fluid05

1FUNDAMENTALS OFFUNDAMENTALS OF FLUID MECHANICSFLUID MECHANICS Chapter 5 Flow Analysis Chapter 5 Flow Analysis Using Control Volume Using Control Volume JyhJyh--CherngCherng ShiehShieh Department of BioDepartment of Bio --Industrial Industrial MechatronicsMechatronics Engineering Engineering National Taiwan UniversityNational Taiwan University 10/19/200910/19/2009 2MAIN TOPICSMAIN TOPICS Conservation of MassConservation of Mass NewtonNewton ’’s Second Law s Second Law ––The Linear Momentum The Linear Momentum EquationsEquations The MomentThe Moment --ofof--Momentum EquationsMomentum Equations First Law of Thermodynamics First Law of Thermodynamics ––The Energy EquationThe Energy Equation Second Law of Thermodynamics Second Law of Thermodynamics ––Irreversible FlowIrreversible Flow 3Learning ObjectsLearning Objects Select an appropriate finite CV to solve a fluid Select an appropriate finite CV to solve a fluid mechanics problem.mechanics problem. Apply basic laws to the contents of a finite CV to get Apply basic laws to the contents of a finite CV to get important answers.important answers. How to apply these basic laws? How to express these basic laws based on CV method? 4Review of Reynolds Transport TheoremReview of Reynolds Transport Theorem dAnVb VbdtdAnVbtB dtdB CS CVCSCV sys   This is the fundamental relation between the rate of This is the fundamental relation between the rate of change of any arbitrary extensive property, B, of a change of any arbitrary extensive property, B, of a system and the variations of this property associated system and the variations of this property associated with a control volume.with a control volume. 5Conservation of Mass Conservation of Mass –– The Continuity Equation The Continuity Equation 1/41/4 Basic Law for Conservation of MassBasic Law for Conservation of Mass For a system and For a system and a fixed, a fixed, nondeformingnondeforming control volumecontrol volume that are coincident at an instant of time, the Reynolds that are coincident at an instant of time, the Reynolds Transport Theorem leads toTransport Theorem leads to  CS CV sysdAnV VdtVddtd  B=M and b =1 Time rate of change Time rate of change of the mass of the of the mass of the coincident systemcoincident systemTime rate of change of the Time rate of change of the mass of the content of the mass of the content of the coincident control volumecoincident control volumeNet rate of flow of Net rate of flow of mass through the mass through the control surfacecontrol surface== ++0dtdM system Vd dm M ) system(V ) system(Msystem    System method下，描述質量守恒的方程式 依據Chapter 4推導Reynolds transport theorem的步驟第一時間 ，兩者一致 6Conservation of Mass Conservation of Mass –– The Continuity Equation The Continuity Equation 2/42/4 System and control volume at three different instances of time. System and control volume at three different instances of time. (a) System and control volume at time (a) System and control volume at time t t ––δδtt. (b) System and . (b) System and control volume at time control volume at time tt, coincident condition. (c) System and , coincident condition. (c) System and control volume at time control volume at time t + t + δδtt..Chapter 4：system與CV在不同時間下的關係 The instant time considered 7Conservation of Mass Conservation of Mass –– The Continuity Equation The Continuity Equation 3/43/4 For a fixed, For a fixed, nondeformingnondeforming control volume, the control control volume, the control volume formulation of the conservation of mass: The volume formulation of the conservation of mass: The continuity equationcontinuity equation 0 dAnV Vdt dtMd CS CVsystem  CS CVdAnV Vdtin out m m   Rate of increase Of mass in CVNet influx of mass當CV是固定且不變形 解讀結果 8Conservation of Mass Conservation of Mass –– The Continuity Equation The Continuity Equation 4/44/4 Incompressible FluidsIncompressible Fluids For Steady flowFor Steady flow0 dAnV Vdt0 dAnV VdtCS CV CS CV    The mass flow rate into a control volume The mass flow rate into a control volume must be equal to the mass flow rate out of must be equal to the mass flow rate out of the control volume.the control volume.0 dAnV CSin out m m  Special case 9Other DefinitionOther Definition Mass Mass flowrateflowrate through a section of control surfacethrough a section of control surface The average velocityThe average velocity in outAm m dAnV Q m   AdAnVVA 衍生定義 10Fixed, Fixed, NondeformingNondeforming Control Volume Control Volume 1/21/2 When the flow is steadyWhen the flow is steady When the flow is steady and incompressible When the flow is steady and incompressible When When the flow is not steadythe flow is not steady 0VdtCV0VdtCV in out m m   in out Q Q  ““++””: the mass of the contents of the control volume is increasing: the mass of the contents of the control volume is increasing ““--””: the mass of the contents of the control volume is decreasing: the mass of the contents of the control volume is decreasing ..Special case 11Fixed, Fixed, NondeformingNondeforming Control Volume Control Volume 2/22/2 When the flow is uniformly distributed over the opening When the flow is uniformly distributed over the opening in the control surface (one dimensional flow)in the control surface (one dimensional flow) When the flow is When the flow is nonuniformlynonuniformly distributed over the distributed over the opening in the control surfaceopening in the control surfaceAV m VA mSpecial case 12Example 5.1 Conservation of Mass Example 5.1 Conservation of Mass –– Steady, Incompressible FlowSteady, Incompressible Flow Seawater flows steadily through a simple conicalSeawater flows steadily through a simple conical --shaped nozzle at shaped nozzle at the end of a fire hose as illustrated in Figure E5.1. If the nozthe end of a fire hose as illustrated in Figure E5.1. If the noz zle exit zle exit velocity must be at least 20 velocity must be at least 20 m/sm/s, determine the minimum pumping , determine the minimum pumping capacity required in mcapacity required in m33/s./s. Figure E5.1Figure E5.1 13Example 5.1 Example 5.1 SolutionSolution Steady flow 2 2 111 2 1 2CS Q Qm m or 0 m m dAnV   0 dAnV VdtCS CVThe continuity equation s/m 0251.0... AV Q Q3 2 2 2 1 2 1 With incompressible condition 求minimum pumping capacityminimum pumping capacity 14Example 5.2 Conservation of Mass Example 5.2 Conservation of Mass –– Steady, Compressible FlowSteady, Compressible Flow Figure E5.2Figure E5.2Air flows steadily between two sections in a long, straight portAir flows steadily between two sections in a long, straight port ion of ion of 44--in. inside diameter as indicated in Figure E5.2. The uniformly in. inside diameter as indicated in Figure E5.2. The uniformly distributed temperature and pre ssure at each section are given. distributed temperature and pre ssure at each section are given. If the If the average air velocity (average air velocity ( NonuniformNonuniform velocity distribution) at section (2) velocity distribution) at section (2) is 1000ft/s, calculate the average air velocity at section (1). is 1000ft/s, calculate the average air velocity at section (1). 求求section (1)section (1) 平均速度平均速度 15Example 5.2 Example 5.2 SolutionSolution Steady flow 22 2 11 11 2 1 2CS VA VAm m 0 m m dAnV  0 dAnV VdtCS CVThe continuity equationThe continuity equation 2 12 1 V V Since A1=A2 s/ft219... VTpTpV2 2112 1   The ideal gas equationThe ideal gas equation RTp假設氣體為理想氣體假設氣體為理想氣體 16Example 5.3 Conservation of Mass Example 5.3 Conservation of Mass –– Two FluidsTwo Fluids Moist air (a mixture of dry air and water vapor) enters a Moist air (a mixture of dry air and water vapor) enters a dehumidifier at the rate of 22 slugs/hr. Liquid water drains outdehumidifier at the rate of 22 slugs/hr. Liquid water drains out of the of the dehumidifier at a rate of 0.5 slugs/hr. Determine the mass dehumidifier at a rate of 0.5 slugs/hr. Determine the mass flowrateflowrate of the dry air and the water vapor leaving the dehumidifier. of the dry air and the water vapor leaving the dehumidifier. Figure E5.3Figure E5.3求出口處的求出口處的 mass mass flowrateflowrate 17Example 5.3 Example 5.3 SolutionSolution 0 dAnV VdtCS CVSteady flow hr/ slugs5.21 hr/ slugs5.0 hr/ slugs22 m m m0 m m m dAnV 3 1 23 2 1CS    The continuity equationThe continuity equation 18Example 5.4 Conservation of Mass Example 5.4 Conservation of Mass –– NonuniformNonuniform Velocity ProfilesVelocity Profiles Incompressible, laminar water flow develops in a straight pipe Incompressible, laminar water flow develops in a straight pipe having radius R as indicated in Figure E5.4. At section (1), thehaving radius R as indicated in Figure E5.4. At section (1), the velocity profile is uniform; the velocity is equal to a constant velocity profile is uniform; the velocity is equal to a constant value value U and is parallel to the pipe axis everywhere. At section (2), tU and is parallel to the pipe axis everywhere. At section (2), t he he velocity profile is velocity profile is axisymmetricaxisymmetric and parabolic, with zero