山东大学：《工程热力学》课程PPT教学课件（双语）第七章 Entropy

Chap6 Summary-1Whyweneed2ndLaw?Allprocessessatisfy1stLawSatisfying1stdoesnotensuretheprocesscanactuallyoccurIntroductionto2ndLawAprocesshasdirectionEnergyhasqualityandquantityHeat SinkHeat SourceHeatengineThermalenergyReservoirWact.ouOReceiveheatQfromahightemperature sourceHMthQHWConvert part Qtowork Wnet.outnetoutOHeatEnginesQTihRejectwaste heatQtoalowtemperature sink92ndlawKelvin-PlanckStatement:Itisimpossibleforanydevicethatoperatesonacycletoreceiveheatfromasinglereservoirandproduceanetamountofwork.Noheatenginecanhaven=100%Refrigerators/heatpump:ThedevicesdriveheatQtransferfromT,toTHW.RefrigeratorThework inputtotherefrigerator/heatpumpnet,inwants QLQHeatQabsorbedfromrefrigeratedspaceTHeatpumpQHHeatQrejectedtohightemperature THwants QHRefrigerator,HeatPumpDesiredoulputuDesired outputQ,AirCOPCOPHCOPW.WreLinRequired inputRequired inputConditionerDCEI2nd law,Clausius Statement:Heatdoesnot,of its ownvolition,transferfromacoldmediumtoawarmerone.（热不能自发地、不付代价地从低温物体传到高温物体）

Chap6 Summary-1 1 Why we need 2nd Law? All processes satisfy 1st Law; Satisfying 1st does not ensure the process can actually occur Heat Engines Refrigerator, Heat Pump Introduction to 2nd Law Refrigerators/heat pump: The devices drive heat Q transfer from TL to TH, Thermal energy Reservoir Receive heat QH from a high temperature source The work input to the refrigerator/heat pump Heat QL absorbed from refrigerated space TL A process has direction Energy has quality and quantity Heat Source Heat Sink Convert part QH to work Wnet,out Reject waste heat QL to a low temperature sink Heat engine 2nd law, Kelvin-Planck Statement: It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce a net amount of work. No heat engine can have η=100% Heat QH rejected to high temperature TH Refrigerator wants QL Heat pump wants QH COP 2nd law, Clausius Statement: Heat does not, of its own volition, transfer from a cold medium to a warmer one. (热不能自发地、不付代价地从低温物体传到高温 物体) Air Conditioner

Chap6 Summary-2SystemAprocesscanbereversedwithouleavinganytraceonthesurroundingsSurroundingsReversible ProcessesInternal RevExternal RevWhy need RevIrreversible:heattransferThe bestknown reversible cycle; four reversible processesIsothermal expansionAdiabatic expansionIsothermalCompressionAdiabaticcompressionCarnotCycleCarnotheatengineReversed CarnotCycleCarnotrefrigerator/heat pumpCarnot Principle 1:Given T,andTh,Nth.irrev<.Nth,revCarnot Principle 2:GivenT,andTh,Nth.all rev=Nth,revThe heat engine operates on the reversible Carnot Cycleirreversibleheat engineMhrevCarnot HeatEngineTLQLreversible heat engineTth,revThTih.revTHOn二impossibleheatengineTih,reyThe refrigerator / heat pump operates on a reversible Carnot CycleCarnotRefrigeratorCOPR.ECOPRIEVirreversiblerefrigeratoTH/T,-1Carnot Heat PumpCOPCOPRrevreversiblerefrigeratorCOPHPeCOPR.revimpossiblerefrigeratorI-TL/TH2

