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3?1?h?12?u?g??z42q?w?z??3?m?h??(2300?3230)?10?43J10?kg?kg3600s?q??0kgm??2.778s?h??2.583?1010?kgkg?2.7783600s?s6J?u120?50Jkg22109?Jkg?u??109?s4J12?m?u?1.65?102sJg??z?m?81.729s6J?h?12?m?u?g??z?m??2.567?102sw6J?6?2.56710???2.56710??W?s??wc??2.583?2.567?100%2.567wc?0.623%

4-2

方法一：

?h?1212?u?g??z22q?wu1??30.0750.2522R??8.314?h??u?wu2???u1u2?0.27

T2??353.15?HT1??593.15Cpmh?T2?T1?HR2?HR1

??P????r0???B0?B1???00????????HR1??R?647.3?Tr??dB0??0.344?dB1?dPr?000??TT????r0?r0??????????0????HR1??576.771

P????r1???B0?B1???1???dB0?1??0.344??dPr?HR2??R?647.3?Tr?????dB1?1????1Tr??1Tr????1?1???????0???HR2??56.91

经计算得

Cpmh?35.03?J?mol?1?K?1

?0.075?3?1体积流速为：V???u1??d/2??3.14?3????0.0132?m?s

?2?22摩尔流速为：n?VV0.0132???4.015?mol?s?1 VmRT/p8.314?593.15/1500000根据热力学第一定律，绝热时Ws = -△H，所以

?HWsn?Cpmh?T2?T1?n?HR2?HR1 ?4.015??)???8.408?10?(?56.91?576.771?3????3.167?10?W4

方法二：

根据过热蒸汽表，内插法应用可查得

35kPa、80℃的乏汽处在过热蒸汽区，其焓值h2=2645.6 kJ·kg-1； 1500 kPa、320℃的水蒸汽在过热蒸汽区，其焓值h1=3081.5 kJ·kg-1；

?w?h?1?22??u2?u1??22645.6?3081.5?4.46410??3?435.904kJ?kg?1

按理想气体体积计算的体积VR?TP8.314593.15??3?3.288?1015000000.0132?m?s3?1m?mol3?1molN??4.015sw435.90418??Nmol?4.015?33?1s3.288?10?m?mol3.15?10W4

4-6 解：

二氧化碳T1??303.15P1??1.5?10Tc??304.2???0.2256R??8.314PaP2??0.1013310?PaPc??7.35710?Cp(T)??45.369?8.68810??366Pa5?2?T?9.61910??T

?1.6?4.2?R?Tc?P1?TT?????11?H1R???0.083?0.139???1.097???0.894????????Pc??Tc??Tc?? ?1.6?4.2?R?Tc?P2??T2??T2???H2RT2???0.083?0.139???1.097???0.894????????Pc??Tc??Tc?? ??通过T2?????TT21H1R?T1迭代计算温度，T2=287.75 K CpmhCpms??Cp(T)dT?H2RT2?H1R?2C(T)p?dT?T?T1T?H??1.822?10?8?J?mol?1?T2?ln??T1?

?2.6?5.2??R?P1?TT?????11?S1R???0.675?????0.722?????Pc???Tc??Tc?? ?2.6?5.2??R?P2?TT?????22?S2R???0.675?????0.722?????Pc???Tc??Tc??

?S????TT2Cp(T)T1?P2??1?1dT?R?ln???mol?K??S2R?S1R21.801J?P1? 4-7 解：

T1??473.15P1??2.5?10Tc??305.4???0.0986R??8.314PaP2??0.20?10Pc??4.88?10Cp(T)??9.403?159.83710?6PaPa?626?3?T?46.23410??T

?2.6?5.2??R?P1?TT?????11?S1R???0.675?????0.722?????Pc???Tc??Tc?? ?2.6?5.2??R?P2?TT?????22?S2RT2???0.675?????0.722?????Pc???Tc??Tc??

??经迭代计算（参考101页例题4-3）得到T2=340.71K。 ?1.6?4.2?R?Tc?P1?TT?????11?H1R???0.083?0.139???1.097???0.894????????Pc??Tc??Tc?? ?1.6?4.2?R?Tc?P2?TT?????22?H2RT2???0.083?0.139???1.097???0.894????????Pc??Tc??Tc?? ??????TT21?HCp(T)dT?H2RT2?H1R???8.32725?10?J?mol3?1。

146页第五章

5-1：b 5-2: c 5-4: a 5-5: a

5-1：

解：可逆过程熵产为零，即?Sg??Ssys??Sf??Ssys?5-2：

解：不可逆过程熵产大于零，即?Sg??Ssys??Sf??Ssys?系统熵变可小于零也可大于零。

5-4：

解：不可逆绝热过程熵产大于零，即?Sg??Ssys??Sf??Ssys?0。所以流体熵变大于零。 5-5：

解：不可逆过程熵产大于零，即?Sg??Ssys??Sf??Ssys??5?0??Ssys?0。 T0?5?5。即?0??Ssys?T0T01010?0??Ssys?。 T0T0

5-3：

解：电阻器作为系统，温度维持100℃，即373.15K，属于放热；环境温度298.15K，属于吸热，根据孤立体系的熵变为系统熵变加环境熵变，可计算如下：

50???(20?A)?2?3600?s?1.44?10J?1.44?10J373.15K?828?1.44?10J298.15K?8?9.707?1041JK

5-6： 解：理想气体节流过程即是等焓变化，温度不变，而且过程绝热，所以系统的熵变等于熵产，计算如下：

?8.314?J?mol?1?K?1?ln??0.09807??24.9011J?Kmol ?1.96?所以过程不可逆。

5-7： 解： 页4-7绝热稳流过程?M所以M?h320?kg?s?1m1?m2??Hm1?h1?m2?h20T3???(90?273.15)?K?30?kg?s50?kg?s?1?1?(50?273.15)?KT3?339.15K?Sg339.15?273.15?66?mj?Sj??mi?Siji?T3??T3?m1?Cpms?ln???m2?Cpms?ln???T1??T2?209.33

查表可得?h1376.92?h2

4.19?kJ?kg?1CpmsCpmh376.92?209.3390?50?K?1 ?Sg?mj?Sj??mi?Siji?T3??T3?m1?Cpms?ln???m2?Cpms?ln??T1???T2?0.345kJ??K?1?1

。

不同温度的S值也可以直接用饱和水表查得。计算结果是0.336。 5-12

解：（1）循环的热效率

?Sg20?4.184?ln??339??30?4.184?ln?339?????363??323??sWNWS,Tur?W4?1 ?T??QHH2?H1(2) 水泵功与透平功之比

H2=3562.38 kJ·kg-1，H3=2409.3 kJ·kg-1，H4=162.60 kJ·kg-1，H5=2572.14 kJ·kg-1，

H4?V??p?H1?162.60??14?0.007??103?0.001?176.6?kJ?kg?1

W4?1WS,TurV??p0.001?(14?0.007)?103???0.012 H2?H33562.38?2409.3?T?H2?H3?H1?H4?0.345

H2?H1