velocity at and parabolic, with zero velocity at the pipe wall and a maximum value of the pipe wall and a maximum value of uumaxmaxat the centerline. How at the centerline. How are U and are U and uumaxmaxrelated? How are the average velocity at section related? How are the average velocity at section (2), , and (2), , and uumaxmaxrelated?related?2V 19Example 5.4 Example 5.4 SolutionSolution Steady flow 2/ u V U2 u0 rdrRr1 u2UA max 2 maxR 02 max 1         0 dAnV VdtCS CV With incompressible conditionWith incompressible conditionThe continuity equationThe continuity equation 0 rdr2u UA 0 dAnV UAR 02 2 1 1A1 1 2   2 1     2 maxRr1 uu 20Example 5.5 Conservation of Mass Example 5.5 Conservation of Mass –– Unsteady Flow Unsteady Flow A bathtub is being filled with water from a faucet. The rate of A bathtub is being filled with water from a faucet. The rate of flow flow from the faucet is steady at 9 gal/min. The tub volume is from the faucet is steady at 9 gal/min. The tub volume is approximated by a rectangular sp ace as indicate Figure E5.5a. approximated by a rectangular sp ace as indicate Figure E5.5a. Estimate the time rate of change of the depth of water in the Estimate the time rate of change of the depth of water in the tub, , in in./min at any instant.tub, , in in./min at any instant. t/h Figure E5.5求水深的時間改變率求水深的時間改變率 21Example 5.5 Example 5.5 SolutionSolution1/21/2 airvolume waterwater water watervolumeairair airCS CV m m VdtVdt0 dAnV Vdt      0 m VdtairForairvolumeairair air The continuity equationThe continuity equation 22Example 5.5 Example 5.5 SolutionSolution2/22/2 )ft10)(ft/gal48.7()ft/.in12 min)(/gal9( )A ft10(Q thmth)A ft10(]A)hft5.1()ft5)(ft2(h[ Vdm VdtwaterFor 2 3 j2waterwater j2 watervolume water j water water watervolume water water water water         2 j ft10 ACV內水的體積 23FilmsFilms Vacuum filter Flow through a contraction Sink flow 24Moving, Moving, NondeformingNondeforming Control VolumeControl Volume1/21/2 When a moving control volume is used, the fluid velocity When a moving control volume is used, the fluid velocity relative to the moving control is an important variable.relative to the moving control is an important variable. WWis the relative fluid velocity seen by an observer is the relative fluid velocity seen by an observer moving with the control volume. moving with the control volume. VVcvcvis the control volume veloci ty as seen from a fixed is the control volume veloci ty as seen from a fixed coordinate system. coordinate system. VVis the absolute fluid velocity seen by a stationary is the absolute fluid velocity seen by a stationary observer in a fixed coordinate system.observer in a fixed coordinate system.CV是移動，不變形 觀察者站在CV上看到的流體速度 站在固定座標 看到的CV移動速度 站在固定 座標看到 的流體速 度（絕對） 25Moving, Moving, NondeformingNondeforming Control VolumeControl Volume2/22/2 CVV WV  0 dAnW Vdt.S.C CV dAnW Vdt dtdM .S.C CVsys   Velocities seen from the control Velocities seen from the control volume reference frame (relative volume reference frame (relative velocities) velocities) 在移動的CV上架設一參考座標系 26Example 5.6 Conservation of Mass Example 5.6 Conservation of Mass --Compressible Compressible Flow with a Moving Control VolumeFlow with a Moving Control Volume An airplane moves forward at speed of 971 km/hr as shown in 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 0.80mFigure E5.6a. The frontal intake area of the jet engine is 0.80m22and and the entering air density is 0.736 kg/mthe entering air density is 0.736 kg/m33. A stationary observer . A stationary observer determines that relative to th e earth, the jet engine exhaust ga determines that relative to th e earth, the jet engine exhaust ga ses ses move away from the engine with a speed of 1050 km/hr. The enginemove away from the engine with a speed of 1050 km/hr. The engine exhaust area is 0.558 mexhaust area is 0.558 m22, and the exhaust gas density is 0.515 kg/m, and the exhaust gas density is 0.515 kg/m33. . Estimate the mass Estimate the mass flowrateflowrate of fuel into the engine in kg/hr.of fuel into the engine in kg/hr. Determine the mass flowrate of fuel into the engine in kg/hr Figure E5.6Figure E5.6CV選在移動的引擎 27Example 5.6 Example 5.6 SolutionSolution 0 dAnW Vdt.S.C CV  hr/kg 9100...)km/m 1000)(hr/km 2021)(m558.0)(m/kg515.0( mhr/km201 hr/km971hr/km 1050 V V WWA WA m0 WA WA m 2 3 in fuelplane 2 21 11 2 2 2 in fuel2 2 2 1 11 in fuel     =0The intake velocity, WThe intake velocity, W11, relative to the moving , relative to the moving control volume. The exhaust velocity, Wcontrol volume. The exhaust velocity, W22, also , also needs to be measured relative to the moving needs to be measured relative to the moving control volume.control volume.The continuity equationThe continuity equation Assuming oneAssuming one --dimensional flowdimensional flow 飛機飛行速度在地面觀察者看到的排氣離開引擎的速度WW11是從是從CVCV觀察到的進氣速度觀察到的進氣速度 WW22是從是從CVCV觀察到的排氣速度觀察到的排氣速度 WW22是從是從CVCV觀察到的排氣速度觀察到的排氣速度 28Example 5.7 Conservation of Mass Example 5.7 Conservation of Mass -- Relative VelocityRelative Velocity Water enters a rotating lawn sprinkler Water enters a rotating lawn sprinkler through its base at the steady rate of through its base at the steady rate of 1000 ml/s as sketched in Figure E5.7. 1000 ml/s as sketched in Figure E5.7. If the exit area of each of the two If the exit area of each of the two nozzle is 30 mmnozzle is 30 mm2 2 , determine the , determine the average speed of the water leaving average speed of the water leaving each nozzle, relative to the nozzle, if each nozzle, relative to the nozzle, if (a) the rotary sprinkler head is (a) the rotary sprinkler head is stationary, (b) the sprinkler head stationary, (b) the sprinkler head rotates at 60 rpm, and (c) the rotates at 60 rpm, and (c) the sprinkler head accelerates from 0 to sprinkler head accelerates from 0 to 600 rpm.600 rpm. Determine the average speed of the water leaving each nozzle, relative to the nozzle…Figure E5.7CV選在轉動的撒水器部分 CV跟著轉動 29 Example 5.7 Example 5.7 SolutionSolution 0 dAnW Vdt.S.C CV  2 22 2 6 3 22 Ws/m7.16) mm30)(2)(liter/ml 1000()m/ mm10)(liter/m001.0)(s/ml 1000( A2QW   =0 The value of WThe value of W22is independent of the speed of rotation of the sprinkler head is independent of the speed of rotation of the sprinkler head and represents the average speed of the water exiting from each and represents the average speed of the water exiting from each nozzle with nozzle with respect to the nozzle for case (a), (b), (c).respect to the nozzle for case (a), (b), (c).The continuity equationThe continuity equation Q m WA2 m0 m m dAnW in 2 2 outout in .S.C   W是相對於轉動的Nozzle的速度 30Deforming Control Volume Deforming Control Volume 1/21/2 A deforming control volume involves changing volume A deforming control volume involves changing volume size and control surface movement.size and control surface movement. The Reynolds transport theorem for a deforming control The Reynolds transport theorem for a deforming control volume can be used for this case.volume can be used for this case. CSV WV dAnW Vdt dtdM .S.C CVsys   VVcscsis the velocity of the control surface as seen by a fixed obseris the velocity of the control surface as seen by a fixed obser ver.ver. W is the relative velocity re ferenced to the control surface. W is the relative velocity re ferenced to the control surface.CV變形，CS當然就出現移動 31Deforming Control Volume Deforming Control Volume 2/22/2 dAnW Vdt dtdM .S.C CVsys   通常不是零。因通常不是零。因 CVCV範圍範圍 與時俱變與時俱變 ，，必須妥善處必須妥善處 理。理。相對於變形的相對於變形的 control volumecontrol volume 表面表面 的速度。的速度。 CVCV表面速度並非每一點都表面速度並非每一點都 相同。因此，相對速度自然較為複相同。因此，相對速度自然較為複 雜。雜。 32Example 5.8 Conservation of Mass Example 5.8 Conservation of Mass –– Deforming Control Volume Deforming Control Volume 1/21/2 A syringe is used to inoculate a cow. The plunger has a face arA syringe is used to inoculate a cow. The plunger has a face ar ea of ea of 500 mm500 mm22. If the liquid in the syringe is to be injected steadily at a . If the liquid in the syringe is to be injected steadily at a rate of 300 cmrate of 300 cm33/min, at what speed should the plunger be advanced? /min, at what speed should the plunger be advanced? The leakage rate past the pl unger is 0.01 times the volume The leakage rate past the pl unger is 0.01 times the volume flowrateflowrate out of the needle.out of the needle. Determine the speed of the plunger be advanced Figure E5.8Figure E5.8Leakage rate 33Example 5.8 Example 5.8 SolutionSolution p 1A A min/mm660...AQ QV0 Q Q VA0 Q m VA VtLet 1leak 2 pleak 2 p1leak 2 p1 p  2 2 Q mThe continuity equationThe continuity equation 0 dAnW Vdt.S.C CV  tA Vdt) V A( Vd 0 Q mVdt 1 CVneedle 1 CV leak 2 CV   外漏加上從針頭注出者 =貫穿 CS進出 CV的量 34The Linear Momentum Equations The Linear Momentum Equations 1/41/4 NewtonNewton ’’s second law for a system moving relative to an inertial s second law for a system moving relative to an inertial coordinate system.coordinate system. systemsys B S sysdtPdVdVdtdF F F    Time rate of change of Time rate of change of the linear momentum of the linear momentum of the the systemsystem=Sum of external forces Sum of external forces acting onacting on the the systemsystem VdV dmV P ) system(V ) system(Msystem       利用System method描述Newton’s second law 35The Linear Momentum Equations The Linear Momentum Equations 2/42/4 External forces acting on system and External forces acting on system and coincident control volumecoincident control volumevolume control coincidenttheof contents sys F FWhen a control volume is coincident with a system at an instant When a control volume is coincident with a system at an instant of of time, the force acting on the system and the force acting on thtime, the force acting on the system and the force acting on th e e contents of the coincident control volume are instantaneously contents of the coincident control volume are instantaneously identical.identical. 