  loss 下游 上游 p V2 p V2 out  out  gz  in  in  gz  loss  2 out  2 in 2 2 p2 V2 p1 V1 因為12有摩擦損失，   gz2 自然而然就低於   gz1  2  2 110 EnergyEnergy EquationEquation && BernoulliBernoulli EquationEquation 3/4

For steady, incompressible flow with friction and shaft work… 有摩擦損失有軸功進來 2 2   pout   pin  Vout  Vin  m uˆ out  uˆ in         gzout  zin   Q net in  W shatf net in        2  在 in  out 注入 p V2 p V2 out  out  gz  in  in  gz  w  (uˆ  uˆ  q )  m  2 out  2 in shaft net in out in net in p V2 p V2 out  out  gz  in  in  gz  w  loss  2 out  2 in shaft net in 在 in  out 注入 p V2 p V2  g out  out  z  in  in  z  h  h  2g out  2g in s L

wshaft W shaft W shaft loss net / in net / in net / in Shaft head hS    Head loss hL  111 g m g Q g EnergyEnergy EquationEquation && BernoulliBernoulli EquationEquation 4/4

p V2 p V2 out  out  z  in  in  z  h  h  2g out  2g in s L in  out 輸出

ForFor turbineturbine h s  h T (h T  0) hT is turbine head in  out 輸入 ForFor pumppump h s  h P hp is pump head

TheThe actualactual headhead dropdrop acrossacross thethe turbineturbine Energy transfer h T  (h s  h L )T 想像：讓loss擴大 TheThe actualactual headhead dropdrop acrossacross thethe pumppump

h p  (h s  h L ) p 想像：讓loss減緩

112 ExampleExample 5.235.23 EnergyEnergy –– EffectEffect ofof LossLoss ofof AvailableAvailable EnergyEnergy

 Compare the volume flowrates associated with two different vent configurations, a cylindrical hole in the wall having a diameter of 120 mm and the same diameter cylindrical hole in the wall but with a well- rounded entrance (see Figure E5.23a). The room pressure is held constant at 0.1 kPa above atmospheric pressure. Both vents exhaust into the atmosphere. As discussed in Section 8.4.2. the loss in available energy associated with flow through the cylindrical bent from the room to the vent exit is 2 0.5V2 /2 where V2 is the uniformly distributed exit velocity of air. The loss in available energy associated with flow through the rounded entrance vent from the 2 room to the vent exit is 0.05V2 /2, where V2 is the uniformly distributed exit velocity of air. 113 ExampleExample 5.235.23 SolutionSolution

For steady, incompressible flow with friction, the energy equation V1=0 No elevation change p V2 p V2 2  2  gz  1  1  gz  loss  2 2  2 1 1 2

2  p1  p2   V2 V2  2 1loss2  1loss2  K L     2

p1  p2 V2  1 K L / 2 2 D2 p1  p2 Q  A2V2  4 1 K L / 2 114 ExampleExample 5.245.24 EnergyEnergy –– FanFan WorkWork andand EfficiencyEfficiency  An axial-flow ventilating fan driven by a motor that delivers 0.4 kW of power to the fan blades produces a 0.6-m-diameter axial stream of air having a speed of 12 m/s. The flow upstream of the fan involves negligible speed. Determine how much of the work to the air actually produces a useful effects, that is, a rise in available energy and estimate the fluid mechanical efficiency of this fan.

115 ExampleExample 5.245.24 SolutionSolution

For steady, incompressible flow with friction and shaft work… 2 2  p2 V2   p1 V1  wshaft net in  loss     gz 2      gz1    2    2 

p1=p2=atmospheric pressure, V1=0, no elevation change V 2 w  loss  2  72.0N  m / kg shaft net in 2 wshaft net in  loss Efficiency   wshaft net in W W w  shaft net in  shaft net in  95.8N  m / kg shaft net in m AV  116 ExampleExample 5.255.25 EnergyEnergy –– HeadHead LossLoss andand PowerPower LossLoss  The pump shown in Figure E5.25 adds 10 horsepower to the water as it pumps water from the lower lake to the upper lake. The elevation difference between the lake surfaces is 30 ft and the head loss is 15 ft. Determine the flowrate and power loss associated with this flow.

117 ExampleExample 5.255.25 SolutionSolution

The energy equation p V2 p V2 A  A  z  B  B  z  h  h  2g A  2g B s L

pA  pB  0 VA  VB  0 The pump head W h  h  z  z  shaft net / in  88.1/ Q s L A B Q

Power loss W loss  QhL  ...

