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 ndA dt t CS bdV bV ndA 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 ndA 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 ndA 0 dt t CV CS 解讀結果 system Rate of increase Net influx of Of mass in CV mass dV V ndA t CV CS 7 ConservationConservation ofof MassMass –– TheThe ContinuityContinuity EquationEquation 4/4 Special case IncompressibleIncompressible FluidsFluids dV V ndA 0 dV V ndA 0 t CV CS t CV CS ForFor SteadySteady flowflow m out m in V ndA 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 ndA 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 ndA 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 ndA 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 ndA 0 t CV CS R A U V ndA 0 A U u 2rdr 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 2u 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 ndA 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 ndA CV dt t CV C.S. dV W ndA 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