Fundamentals-of-thermody教学讲解课件.ppt

1、Fundamentals of thermodynamics Study what rules must follow when the macroscopic parameters of thermodynamic system change.7-1 7-1 Internal Energy Heat&Workits surroundings1.Two basic concepts:is a system that can interact with its surroundings in at least two ways,one of which must be heat transfer

2、.Thermodynamic system:everything outside the system.Surroundings(exterior)Thermodynamic systemThermodynamic systems in engineering:Thermodynamic systemworking substanceGas:such as air.Vapor:such as steam.Liquid:such as freon(氟利昂氟利昂).Solid:such as semiconductor.They are also called as working substan

3、ceDepends on T of the system only.2.Internal energy Internal energyIdeal gasRTiTE2M)(jipijikiE,state quantitydlSFdldW pSdl pdV 21VVpdVWpVdVpF1V2V12SFSFSFSFSFSFWork is process quantitythe molecules of the system and the molecules of the environment exchange their thermal kinetic energies.Process quan

4、tity Joule proved that 1 cal of heat causes the same temperature increment as that of 4.18 joule of work does.i.e.1cal=4.186 J system does work on exterior.exterior does work on the system.WEEQ 12WE 7-2 The First Law of ThermodynamicsNote The FLT can be used for any thermodynamic process of any ther

5、modynamic system.-Conservation law of energy including the heat phenomena and thermal motion.differential form：for the equilibrium process of ideal gas：21)(212VVpdVTTRiMQ dWdEdQ pdVRdTiMdQ 2 There is not any system(machine)that can do work forever without the supply of heat.It is impossible to make

6、out the first kind of perpetual motion machine.Or Other description of the FLT:HeatWorksystem“Heat”and“Work”can not be transformed without thermodynamic system.0 dVpdVdW 0 WEQ Isochoric(Constant volume)process0)(212TTRiM ABpV00 Q7-3 Application of the First Law to isochoric,isobaric&isothermal Proce

7、ssesVPPi)(212 internal energy.Isothermal(Constant temperature)process 0 dT0 EWQ 21VVdVVRTM 12lnVVRTM Or 2111lnppVpQ 21VVpdV ABpV0The isothermal expansion process0 QIsobaric(Constant pressure)process 21VVdVpEQ)()(21212TTRMTTRiM )(2212TTRiM Or)(2212VVpiQ )(12VVpE pV0ABIsobaric expansion0 Q2 iiQE22 iQW

8、FurthermoreConstant volumeRdTidEdQ2 pdVdQ RdTidQ22 Pa)10013.1(5p)10(33mVabcd112233adEEE )(2adTTRi )(2aaddVpVpi 0)(abapVVpW 3510110013.13 J304 bcbTVVRTWln bcbbVVVpln J246 0 VWVTpWWWW J550 WEQ J550 7-4 Heat Capacities of an Ideal GasdTdQC VVdTdQC dEdQV)(RiCV2 RdTi2 Under constant volume process,Absorb

9、 heat for increasing internal energy only 21)(212VVpdVTTRiMQ 21)(12VVVpdVTTCM ppdTdQC RdTidQp22)(RCCVp Under constant pressure process,Absorb heat not only for increasing internal energy but for working for exterior.)(2212TTRiMQp VpCC ii2 1)(12TTCMp For Mkg gaspdVdEdQ dTdQC aT VdVRTdTCadTV bVRTCaTV

10、lnln.consteVTRaTRCV dVVRTdTCaTdTV No heat transfer between a system and exterior.7-5 Application of the first Law of thermodynamics to Adiabatic Process.0 WEEW thennkTp pdVdEdQ 0 RTMpV RdTMVdppdV pdVdTCMV ：0)(RpdVVdppdVCV0)(VdpCpdVRCVV0 pdpCVdVCVp0 pdpVdV.ConstpV .1ConstTV .1ConstTp ABCpVO CpV 0 Vdp

11、pdVAATVpdVdp )(CpV 01 dVpVdpV AAQVpdVdp )(ABCpVO nkTp pVIKJAB0 IQIWE JJWEQ JIWW 0 KKWEQ KIAA 0 pVOCDAB0V02V0 E0 E321WWW WEQ 321QQQ TCMEV T AABBVTVT AABBTVVT AT2 0 EAABTTTT DDAATVTV11 ADADTVVT1)(AT1)21(AADTTTT)211(1 AT ABRemoving baffleMolecules AB quicklyNon-equilibrium process Q=0W=0 E=0When the sy

