山东大学：《工程热力学》课程PPT教学课件（双语）第四章 Energy analysis of closed systems

Chap.3 SummaryAsinglechemicalelementorcompoundAsubstancethat hasPure substanceHomogeneousmixtureofvariouschemicalelementsorcompoundsafixedchemicalcompositionmixtureoftwoormorephasesapuresubstanceCompressed liguid orsubcooled liquidliquidSaturated liquidSatruated liquid-vapor mixturePhase changeprocessSatruated vaporvaporSuperheatedvapor+Psat,Tsat,Latentheatoffusion/vaporization,sublimationsolidCritical pointSaturated liquid/vapor lineSuperheatedvaporregionT-v,P-v,P-T, P-v-T diagramsTripple pointCompressedliquid regionSaturated liquid-vaporregion1Enthalpy/焰:acombinationpropertyh=u+Pv(kJ/kg);Entropy/PropertiestablesSaturated liguid/vaporormixture.ht,ha,haQuality x=[0,1]SuperheatedvaporCompressedliquid P,TReferencestatevalues1PV=RTPV=mRTR,RuIdeal gasequationofstateIdeal-gas/realgasZ=Pv/RTCompressibilityfactorReducedPReducedTequationofstateGeneralizedVanderwallsPrincipleofcorresponding statescompressibility chartequationofstate

Chap.3 Summary 1 A substance that has a fixed chemical composition Phase change process A single chemical element or compound Enthalpy/焓: a combination property h=u+Pv (kJ/kg); Entropy/熵 Critical point T-v, P-v,P-T, P-v-T diagrams Properties tables Ideal-gas / real gas equation of state Pure substance liquid vapor solid Homogeneous mixture of various chemical elements or compounds mixture of two or more phases a pure substance Saturated liquid Satruated liquid-vapor mixture Satruated vapor Superheated vapor Compressed liquid or subcooled liquid Psat, Tsat, Latent heat of fusion/vaporization, sublimation Tripple point Saturated liquid/vapor line Superheated vapor region Compressed liquid region Saturated liquid-vapor region Saturated liquid/vapor or mixture. hf ,hg,hfg Quality x=[0,1] Superheated vapor Compressed liquid P, T Reference state/values Ideal gas equation of state Pv=RT PV=mRT R, Ru Compressibility factor Z=Pv/RT Reduced P Reduced T Principle of corresponding states Generalized compressibility chart Van der walls equation of state

Chap4 SummaryGeneral:Ow,=Pdy,P-VdiagramThework associatedBoundary workwithexpansionandIsobaricprocess,Pvn=const PolytropicprocesscompressionisothermalprocessofanidealgasW,=P,V,ln(V2/V)AE.(KJ)E-EouEnergyBalancesyatenNet enengy tnansferChange in intemal,kioctic.foranysystemandanyprocessbyhealwortandmasspoteatial,tc,enengis1stlawforclosedsystem=WWWWQ=Qnetin=0-0aet.ououtEnergyBalanceforclosedsystemO-W=AEThe energyrequired to raise T of 1kg ofa substanceby 1K,kJ/(kg.k)ahSpecific heatCv atconstant1Cp atconstantPCp22aTIdealgases:Cp=Cv+R;k=Cp/CvForsolidsand liquids,Cp=Cv=cU,h fromtable.Ah=h-hc,(T)dTepavg(T-T)Changes ofu, h for ideal gasesByusingCvorCpUsingaveragespecificheats.Cv,avg=Cv(T2/2+T1/2)Cp,avg=Cp(T2/2+T1/2)Forimcompressiblec(T)dT=Cav(T2-T)Ah=Au+VAPChanges of u,h for solid/fluidsubstances

Chap4 Summary 2 The work associated with expansion and compression 1st law for closed system General: δWb=Pdv, P-V diagram U,h from table. Specific heat Changes of u, h for ideal gases Changes of u, h for solid/fluid Boundary work Energy Balance for any system and any process Isobaric process, PVn=const Polytropic process isothermal process of an ideal gas Wb=P1V1ln(V2/V1) The energy required to raise T of 1kg of a substance by 1K, kJ/(kg.k) By using Cv or Cp Using average specific heats. Cv,avg=Cv(T2/2+T1/2) Cp,avg=Cp(T2/2+T1/2) Energy Balance for closed system Cv at constant V Cp at constant P Ideal gases: Cp=Cv+R; k=Cp/Cv For solids and liquids, Cp=Cv=C For imcompressible substances

Chapter 4 Energy analysis of closed systemsmoving boundary work or P dVwork general energy balance applied to closed systemsCp,Cy -specific heat at constant pressure/volumeincompressible substances , the changes in their internalenergy and enthalpy Solve energy balance problems for closed (fixed mass)systems that involve heat and work interactions forgeneral pure substances, ideal gases, andincompressiblesubstances