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） 36Review of Reynolds Transport TheoremReview of Reynolds Transport Theorem dAnVb VbdtdAnVbtB dtdB CS CVCSCV sys   This is the fundamental relation between the rate of This is the fundamental relation between the rate of change of any arbitrary extensive property, B, of a change of any arbitrary extensive property, B, of a system and the variations of this property associated system and the variations of this property associated with a control volume.with a control volume. 37The Linear Momentum Equations The Linear Momentum Equations 3/43/4 For the system and a fixed, For the system and a fixed, nondeformingnondeforming control volume that are control volume that are coincident at an instant of time, the Reynolds Transport Theorem coincident at an instant of time, the Reynolds Transport Theorem leads toleads to   CS CV sysdAnVV VdVtVdVdtd    B=P and B=P and Vb   CS CV sysdAnVV VdVtVdVdtd    Time rate of change Time rate of change of the linear of the linear momentum of the momentum of the coincident systemcoincident systemTime rate of change of the Time rate of change of the linear momentum of the linear momentum of the content of the coincident content of the coincident control volumecontrol volumeNet rate of flow of Net rate of flow of linear momentum linear momentum through the control through the control surfacesurface== ++依據Chapter 4推導Reynolds transport theorem的步驟 解讀sysFvolume control coincidenttheof contentsFmass flow ratemass flow rate 因為質量進出所 引發的動量進出 38The Linear Momentum Equations The Linear Momentum Equations 4/44/4 For a For a fixed and fixed and nondeformingnondeforming control volume, control volume, the control the control volume formulation of Newtonvolume formulation of Newton ’’s second laws second law  F dAnVV VdVtCS CV   Contents of the coincident control volume Linear momentum equationLinear momentum equation對固定、不變形的CV而言 這是基於這是基於 CV methodCV method ，描述牛頓第二定律的方程式，描述牛頓第二定律的方程式mass flow ratemass flow rate 39Linear momentum equation written for a moving control volume 40Moving, Moving, NondeformingNondeforming Control VolumeControl Volume1/31/3 CVVWV  CS CV sysdAnWV VdVtVdVdtd     F dAnWV VdVtCS CV   Contents of the coincidentContents of the coincident control volumecontrol volume   F dAnW)V W( Vd)V W(tCSCVCVCV   Contents of the coincident control volumedAnWb Vdbt dtdB CS CVsys  Chapter 4: Reynolds transport Chapter 4: Reynolds transport equation for a control volume equation for a control volume moving with moving with constant velocityconstant velocity isis 對移動、不變形的CV而言 mass flow ratemass flow rate 41Moving, Moving, NondeformingNondeforming Control VolumeControl Volume2/32/3     CSCVCS CSCV dAnW V dAnWW dAnW V W   0Vd V WtCVCVFor a constant control volume velocity, For a constant control volume velocity, VVcvcv, and , and steady steady flowflow in the control volume reference framein the control volume reference frame For steady flowFor steady flow , , continuity equationcontinuity equation=0=0 0 dAnW Vdt dtdM .S.C CVsys0 dAnW.S.C STEADY FLOW STEADY FLOW特例特例 42Moving, Moving, NondeformingNondeforming Control VolumeControl Volume3/33/3   F dAnWW CS  For For a movinga moving , , nondeformingnondeforming control volume, the control volume, the linear momentum equation of linear momentum equation of steady flowsteady flow Contents of the coincident control volume mass flow ratemass flow rate 43Vector Form of Momentum EquationVector Form of Momentum Equation TheThesum of all forcessum of all forces (surface and body forces) acting on a (surface and body forces) acting on a NonNon--accelerating control volume is equal to theaccelerating control volume is equal to the sum of the sum of the rate of change of momentum inside the control volume rate of change of momentum inside the control volume and the net rate of flux of momentum out through the and the net rate of flux of momentum out through the control surfacecontrol surface ..   ASCVB Adp- FVdB dmB F    Where the velocities are measuredWhere the velocities are measured Relative to the control volume.Relative to the control volume.   CS CVB Svolume control coincidenttheof contents dAnVV VdVtF F F   解讀 44FILMSFILMS 煙囪雲煙 Smokestack plume momentum 船舶推力Marine propulsion Force due to a water jet Running onwater Fire hose Jelly fish水母 45Linear Momentum Equations Linear Momentum Equations 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 ，，（＋（＋ for flow for flow out of the CVout of the CV ，－，－ for flow into the for flow into the CVCV））。。 nV  46Application for FIXING CVApplication for FIXING CV 5.10~5.165.10~5.16 47Example 5.10 Linear Momentum Example 5.10 Linear Momentum ––Change in Change in Flow DirectionFlow Direction As shown in Figure E5.10a, a horizontal jet of water exits a nozAs shown in Figure E5.10a, a horizontal jet of water exits a noz zle zle with a uniform speed of Vwith a uniform speed of V11=10 ft/s, strike a vane, and is turned =10 ft/s, strike a vane, and is turned through an anglethrough an angle θθ. Determine the anchoring force needed to hold . Determine the anchoring force needed to hold the vane stationary. Neglect gravity and viscous effects.the vane stationary. Neglect gravity and viscous effects. Determine the anchoring force needed to hold the vane stationary. 48Example 5.10 Example 5.10 SolutionSolution     zCS CVxCS CV F dAnVw VdwtF dAnVu Vdut The x and z directionThe x and z direction components of linear momentum equationcomponents of linear momentum equation lb sin64.11... sinAV Flb) cos1(64.11 ..) cos1(AV FF A)V( sinV A)V()0(F A)V( cosV A)V(V 112 Az112 AxAz 2 1 1 1 1Ax 2 1 1 1 1 1     kwiuV  49Example 5.11 Linear Momentum Example 5.11 Linear Momentum ––Weight, Weight, pressure, and Change in Speedpressure, and Change in Speed Determine the anchoring force required to hold in place a conicaDetermine the anchoring force required to hold in place a conica l l nozzle attached to the end of a laboratory sin faucet when the wnozzle attached to the end of a laboratory sin faucet when the w ater ater flowrateflowrate is 0.6 liter/s. The nozzle mass is 0.1kg. The nozzle inlet and 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 iexit diameters are 16mm and 5mm, respectively. The nozzle axis i s s vertical and the axial distance between section (1) and (2) is 3vertical and the axial distance between section (1) and (2) is 3 0mm. 0mm. The pressure at section (1) is 464 The pressure at section (1) is 464 kPakPa..to hold the vane stationary. to hold the vane stationary. Neglect gravity and viscous effects.Neglect gravity and viscous effects. 50Example 5.11 Example 5.11 SolutionSolution1/31/3 51Example 5.11 Example 5.11 SolutionSolution2/32/3 dAw dAnVAp W Ap W F dAnVw Vdwt2 2 w 11 n ACS CV  The z direction component of linear moment equationThe z direction component of linear moment equation With the With the ““++””used for flow out of the control volume and used for flow out of the control volume and ““--””used used for flow in.for flow in. 2 2 w 11 n 2 1 A2 2 w 11 n 2 2 1 1 Ap W Ap W)w w(m FAp W Ap W )w(m)w)(m(    m m m2 1 s/kg599.0...Q Aw m m m11 2 1  52Example 5.11 Example 5.11 SolutionSolution3/33/3   N 0278.0...gV)DD D D(h121gV WN981.0)s/m81.9)(kg1.0(gm Wm6.30... 4/DQ AQws/m98.2... 4/DQ AQw w 2122 12 w w2 n n2 2 222 1 11      N8.77... )(...)s/kg599.0(Ap W Ap W)w w(m F2 2 w 11 n 2 1 A   53Example 5.12 Linear Momentum Example 5.12 Linear Momentum ––Pressure , Pressure , Change in Speed, and FrictionChange in Speed, and Friction Water flows through a horizontal, 180Water flows through a horizontal, 180 °°pipe bend. The flow crosspipe bend. The flow cross -- section area is constant at a value of 0.1ftsection area is constant at a value of 0.1ft22through the bend. The through the bend. The magnitude of the flow velocity everywhere in the bend is axial amagnitude of the flow velocity everywhere in the bend is axial a nd nd 50ft/s. The absolute pressure at the entrance and exit of the be50ft/s. The absolute pressure at the entrance and exit of the be nd are nd are 30 30 psiapsia and 24 and 24 psiapsia, respectively. Calculate the horizontal (x and y) , respectively. Calculate the horizontal (x and y) components of the anchoring force required to hold the bend in components of the anchoring force required to hold the bend in place.place. 