(3) 提供1 kw电功的蒸气循环量
m?10001000??0.857?g?s?1 WN1167.08
5-15题：
4?15T0THWQHQLWTLT0?TL?C1???TQLQH?C?60%??T??????20%??1????TL????T?T??0.6?0.20.555TH??0L?T0?T0QL??ir??ir??c?60%???20%???1?TQHH???TL?????T?T???0.6?0.2??0L?
21?273.15??6?273.15????1?????0.12?0.555118?273.1521?6????
194页第六章
6-1：
解：水蒸气的摩尔流量为：
nmM?360016801000?18?360025.926mol??s?1
??T??35?2???H(T)??n?8.314??3.47?1.45?10?T?0.12110??T?dT??703??? ??35?28.314?3.47?1.45?10?T?0.121?10?T??p??S(T?p)??n?dT?n?8.314?ln???T3.727???703T?? （a）通过内插法求出0.1049MPa时对应的温度，如下
100?t101.325?104.9t100.964 100?95101.325?84.550.298 Wid??H(374.114)?T0??S(374.1140.1049?)3.607?10?J?s5?1 Ws??H(374.114)3.046?10?J?s5?1 ?aWsWid0.844 5?1（b）
Wid??H(333)?T0??S(333?0.0147)4.926?10?J?s
Ws?a??H(333)WsWid3.408?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??0.044kJ??kg?1?K?1根据热力学第一定?律热损失为Q?H?83.94?kJ?kg?1或Q?1.511?10?J?mol3?1功损失为WLT0??Sg13.1?kJ?kg?1或WL235.8J??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??7.42?103?J?mol?1??????1.96?? ??298?8.314?ln?0.09807??7.42?103?J?mol?1??????1.96??
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：
W1 0.21 O2 0.98 N2 分离 Wid 纯 O2 纯 N2 混合 W2 98% N2
50% O2
总的功
W?Wid?W1?W2
W1??T0R?0.98ln0.98?0.02ln0.02???298.15?8.314???0.098??242.924?J?mol?1
W2??T0R?0.5ln0.5?0.5ln0.5???298.15?8.314???0.693? ?1.718?kJ?mol?1
?W??1.274?242.924?10?3?1.718?0.687kJ?mol
6-12： 解：
查表得?HfS???1H20H2O?285.84NH3?46.19O20CH3OH?238.64CO2?393.51N20130.5969.94192.51205.03126.8213.64191.49
H21H2??O2???H2?O(l)2?H?285.84?kJ?mol?1?SEXCH269.94?130.59?0.5??205.03?8.314ln????H?T0??S285.84?298169.785?1000???0.0206105?????0.10133??235.244kJ??mol?169.785?J?mol?K?1?1?1
??
NH313N2?H2????NH322EXC(NH3)3?117.61?1?0.335?1?16.63336.535?kJ?mol?1
CH3OHEXC(CH3OH)4?117.61?1?1.966?410.54?166.31716.636kJ??mol?1
6-13 解：
?HQ1?Q20Q1m1?h3?h10??Q2m2?h3?h2??
m1?h3?h1?m2?h3?h2???? 72000?1m1??kg?s3600h1??376.92kJ?kg
108000?1m2??kg?s3600
?1?1
?1S1??1.1925kJ?kg?K?1
h2??209.3kJ3?kg?1
S2??0.7038kJ?kg?K?1
m1?h3?h1?m2?h3?h2????0 h3276.366?kJ?kg?1 m?C?t?t?m2?Cpmh?t3?t2由1pmh31????0t，可得366.03?℃
使用内插法可求得66.03℃时的熵值，
S3?0.893566.03?650.9549?0.893570?65S30.906kJ?kg?1?K?1
（1）利用熵分析法计算损耗功，
WLT0??SgT0??SsysT0???m1?S3?S1?m2?S3?S2??????100.178?kJ?s?1
（2）利用火用分析法：
h0??104.89S0??0.3674M??m1?m2
EX1???m1?h0?h1?m1?T0?S0?S1EX3???M?h0?h3?M?T0?S0?S3WL100.178kJ??s?1????EX2???m2?h0?h2?m2?T0?S0?S2WL??EX1?EX2?EX3????????
?1或者
WL3.606?10?kJ?h5
241页第七章 7-2
解：假设需水m kg，则
1000?96%?56%?m?714.3kg
1000?m产品酒中含水
?1000?714.3???1?56%??754.3kg
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：
VV1P?109.4?16.8?x?2.64?x2??cm3?mol?111??V?x2???d?dx2?V??V?x2???d?dx1?V??V2PV?x1???d?dx1?V??2.64?x1?109.42
V1Px1x1?V2.64?x1?5.28?x1?92.61V1P0V2PV?V前V1V289.96cm??mol109.4cm??mol33?1?122109.4?16.8?x1?2.64?x1?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
?xdMiii?0
i则有
?xdVii?0，所以
x1dV1?x1dV1?x1??b?a??2bx1?dx1?x2??b?a??2bx2?dx2?dx1??dx22?dx1?x1dV1?x1dV1??x1?x2??b?a??2b?x12?x2??
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。