12、stem becomes equilibrium state,T1=T2 Use state equation 1221pp Note(1)It is not adiabatic process though Q=0It is not isothermal process though T1=T2(2)For real gas,-Non-equilibrium processKeep p1 p2 when moving the pistonsThe gas through 多孔塞多孔塞 adiabatic free expansion-Joule-Thomson effectFp1p2多孔塞多

13、孔塞pistonpiston0 EabpVO7-6 Cyclical Processesp、V、T return their initial values pVabOA1Q2QQA 21QQQ 21QQ Q1,Q2 are all positiveA simplified heat engine:1QWCold reservoirat T2Heat engineHot reservoirat T12Q 1QW 121QQ We hope that the work output from heat engine is as large as possible and the heat thro

14、wn away is as small as possible.In 17941840,was about 3%4%.So low!Now,is about 30%40%.pVOABC1V2V1p2pCAABQQQ 112lnVVRTA)(CAVTTC )(2BCpTTCQ )(271222VpVp )(25ln12111211VpVpVVVp )(25ln)(27121112111222ppVVVVpVpVp 121QQ bc,da adiabatic process2T1TabcdpVO1.Carnot cycle:Carnot answered this question.Consist

15、s of two isothermal and two adiabatic processesabVVRTQln11 1Q1Tabcd2TpVOdcVVRTQln22 2Q121QQc abdcVVTVVT/ln/ln112 1QCold reservoirat T2Heat engineHot reservoirat T12Q21QQW dcabVVVV 121TT abdccVVTVVT/ln/ln112 1T2QabcdpVO1Q2T1211 cbVTVT1211 daVTVT2.Reverse cycle RefrigeratorA heat engine operating in r

16、everse.It takes heat from a cold place and give it off to a warmer place.Performance coefficient（致冷系数）（致冷系数）2122QQQWQK 1Q2QWHot reservoirat T1Cold reservoirat T2Refrigerator212TTTKc 1T2QabcdpVO1Q2Ti.e.Q2=0 in?121QQ i.e.W=0 in?WQK2 Hot reservoir T1Cold reservoir T21QW1.“Engine”statement (Kelvin state

17、ment开尔文说法）开尔文说法）Note:This statement lies in the difference between the nature if internal energy and that of macroscopic mechanical energy.Random motion of moleculesCoordinated macroscopic motion Cold reservoirT2Hot reservoir T12Q1QCan realize by supplying some work to the system T2T12Q2QT1T221QQ W2

18、QT1T21QW2QT1T22Q1QWequivalenteffecti.e.Violate the“engine”Statement T1T21Q1QW T2T12Q21QQ equivalenteffectT2T12Q2QViolate the“refrigerator”Statement Any directional process can give a statement of the second law of thermodynamicsWhat is the essence of the second law of thermodynamics?All natural proc

19、ess always proceeds spontaneously along the direction in which the irregularity of molecular thermal motion is increasing.7-9 Reversible process&Irreversible process Carnot Theorem1.Reversible Process&Irreversible ProcessReversible Process:Process P(AB)if there is a way to make system back to the in

20、itial state without other effects.AP(AB)There is a way(other processes)BAP(AB)There is no way(other processes)BIrreversible processIrreversible Processes:examples（1）The process in which work transfers into heat is irreversibleworkspontaneouslyheatCant be spontaneously (2)The process in which heat fl

21、ows from a hotter body to a colder one is irreversible.hotcoldHigh T spontaneouslyLow TCant be spontaneously(3)The free expansion of gas is irreversible;AB(4)The rapid expansion of gas is irreversible;ABV.Conclusion:All the real processes that relate with thermal phenomenon are irreversibleReversibl

22、e Processes must be satisfy:(1)must be equilibrium process(2)No friction in each step of process.(1)All Carnot engine operating between the same two temperatures T1 and T2 have the same efficiency,irrespective of the nature of the working substance.2.Carnot Theorem121TT (2)No real engine operating b

23、etween two given temperatures T1 and T2 can have a greater efficiency than that of Carnot engine operating between the same two temperaturesCarnot theorem points out the way to enhance the efficiency of heat engine.121TT Make the process near equilibrium process as possible.Increase the temperature