54Example 5.12 Example 5.12 SolutionSolution1/21/2 The x direction component of linear moment equationThe x direction component of linear moment equation AxCS CVF dAnVu Vdut 2 2 11 AyCS CVAp Ap F dAnVv VdvtAt section (1) and (2), the flow is in the y direction and thereAt section (1) and (2), the flow is in the y direction and there fore fore u=0 at both sections.u=0 at both sections. 0 FAx The y direction component of linear moment equationThe y direction component of linear moment equation 55Example 5.12 Example 5.12 SolutionSolution2/22/2 For oneFor one --dimensional flowdimensional flow 2 2 11 Ay 2 2 1 1 Ap Ap F)m)(v()m)(v(    2 2 11 Ay 2 1 Ap Ap F)vv(m  s/ slugs70.9... Av m m m11 2 1 lb 1324 ... Ap Ap)vv(m F2 2 11 2 1 Ay  56Example 5.13 Linear Momentum Example 5.13 Linear Momentum ––Weight, Weight, pressure, and Change in Speedpressure, and Change in Speed Air flows steadily between two cross sections in a long, straighAir flows steadily between two cross sections in a long, straigh t t portion of 4portion of 4 --in. inside diameter pipe as indicated in Figure E5.13, in. inside diameter pipe as indicated in Figure E5.13, where the uniformly distributed temperature and pressure at eachwhere the uniformly distributed temperature and pressure at each cross section are given, If the av erage air velocity at section cross section are given, If the av erage air velocity at section (2) is (2) is 1000 ft/s, we found in Example 5.2 that the average air velocity 1000 ft/s, we found in Example 5.2 that the average air velocity at at section (1) must be section (1) must be 219 ft/s219 ft/s . Assuming uniform velocity . Assuming uniform velocity distributions at sections (1) and (2), determine the frictional distributions at sections (1) and (2), determine the frictional force force exerted by the pipe wall on the air flow between sections (1) anexerted by the pipe wall on the air flow between sections (1) an d (2).d (2). 57Example 5.13 Example 5.13 SolutionSolution1/21/2 The axial component of linear moment equationThe axial component of linear moment equation 2 2 11 xCS CVAp Ap R dAnVu Vdut 2 2 11 x 2 2 1 1 Ap Ap R )m)(u()m)(u(    )pp(A R )uu(m2 1 2 x 1 2  s/ slugs297.0... u4D RTpm m m222 22 2 1        )u u(m)pp(A R1 2 2 1 2 x   58Example 5.13 Example 5.13 SolutionSolution2/22/2 4DARTp 22 222 2 )u u(m)pp(A R1 2 2 1 2 x   lb793...)u u(m)pp(A R1 2 2 1 2 x   59Example 5.14 Linear Momentum Example 5.14 Linear Momentum –– Weight, Pressure,Weight, Pressure, …… If the flow of If the flow of Example 5.4Example 5.4 is is vertically upward, develop an vertically upward, develop an expression for the fluid pressure drop expression for the fluid pressure drop that occurs between sections (1) and that occurs between sections (1) and (2).(2). 60Example 5.14 Example 5.14 SolutionSolution 2 2 z 11CS2 2 2 1 12 2 z 11CS CV Ap W R Ap)dAw()w( )m)(w(Ap W R Ap dAnVw Vdwt   The axial component of linear moment equationThe axial component of linear moment equation     2 1 2Rr1w2 w  3Rw4 rdr2w )dAw()w()R/r(1w2 w 2 2 1R 02 2CS2 2 22 1 2    1 1z2 1 2 12 2 z 112 12 2 1 AW AR 3wppAp W R ApR w34R w  61 Example 5.15 Example 5.15 Linear Momentum Linear Momentum --TrustTrust A static thrust as sketched in Figure E5.15 is to be designed foA static thrust as sketched in Figure E5.15 is to be designed fo r r testing a jet engine. The following conditions are known for a testing a jet engine. The following conditions are known for a typical test: Intake air velocity = 200 typical test: Intake air velocity = 200 m/sm/s; exhaust gas velocity= = ; exhaust gas velocity= = 500 500 m/sm/s; intake cross; intake cross --section area = 1msection area = 1m22; intake static pressure = ; intake static pressure = -- 22.5 22.5 kPakPa=78.5 =78.5 kPakPa(abs); intake static temperature = 268K; exhaust (abs); intake static temperature = 268K; exhaust static pressure =0 static pressure =0 kPakPa=101 =101 kPakPa(abs). Estimate the normal trust for (abs). Estimate the normal trust for which to design.which to design. 62Example 5.15 Example 5.15 SolutionSolution N 83700...)uu(m A A FF Ap Ap)uu(muA muA mmF A)pp( A)pp()m)(u()m)(u()A A(p Ap F Ap dAnVu Vdut 1 2 2 2 11 thth 2 2 11 1 222 2 2 111 1th 2 atm 2 1 atm 1 2 2 1 12 1 atm 2 2 th 11CS CV    The x direction component of linear moment equationThe x direction component of linear moment equation s/kg204...uA mRTp 111 11 1    63 Example 5.16 Linear Momentum Example 5.16 Linear Momentum –– NomuniformNomuniform Pressure Pressure A sluice gate across a A sluice gate across a channel of width b is shown channel of width b is shown in the closed and open in the closed and open position in Figure 5.16a and position in Figure 5.16a and b. Is the anchoring force b. Is the anchoring force required to hold the gate in required to hold the gate in place larger when the gate is place larger when the gate is closed or when it is open?closed or when it is open? 64Example 5.16 Example 5.16 SolutionSolution When the gate is open, the horizontal forces acting on the contents of the control volume are identified in Figure E5.16d. hbu Fbh21bH21R u uandh HForFbh21RbH21hbu HbuFbh21RbH21dAnVu 2 2 f2 2 x 2 1f2 x2 2 22 1f2 x2 CS  When the gate is closed, the horizontal forces acting on the contents of the control volume are identified in Figure E5.16c. bH21R RbH21dAnVu2 x x2 CS 65Application for MOVING CVApplication for MOVING CV 5.175.17 66..Example 5.17 Linear MomentumExample 5.17 Linear Momentum -- Moving Control Volume Moving Control Volume 1/21/2 A vane on wheels move with a constant velocity VA vane on wheels move with a constant velocity V00when a stream when a stream of water having a nozzle exit velocity of Vof water having a nozzle exit velocity of V11is turned 45is turned 45 °°by the vane by the vane as indicated in Figure E5.17a. Note that this is the same movingas indicated in Figure E5.17a. Note that this is the same moving vane considered in Section 4.4.6 earlier . Determine the magnituvane considered in Section 4.4.6 earlier . Determine the magnitu de de and direction of the force, F, exerted by the stream of water onand direction of the force, F, exerted by the stream of water on the the vane surface. The speed of the wa ter jet leaving the nozzle is 1 vane surface. The speed of the wa ter jet leaving the nozzle is 1 00ft/s, 00ft/s, and the vane is moving to the right with a constant speed of 20 and the vane is moving to the right with a constant speed of 20 ft/s.ft/s. CV移動速度Vo V1是流體離開Nozzle的絕對速度 67..Example 5.17 Linear MomentumExample 5.17 Linear Momentum -- Moving Control Volume Moving Control Volume 2/22/2 移動的CV 68..Example 5.17 Example 5.17 SolutionSolution1/21/2 x 2 2 1 1xCSx R )m)(45cosW()m)(W(R dAnWW     2 2 2 2 11 1 1 AW m AW m    The x direction component of linear moment equationThe x direction component of linear moment equation w z 2 2W zCSz W R)m)(45sinW(W R dAnWW    ... VV W WAW m AW m 0 1 2 12 2 2 2 11 1 1   The z direction component of linear moment equationThe z direction component of linear moment equation 69..Example 5.17 Example 5.17 SolutionSolution2/22/2 xz 12 z2 xw 12 1 z12 1 x RRtanlb3.57... R R Rlb53... W 45sinAW Rlb8.21...)45cos1(AW R   1 w gA W 70From the Proceeding ExamplesFrom the Proceeding Examples A flowing fluid can be forcedA flowing fluid can be forced to change direction, to change direction, Speed up or slow down, have a velocity profile change, do Speed up or slow down, have a velocity profile change, do only some or all of the above, do only some or all of the above, do nobenobe of the above.of the above. A net force A net force on the on the fluidisfluidis required for achieving any or required for achieving any or all of the first four above. The forces on a flowing fluid all of the first four above. The forces on a flowing fluid balance out with no net force for the fifth.balance out with no net force for the fifth. Typical forceTypical force considered include pressure, considered include pressure, friction,friction, weight.weight. 