Cid?26.75?p?Hp??Sidp42.2?58?31T0?? 0(1?4.25T1?62)id??397.15315.15id-1CpdT?3279.6 kJ?kmol id?397.15CpT315.15dT?Rln15.7928.053?3.635 kJ?kmol?K-1?1
过程③ 用普压法：
H2RRTc=(H2)RTcR0+?(H2)RTcR'??1.72?0.225?0.20??1.78
S2RR?(S2)RR0??(S2)RR'??0.95?0.225?0.20??1.02
R-1?H2??1.78?8.314?304.2??4501.8 kJ?kmol
R-1-1S2??1.02?8.314??8.48kJ?kmol?K
?H= (?H1)?(H2)??HRidRRRid?6424?3279.6?4501.8?5202 kJ?kmol-1-1-1
?S= (?S1)??Sp?(S2)?13.14?3.635?8.48?8.3 kJ?kmol?K
-1?U??H??pV?5202?115.792?149.9?8.053?219.3?4601 kJ?kmol
9 解：(1)常压沸点时的蒸发热可用下式计算
?HnTnvap?2.17(lnpc?1)?4.184(0.930?Trn)
?Tn?273.15?(?0.5)?272.65 K由表中数据知，正丁烷的临界压力，临界温度，如下：
Pc?3.786MPa?38atmTc?425.13K??Hnvap
?2.17(ln37.86?1）5.7236-1?4.184? J?mol272.650.28880.93?425.13-1?Hn?19.8205 kJ?mol（2）T?137.78?27.3.15?410.93K 蒸发热可按watson经验式计算：
?H1r?H2rT2r?T1r?vapvap?(1?T1r1?T2r0.38)
272.65425.13410.93425.13?0.6413?0.96661?0.96661?0.64130.38?H410.93K?19.8205?(vap)?8.0415 kJ?kmol-1（3）由Clapeyron方程知，
11
dpvapdT?Hvap??Hvap
vapT?vdpdT?T?v由数据拟合lnpvap?1得到lnpvap?21.752?2808.4/T
T1pvapdpvapdTvap?2808.4T2?dPvapdT?2808.4PT62vap
?3.009?10?8.988 kJ?mol6-1??H?411?(10.301?2.764)?10?8.98?88.0415?100%?8.9882808.44112相对误差：
10.5 3解： 1）查饱和蒸气表得：70℃时
Hl?292.98 kJ?kg-1-1?1?1
Hg?2626.8 kJ?kg-1Sl?0.9549 kJ?kg?KSg?7.7553 kJ?kg?K-1?H?0.95Hg??1?0.95?Hl?0.95?2626.8?(1?0.95)?292.68?2510.1 kJ?kg-1-1
-1S?0.95Sg?l?1?0.95?Sl?0.95?7.7553?(1?0.95)?0.9549?7.4153 kJ?kg?K2) 由表可知4.0MPa时，Sg?6.0701kJ?kg?K-1-1?S?6.4kJ?kg?K-1-1
故水蒸气处于过热状态。
由表可知4.0MPa ，280℃时，H=2901.8 kJ·kg-1, S=6.2568 kJ·kg-1·K-1 320℃时，H=30154.4 kJ·kg-1, S=6.4553 kJ·kg-1·K-1
由内插法求得，S=6.4 kJ·kg-1·K-1时，对应的温度T=308.9℃，H=2983.75 kJ·kg-1 3）查饱和蒸气表得，4.0MPa时，
Hl?1087.3 kJ?kg-1-1-1
Hg?2801.4 kJ?kg-1Sl?2.7964 kJ?kg?KSg?6.070 kJ?kg?K-1-1由于Sl处于Sg之间，故水蒸气为湿蒸气，且满足
S?xSg??1?x?Sl 即
5.8?6.07x0?1??x1??，得 2.7964x?0.9175
故 H?xH??1?x?H?0.9175?2801.4??1?0.9175??1087.3?2660 kJ?kg-1
gl对应的温度为T=250.4℃
11解：查氨的饱和蒸气表，通过内插法得，0.304 MPa时，Sl=8.8932 kJ·kg-1·K-1
绝热可逆膨胀到0.1013MPa后，S不变，通过内插得
-1-1Sl?3.5855 kJ?kg?K
12
Sg?9.2751 kJ?kg?K-1-1
而S位于Sl与Sg之间，设干度为x, 则
S?xSg??1?x?Sl
即 8.893?2解得：x?0.9329
9.27x?511???x? 58553.12 解：查表知0.1985MPa（饱和）的饱和蒸汽压H=2706.3 kJ·kg-1
因为节流膨胀是等焓过程，过1.906 MPa时湿蒸气的H=2706.3 kJ·kg-1 查表知，1.906 MPa时，H?2798.1 kJ?kg-1 ，Hl?896.8 kJ?kg-1
g干度为x， 则
H?xHg??1?x?Hl
即
2706.?3279x8?.11??x? 96.88解得 x?0.951 73313解： 1）v?0.04166 m3?kg?1 vf?1.2088?10?3m3?kg?1 vg?71.5?8?10?m?1 gkv?vf(1?x)?vg?xx?v?vfvg?vf?0.04166?1.2088?10171.58?1.2088?10?3
?0.5748?3 2)
Hf?990.12 kJ?kg?11， Hg?2804.0 ? k?J kg?1H?Hf(1?x)?Hg?x?990.12?(1?0.5748)?2804.0?0.5748?2032.7 kJ?kg