Chap6 Summary-2 2 A process can be reversed without leaving any trace on the surroundings. Carnot Cycle Carnot Refrigerator Carnot Heat Pump Reversible Processes The heat engine operates on the reversible Carnot Cycle The best known reversible cycle; four reversible processes Carnot heat engine Carnot Principle 1: Given TL and TH, ηth,irrev < ηth,rev System Surroundings Internal Rev External Rev Why need Rev Irreversible: heat transfer Isothermal expansion Isothermal Compression Adiabatic compression Reversed Carnot Cycle Carnot refrigerator /heat pump Carnot Principle 2: Given TL and TH, ηth,all rev = ηth,rev Carnot Heat Engine The refrigerator / heat pump operates on a reversible Carnot Cycle Adiabatic expansion

Chap7 SummaryClausiusinequality:thecyclicintegral ofisalways≤zeroEntropy（熵）60QAS= S, - SiASEntropyIsoTTotintrevEntropychangeofaclosedsystemASm=S,-S,ReversibleprocessIrreversibleprocess2ASm-S1-S1-JSm= S, - S,>IncreaseofEntropySgen≥0Principle(增原理Increaseofentropyprinciple（孤立系统炳增原理，简称摘增原理）：theentropyofanisolatedsystemduringaprocessalwaysincreaseor,inthelimitingcaseofareversibleprocessremainsconstant.（孤立系统的焰可以增大，或保持不变，但不可能减少）AS=mAS=m(S.EntropychangeofpuresubstancesS0,0,ASsys=S.S2,IsentropicprocessadibaticReversibleas=03,T-S,h-sdiagramsdsAeSomeremarks4,The3rd lawof thermodaynamics:The entropy ofapurecrystallinesubstance atabsolutezerotemperatureiszerodhvdpPdvdudsas5,Tds relationsTTTT6,reversibleworkoutputAkeADC

Chap7 Summary 3 Clausius inequality: the cyclic integral of is always ≦ zero Increase of Entropy Principle(熵增原理) Entropy (熵) 1, Entropy change of pure substances: Entropy change of a closed system: Increase of entropy principle (孤立系统熵增原理，简称熵增原理)：the entropy of an isolated system during a process always increase or , in the limiting case of a reversible process remains constant. (孤立系统的熵可以增大，或保持不变，但不可能减少) Entropy Some remarks Iso T Reversible process Irreversible process ≥ 0 sf at 0.01℃=0 kJ/kg.k 2, Isentropic process 0, adibatic 0, Reversible 3, T-S, h-s diagrams = 4, The 3rd law of thermodaynamics: The entropy of a pure crystalline substance at absolute zero temperature is zero 5, T ds relations: 6, reversible work output >

Chapter 7 Entropy

Chapter 7 Entropy 4

7-1 EntropyThe 2nd law of thermodynamics leads toexpressions that involve inequalitiesHeat engine:Nth,irrev< Nth,rev- Heat pump: COPHP,irev< COPHP,rev COPR,irev < COPR,rev :Clausius inequality:Thecyclicintegralofisalwayslessthanorequal tozero.Stated byR.J.E.Clausius.Valid for all cycles, reversible or irreversibleThecyclicintegral ofcanbeviewed asthe sumof all thesedifferential amounts of heat transfer divided by the temperature of theboundary

7-1 Entropy • The 2nd law of thermodynamics leads to expressions that involve inequalities. – Heat engine:ηth,irrev< ηth,rev – Heat pump: COPHP,irrev < COPHP,rev COPR,irrev < COPR,rev ; • Clausius inequality: – The cyclic integral of is always less than or equal to zero. – Stated by R.J.E. Clausius. – Valid for all cycles, reversible or irreversible – The cyclic integral of can be viewed as the sum of all these differential amounts of heat transfer divided by the temperature of the boundary. 5

