(3) 提供1 kw电功的蒸气循环量
m?1000W?1000.08?0.857?g?s?1 N1167
5-15题：
4?15T?0WQLTLC1?T?HQHWT0?TL?T?C?60%????20%QL?T0Q??HT???1???T?T????L??TL??0.6?0.20.555HT0??QL???T0?Qir??ir??c?60%???20%?????H??1?T????TL?H?T?TL???0.6?0.20???21?273.15??6?273?1?118?273.15???.15??21?6???0.12?0.555
194页第六章
6-1： nm1680?10001解：水蒸气的摩尔流量为：M?360018?360025.926mol??s? ??H(T)??n??T?35?2???8.314??3.47?1.45?10?T?0.121?10?T?d???T703?? ?T?S(T?p)??n??8.314??3.47?1.45?10?3?T?0.121?105?T?2???p??TdT?n?8.314?ln?703?3.727??（a）通过内插法求出0.1049MPa时对应的温度，如下
100?t100?95101.325?104.9101.325?84.550.298 t100.964 Wid??H(374.114)?T0??S(374.114?0.1049)3.607?105?J?s?1
Ws??H(374.114)3.046?105?J?s?1 41
?aWsWid0.844 5?1（b）
WidWs?a??H(333)?T0??S(333?0.0147)??H(333)WsWid4.926?10?J?s
3.408?10?J?s5?1 0.692 292.98?376.92?83.946-3
?H?Ssur?Sgh2?h1??HT0?Ssys??Ssur?0.9549?1.1925??83.94???298??0.044?kJ?kg?1?K?1根据热力学第一定律?热损失为Q功损失为WLT0??Sg13.1?kJ?kg?1?H?83.94?kJ?kg?1或Q?1.511?10?J?mol3?1或WL235.8?J?mol?1
6-6：
解：理想气体经一锐孔降压过程为节流过程，?H?0,且Q?0，故WS?0，过程恒温。 则绝热膨胀过程的理想功和损耗功计算如下： WidWL??H?T0??SWidT0??SgT0??S??298?8.314?ln?0.09807????????1.96??7.42?10?J?mol3?13?1 ??298?8.314?ln?0.09807????????1.96??7.42?10?J?mol 6-8： 解：（1）产品是纯氮和纯氧时，
Wid?RT0(y1lny1?y2lny2)?8.314?298.15?(0.21ln0.21?0.79ln0.79)??1.274kJ?mol?1
（2）产品是98% N2和50% O2的空气时，设计计算流程如下：
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W1 98% N2
0.21 O2 分离 纯 O2 混合 0.98 N2 Wid 纯 N2 W2 50% O2
总的功
W?Wid?W1?W2
W1??T0R?0.98ln0.98?0.02ln0.02?W2??T0R?0.5ln0.5?0.5ln0.5???298.15?8.314???0.098?
??298.15?8.314???0.693? ?242.924?J?mol?1?1.718?kJ?mol?1?W??1.274?242.924?10?3?1.718?0.687kJ?mol?1
6-12： 解：
查表得H2H2ONH3O2CH3OHCO2N2?H?f0?285.84?46.190?238.64?393.510S?130.5969.94192.51205.03126.8213.64191.49
H2H?122?O2???H2?O(l)?H?285.84?kJ?mol?1?S69.94?130.59?0.5???205.03?8.314?ln??0.0206105???0.10133??????169.785?J?mol?1?K?1
EXC?H2???H?T0??S285.84?298?169.7851000235.244kJ??mol?1 NH312N2?32H2????NH3EXC(NH3)3?117.61?1?0.335?1?16.63336.535kJ??mol?1
CH3OHEXC(CH3OH)4?117.61?1?1.966?410.54?166.31716.636kJ??mol?1
6-13 解：
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?HQ1?Q20Q1m1?h3?h10??Q2m2?h3?h2??
m1?h3?h1?m2?h3?h2????
kg?s?172000?1m1??kg?s3600
m2??1080003600
kJ?kg?1h1??376.92kJ?kg?1S1??1.1925?K?1h2??209.33kJ?kg?1
S2??0.7038kJ?kg?1?K?1
m1??h3?h1??m2??h3?h2?0
h3276.366kJ??kg?1
由m1?Cpmh??t3?t1??m2?Cpmh??t3?t2?0，可得t366.03?℃ 使用内插法可求得66.03℃时的熵值，
S3?0.89350.9549?0.8935166.03?6570?65S30.906kJ?kg??K?1
（1）利用熵分析法计算损耗功，
WLT0??SgT0??SsysT0???m1??S3?S1??m2??S3?S2???100.178kJ??s?1 （2）利用火用分析法：
h0??104.89S0??0.3674M??m1?m2
EX1???m1??h0?h1??m1?T0??S0?S1?EX2???m2??h0?h2??m2?T0??S0?S2?EX3???M??h0?h3??M?T0??S0?S3?WL??EX1?EX2?EX3WL100.178kJ??s?1
或者
WL3.606?105?kJ?h?1
241页第七章 7-2
解：假设需水m kg，则
1000?9600?m?56%?m?714.3kg
产品酒中含水
?1000?714.3???1?56%??754.3kg
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产品酒中含醇
?1000?714.3??56%?960kg
所以酒的体积
V??miVi?754.3?0.953?960?1.243?1912?103cm3?1.912m3
7-3 解：
V??109.4?16.8?x1?2.64?x12???cm3?mol?1V1PV?x2???dV?dx2???V?x2???d??dxVV1???2PV?x1??d2.64?x?dxV1???12?109.4
V1P2.64?x12?5.28?x1?92.6x11V1PV189.96?cm3?mol?1x10V2PV2109.4?cm3?mol?1?VV?V前109.4?16.8?x1?2.64?x12??x1?V1?x2?V2?2.64?x1??1?x1?
7-4
解：根据吉布斯-杜亥姆公式，恒温恒压时
?xidMi?0
i则有
?xidVi?0，所以
ix1dV1?x1dV1?x1??b?a??2bx1?dx1?x2??b?a??2bx2?dx2?dx1??dx2?xdV?21dV1?x11??x1?x2??b?a??2b?x21?x2??dx1
把x1?1?x2代入上式，得x1dV1?x1dV1???3b?a??2?a?b?x2?dx1?0所以设计的方程不合理。
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