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化工热力学第三版习题答案
《化工热力学》（第三版习题参考答案
58页第2章
2-1求温度673.15K、压力4.053MPa的甲烷气体摩尔体积。
解:
（a理想气体方程
1
33
6
10
381.110
053.415.673314.8--⋅⋅⋅=⋅⋅=
=
⇒=mol
mpRTVRTpV
（b用R-K方程
①查表求cT、cp;②计算a、b;③利用迭代法计算V。
（
（（1
3
3
11
3
3
0110
3896.110381.1--+--+⋅⋅⋅=⋅
⋅⋅⋅⋅⋅=+⋅⋅--
+=
+⋅⋅-
-=
mol
mVmol
mVbVVTbVabpRTVbVVTabVRTpiiiii
（c用PR方程
步骤同（b,计算结果:
1
331103893.1--+⋅⋅⋅=mol
mVi。
（d利用维里截断式
2
.41
6
.10
1
172.0139.0422.0083.0111r
r
r
rr
rr
rc
cTBTB
TpBTpBTpRTBpRTBpRT
pVZ-
=-=⋅
⋅+⋅
+=⋅
+
=+
==ω
查表可计算rp、rT、0B、1B和Z由1
33
10
391.1--⋅⋅⋅==
⇒=
mol
mp
ZRTVRT
pVZ
2-4V=1.213m3,乙醇45.40kg,T=500.15K,求压力。
解:
（a理想气体状态方程
MPaV
RTM
mV
nRTp⋅=⋅⋅
=⋅==
383.3213
.115
.500314.846
40.45
（b用R-K方程
a0.42748R
2
⋅TC
2.5
⋅
PC
28.039
b0.08664R⋅TC
⋅
PC
0.058（MPa
p
kmolm
n
V
V
bVV
a
b
V
RT
p
m
m
m
m
⋅
=
⋅
⋅
=
=
=
+
⋅
⋅
-
-
=
-759
.2
229
.1
46
/
40
.
45
213
.1
13
（c用
SRK方程计算
（d用
PR方程计算
（e用三参数普遍化关联
（MPa
B
VRTpBBB
pT
RB
BBBBRTBpB
VRTpRT
BpRT
pVZkmolmnVVkmolMmnmc
c
ccmm
m⋅=-=
∴-=∴=-=-=⋅⋅⋅⋅+=⇒⋅+=-=
⇒+
==⋅⋅==
=⋅===
--779.2267.0635
.0,057.0,361.01229.1987.0213.1987.046
4.451
1
1
1
1
3
ωωω
2-7计算T=523.15K,p=2MPa的水蒸气的Z和V解:
（a用维里截断式2
2
1pV
CRTpV
BRTp
RTVV
CVBRT
pVZ+
+
=
⇒++
≈=
采用迭代法计算V=2.006
之后求得Z=0.923
（d利用维里截断式
2
.41
6
.10
1
172.0139.0422.0083.0111r
r
r
rr
rr
rc
cTBTB
TpBTpBTpRTBpRTBpRT
pVZ-
=-=⋅
⋅+⋅
+=⋅
+
=+
==ω
查表可计算rp、rT、0B、1
B可得到Z=0.932;
由1
33
10
025.2--⋅⋅⋅==
⇒=
mol
mp
ZRTVRT
pVZ
（c水蒸气表9223
.015
.523314.800592.2200000592.21811144.011144.01
31
3
=⋅⋅=
=
∴⋅⋅=⋅≈⋅⋅=--RT
VpZkmol
mkgmV
92页第三章3-4
丁二烯
1-3
-R8.314
:
=T1127273.15+:
=T2227273.15+:
=P12.53106
⋅:
=PaP212.67106
⋅:
=Pa
Tc425
:
=Pc4.32610
