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