3?1?h?1?u?g??z2q?w?z??3?m)?10??h??(2300?323043J10?kg?q??02kg3600s?m??2.778kg104?kgkgs3600s??2.778s?u1202?502109?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?106J2sw???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?HR?647.3?T??B00??B1???1??Rr?????0??0???????dB00?T?0.344?dB1?dP?HR1??576.771?r??0Tr??r????00????0????
???Pr1HR??B0??B1????2??R?647.3?Tr???1???????dB01?1?T?0.344??dB1?1??dP?r??1Tr??r?HR2??56.91??????1??1??0??
经计算得
36
Cpmh?35.03?J?mol?1?K?1
0.0752体积流速为：V???u2.14?3????3?11??d/2??3?2???0.0132?m?s
摩尔流速为：n?V?V?0.0132?4.015?mol?s?1V mRT/p8.314?593.15/1500000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?m3?mol?1N??4.015mol0.0132m?3?s?1mols3.288?10?3?4.015?m3?mol?1s
w435.90418??N3.15?104W
4-6 解：
二氧化碳T1??303.15R??8.314P1??1.5?106PaP2??0.1013310?6PaTc??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???1.6?4.2?2R?T2???P??T?0.083?0.139???1.097?2??T2??c???Tc??0.894??????Tc????
?T2通过TH?Cp(T)2?1RC?T1迭代计算温度，T2=287.75 K
?pmh?TdTTC??T12pms?H???C?p(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?P?2.6?S2?2R????T2??T2?5.2??P?c?0.675????Tc????0.722????Tc????
?T2C?S?p(T)?P???S2R?S21.801J??mol?1?K?1?TdT?R?ln?2?T1?P1?1R 4-7
解：
T1??473.15R??8.314P1??2.5?106PaP2??0.20?106PaTc??305.4Pc??4.88?106Pa???0.098Cp(T)??9.403?159.83710??3?T?46.23410??6?T2
?R?P?2.6?5.2?S1??T??T1R??P???0.675?11??c?????0.722??Tc????Tc???? ???R?PS2?T?2.6?5.2?2RT2??P???c?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???? 38
?1.6?4.2?R?Tc?P2?TT?????22?H2RT2????0.083?0.139???1.097???0.894???????Pc??Tc??Tc??
??????T2?HCp(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?0??Ssys?。即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
8828
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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?1?1?0.09807?1?8.314?J?mol?K?ln??1.96???24.901KmolJ 所以过程不可逆。
5-7： 页4-7解： 绝热稳流过程?Mm1?m2??H0所以M?h3m1?h1?m2?h220?kg?s?1?1T?273.15)?K?30?kg?s?(50?273.15)?K3???(9050?kg?s?1T3?339.15K339.15?273.15?66?Sg?mj?Sm?C?T3?m?T3?j?j?mi?Si1pms?ln??T1??i?2?Cpms?ln??T2??
查表可得?h1376.92?h2209.33
CpmsC376.92?209.33pmh190?504.19?kJ?kg??K?1 ?Sg?mj?Sj??mm?ln??T3??T??mC?T3?i?Si1?Cpms1?2?pms?ln?ji?T2??
?Sg20?4.184?ln??339??363???30?4.184?ln??339??323??0.345?kJ?K?1?s?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
??WNWS,Tur?W4?1TQ?
HH2?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?1V??p0.001?(14?0.007)?103W?S,TurHH?38?2409.3?0.012 2?33562.?1?H4T?H2?H3?HH?H?0.345
21 40