6
⋅:
=Pa
ω0.181
:
=CpT（22.738222.79610
3
-⋅T⋅+73.87910
6
-⋅T
2
⋅-:
=
利用三参数压缩因子计算方法,查图表,得到压缩因子:
Tr1T1Tc
:
=
Tr2T2Tc
:
=
Tr10.942
=Tr21.177
=Pr1P1Pc
:
=
Pr2P2Pc
:
=
Pr10.585=Pr22.929
=Z10.677
:
=Z20.535
:
=∆VZ2R⋅T2⋅P2
Z1R⋅T1⋅P1
-:
=∆V7.146-10
4
-⨯=m3mol
1
-⋅
H2RRTc⋅P2⋅Pc0.0830.139ω⋅+1.097T2Tc
⎛
⎝⎫
⎪⎭
1.6
-⋅-0.894ω⋅T2Tc
⎛
⎝
⎫
⎪⎭
4.2
-⋅-⎡⎢
⎢⎣⎤⎥⎥⎦
⋅:
=H2R8.475-10
3
⨯=H1RRTc⋅P1⋅Pc0.0830.139ω⋅+1.097T1Tc⎛⎝⎫⎪⎭1.6-⋅-0.894ω⋅T1T
c⎛⎝⎫⎪⎭4.2-⋅-⎡⎢⎢⎣⎤
⎥
⎥⎦
⋅:
=H1R2.704-10
3
⨯=∆HT
1
T2
TCpT（⎛
⎜
⎜⎠dH2R+H1R
-:
=∆H5.02810
3
⨯=Jmol
1
-⋅
S2RR-P2⋅Pc0.675T2Tc⎛⎝⎫⎪⎭2.6-⋅ω0.722⋅T2T
c⎛⎝⎫⎪⎭5.2-⋅+⎡⎢⎢⎣⎤
⎥
⎥⎦
⋅:
=S2R12.128
-=S1RR-P1⋅Pc0.675T1Tc⎛⎝⎫⎪⎭2.6-⋅ω0.722⋅T1T
c⎛⎝⎫⎪⎭5.2-⋅+⎡⎢⎢⎣⎤
⎥
⎥⎦
⋅:
=S1R4.708
-=∆ST
1
T2
TCpT（T
⎛
⎜
⎜⎠dRlnP2P1
⎛
⎝
⎫
⎪⎭
⋅-S2R+S1R-:
=∆S3.212=Jmol
1
-⋅K
1
-⋅
3-7:
解:
（1
1
3
3
3
3
261.510
261.5381200010
551.110
095.2SV,,TVV121
-----⋅⋅⋅-=⋅⋅⋅
⨯-=-⨯⨯⨯⨯-=⋅-=∆⇒⋅-=⎪⎪⎭⎫
⎝⎛∂∂⎪⎭⎫
⎝⎛∂∂-=⎪
⎪⎭⎫⎝⎛∂∂⎪⎭⎫⎝⎛∂∂=
⎰
K
kg
JK
kgkPamVdppSTVpSppTp
Tpβββ
注意:
JkPam⋅=⋅3
3
10
（1
6
3
091.110
619.110
551.1261.5270--⋅⋅=⨯⨯⨯+-⨯=∆⋅+∆⋅=∆kg
kJpVSTH
或者
（（（1
5
6
3
3
6.109010
81.310210
551.127010095.2112
1
---⋅=⋅-⋅⋅⋅⋅⋅⋅-=-=
∆⎰kg
JdpVTHL
pp
β
3-9
解:
乙腈的Antonie方程为
（kPactps
85
.241/24.32717258.14ln+︒-
=
（160℃时,乙腈的蒸气压kPa
ppss
⋅=∴=+-
=813.48888
.385
.2416024.32717258.14ln
（2乙腈的标准沸点
c
tct︒⋅=∴=+︒-=375.81605
.485
.241/24.32717258.14100ln
（320℃、40℃和标准沸点时的汽化焓
（（（（（mol
kJcHmolkJcHmolkJcHtT
HRT
HtRT
HdTpds
/72.32375.81;/57.3340;/09.342085.241314.824.327185.24124
.3271ln2
2
2
2
2
=︒∆=︒∆=︒∆+⨯⨯=
∆∴∆=
+⇒
∆=
117页第四章4-1
31
-∆h12∆u
2
+
g∆z⋅+qw-∆z3m⋅:
=∆h23003230-（103
⋅Jkg⋅
104
kg
⋅3600s
⋅⋅
:
=q0
:
=m2.778
kgs
:
=104
kg⋅3600s