118 ApplicationApplication ofof EnergyEnergy EquationEquation toto NonuniformNonuniform FlowsFlows 1/2 IfIf thethe velocityvelocity profileprofile atat anyany sectionsection wherewhere flowflow crossescrosses thethe controlcontrol surfacesurface isis notnot uniformuniform…… 2 V   當進出CS的流體速度分布不是uniform  V  ndA ???? C.S. 2 For one stream of fluid entering and leaving the control volume…. 以平均值取代 ~ ~ V2    V2  V2  V  ndA  m out out  in in   CS    2  2 2  2  V    Where  is the kinetic energy   V  ndA A 2 coefficient and V is the      1 新增參數 2 average velocity  V  m    2  119 ApplicationApplication ofof EnergyEnergy EquationEquation toto NonuniformNonuniform FlowsFlows 2/2

ForFor nonuniformnonuniform velocityvelocity profileprofile………….. p  V2 p  V2 out  out out  gz  in  in in  gz  w  loss  2 out  2 in shaft net in   V2  V2 p  out out  z  p  in in  z  w  (loss) out 2 out in 2 in shaft net in  g 2 2 w pout outVout pin inVin shaft net in   zout    zin   hL  2g  2g g 120 ExampleExample 5.265.26 EnergyEnergy –– EffectEffect ofof NonuniformNonuniform VelocityVelocity ProfileProfile 1/21/2  The small fan shown in Figure E5.26 moves air at a mass flowrate of 0.1 kh/min. Upstream of the fan, the pipe diameter is 60 mm, the flow is laminar, the velocity distribution is parabolic, and the kinetic

energy coefficient, α1, is equal to 2.0. Downstream of the fan, the pipe diameter is 30 mm, the flow is turbulent, the velocity profile is

quite uniform, and the kinetic energy coefficient, α2 , is equal to 1.08. If the rise in static pressure across the fan is 0.1 kPa and the fan motor draws 0.14 W, compare the value of loss calculated: (a) assuming uniform velocity distributions, (2) considering actual velocity distribution.

121 ExampleExample 5.265.26 EnergyEnergy –– EffectEffect ofof NonuniformNonuniform VelocityVelocity ProfileProfile 2/22/2

122 ExampleExample 5.265.26 SolutionSolution1/21/2

The energy equation for nonuniform velocity profile……. p  V2 p  V2 2  2 2  gz  1  1 1  gz  w  loss  2 2  2 1 shaft net / in 2 2  p2  p1  1V1 2V2 loss  wshaft net in         2 2 power to fan motor wshaft net / in  m (0.14W)[(1N  m / s) / W]  (60s / min)  84.0N  m / kg 0.1kg / min m m V1   ...  0.479m / s V2   ...  1.92m / s A1 A2 123 ExampleExample 5.265.26 SolutionSolution1/21/2

2 2  p2  p1  1V1 2V2 loss  wshaft net / in         2 2

 0.975N  m / kg1  2  1

2 2  p2  p1  1V1 2V2 loss  wshaft net / in         2 2

 0.940N  m / kg1  2,2  1.08

124 ExampleExample 5.285.28 EnergyEnergy –– FanFan PerformancePerformance  For the fan of Example 5.19, show that only some of the shaft power into the air is converted into a useful effect. Develop a meaningful efficiency equation and a practical means for estimating lost shaft energy.

125 ExampleExample 5.285.28 SolutionSolution1/21/2

p V2 p V2 2  2  gz  1  1  gz  w  loss (1)  2 2  2 1 shaft net in useful effect  wshaft net / in  loss 2 2  p2 V2   p1 V1  (2)     gz2      gz1    2    2 

wshaft net in  loss Efficiency   (3) wshaft net in

wshaft net in  U2V2 (4) 126 ExampleExample 5.285.28 SolutionSolution2/22/2

(2)+(3)+(4)

2 2   {[(p2 /)  (V2 / 2)  gz2 ]  [(p1 /)  (V1 / 2)  gz1]}/ U2V2

(2)+(4)

2 2   U2V2  [(p2 /  V2 / 2  gz2 )  (p1 /  V1 / 2  gz1)]

127 FirstFirst LawLaw ofof ThermodynamicsThermodynamics –– ForFor SemiSemi--infinitesimalinfinitesimal CVCV 1/21/2 ApplyingApplying thethe oneone--dimensional,dimensional, steadysteady flowflow energyenergy equationequation toto thethe contentcontent ofof aa semisemi--infinitesimalinfinitesimal controlcontrol volumevolume  2 2   pout   pin  Vout  Vin m uˆ out  uˆ in         gzout  zin   Q net in        2    p   V2   ˆ m du  d   d   gdz   Q net in      2   semi-infinitesimal control volume