3)
Sf?2.6099 kJ?kg?K-1-1 Sg?6.2146?kJ-1k?g- 1k-1-1S?Sf?(1?x)?Sg?x?2.6099?(1?0.5748)?6.2146?0.5748?4.682 kJ?kg?k
14 解： A、B两物流进行绝热换热，终态为c 状态 A股湿蒸汽
pA?0.5MPa?1gs 8 GA?1 k? xA?0.9
B股过热蒸汽
a TB?473.1pA?0.5MPK5
?0.5MPa?1
① 查饱和水蒸气表知：pA时，
Hg,A?2747.5 kJ?kgHl,A?640.12 kJ?kg?1
13
HA?xA?Hg,A?(1?xA)Hl,A ?0.98?2747.5?(1?0.98)?640.12
?2705.35 kJ?kg?1
查过热蒸汽表知：pA?0.5MPa?1 TB?473.1K5 HB?2855.1 ? kJ kg0.5 Mpa的饱和蒸汽: ② 绝热混合过程 QA?QB
Hc?2747.5KJ/Kg
mB(Hc?HB)?mA(HA?Hc)
∴mB?mA(HA?HC) Hc?HB
∴B股蒸汽的流量mB
15 解：（1）查423.15K，饱和水蒸汽
hl?632.15 kJ?kg-1-1-1?1?(2705.35?2747.5)2747.5?2855.1?0.392Kg/s
Sl?1.8416 kJ?kg?Khg?2745.4 kJ?kg-1Sg?6.8358 kJ?kg?K-1-1 V?0.3924 m3?kg?1
查423.15K, 0.1MPa,过热蒸汽，
SP?0.475MPa
h'?2776.3 kJ?kg-1-1-1
S'?7.6137 kJ?kg?Kv'?1.936 m?kg3-1423.15K, 0.15MPa, 过热蒸汽
h''?2760.75 kJ?kg-1-1-1
S''?7.4164 kJ?kg?Kv''?1.221 m?kg3-1（2）423.15K,0.14MPa, 过热蒸汽焓熵由线性内插求出：
h?h'0.14?0.1h??h''?h'0.15?0.1(0.14?0.1)?2776.3-1
h''?h'0.15?0.1?2763.86kJ?kgS?S''?S'0.15?0.1(0.14?0.1)?7.6137?-1-17.4164?7.61370.15?0.1(0.14?0.1)?7.6137
?7.4559 kJ?kg?K 14
(3) ?等T可逆过程，
QR?m?TdS?mT?S1?S2?s2s1
?8?423.15(Sg?S)?8?423.15(6.8358?7.4559)??2099.03 kJ体系放出热量为2099.03Kj
由热力学第一定理知， ?u?q?w
?u?h?Pv?u1?h1?P1v1?2763.86?0.14?10?1.221?106-16?3-1-1?2592.92 kJ?kg?3
u2?h2?P2v2?2745.4?0.476?10?0.3924?10 ?u?u2?u1??34.3024 kJ?kg?2558.62 kJ?kg w?2099.03?8?(?34.3024)?1824.61 kJ体系获得功量为1824.061kJ
3-16 解： ①
4MPa 饱和水蒸汽查表知：t=250.33℃
状态1过热度为150℃. 即 t=250.33+150=400.33℃ 对应此状态点的焓、熵分别为h1、S1
查过热水蒸汽表知：400.33℃下
2.5MPa时 h’=3240.7kg·kg-1 S’=7.0178 kJ·kg-1·k-1 V’=0.12004m·kg
5MPa 时 h”=3198.3 kg·kg-1 S”=6.65 kJ·kg-1·k-1 V”=0.05779 m3·kg-1
内插求出4MPa ，400.33℃下 对应的h 和s分别为：
h?3240.74?2.5?3198.3?3240.75?2.53
-1
?1h 1?3198.3?3240.75?2.5(4?2.5)?3240.7?3215.26 kJ?kg
s?7.01784?2.5s1?v??6.6508?7.01785?2.5
?16.6508?7.01785?2.50.05779?0.120045?2.5(4?2.5)?7.0178?6.7976 kJ?kg3?K?1
(5?2.5)?0.12004?0.08269 m?kg?1
② 经过绝热可逆过程s2?s1?6.7976 kJ?kg?1?K?1， P2=50KPa 对应该压力下其饱和热力学参数查表知：
S2g=7.5947 kJ·kg-1·K-1 H2g=2646.0 kJ·kg-1
S2l=1.0912 kJ·kg
3
-1-1-1·K
H2l=340.56 kJ·kg
3·kg
-1
-1
v2g=3.24 m·kg v2l=0.0010301 m∵ S2＜S2g
∴ 终态为湿蒸汽状态
S2?Sl,2?(1?x)?Sg,2?x
15
H2?Hl,2?(1?x)?Hg,2?x
即有：x?S2?Sl,2?6.7976?1.0912?5.7064?0.8774
Sg,2?Sl,27.5947?1.09126.5035v2?0.8774?3.24?(1?0.8774)?0.0010301?2.8429 m?kg3?1
H2?0.8774?2646.0?(1?0.8774)?340.56?2363.35 kJ?kg?1③ 作图
T 4MPa 1 50kPa 2 S ④ 若过程在封闭体系进行。由热力学第一定律 ?u?q?w
∵q=0
w??u??(h?pv)??h??(pv)∴
?(2363.35?3215.26)?(50?10?2.8429?4?10?0.08269)?10 ??851.91?188.615 ??663.30 kJ?kg?136?3
即：体系从外界获功663.30 kJ·kg
-1
16