⋅2.778
kgs
=∆u120
2
502
-109Jkg
⋅
∆u109Jkg
⋅
:
=∆h2.583-10
6⨯J
s
=12
m∆u
2
⋅1.6510
4⨯J
s
=g∆z⋅m⋅81.729
Js
=∆h12
m∆u
2
⋅+
g∆z⋅m⋅-2.567-10
6⨯J
s
=w
2.567106⋅Js⋅⎛⎝
⎫⎪
⎭2.567106
⋅W
⋅wc2.5832.567
-2.567
100⋅%
:
=
wc0.623%
=
4-2
方法一:
∆h12∆u
2
+
g∆z⋅+qw
-u13
:
=R8.314
:
=∆h12∆u
2
+w
-u20.0752
0.25
2
u1⋅:
=
u20.27
=
T2353.15:
=T1593.15
:
=∆H
CpmhT2T1-（
⋅HR2+HR1
-
HR1R647.3⋅Tr0⋅0
Pr0
PrdB00
B0
Tr0-
⎛⎝
⎫⎪⎪⎭0.344dB10
B1
Tr0-⎛⎝
⎫⎪⎪⎭⋅+⎡⎢⎢⎣⎤⎥⎥⎦
⎛⎜
⎜
⎜⎠d-
⎡⎢⎢⎢⎢⎣⎤⎥⎥⎥⎥⎦
⋅:
=HR1576.771
-=
HR2R647.3⋅Tr1⋅0
Pr
1
PrdB01
B0
1Tr1-
⎛⎝
⎫⎪⎪⎭0.344dB11
B1
1
Tr1-⎛⎝
⎫⎪⎪⎭⋅+⎡⎢⎢⎣⎤⎥⎥⎦
⎛⎜
⎜
⎜⎠d-
⎡⎢⎢⎢⎢⎣⎤
⎥⎥⎥⎥⎦
⋅:
=HR256.91
-=
经计算得
1
1
03.35--⋅⋅⋅=K
mol
JCpmh
体积流速为:
（
1
32
2
10132.02075.0314.32/-⋅⋅=⎪⎭
⎫⎝⎛⋅⋅=⋅⋅=smduVπ
摩尔流速为:
1
015.41500000
/15.593314.80132
.0/-⋅⋅=⋅===s
molp
RTVVVnm
根据热力学第一定律,绝热时Ws=-△H,所以
∆H
nCpmh⋅T2T1-⋅nHR2HR1
-⋅+
Ws
4.015-8.408-10⋅56.91-576.771+（+⎣⎦
⋅3.16710⨯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=2645.6kJ·kg-1
;1500kPa、320℃的水蒸汽在过热蒸汽区,其焓值h1=3081.5kJ·kg-1;
w
-∆h12u22u12-⎛⎝
⎫
⎭⋅-
2645.63081.5-4.46410
3
-⋅-435.904-kJkg
1
-⋅
按理想气体体积计算的体积VRT⋅P
8.314593.15⋅1500000
3.288103
-⨯=m3
mol
1
-⋅N4.015
mols
:
=0.0132m3⋅s1
-⋅3.28810
3
-⨯m3
⋅mol1
-⋅4.015
mols
=
w435.90418⋅N⋅3.1510
4
⨯W
4-6解:
二氧化碳
T1303.15:
=R8.314
:
=P11.5106
⋅:
=Pa
P20.10133106
⋅:
=Pa
Tc304.2
:
=Pc7.35710
6
⋅:
=Pa
ω0.225
:
=CpT（45.3698.68810
3
-⋅T⋅+9.619105⋅T
2
-⋅-:
=
H2RT2（
RTc⋅P2⋅Pc0.0830.139ω⋅+1.097T2Tc⎛⎝⎫⎪⎭⋅-0.894ω⋅T2T
c⎛⎝⎫⎪⎭⋅-⎢⎢⎣⎥
⎥⎦⋅:
=
通过112TCHTpmh
R+=
迭代计算温度,T2=287.75K
T
1
T2
T
CpT（T
⎛
⎜⎜⎠dlnT2T1
⎛⎝
⎫⎭
:
=
∆H
T
1
2
TCpT（⎛⎜⎜⎠dH2RT2（
+H1R
-1.82210
8
-⨯J⋅mol
1
-⋅
4-7
解:
T1473.15:
=R8.314
:
=P1
2.5106
⋅:
=Pa
P20.20106
⋅:
=Pa
Tc305.4
:
=Pc4.8810
6
⋅:
=Pa
ω0.098
:
=CpT（9.403159.83710
3
-⋅T⋅+46.23410
6
-⋅T
2
⋅-:
=
S2RT2（
R-P2⋅Pc
0.675
T2Tc⎛⎝⎫⎪⎭⋅ω0.722⋅T2Tc⎛⎝⎫
⎪
⎭
⋅+⎢
⎢⎣
⎥
⎥⎦⋅:
=
经迭代计算（参考101页例题4-3得到T2=340.71K。
H2RT2（
RTc⋅P2⋅Pc0.0830.139ω⋅+1.097T2Tc
⎛
⎝⎫⎪⎭
⋅-0.894ω⋅T2Tc⎛⎝⎫
⎪
⎭
⋅-⎢
⎢⎣⎥
⎥⎦⋅:
=
∆H
T
1
2
TCpT（⎛
⎜⎜⎠dH2RT2（
+H1R
-8.32725-103
⨯J⋅mol
1
-⋅。
146页第五章
5-1:
b5-2:
c5-4:
a5-5:
a
5-1:
解:
可逆过程熵产为零,即0050
<∆⇒=--∆=∆-∆=∆syssysfsysgSTSSSS。
5-2:
解:
不可逆过程熵产大于零,即0
505TSTSSSSsyssysfsysg->
∆⇒>--∆=∆-∆=∆。
即
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0>∆=∆-∆=∆sysfsysgSSSS。
所以流体熵变大于零。
5-5:
解:
不可逆过程熵产大于零,即0
10010TSTSSSSsyssysfsysg>
∆⇒>-∆=∆-∆=∆。
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:
50Ω⋅20A⋅（2
⋅2⋅3600⋅s⋅1.44108
⨯J
=1.44-108
⨯J373.15K
⋅1.44108
⨯J298.15K
⋅+
9.70710
4⨯1
K
J
=
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:
所以过程不可逆。
5-7:
解:
页4-7
绝热稳流过程M
m1m2+,∆H0
所以
Mh3⋅m1h1⋅m2h2
⋅+T320kg⋅s
1
-⋅90273.15+（⋅K⋅30kg⋅s
1
-⋅50273.15+（⋅K
⋅+50kg⋅s
1
-⋅:
=
T3339.15K
=339.15273.15-66
=∆Sg
j
mjSj⋅∑i
miSi
⋅∑-m1Cpms⋅lnT3T1
⎛
⎝
⎫
⎪⎭⋅m2Cpms⋅lnT3T2
⎛
⎝
⎫⎪
⎭
⋅+
Cpms
Cpmh
9050
-4.19kJ⋅kg
1
-⋅K
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
1
21
4,HHWWQWTurSH
NT-+==
-η
（2水泵功与透平功之比
H2=3562.38kJ·kg-1,H3=2409.3kJ·kg-1,H4=162.60kJ·kg-1,H5=2572.14kJ·kg-1,
（1
3
14
6.176001.010007.01460.162-⋅⋅=⋅⋅-+==∆⋅-kg
kJHpVH
012.03
.240938.356210
007.014（001.03
3
2,14=-⋅-⋅=
-∆⋅=
-HHpVTur
S
345.01
24
132=--+-=
HHHHHHTη
（3提供1kw电功的蒸气循环量
1
857.008
.116710001000-⋅⋅==
=
s
gWmN
5-15题:
415
-ηC
1T0TH
-
WQH
ξ
QLW
TLT0TL
-ηTηC60⋅%
ξ
−ξ20⋅%
QLQH
ηTξ
−⋅1T0TH-
⎛⎝
⎫⎪⎭TLT0TL-⎛
⎝⎫
⎪⎭
⋅0.6⋅0.2⋅0.555