24、difference between two reservoirs121QQ T2T11Q2QA甲1Q2QA乙112QQ -违反热二律违反热二律 T2T11Q2QA甲1Q2QA乙11 QQ AA T2T1 Q2=Q2 121211QQQQ 11QQ 可可不不 121TT 不不 不不可可 可可不不 不不可可 T2T11Q2QA甲1Q2QA乙there are N=3 molecules in A vessel at beginning.When the plank is removed,the three molecules may move in A+B vessel.AB7-10 Stati

25、stical Meaning of the Second Law of ThermodynamicsTake the free expansion of gas as an example.AB -It expresses the irregularity of molecular thermal motions2.Thermodynamic Probability 热力学概率热力学概率 (A3B0)=1(A2B1)=3(A1B2)=3(A0B3)=1Statistical mechanics assumes that the probability for each microstate i

26、s equal.the position distribution for N=20 moleculesMacrostateL 20R 01L 18R 2190L 15L 11L 10L 9L 5L 2L 0R5R 9R 10R 11R 15R 18R 2015504167960184765167960155041901 is different for different macrostate.The greater the number of molecules,the greater Rules:The macrostate that contains great number of m

27、icrostates(is great)has great appearing probability.When N is much great,the macrostate corresponding to small nearly does not appear.For a isolated system with huge N,its equilibrium state corresponds to the macrostate that gets maximum value.If the system is not at equilibrium state,its does not g

28、et maximum value.-non-equilibrium stateThen it must take place such process:The state of the system changes spontaneously from the state with smaller to the one with biggest with the time extending.gets maximumequilibrium stateStable stateThe proceeding direction of natural process is always along t

29、he direction that increases.Small corresponds non-equilibrium state With time increasingbiggest .i.e.equilibrium state 玻尔兹曼熵玻尔兹曼熵 lnkSto describe the irregularity(disorder)of the system quantitatively initial final21lnlnln kkkS21SS i.e.Natural process always proceeds spontaneously in the direction o

30、f increasing entropy.-the principle of the increase of entropy7-11 The formula of Clausius entropy121121TTTQQQ 02211 TQTQpVOreversible 0 iiTQ0 iiTQ reversible 0 TdQreversibleTdQpVOABinitialfinalFor any reversible cycle,from initial state to final state,iBfiAfTdQTdQ fiTdQFor any reversible process fr

31、om initial state to final state ifSS S Clausius entropy,its a state function of the thermodynamic system.J/KClausius entropy.Clausius entropyClausius entropy.i.e.the formula fiTdQifSSS can calculate S of the system form i state to f state no matter the process is reversible or not.S),(12VT TdQS1 21T

32、TVTdTC12lnTTCV VpO),(11VT),(22VT12 TdQS2 TpdV 21VVVdVR12lnVVR 21SSS 1212lnlnVVRTTCV 说明：说明：若此系统若此系统非孤立非孤立，S可正、可负、可可正、可负、可=0 121121TTTQQQ 不不 02211 TQTQ 0)(不不TdQ pVO12a 21120)()(abTdQTdQ可可不不b 21)(bTdQ可可 12SS 2112)(不不TdQSS不不)(TdQdS 0 dSAB1V2VAB21VV 0 Q0 A0 E TdQSS12 dQT1121ln1VVVRTT 121lnVVVR 0 熵增加熵增加 2

33、1)(可可TdQS lnkS21lnlnln kkkS21SS 用宏观状态量表示熵用宏观状态量表示熵只适用于平衡态只适用于平衡态对对 mol单原子理气：单原子理气：平衡态可用平衡态可用V、T 确定，确定，设此时设此时),(TVSS =？先求先求),(TV =？由分子的位置和速度确定由分子的位置和速度确定vp 分子按位置分布的可能微观状态数分子按位置分布的可能微观状态数分子按速度分布的可能微观状态数分子按速度分布的可能微观状态数一个一个分子按分子按位置位置分布的可能微观状态数分布的可能微观状态数 x,y,z=V一个一个分子按分子按速度速度分布的可能微观状态数分布的可能微观状态数 Mol 气体：气