71MomentMoment --ofof--Momentum EquationMomentum Equation1/41/4 Applying NewtonApplying Newton ’’s second law of motion to a particle of fluids second law of motion to a particle of fluid Taking moment of each side with respect to the origin of an ineTaking moment of each side with respect to the origin of an ine rtial rtial coordinate systemcoordinate systemparticleF)VV(dtd    particleFr)VV(dtdr   )VV(dtdrVVdtrdV)Vr(dtdVdtrd  0VVThe velocity measured in an inertial reference system particleFr V)Vr(dtd   牛頓第二定律 針對一個Particle對 座 標 原 點 的 力 矩r是質點與座標原點的距離 Acting on the particle 72MomentMoment --ofof--Momentum EquationMomentum Equation2/42/4 particleFr V)Vr(dtd   sys sys)Fr( V)Vr(dtd        sys syssys sys )Fr( Vd)Vr(dtdV)Vr(dtdVd)Vr(dtd    The time rate of change of theThe time rate of change of the MomentMoment --ofof--momentum of the systemmomentum of the systemSum of external torquesSum of external torques Acting on systemActing on system適用到所有particles 積分 Based on system method適用所有質點，擴及整個適用所有質點，擴及整個 SystemSystem 73MomentMoment --ofof--Momentum EquationMomentum Equation3/43/4 When a control volume is coincident with a system at an When a control volume is coincident with a system at an instant of time, the torque ac ting on the system and the instant of time, the torque ac ting on the system and the torque acting on the contents of the coincident control torque acting on the contents of the coincident control volume are instantaneously identicalvolume are instantaneously identical For fixed and For fixed and nondeformingnondeforming control volume, the momentcontrol volume, the moment -- ofof--momentum equation:momentum equation:     )Fr( dAnV)Vr( Vd)Vr(tCS CV   Contents of the coincidentContents of the coincident control volumecontrol volume cv sys )Fr( )Fr(  在時間 t 的瞬間，作用在CV上的 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…… Based on Control VolumeBased on Control Volume 74Review of Reynolds Transport TheoremReview of Reynolds Transport Theorem dAnVb VbdtdAnVbtB dtdB CS CVCSCV sys   This is the fundamental relation between the rate of This is the fundamental relation between the rate of change of any arbitrary extensive property, B, of a change of any arbitrary extensive property, B, of a system and the variations of this property associated system and the variations of this property associated with a control volume.with a control volume. 75MomentMoment --ofof--Momentum EquationMomentum Equation4/44/4 For the system and the contents of the coincident control For the system and the contents of the coincident control volume that is fixed and volume that is fixed and nondeformingnondeforming , The Reynolds , The Reynolds transport theorem leads totransport theorem leads to    CS CV sysdAnV)Vr( Vd)Vr(tVd)Vr(dtd    Time rate of change Time rate of change of the of the momentmoment --ofof-- momentum of the momentum of the systemsystemTime rate of change of the Time rate of change of the momentmoment --ofof--momentummomentum of of the content of the the content of the coincident control volumecoincident control volumeNet rate of flow of Net rate of flow of momentmoment --ofof--momentummomentum through the control through the control surfacesurface== ++ mVr bmB Vrb  依據 Chapter 4 推導 Reynolds transport theorem 的步驟 選擇一固定且不變形的CV = SYSTEM 解 讀 76ApplicationApplication1/81/8 Consider the rotating sprinkler.Consider the rotating sprinkler. The flows are oneThe flows are one --dimensional.dimensional. The flows are steady or steadyThe flows are steady or steady --inin--thethe--mean.mean. Using the axial component of the momentUsing the axial component of the moment --ofof--momentum momentum equation to analyze this flowequation to analyze this flow Using the fixed and nonUsing the fixed and non --deforming deforming control volume which contains control volume which contains within its boundaries the spinning within its boundaries the spinning or stationary sprinkler head and or stationary sprinkler head and the portion of the water flowing the portion of the water flowing through the sprinkler contained in through the sprinkler contained in the control volume.the control volume.0Vd)Vr(tCVCV是固定的 ，注意所涵蓋範圍 CV CV假 設條 件 77ApplicationApplication2/82/8  CSdAnV)Vr( 0VrAt section (1)At section (1) At section (2)At section (2)2 2VrVr rr22is the radius from the axis of rotation to the nozzle centerlinis the radius from the axis of rotation to the nozzle centerlin e and Ve and V22is the is the tangential component of the velocity of the flow exiting each notangential component of the velocity of the flow exiting each no zzle as zzle as observed from a frame of reference attached to the fixed and observed from a frame of reference attached to the fixed and nondeformingnondeforming control volume.control volume.    )Fr( dAnV)Vr( Vd)Vr(tCS CV   This term can be nonzero only where fluid is This term can be nonzero only where fluid is crossing the control surface. Everywhere else on crossing the control surface. Everywhere else on the control surface this term will be zero becausethe control surface this term will be zero because There is no axial momentThere is no axial moment --ofof-- momentum flow in momentum flow in section (1)section (1)解 構 此項 的 內 容轉軸到噴嘴中心的距離 流體流出噴嘴的絕對切線速度 78ApplicationApplication3/83/8 U is the velocity of the moving nozzle as measured relative to U is the velocity of the moving nozzle as measured relative to the fixed control surface.the fixed control surface. W is relative velocity of exit flow as viewed from the nozzleW is relative velocity of exit flow as viewed from the nozzle V is the absolute velocity of exit flow relative to a fixed contV is the absolute velocity of exit flow relative to a fixed cont rol rol surface.surface.U WV U：Moving nozzle相對於固定座標的速度W：流體相對Moving nozzle的相對速度 V：流體流出噴嘴的絕對速度 79ApplicationApplication4/84/8 m)Vr( dAnVρ)Vr(θ2 2axialCS   Where m is the total mass Where m is the total mass flowrateflowrate through both nozzles. The through both nozzles. The mass mass flowrateflowrate is the same whether the sprinkler rotates or not.is the same whether the sprinkler rotates or not.nV ““--””for flow intofor flow into ““++””for flow outfor flow out Vr““++””or or ““--””ascertained by ascertained by using the rightusing the right --hand rulehand rule  CSdAnVρ)Vr( Vr如何判斷正負 右手定則 80ApplicationApplication5/85/8 The correct algebraic sign of the axial component of The correct algebraic sign of the axial component of can be easily remembered in the following way:can be easily remembered in the following way: If VIf Vθθand U are in the same direction, use +and U are in the same direction, use + If VIf Vθθand U are in opposite direction, use and U are in opposite direction, use --VrmVr T )Fr(2 2 shaft axialCVtheof content  The torque termThe torque termvolume controltheof content)Fr( Acting on the shaftActing on the shaft 『『負負』』表示表示TorqueTorque與旋轉方向相反與旋轉方向相反Turbine（渦輪機）的Torque 為『負』 Vr 81ApplicationApplication6/86/8 Negative shaft work is work out of the control volume, that is, Negative shaft work is work out of the control volume, that is, work done by the fluid on the rotor and thus its shaft.work done by the fluid on the rotor and thus its shaft.Shaft power?Shaft power? mVr T W2 2 shaft shaft 2 2 shaft shaft VU m W w 2rU Sprinkler speedSprinkler speed TorqueTorque與旋轉方向相反與旋轉方向相反 『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！Acting on the CV 82ApplicationApplication7/87/8 outθ out out inθ in in shaft Vr m Vr m T       )Fr( dAnV)Vr( Vd)Vr(tCS CV   The The ““--””is used with mass is used with mass flowrateflowrate into the control into the control volume, mvolume, minin, and the , and the ““++””is used with mass is used with mass flowrateflowrate out out of the control volume, of the control volume, mmoutout, to , to acountacount for the sign of the for the sign of the dot product .dot product . nV  The The ““++””or or ““--””is used with the is used with the rVrVproduct depends product depends on the direction ofon the direction of axialVr If VIf Vθθand U are in the same direction, use +and U are in the same direction, use + If VIf Vθθand U are in opposite direction, use and U are in opposite direction, use -- Contents of theContents of the Control volumeControl volume General case General case 83ApplicationApplication8/88/8 The shaft powerThe shaft power       out out out in in in shaftout out out in in in shaft shaft VU U W) )( () )( ( T W θ θθ θ m V mVr m Vr m       out inm mm  out out in in shaft VU U wθ θV 2rU 質量守衡質量守衡 When shaft torque and shaft rotation are in the same When shaft torque and shaft rotation are in the same (opposite) direction, power is into (out of ) the fluid.(opposite) direction, power is into (out of ) the fluid. 84判斷 rVrV正負 A simple way to determine the A simple way to determine the sign of the sign of the rVrVproductproduct is is to compare the direction of to compare the direction of VVand the blade speed U.and the blade speed U. If VIf Vand U are in the same direction, the product and U are in the same direction, the product rVrVis positive.is positive. If VIf Vand U are in opposite direction, the product and U are in opposite direction, the product rVrVis negative.is negative. 85..………………………………Example 5.18 Moment of Example 5.18 Moment of Momentum Momentum ––Torque Torque 1/21/2 Water enters a rotating lawn sprinkler through its base at the sWater enters a rotating lawn sprinkler through its base at the s teady teady rate of 1000 ml/s as sketched in Figure E5.18. The exit area of rate of 1000 ml/s as sketched in Figure E5.18. The exit area of each each nozzle is in the tangential direction. The radius from the axis nozzle is in the tangential direction. The radius from the axis of of rotation to the centerline of each nozzle is 200mm. 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. 86..………………………………Example 5.18 Moment of Example 5.18 Moment of Momentum Momentum ––Torque Torque 2/22/2 87..