555
.012.062115.273615.27311815.2732112.06.01%20%6000=⋅⎪⎭
⎫⎝⎛-+⎪⎭⎫⎝⎛
++-=⋅⋅⎪⎪⎭
⎫
⎝⎛-⎪⎪⎭⎫⎝⎛-=⋅⋅⋅=⋅=LLHcir
irH
LTTTTTQQξηξη
194页第六章
∆STp,（n703
T8.3143.471.4510
3
-⋅T⋅+0.121105⋅T
2
-⋅+（
⋅T
⎛⎜
⎜⎠
d⋅
n8.314⋅lnp
3.727
⎛
⎝⎫⎪
⎭
⋅-:
=
Wid∆H374.114（-T0∆S374.1140.1049,（⋅+3.60710⨯J⋅s
⋅
（b
Wid∆H333（-T0∆S3330.0147,（⋅+4.92610⨯J⋅s
⋅
Ws
∆H333（-3.40810⨯J⋅s
⋅
6-3
∆H
h2h1-292.98376.92-83.94
-∆Ssur
∆H-T0∆Sg
∆Ssys∆Ssur
+0.95491.1925-83.94-298-⎛
⎝⎫⎪
⎭0.044kJ⋅kg
1
-⋅K
1
-⋅根据热力学第一定律
热损失为
Q∆H83.94-kJ⋅kg1
-⋅或Q1.511-103
⨯J⋅mol
1
-⋅功损失为
WL
T0∆Sg
⋅13.1kJ⋅kg
1
-⋅或
WL
235.8J⋅mol
1
-⋅
6-6:
解:
理想气体经一锐孔降压过程为节流过程,0=∆H,且0=Q,故0=SW,过程恒温。
WL
Wid
T0∆Sg
⋅298-8.314⋅ln1.96⎝⎪⎭⋅⎝⎪⎭
7.42103
⨯J⋅mol
1
-⋅
6-12:
解:
191.49
213.64126.8205.03192.5169.94130.59SΘ0
393.51
-238.64
-0
46.19
-285.84
-0
∆Hf
Θ
N2CO2CH3OHO2NH3H2OH2查表得
H2H212
O2⋅+
H2--O⋅l（
-∆H285.84-kJ⋅mol
1
-⋅∆S
69.94130.59-0.5205.038.314ln0.02061050.10133⎛
⎝⎫⎪⎭⋅-⎛⎝⎫
⎪
⎭
⋅-169.785-J⋅mol
1
-⋅K
1
-⋅
EXCH2
（
∆H-T0∆S
⋅+285.841000
-
235.244kJ⋅mol
1
-⋅
NH312
N232
H2+NH3
----EXCNH3（
3117.61⋅10.335⋅+116.63⋅-336.535kJ⋅mol
1
-⋅
CH3OH
EXCCH3OH（
4117.61⋅11.966⋅+410.54+166.31
-716.636kJ⋅mol
1
-⋅
6-13解:
∆H
Q1Q2
+0
Q1
m1h3h1
-（
⋅Q2m2h3h2
-（
⋅
131232
m1720003600
:
=
kgs
1
-⋅
m21080003600
:
=
kgs
1
-⋅
h1376.92:
=kJkg
1
-⋅
S11.1925:
=kJkg
1
-⋅K
1
-⋅
h2209.33:
=kJkg
1
-⋅
S20.7038:
=kJkg
1
-⋅K
1
-⋅
131232
h3
276.366kJ⋅kg
⋅
由
1pmh31
2pmh32
使用内插法可求得66.03℃时的熵值,
S30.8935-66.0365
-0.95490.8935
-7065
-S3
0.906
kJkg
1
-⋅K
1
-⋅
（1利用熵分析法计算损耗功,
WL
T0∆Sg
⋅T0∆Ssys
⋅T0m1S3S1-（⋅m2S3S2-（
⋅+⎡⎣⎤
⎦
⋅100.178kJ⋅s
⋅
（2利用火用分析法:
h0104.89
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