CV非有限大，也非無限小，故將 in 與 out 間的變化以 difference來表達

128 FirstFirst LawLaw ofof ThermodynamicsThermodynamics –– ForFor SemiSemi--infinitesimalinfinitesimal CVCV 2/22/2

  1  For all pure substances including common  Tds  du  pd  engineering working fluids, such as air, water,    oil, and gasoline   1   p   V2   m Tds  pd   d   d   gdz   Q net in         2   SemiSemi--infinitesimalinfinitesimal controlcontrol volumevolume statementstatement ofof thethe energyenergy equationequation

dp  V2   d   gdz  (Tds  q )   2  net in   129 SecondSecond LawLaw ofof ThermodynamicsThermodynamics –– IrreversibleIrreversible FlowFlow 1/31/3 AA generalgeneral statementstatement ofof thethe secondsecond lawlaw ofof thermodynamicsthermodynamics The time rate of increase of Sum of the ratio of net heat transfer rate into the entropy of a system ≥≥ system to absolute temperature for each particle of mass in the system receiving heat d Q net in  sdV    from surroundings dt  sys  T  sys Based on system method 依據 Chapter 4 推導 Reynolds transport theorem 的步驟 ForFor thethe systemsystem andand thethe contentscontents ofof thethe coincidentcoincident controlcontrol volumevolume thatthat isis fixedfixed andand nondeformingnondeforming ---- ReynoldsReynolds TransportTransport TheoremTheorem leadsleads toto 選擇一固定且不變形的CV = SYSTEM d   sdV  sdV  sV  ndA  sys  CV  C.S. dt t 130 SecondSecond LawLaw ofof ThermodynamicsThermodynamics –– IrreversibleIrreversible FlowFlow 2/32/3 ForFor thethe systemsystem andand controlcontrol vovolumelume atat thethe instantinstant whenwhen systemsystem andand controlcontrol volumevolume areare coincidentcoincident

Q net in  Q net in  在時間 t 的瞬間，進出 CV 的      Q 等於進出 system 者。透過 T T 這層連結，才能得到……  sys  cv TheThe controlcontrol volumevolume formulaformula forfor thethe secondsecond lawlaw ofof thermodynamicsthermodynamics

   Q net in  sdV  sV  ndA    t  CV  CS  T   CV Based on Control Volume 131 SecondSecond LawLaw ofof ThermodynamicsThermodynamics –– IrreversibleIrreversible FlowFlow 3/33/3 ForFor oneone streamstream ofof fluidfluid enteringentering andand leavingleaving thethe controlcontrol volumevolume…….. Special & simple case Q net in Steady flow m (sout  sin )   T

Q net in Semi-infinitesimalSemi-infinitesimal thinthin CVCV m ds   T

UniformUniform temperaturetemperature Tds  q net in  0

132 FirstFirst andand SecondSecond LawLaw ofof ThermodynamicsThermodynamics 1/41/4 Semi-infinitesimal control volume statement of the CV Semi-infinitesimal control volume statement of the 非 energyenergy equationequation 有 2 限 dp  V  大  d   gdz  (Tds  qnet in ) ，   2  也 非 SemiSemi--infinitesimalinfinitesimal CVCV ofof thethe secondsecond lawlaw ofof 無 thermodynamicsthermodynamics 限 小 Tds  q net in  0 dp  V2      d   gdz  0   2    133 FirstFirst andand SecondSecond LawLaw ofof ThermodynamicsThermodynamics 2/42/4

dp  V2      d   gdz  (loss)  (Tds  qnet in )    2  

For steady frictionless flow dp  V2      d   gdz  0    2   The shaft work is involved dp  V2      d   gdz  (loss)  wshaft net in   2    134 FirstFirst andand SecondSecond LawLaw ofof ThermodynamicsThermodynamics 3/43/4

dp  V2      d   gdz  (loss)  (Tds  qnet in )    2     1   Tds  du  pd    

  1  du  pd   qnet in  (loss)     For incompressible flow du  qnet in  (loss)

135 FirstFirst andand SecondSecond LawLaw ofof ThermodynamicsThermodynamics 4/44/4

WhenWhen controlcontrol volumevolume isis finitefinite

  out  1  uout  uin  pd   qnet in  loss in        ForFor incompressibleincompressible flowflow uout  uin  qnet in  loss

136 ApplicationApplication ofof thethe LossLoss FormForm1/21/2

dp  V2      d   gdz  (loss)  wshaft net in    2   Frictionless, loss=0, no shaft work, incompressible p V2 p V2 IntegratingIntegrating 2  2  gz  1  1  gz Bernoulli equation  2 2  2 1 Frictionless, loss=0, no shaft work, compressible

2 2 2 IntegratingIntegrating dp V2 V1   gz2   gz1 1  2 2

137 ApplicationApplication ofof thethe LossLoss FormForm2/22/2

For adiabatic flow of an ideal gas p  cons tan t k 2 dp k  p2 p1  IntegratingIntegrating     1    k 1 2 1 

2 2 k p2 V2 k p1 V1   gz2    gz1 k 1 2 2 k 1 1 2

138