………………………………Example 5.18 Example 5.18 SolutionSolution1/21/2 7.5 /7.162 222 2 2 2 Example fromsm V where mVr TV V mVr T shaftshaft      mN34.3)liter/ml 1000()]s/m/()kg/N(1)[s/kg999.0)(s/m7.16)(mm200(Ts/kg999.0)liter/ml 1000()m/kg999)(liter/m10)(s/ml 1000(Qm 2 shaft3 3 3    (a) (b) s/m2.6min)/s60)(m/mm 1000()rev/rad2 min)(/rev500)(mm200(s/m7.16 Vr Us/m7.16 W whereU W V 22 2 22 2 2    88..………………………………Example 5.18 Example 5.18 SolutionSolution2/22/2 mN24.1)liter/ml 1000()]s/m/()kg/N(1)[s/kg999.0)(s/m2.6)(mm200(T2 shaft   (c) rpm797s/rad5.83)mm200()m/mm 1000)(s/m7.16( rW0m)r W(r T 222 2 2 shaft    mVr T22 shaft 89..………………………………Example 5.19 Moment of Example 5.19 Moment of Momentum Momentum ––Power Power 1/21/2 An air fan has a bladed rotor of 12An air fan has a bladed rotor of 12 --in. outside diameter and 10in. outside diameter and 10 --in. in. inside diameter as illustrated in Figure E5.19a. The height of einside diameter as illustrated in Figure E5.19a. The height of e ach ach rotor is constant at 1 in. from blade inlet to outlet. The rotor is constant at 1 in. from blade inlet to outlet. The flowrateflowrate is is steady, on a timesteady, on a time --average basis, at 230 ftaverage basis, at 230 ft33/min, and the absolute /min, and the absolute velocity of the air at blade inlet, Vvelocity of the air at blade inlet, V11, is radial. The blade discharge , is radial. The blade discharge angle is 30angle is 30 °°from the tangential direction. If the rotor rotates at a from the tangential direction. If the rotor rotates at a constant speed of 1725 rpm, constant speed of 1725 rpm, estimate the power required to run the estimate the power required to run the fan.fan. 90..………………………………Example 5.19 Moment of Example 5.19 Moment of Momentum Momentum ––Power Power 2/22/2 91..………………………………Example 5.19 Example 5.19 SolutionSolution  2θ2 2 1θ1 1 shaft VU m VU m W    0 (V0 (V11is radial)is radial) hp972.0... VUm Ws/ft3.29...)30sinhr2/(m WhVr2 VA Q mV 30cosW 30cosW U V U W Vs/ft3.90min)/s60)(ft/.in12()rev/rad2)(rpm 1725.)(in6(r Us/slug 00912.0...Q m 2 2 shaft2 22r 2 2r22r 2 2 2 2 2 2 22 2        92First Law of Thermodynamics First Law of Thermodynamics –– The Energy EquationThe Energy Equation1/51/5 The first law of thermodynamics The first law of thermodynamics for a systemfor a system isis Time rate of increase Time rate of increase of the total stored of the total stored energy of the systemenergy of the systemNet time rate of energy Net time rate of energy addition by heat transfer addition by heat transfer into the systeminto the systemNet time rate of energy Net time rate of energy addition by work addition by work transfer into the systemtransfer into the system= +      gz2VuˆeW Q VdedtdorW Q W W Q Q Vdedtd 2sysinnet innet syssysin/net in/netsysout insysout in sys         Total stored energy per unit Total stored energy per unit mass for each particle in the mass for each particle in the systemsystem““++””going into systemgoing into system ““--””coming out coming out The net rate of heat transfer into the systemThe net rate of heat transfer into the systemThe net rate of work transfer The net rate of work transfer into the systeminto the systemBased on system methodBased on system method 注意『正』、『負』 93First Law of Thermodynamics First Law of Thermodynamics –– The Energy EquationThe Energy Equation2/52/5 For the control volume that is coincident with the system For the control volume that is coincident with the system at an instant of time.at an instant of time. volume control coincidentinnet innet sysinnet innet ) W Q( ) W Q(    在時間在時間t t 的瞬間，進出的瞬間，進出 CV CV 的的Q Q 與與W W 等於進等於進 出出system system 者。透過這層連結，才能得到者。透過這層連結，才能得到 ………… 94Review of Reynolds Transport TheoremReview of Reynolds Transport Theorem dAnVb VbdtdAnVbtB dtdB CS CVCSCV sys   This is the fundamental relation between the rate of This is the fundamental relation between the rate of change of any arbitrary extensive property, B, of a change of any arbitrary extensive property, B, of a system and the variations of this property associated system and the variations of this property associated with a control volume.with a control volume. 95First Law of Thermodynamics First Law of Thermodynamics –– The Energy EquationThe Energy Equation3/53/5 For the system and the contents of the coincident control volumeFor the system and the contents of the coincident control volume that is fixed and that is fixed and nondeformingnondeforming ----Reynolds Transport Theorem Reynolds Transport Theorem leads toleads to dAnVe VdetVdedtd .S.C CV sys   Time rate of increase Time rate of increase of the total stored of the total stored energy of the systemenergy of the systemNet time rate of increase Net time rate of increase of the total stored energy of the total stored energy of the contents of the of the contents of the control volumecontrol volumeThe net rate of flow of the The net rate of flow of the total stored energy out of total stored energy out of the control volume through the control volume through the control surfacethe control surface= +依據 Chapter 4 推導 Reynolds transport theorem 的步驟 解 讀選擇一固定且不變形的CV = SYSTEM meB eb   96First Law of Thermodynamics First Law of Thermodynamics –– The Energy EquationThe Energy Equation4/54/5 CV CSinnet innet cv ) W Q( dAnVe VdetThe control volume formula for the first law of The control volume formula for the first law of thermodynamics:thermodynamics: Based on Control VolumeBased on Control Volume NEXT PAGE 97Rate of Work done by CVRate of Work done by CV Shaft work : the rate of work transferred into thro Shaft work : the rate of work transferred into thro ugh ugh the CS by the shaft work ( negative for work transferred out, the CS by the shaft work ( negative for work transferred out, positive for work input required) positive for work input required) Work done by normal stresses at the CS:Work done by normal stresses at the CS: Work done by shear stresses at the CS:Work done by shear stresses at the CS: Other work Other work other shear normal Shaft W W W W W     ShaftW   CS CSnn normal normal dAnVp dAnV V F W   dAnV W CSshear     CSinnet shaft innetCScv dAnVp W Q dAnVe Vdet Negligibly smallNegligibly small藉由shaft傳遞的功 +輸入系統者，-輸出系統者 98First Law of Thermodynamics First Law of Thermodynamics –– The Energy EquationThe Energy Equation5/55/5 in/ Shaft in/netCS2 CVW Q dAnV)gz2Vpuˆ( Vdet    CSinnet Shaf innetCS CVdAnVp W Q dAnVe Vdet Energy equationEnergy equation gz2Vuˆe2  99Application of Energy EquationApplication of Energy Equation1/31/3 0VdetCV mgz2Vpuˆ mgz2Vpuˆ dAnV gz2Vpuˆ in2 out2 2 CS            in in2 out out22 CS m gz2Vpuˆ m gz2VpuˆdAnV gz2Vpuˆ          When the flow is steadyWhen the flow is steady The integral of The integral of dAnV gz2Vpuˆ2 CS  ?????? Uniformly distribution Only one stream entering and leaving Only one stream entering and leavingSpecial & simple case Special & simple case 100Application of Energy EquationApplication of Energy Equation2/32/3  innet shaft innetin out2 in2 out in outin out W Qz zg2V V p puˆ uˆm              puˆhˆ in/net shaft in/net in out2 in2 out in out W Q z zg2V Vhˆ hˆm    If shaft work is involvedIf shaft work is involved …….. OneOne--dimensional energy equation dimensional energy equation for steadyfor steady --inin--thethe--mean flowmean flow EnthalpyEnthalpy The energy equation is written in terms The energy equation is written in terms of enthalpy.of enthalpy.當 shaft work 包括進來 101Application of Energy EquationApplication of Energy Equation3/33/3  innetin out2 in2 out in outin out Qz zg2V V p puˆ uˆm            in/net in out2 in2 out in out Q z zg2V Vhˆ hˆm    If shaft work is zeroIf shaft work is zero …….. OneOne--dimensional energy equation dimensional energy equation for steadyfor steady --inin--thethe--mean flowmean flow沒有shaft work The The PeltonPelton wheelwheel is among the most is among the most efficient types of water efficient types of water turbinesturbines . It . It was invented by Lester Allan was invented by Lester Allan PeltonPelton (1829(1829 --1908) in the 1870s 1908) in the 1870s 102Example 5.20 Energy Example 5.20 Energy ––Pump Power Pump Power 1/21/2 A pump delivers water at a steady rate of 300 gal/min as shown iA pump delivers water at a steady rate of 300 gal/min as shown i n n Figure E5.20. Just upstream of the pump [section(1)] where the pFigure E5.20. Just upstream of the pump [section(1)] where the p ipe ipe diameter is 3.5 in., the pressure is 18 diameter is 3.5 in., the pressure is 18 psipsi. Just downstream of the . Just downstream of the pump [section (2)] where the pipe diameter is 1 in., the pressur pump [section (2)] where the pipe diameter is 1 in., the pressur e is e is 60 60 psipsi. The change in water elevatio n across the pump is zero. The . The change in water elevatio n across the pump is zero. The rise in internal energy of water, urise in internal energy of water, u22--uu11, associated with a temperature , associated with a temperature rise across the pump is 3000 rise across the pump is 3000 ftft··lblb/slug. If the pumping process is /slug. If the pumping process is considered to be adiabatic, dete rmine the power (hp) required by considered to be adiabatic, dete rmine the power (hp) required by the the pump.pump. 