3?1?h?1?u?g??z2q?w?z??3?m?h??(2300?3230)?10?43J10?kg?q??02kg3600s?m??2.778kg104?kgkg2s3600s??2.778s?u1202?50109?Jkg?u??109?J?h??2.583?106Jkgs1?m?u2?1.65?104J2sg??z?m?81.729Js?h?1?m?u2?g??z?m??2.567?106J?2sw??2.56710?6?J?s??2.56710?6?Wwc??2.583?2.5672.567?100%wc?0.623% 4-2
方法一：
?h?12?u2?g??zq?wu1??3R??8.314?h?12?u2?wu0.07522???u1u2?0.270.252
T2??353.15T1??593.15?HCpmh??T2?T1??HR2?HR1
???Pr0?HR1??R?647.3?Tr?????B00????B10?????0???????dB00?T?0.344?dB10?dPr?HR1??576.771??r??T?????00????r0????
???Pr1HR2??R?647.3?Tr??????B01????dB0?1??B1?0.344????dB11?1??????dPr????1TTr???HR2??56.91?r????1???1??0??
经计算得
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Cpmh?35.03?J?mol?1?K?1
0.0752体积流速为：V???u214?3????3?11??d/2??3.?2???0.0132?m?s
摩尔流速为：n?V?Vp?0.01328.314?593.15/1500000?4.015?mol?s?1V mRT/根据热力学第一定律，绝热时Ws = -△H，所以
?Hn?Cpmh??T2?T1??n??HR2?HR1? Ws?4.015????8.408?103?(?56.91?576.771)??3.167?104?W
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???u22?u12?2?2645.6?3081.5?4.464?10?3?435.904kJ?kg?1
按理想气体体积计算的体积VR?T8.314593.15?3P1500000?3.288?10?3m?mol?1N??4.015mol0.0132?m3?s?1s?mol3.288?10?3?m3?mol?14.015s
w435.904?18?N3.15?104W
4-6 解：
二氧化碳T1??303.15R??8.314P1??1.5?106PaP2??0.10133?106PaTc??304.2Pc??7.357?106Pa???0.225Cp(T)??45.369?8.688?10?3?T?9.619?105?T?2
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R?THc?P1?P???T??1.6T?4.2?c?0.083?0.139???1.097?1?1?1R??????Tc??0.894??????Tc???? ??R?THc?P2??T?1.6?4.2?2RT2??P???0.083?0.139???1.097?2??T2??c??T?0.894??c??????Tc????
通过TH?T2CR?p(T)12?C?T1迭代计算温度，T2=287.75 K ??TdTpmh?H?TCT12pms????Cp(T)dT?H2R?T2??H1R1.822?10?8?J?mol?1ln??T2??T?T1?1 ?R?PS1??T??2.6?T??5.2?1R??P??c?0.675???1???0.722??Tc???1??Tc???? ?R?P2??T?2.6?5.2?S2R???2??T2??P?c?0.675????Tc????0.722????Tc???? ?T2?CT)?Sp(dT?R?ln??P2???S2R??1?21.801J??mol?1?K?TST1?P1?1R 4-7
解：
T1??473.15R??8.314P1??2.5?106PaP2??0.2010?6PaTc??305.4Pc??4.8810?6Pa???0.098Cp(T)??9.403?159.837?10?3?T?46.23410??6?T2
?R?P?2.6?5.2?S1??T?T1R??P??1?1??c?0.675??????0.722??Tc????Tc???? ???R?PS2??2.6?5.2?2RT2??P???Tc?0.675?2??T2?????Tc????0.722????Tc????
经迭代计算（参考101页例题4-3）得到T2=340.71K。
R?THc?P1?1R????T1??1.6?T1??4.2?P?c?0.083?0.139???1.097????0.894????Tc?????Tc????
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?1.6?4.2?R?Tc?P2?TT?????22?H2RT2????0.083?0.139???1.097???0.894???????Pc??Tc??Tc??
???HT?2?Cp(T)dT?H2RT2?H1R?????8.32725?10?J?mol3?1T1。
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?10J1.44?10J41??9.707?10J373.15K?298.15K?K
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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：
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