Chapter 4 Energy analysis of closed systems • moving boundary work or P dV work • general energy balance applied to closed systems. • Cp ,Cv -specific heat at constant pressure/volume • incompressible substances , the changes in their internal energy and enthalpy. • Solve energy balance problems for closed (fixed mass) systems that involve heat and work interactions for general pure substances, ideal gases, and incompressible substances. 3

4-1 Moving Boundary WorkMovingboundarywork:oneformofmechanicalworkassociatedwiththeexpansionorcompressionofagasinadevicelikecarengines. PdVExpansionworkCompressionworkThemost common formofmechanical workFor real engines (compressors),to analyze the moving boundarywork.Piston moves at very higherspeed.StatesarehardtodefineProcessesPathWork-pathfunctionGAS

4-1 Moving Boundary Work • Moving boundary work: one form of mechanical work associated with the expansion or compression of a gas in a device like car engines. PdV – Expansion work – Compression work – The most common form of mechanical work • For real engines (compressors), to analyze the moving boundary work. – Piston moves at very higher speed. – States are hard to define – Processes – Path – Work- path function 4

4-1 Moving Boundary Work:Aquasi-equilibriumprocess:thesystemremainsnearlyinequilibriumat all times of a process.ApproximationMaximum work output(engines)minimumworkinput (compressors)ReversibleprocessSW.=Fds=PAds=PdVP-absolutepressure, positivedVpositivefor expansionprocess,negativeforcompressionprocessGASThis explains why the moving boundary work is called PdV work

4-1 Moving Boundary Work • A quasi-equilibrium process: the system remains nearly in equilibrium at all times of a process. – Approximation – Maximum work output (engines) – minimum work input (compressors) • Reversible process • This explains why the moving boundary work is called PdV work 5 P-absolute pressure, positive dV—positive for expansion process, negative for compression process

Differentialwork:SW=Fds=PAds=PdVTotal boundary work for process fromstate 1 to state 2 (for reversible process):Pav(kJ)

• Differential work: • Total boundary work for process from state 1 to state 2 (for reversible process):

ProcesspathArea=dA=PdVV2-dPav(kJ)十1H十十1Pthe area underthe process curve on a P-V diagramis equal, inmagnitude, to the work done during a quasi-equilibrium expansionorcompressionprocessofaclosedsystem.(On the P-v diagram, it represents the boundary work done per unit mass

• the area under the process curve on a P-V diagram is equal, in magnitude, to the work done during a quasi-equilibrium expansion or compression process of a closed system. • (On the P-v diagram, it represents the boundary work done per unit mass.) 7

Work is a path function (depends on the paths followedas well as the end states)W=10kJWB=8kJWc=5kJIf work is not path function ,but states function. What willhappen for cyclic devices. (car engines, power plants..)- Wb=0WNet work done in cycle:Expansionprocess:WbispositiveSCompressionprocess:WbisnegativeV2Vi

• Work is a path function (depends on the paths followed as well as the end states) • If work is not path function ,but states function. What will happen for cyclic devices. (car engines, power plants.) – Wb=0 • Net work done in cycle: – Expansion process: Wb is positive – Compression process: Wb is negative. 8

RVQW正向循环：热机WQ逆向循环：制冷、热泵QuO2Tu=const.ATu=constWeLatWeeLinTiaconsteconstOQrLCarnot cycleReversed Carnot cycle

Carnot cycle Reversed Carnot cycle P V V P Q➔W 正向循环：热机 W➔Q 逆向循环：制冷、热泵

Apiston-cylinderdeviceinitiallycontains0.4m3ofairat100kPaand80°CThe airis nowcompressed to O.1 m3 in such awaythatthetemperatureinside the cylinder remains constant.Determine the work done during thisprocess. Example 4-1:Solution Air in a piston-cylinder device is compressed isothermally.Theboundarywork doneis to be determined.Analysis Asketch of the system and the P-V diagram of the process areshown inFig.4-8.Assumptions1Thecompression processisquasi-equilibrium.2Atspecifiedconditions,aircan beconsidered to bean idealgassince it isata high temperatureand low pressure relativeto its critical-pointvaluesAnalysisForanideal gasatconstanttemperature ToCPV=mRT.=CorCwhereCisaconstant.Substitutingthisinto Eg.4-2,wehaveV2dyV2W.=PV,In(4-7)1V.UTo=80°C=const.InEq.4-7,PVrcanbereplaced by PzV2or mRTg.Also,VaV,canbeAIRreplacedbyP/P,forthiscasesincePV,=P,V2V,=0.4m3Substituting the numerical values into Eq.4-7yieldsP=100kPaTo=80°C=constIKJ01W,=(100kPa)(0.4m)V.m0.10.4041kPa·m=-55.5kJDiscussion The negative sign indicates that this work is done on the system(aworkinput),whichisalwaysthecaseforcompressionprocesses.10

• Example 4-1: 10