絕熱條件：沒有 Q 進出 計算輸入功 103Example 5.20 Energy Example 5.20 Energy ––Pump Power Pump Power 2/22/2 104Example 5.20 Example 5.20 SolutionSolution  in/net shaft in/net1 22 12 2 1 21 2 W Qzzg2V V p puˆ uˆm               hp3.32 ....)s/ slugs30.1( Ws/ft123...AQV s/ft0.10 .....AQV4/DQ AQVs/ slugs30.1min)/s60)(ft/gal48.7(min)/gal300)(ft/slug94.1(Q m innet shaft22 11 233     OneOne--dimensional energy equation for steadydimensional energy equation for steady --inin--thethe--mean flowmean flow =0(Adiabatic flow) 105Example 5.21 Energy Example 5.21 Energy ––Turbine Power Turbine Power per Unit Mass of Flowper Unit Mass of Flow Steam enters a turbine with a velocity of 30m/s and enthalpy, hSteam enters a turbine with a velocity of 30m/s and enthalpy, h11, of , of 3348 kJ/kg. The steam leaves the turbine as a mixture of vapor a3348 kJ/kg. The steam leaves the turbine as a mixture of vapor a nd nd liquid having a velocity of 60 liquid having a velocity of 60 m/sm/sand an enthalpy of 2550 kJ/kg. If and an enthalpy of 2550 kJ/kg. If the flow through the turbine is ad iabatic and changes in elevati the flow through the turbine is ad iabatic and changes in elevati on on are negligible, are negligible, determine the work output involved per unit mass of determine the work output involved per unit mass of steam throughsteam through --flowflow..絕熱條件：沒有 Q 進出 計算輸出功 106Example 5.21 Example 5.21 SolutionSolution in/net shaft in/net 1 22 12 2 1 2 W Q zzg2V Vhˆ hˆm     kg/kJ797...2V Vhˆ hˆ ww w2V Vhˆ hˆ mW w 2 22 1 2 1 outnet shaftinnet shaft outnet shaft2 12 2 1 2innet shaft innet shaft     The energy equation in terms of enthalpy.The energy equation in terms of enthalpy. =0(Adiabatic flow) 107Example 5.22 Energy Example 5.22 Energy ––Temperature Temperature ChangeChange A 500A 500 --ft waterfall involves steady flow from one large body of ft waterfall involves steady flow from one large body of water to another. Determine the temperature change associated wiwater to another. Determine the temperature change associated wi th th this flow.this flow. 108Example 5.22 Example 5.22 SolutionSolution innet 1 22 12 2 1 2 1 2 Q zzg2V V p puˆ uˆm            waterof heat specifictheis)R lbm/(Btu1c wherecuˆ uˆTT1 2 1 2    V2=V1 R 643.0)]slb/()ft lbm(2.32)][R lbm/(lbft778[)ft500)(s/ft2.32( c)zz(gTT22 1 2 1 2   The temperature change is related to the change of internal energy of the water OneOne--dimensional energy equation for steadydimensional energy equation for steady --inin--thethe--mean flow mean flow without shaft workwithout shaft work =0(Adiabatic flow) 109Energy Equation vs. Bernoulli Equation Energy Equation vs. Bernoulli Equation 1/41/4  innet in out in2 in in out2 out outq uˆ uˆ gz2V pgz2V pinnet in out2 in2 out in out in out Q z zg2V V p puˆ uˆm             mFor steady, incompressible flowFor steady, incompressible flow ……OneOne--dimensional energy equationdimensional energy equation m/ Q qinnet innet in2 in in out2 out out z2Vp z2Vp  0 q uˆ uˆinnet in out  wherewhere For steady, incompressible, For steady, incompressible, frictionless flowfrictionless flow …… Bernoulli equationBernoulli equation Frictionless flowFrictionless flow ……沒有 shaft work 有normal stress做的功innet shaft innet in out2 in2 out in outin out W Q z zg2V V p puˆ uˆm               gz2Vuˆe2  沒有摩擦損失參照 chapter 3 110Energy Equation & Bernoulli Equation Energy Equation & Bernoulli Equation 2/42/4 For steady, incompressible, For steady, incompressible, frictional flowfrictional flow …… 0 q uˆ uˆinnet in out  loss q uˆ uˆinnet in out  loss gz2V pgz2V p in2 in in out2 out out Defining “useful or available energy”… gz2Vp2  Defining “loss of useful or available energy”…Frictional flowFrictional flow …… 因為12有摩擦損失 ，22 2 2gz2V p自然而然就低於12 1 1gz2V p 下游 上游Loss發生在in out過程中 111Energy Equation & Bernoulli Equation Energy Equation & Bernoulli Equation 3/43/4 innet shatf innet in out2 in2 out in out in out W Q z zg2V V p puˆ uˆm             m ) q uˆ uˆ( w gz2V pgz2V p innet in out innet shaft in2 in in out2 out out For steady, incompressible flow with friction and shaft workFor steady, incompressible flow with friction and shaft work …… loss w gz2V pgz2V p innet shaft in2 in in out2 out out  gL s in2 in in out2 out outhh zg2V pzg2V p QW gmW gw hin/netshaftin/netshaftin/netshaft S    glosshL Head lossHead loss Shaft headShaft head有摩擦損失有軸功進來 在 in out 注入在 in out 注入 112Energy Equation & Bernoulli Equation Energy Equation & Bernoulli Equation 4/44/4 For turbineFor turbine For pumpFor pump The actual head drop across the turbineThe actual head drop across the turbine The actual head drop across the pumpThe actual head drop across the pump)0 h(h hT T s  P sh h hhppis pump headis pump headhhTTis turbine headis turbine head TL s T )h h( h  pL s p )h h( hL s in2 in in out2 out outhh zg2V pzg2V p in out 輸入in out 輸出 想像：讓loss擴大 想像：讓loss減緩 Energy transfer 113Example 5.23 Energy Example 5.23 Energy ––Effect of Loss Effect of Loss of Available Energyof Available Energy Compare the volume Compare the volume flowratesflowrates associated with two associated with two different vent configurations , a cylindrical hole in the different vent configurations , a cylindrical hole in the wall having a diameter of 120 mm and the same wall having a diameter of 120 mm and the same diameter cylindrical hole in the wall but with a well diameter cylindrical hole in the wall but with a well -- rounded entrance (see Figure E5.23a). The room rounded entrance (see Figure E5.23a). The room pressure is held constant at 0.1 pressure is held constant at 0.1 kPakPaabove above atmospheric pressure. Both vents exhaust into the atmospheric pressure. Both vents exhaust into the atmosphere. As discussed in Section 8.4.2. the loss in atmosphere. As discussed in Section 8.4.2. the loss in available energy associated with flow through the available energy associated with flow through the cylindrical bent from the ro om to the vent exit is cylindrical bent from the ro om to the vent exit is 0.5V0.5V2222/2 where V/2 where V22is the uniformly distributed exit is the uniformly distributed exit velocity of air. The loss in available energy associated velocity of air. The loss in available energy associated with flow through the rounde d entrance vent from the with flow through the rounde d entrance vent from the room to the vent exit is 0.05Vroom to the vent exit is 0.05V2222/2, where V/2, where V22is the is the uniformly distributed exit velocity of air.uniformly distributed exit velocity of air. 114Example 5.23 Example 5.23 SolutionSolution 2 112 1 1 22 2 2loss gz2V pgz2V pFor steady, incompressible flow with friction, the energy equation V1=0 No elevation change   2/K1pp 4DVAQ2/K1ppV2VK loss losspp2 V L2 12 2 22L2 1 22 2 L 2 1 2 12 1 2           115Example 5.24 Energy Example 5.24 Energy ––Fan Work and Fan Work and EfficiencyEfficiency An axialAn axial --flow ventilating fan driven by a motor that delivers 0.4 kW flow ventilating fan driven by a motor that delivers 0.4 kW of power to the fan blades produces a 0.6of power to the fan blades produces a 0.6 --mm--diameter axial stream diameter axial stream of air having a speed of 12 of air having a speed of 12 m/sm/s. The flow upstream of the fan . The flow upstream of the fan involves negligible speed. Determine how much of the work to theinvolves negligible speed. Determine how much of the work to the air actually produces a useful eff ects, that is, a rise in avail air actually produces a useful eff ects, that is, a rise in avail able able energy and estimate the fluid mechanical efficiency of this fan.energy and estimate the fluid mechanical efficiency of this fan. 116Example 5.24 Example 5.24 SolutionSolution       12 1 1 22 2 2 innet shaft gz2V pgz2V ploss wFor steady, incompressible flow with friction and shaft workFor steady, incompressible flow with friction and shaft work …… pp11=p=p22=atmospheric pressure, V=atmospheric pressure, V11=0, no elevation change=0, no elevation change kg/mN0.722Vloss w2 2 innet shaft  innet shaftinnet shaft wloss w  EfficiencyEfficiency kg/mN8.95AVW mW winnet shaft innet shaft innet shaft    117Example 5.25 Energy Example 5.25 Energy ––Head Loss Head Loss and Power Lossand Power Loss The pump shown in Figure E5.25 adds 10 horsepower to the water 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 as it pumps water from the lower lake to the upper lake. The elevation difference between the lake surfaces is 30 ft and the elevation difference between the lake surfaces is 30 ft and the head head loss is 15 ft. Determine the loss is 15 ft. Determine the flowrateflowrate and power loss associated with and power loss associated with this flow.this flow. 118Example 5.25 Example 5.25 SolutionSolution 0 V V 0 p phh zg2V pzg2V p B A B AL s B2 B B A2 A A  The energy equationThe energy equation Q/1.88QW z z h hin/net shaft B A L s The pump headThe pump head Power lossPower loss ... Qh WL loss  119Application of Energy Equation to Application of Energy Equation to NonuniformNonuniform Flows Flows 1/21/2   dAnV2V2 .S.C     2V~ 2V~ m dAnV2V2 inin2 out out2 CS 1 2VmdAnV2V 2A2           If the velocity profile at any section where flow crosses the If the velocity profile at any section where flow crosses the control surface is not uniformcontrol surface is not uniform …… For one stream of fluid entering and For one stream of fluid entering and leaving the control volumeleaving the control volume …….. Where Where is the kinetic energy is the kinetic energy coefficient and V is the coefficient and V is the average velocityaverage velocity????????