34、体：0NpV 速度立方盒的体积速度立方盒的体积很大很大可取可取ppivorvv1000100zyxi,而而21Tvp 23T 23T Mol 气体：气体：0)(23NvT 0)(23NvpVT 令令0)(23NVTC 比例常数比例常数则则 lnkSCkTkNVkNlnln23ln00 R0lnlnSTCVRSv R23单原子理气的单原子理气的vC以相应单位表示的相应物理量的数值以相应单位表示的相应物理量的数值-S的的对双原子分子对双原子分子0)(25NvT 转动速度转动速度与与平动速度平动速度对对 v的贡献等价的贡献等价对多原子分子对多原子分子0)(26NvT 对对S的的TdTCVdVRdSv

35、 dTCVdVRTTdSv 对理气对理气=pVdTCpdVv 对可逆过程对可逆过程=dAdQ TdQdS 可逆)(2112 TdQSS克劳修斯熵公式克劳修斯熵公式对于孤立系统、可逆过程：对于孤立系统、可逆过程：0 dQ0 dS适用于任意系统的适用于任意系统的可逆过程可逆过程 lnkS 2112)(可TdQSSS1212ln kSSS21 、解：解：水冰TdQSTQ KJTm/1022.13 12ln kSKSe 12251084.8 e解：解：21TTTdQS水 21TTTcmdT12lnTTcm)/(1018.43KkgJc 22TTTdQS库2TQ 212TTTcm 库水SSS )/(18

36、4KJ 21TTTTTdQTdQS水水S 22TQTQS 库 221TTTcmTTTcm 库S 库水SSS )/(97KJ 结论：结论：中间热库越中间热库越多多，S越小，越小，中间热库中间热库多，多，S 0，-可逆过程可逆过程),(12VT TdQS1 21TTVTdTC12lnTTCV VpO),(11VT),(22VT12 TdQS2 TpdV 21VVVdVR12lnVVR 21SSS 1212lnlnVVRTTCV 说明：说明：若此系统若此系统非孤立非孤立，S可正、可负、可可正、可负、可=0 121121TTTQQQ 不不 02211 TQTQ 0)(不不TdQ pVO12a 2112

37、0)()(abTdQTdQ可可不不b 21)(bTdQ可可 12SS 2112)(不不TdQSS不不)(TdQdS 0 dSAB1V2VAB21VV 0 Q0 A0 E TdQSS12 dQT1121ln1VVVRTT 121lnVVVR 0 熵增加熵增加TATB BATT 21平衡后温度平衡后温度 BATTT 21熵变熵变BASSS TTTTBATdQTdQ TTvTTvBATdTCTdTC BAvTTTC2ln 0 熵增加熵增加 21)(可可TdQS lnkS由由可可)(TdQdS TdSdQ 得可逆过程系统吸热得可逆过程系统吸热 TdSQSTAB温熵图温熵图Q绝热线绝热线循环过程：循环过

38、程：STABCDMNAC过程：过程：BdS0吸热过程吸热过程吸热吸热Q1=SABCNMACA过程：过程：DdS0放热过程放热过程放热放热Q2=SCDAMNC功功A=SABCDAABCNMAABCDASS 卡诺循环：卡诺循环：STabcdmnT1T2abcnmaabcdaSS 卡 amad 121TT 由图可知：由图可知：T1、T2不变，不变，改变等温过程长度时，改变等温过程长度时，卡 不变不变可逆可逆卡 只与只与T1、T2有关有关不可逆不可逆 TB，A B，dQdQ在在A中时，可中时，可做功的最大值：做功的最大值：dQA卡 1dQTTA)1(0 dQ传到传到B中后，可中后，可做功的最大值：做功

39、的最大值：dQA卡 2dQTTB)1(0 热传导前后：热传导前后：21AAEd dQTTTAB0)11(熵变：熵变：BAdSdSdS BATdQTdQ dSTEd0 3.理气绝热自由膨胀理气绝热自由膨胀 mol 理气，温度理气，温度T，体积体积21VV 对外做功对外做功=0若使其回到若使其回到1V则需外界对其做功则需外界对其做功121lnVVRTA 变为热变为热QQ再转变为有用功再转变为有用功QA卡 210)1ATT （21AAEd 120lnVVRT 而而TQS （设为等温压缩）（设为等温压缩）12lnVVR STEd 0不可逆过程中退降的能量不可逆过程中退降的能量Ed与系统熵的增量成正比与系统熵的增量成正比结论：结论：