ThermalreseelkIWhyWOombinedsyste(system and eyclic device)Considera systemconnected to a thermal energy reservoir at aconstanttemperatureofTthroughareversiblecyclicdeviceThecyclicdevicereceivesheatSQfromreservoir;suppliesheat to the system sg , and producing work swrey.The system produces work sw as a result of heat transfer,andtheboundarytemperatureisT(variable)Wehave a combined systemas the dashed linesyields:SORSQ8Q(SWrey+SWsys)=SWc=8QR-dEcSWC=TRdEcTTRTKelvinPlanck statement:nosystem canproduce a net amount ofwork while operatingina cycle and exchanging heat with a singleSOthermal energy reservoir.For a cycleWc=Wccan not be a work output, can not be a6positivequantity

• Why • • Consider a system connected to a thermal energy reservoir at a constant temperature of TR through a reversible cyclic device. • The cyclic device receives heat from reservoir; supplies heat to the system , and producing work • The system produces work as a result of heat transfer， and the boundary temperature is T (variable) • We have a combined system as the dashed lines yields: 6 For a cycle Kelvin Planck statement: no system can produce a net amount of work while operating in a cycle and exchanging heat with a single thermal energy reservoir. WC can not be a work output, can not be a positive quantity

0: For reversible cycles: all quantities have thesame magnitude but the opposite sign, thus800三Tintrev The equality in the Clausius inequality holds forreversible cyclesThe inequality for the irreversible cycles

• For reversible cycles: all quantities have the same magnitude but the opposite sign, thus: • The equality in the Clausius inequality holds for reversible cycles • The inequality for the irreversible cycles

. Entropy'(%)?intrevCyclic integral is zero:- W, no;- Q, no;- net changein volume,yes.-? Depends on state only, not the process path.So,itisaproperty,statefunctionIn 1865,Clausius named this property as ENTROPY, S.theentropychangeisdefinedasAds(KJ/K)ntrev

• Entropy: • Cyclic integral is zero: – W, no; – Q, no; – net change in volume, yes. – ? • Depends on state only, not the process path. – So, it is a property, state function – In 1865, Clausius named this property as ENTROPY, S. – the entropy change is defined as ?

橘名称的由来克劳修斯（1865）创造entropy()的由来：因为它具有能量的含义，所以用字头en，又因它代表热功转换能力，则用希腊字转变的tropy作字段，所以构成新字en-tropy1923年普朗克到南京作报告《热力学第二定律及之观念》讲 entropy,物理学家胡刚复教授任翻泽，由于它概念太复杂，就想了个简单的办法，根据dS=dQ/T，认为S是热“滴”量与温度之商，而且与火有关，从而构成新字

熵名称的由来 • 克劳修斯(1865)创造 entropy(熵)的由来：因为它具有能 量的含义，所以用字头 en,又因它代表热功转換能力，则 用希腊字转变的 tropy 作字段, 所以构成新字 en-tropy • 1923年普朗克到南京作报告《热力学第二定律及熵之观念 》讲 entropy,物理学家胡剛复教授任翻泽，由于它概念太 复杂，就想了个简单的办法,根据 dS=dQ/T，认为S 是热 量与温度之商，而且与火有关，从而构成新字 “熵” 9

(kJ/K): Then, the entropy change of a system during aprocess can be determined by integrationbetween the initial and the final states:(kJ/K)AS=S2-STintrevWeconcernmoreonchangeofS,butnot S itself,sowehaveareferencestate(S=0),The relation of Q and T is often not available, so we get entropy mostly fromtables.Entropyisaproperty,isfixedatfixedstates.Theentropychange betweentwospecifiedstatesisthesamenomatterwhatprocesspathitisfollowed.Thecalculationisonly validforreversiblepath betweenthetwo states;and usedascomparedvalueforirreversibleprocess.10

• Then, the entropy change of a system during a process can be determined by integration between the initial and the final states: – We concern more on change of S, but not S itself, so we have a reference state (S=0). – The relation of Q and T is often not available, so we get entropy mostly from tables. – Entropy is a property, is fixed at fixed states. The entropy change between two specified states is the same no matter what process path it is followed. – The calculation is only valid for reversible path between the two states; and used as compared value for irreversible process. 10