當進出CS的流體速度分布不是uniform 以平均值取代 新增參數 120Application of Energy Equation to Application of Energy Equation to NonuniformNonuniform Flows Flows 2/22/2 loss w gz2V pgz2V p innet shaft in2 in in in out2 out out out For For nonuniformnonuniform velocity profilevelocity profile ………… ..  )loss( w z2Vp z2Vpinnet shaft in2 in in in out2 out out out   g Linnet shaft in2 in in in out2 out out outhgw zg2V pzg2V p  121Example 5.26 Energy Example 5.26 Energy ––Effect of Effect of NonuniformNonuniform Velocity Profile Velocity Profile 1/21/2 The small fan shown in Figure E5.26 moves air at a mass The small fan shown in Figure E5.26 moves air at a mass flowrateflowrate of 0.1 of 0.1 khkh/min. Upstream of the fan, the pipe diameter is 60 mm, the /min. Upstream of the fan, the pipe diameter is 60 mm, the flow is laminar, the velocity di stribution is parabolic, and the flow is laminar, the velocity di stribution is parabolic, and the kinetic kinetic energy coefficient, energy coefficient, αα11, is equal to 2.0. Downstream of the fan, the , is equal to 2.0. Downstream of the fan, the pipe diameter is 30 mm, the flow is turbulent, the velocity profpipe diameter is 30 mm, the flow is turbulent, the velocity prof ile is ile is quite uniform, and the kinetic energy coefficient, quite uniform, and the kinetic energy coefficient, αα22, is equal to , is equal to 1.08. If the rise in static pressure across the fan is 0.1 1.08. If the rise in static pressure across the fan is 0.1 kPakPaand the and the fan motor draws 0.14 W, compare the value of loss calculated: (afan motor draws 0.14 W, compare the value of loss calculated: (a ) ) assuming uniform velocity distributions, (2) considering actual assuming uniform velocity distributions, (2) considering actual velocity distribution.velocity distribution. 122Example 5.26 Energy Example 5.26 Energy ––Effect of Effect of NonuniformNonuniform Velocity Profile Velocity Profile 2/22/2 123Example 5.26 Example 5.26 SolutionSolution1/21/2 loss w gz2V pgz2V p in/net shaft 12 11 1 22 22 2 The energy equation for The energy equation for nonuniformnonuniform velocity profilevelocity profile ………… .. s/m92.1...AmV s/m479.0...AmVkg/mN0.84 min)/s60(min/kg1.0]W/)s/mN1)[(W14.0(mmotorfanto powerw 22 11in/net shaft    2V 2V ppw loss2 222 11 1 2 innet shaft     124Example 5.26 Example 5.26 SolutionSolution1/21/2  1 kg/mN975.02V 2V ppw loss 2 12 222 11 1 2 in/net shaft        08.1 ,2 kg/mN940.02V 2V ppw loss 2 12 222 11 1 2 in/net shaft       125Example 5.28 Energy Example 5.28 Energy ––Fan Fan PerformancePerformance For the fan of Example 5.19, show that only some of the shaft poFor the fan of Example 5.19, show that only some of the shaft po wer wer into the air is converted into a useful effect. Develop a meanininto the air is converted into a useful effect. Develop a meanin gful gful efficiency equation and a practical means for estimating lost shefficiency equation and a practical means for estimating lost sh aft aft energy.energy. 126Example 5.28 Example 5.28 SolutionSolution1/21/2 loss w gz2V pgz2V p innet shaft 12 1 1 22 2 2          12 1 1 22 2 2in/net shaft gz2V pgz2V ploss w effect useful innet shaftinnet shaft wloss w  EfficiencyEfficiency 2 2 innet shaft VU w(1) (2) (3) (4) 127Example 5.28 Example 5.28 SolutionSolution2/22/2 )]gz2/V /p()gz2/V /p[( VU12 1 1 22 2 2 2 2 (2)+(3)+(4)(2)+(3)+(4) 2 2 12 1 1 22 2 2 VU/]}gz)2/V()/p[(]gz)2/V()/p{[(  (2)+(4)(2)+(4) 128First Law of Thermodynamics First Law of Thermodynamics ––For For SemiSemi --infinitesimal CV infinitesimal CV 1/21/2 Applying the oneApplying the one --dimensional, steady flow energy dimensional, steady flow energy equation to the content of a semiequation to the content of a semi --infinitesimal control infinitesimal control volumevolume CV非有限大 ，也非無限小，故將 in 與 out 間的變化以 difference 來表達innet in out2 in2 out in out in out Q z zg2V V p puˆ uˆm                innet2 Q dzg2Vdpduˆdm              semisemi --infinitesimal control volumeinfinitesimal control volume 129First Law of Thermodynamics First Law of Thermodynamics ––For For SemiSemi --infinitesimal CV infinitesimal CV 2/22/2     1pdud TdsFor all pure substances including common For all pure substances including common engineering working fluids , such as air, water, engineering working fluids , such as air, water, oil, and gasolineoil, and gasoline innet2 Q dzg2Vdpd1pd Tdsm                  ) q Tds( gdz2Vddp innet2    SemiSemi --infinitesimal control volume statement of infinitesimal control volume statement of the energy equationthe energy equation 130Second Law of Thermodynamics Second Law of Thermodynamics –– Irreversible Flow Irreversible Flow 1/31/3 A general statement of the second law of thermodynamicsA general statement of the second law of thermodynamics For the system and the contents of the coincident control For the system and the contents of the coincident control volume that is fixed and volume that is fixed and nondeformingnondeforming ----Reynolds Reynolds Transport Theorem leads toTransport Theorem leads to     sysinnet sysTQ VdsdtdThe time rate of increase of The time rate of increase of the entropy of a systemthe entropy of a systemSum of the ratio of net h eat transfer rate into Sum of the ratio of net h eat transfer rate into system to absolute temperature for each system to absolute temperature for each particle of mass in th e system receiving heat particle of mass in th e system receiving heat from surroundingsfrom surroundings≥≥ dAnVs VdstVdsdtd .S.C CV sys  Based on system method 依據 Chapter 4 推導 Reynolds transport theorem 的步驟 選擇一固定且不變形的CV = SYSTEM 131Second Law of Thermodynamics Second Law of Thermodynamics –– Irreversible Flow Irreversible Flow 2/32/3 For the system and control vo lume at the instant when For the system and control vo lume at the instant when system and control volume are coincidentsystem and control volume are coincident The control volume formula for the The control volume formula for the second law of second law of thermodynamicsthermodynamics CVinnet CS CVTQ dAnVs Vdst                 cvinnet sysinnet TQ TQ  在時間在時間t t 的瞬間，進出的瞬間，進出 CV CV 的的 Q Q 等於進出等於進出 system system 者。透過者。透過 這層連結，才能得到這層連結，才能得到 ………… Based on Control VolumeBased on Control Volume 132Second Law of Thermodynamics Second Law of Thermodynamics –– Irreversible Flow Irreversible Flow 3/33/3  TQ dsminnet For one stream of fluid entering and leaving the control For one stream of fluid entering and leaving the control volumevolume ……..  TQ )s s(minnet in out  0 q TdsinnetSemi-infinitesimal thin CV Semi-infinitesimal thin CV Uniform temperature Uniform temperatureSteady flowSteady flowSpecial & simple case 133First and Second Law of First and Second Law of Thermodynamics Thermodynamics 1/41/4 ) q Tds( gdz2Vddp innet2    SemiSemi --infinitesimal control volume statement of the infinitesimal control volume statement of the energy equationenergy equation SemiSemi --infinitesimal CV of the second law of infinitesimal CV of the second law of thermodynamicsthermodynamics 0 q Tdsinnet 0 gdz2Vddp2       CV 非 有 限 大 ， 也 非 無 限 小 134First and Second Law of First and Second Law of Thermodynamics Thermodynamics 2/42/4 ) q Tds()loss( gdz2Vddp innet2        For steady frictionless flowFor steady frictionless flow 0 gdz2Vddp2        The shaft work is involvedThe shaft work is involved innet shaft2 w)loss( gdz2Vddp       135First and Second Law of First and Second Law of Thermodynamics Thermodynamics 3/43/4 ) q Tds()loss( gdz2Vddp innet2            1pdud Tds )loss( q1pdudinnet     For incompressible flowFor incompressible flow )loss( q udinnet 136First and Second Law of First and Second Law of Thermodynamics Thermodynamics 4/44/4 loss q1pd u uinnetout inin out     When control volume is finiteWhen control volume is finite For incompressible flowFor incompressible flow loss q u uinnet in out  137Application of the Loss FormApplication of the Loss Form1/21/2 innet shaft2 w)loss( gdz2Vddp       Frictionless, loss=0, no shaft work, incompressible Integrating Integrating 12 1 1 22 2 2gz2V pgz2V pBernoulli equation Frictionless, loss=0, no shaft work, compressible Integrating Integrating 12 1 22 22 1gz2Vgz2V dp 138Application of the Loss FormApplication of the Loss Form2/22/2 Integrating IntegratingFor adiabatic flow of an ideal gas ttan consp k      11 222 1p p 1kk dp 12 1 11 22 2 22gz2V p 1kkgz2V p 1kk