化工基础课后习题答案讲解（武汉大学版）南京廖华答案网

化工基础课后习题答案讲解（武汉大学版）下载本文

文章发布时间 : 2024/11/19 13:16:58星期二

化工

第二章．P69 2.解：

?de?4?44?(d1?d2)?70 ?d1??d2d12?习题

?2d23.解：对于稳定流态的流体通过圆形管道，有

u2u1?2d 1d 22

若将直径减小一半，即d 1?u2u1?4

d 2即液体的流速为原流速的4倍. 4.解：

Lu2 Hf????d2gL1u12

Hf1??1??

d12g

Hf2

L2u22

??2??

d22g

Hf2Hf1L2u22?2??d22g?L1u12?1??d12g64?du???64Re?u2?4u1,L1?L2,d1?2d264?L2u22??du?d22g?2264?L1u12??d1u1?d12g64?L2u22??d2u2?d22g?1(u2)2L64??1?412g2d2u2?2d24?111??12?21641?16Hf2Hf1Hf2Hf1Hf2Hf1Hf2?16Hf1

即产生的能量损失为原来的16倍。

6.解：1）雷诺数Re??ud ?其中??1000kg?m?3，u?1.0m?s?1

d?25mm?25?10?3m

??1cp?10?3Ps?s

故Re??ud ?1000?1.0?25?10?3?

10?3?25000

故为湍流。

2）要使管中水层流，则Re?2000

1000?25?10?3m?u?2000 即Re?10?3解得u?0.08m?s?1

7.解：取高位水槽液面为1-1′， A-A′截面为2-2′截面，由伯努利方程

22p1u1p2u1z1???z2???Hf ?g2g?g2g其中z1?10m,z2?2m;

p1?p2;u1?0;Hf??hgf

2u216.15u2?则10?2? 2?9.89.8解得

1）A-A′截面处流量u?u2

u?2.17m?s?1

2）qv?Au?

其中A??d2??3.14?(100?10?3)2

```0

u?2.17m?s?1

1414 3

qv?7.85?10?3?2.17?3600?61.32m3h

15.解：选取贮槽液面为1-1′截面，高位槽液面为2-2′截面，

由伯努利方程得

22p1u1p2u1z1???He?z2???Hf/ ?g2g?g2g其中：z1?2m,z2?10m;u1?u2?0

p1?pvac??100mmHg??13.6?103?9.8?0.1??13332.2pa p2?02??13332.219.61000?He?10?(?4?)?g9.8980

19.613332.2He?12.08???14.08?1.388?15.4689.8980?gP?He?qV???10215.468?2??4?(53?10?3)2?980102?0.655kw

17.解：取水池液面为1-1′截面，高位截面为2-2′截面，

由伯努利方程得

22p1u1p2u1z1???He?z2???Hf ?g2g?g2g其中：z1?0,z1?50m;p1?p2?0

Hf?209.8

He?50?20?52.05 9.8P?He?qV??52.05?36?1000??8.05kw 102?102?0.6?360019.解：取贮槽液面为1-1′截面，

蒸发器内管路出口为2-2′截面，由伯努利方程得

22p1u1p2u1z1???He?z2???Hf ?g2g?g2g其中，z1?0,z1?15m;

p1?0,p2??200?10?3?13.6?103?9.8??26656pa 9.812026656He?15???24.97

9.89.8?1200H?q??24.97?20?1200P?eV??1.632kw

102102?3600Hf?12020.解：1）取贮水池液面为1-1′截面，

出口管路压力表所在液面为2-2′截面，由伯努利方程得

22p1u1p2u1z1???He?z2???Hf ?g2g?g2g其中，z1?0,z2?5.0m;

p1?0,p2?2.5kgf.cm?2

?2.5?9.85?2.45?10pa?40.01忽略出水管路水泵至压力表之间的阻力损失，则：衡算系统的阻力损失主要为吸入管路的阻力损失：

Hf?0.2u?9.8

36?2.2

3600??(76?10?3)242.45?1052.220.2He?5.0???1000?9.82?9.89.8

?5.0?25?0.25?0.02?30.27P?He?qV??30.27?36?1000??3.0kw 102102?36005

2）P?He?qV??3.0??4.3kw 102?0.73）取贮槽液面为1-1′截面，

水泵吸入管路上真空表处液面为2-2′截面，由伯努利方程得

22p1u1p2u1z1???z2???Hf ?g2g?g2g其中：z1?0,z2?4.8m;

p1?0,p2??

忽略进水管路水泵中真空表至水泵之间的阻力损失，则：衡算系统的阻力损失为吸入管路的阻力损失：

Hf?0.29.8

2.220.2p2??(4.8??)?1000?9.8??49600pa 2?9.89.8得真空表的读数为Pvac?49600Pa 23.解：1）取低位槽液面为1-1′截面

高位槽液面为2-2′截面由伯努利方程得

22p1u1p2u1z1???He?z2???Hf ?g2g?g2g其中，z1?0,z2?20m;p1?0,p2?0

Hf?5,He?20?5?25 w e?25?9.8?245J/kg

2）在管路A、B截面间列伯努利方程得：

2pAu2pBuBAzA???zB???Hf?g2g?g2gpApBLu2??zB?zA???g?gd2gu2?2g?ppd?(A?B?zB?zA)L??g?gpA?pB?(?Hg??H2O)?gR??H2O?g?6pA?pB(?Hg??H2O)?R??H2O?6??H2Og?H2O(???H2O)?R??H2O?6d?(Hg?6)L??H2Ou?2g? ?2g?0.05?0.504?2.03m?s?16?0.023）P?25?2.03??4102?0.052?1000?0.976kw

4）根据静力学基本方程式：

pB??H2Og(6?H)??HggR'?p0?pB??HggR'?p0??H2Og(6?H)pA??H2Ogh?pB??H2Og6??H2Og(h?R)??HggR?pA?pB??H2Og6??H2OgR??HggR?pA??HggR'?p0??H2Og(6?H)??H2Og6??H2OgR??HggR?pA?p0??HggR'??H2OgH?(?Hg??H2O)gR?[13.6?1.2?1?1?(13.6?1)?0.04]?9.8?1000?1.55?105pa

第三章传热过程

p105 ex1解：

q??tt1?t41150?30??0.20.10.006?R?1??2??3???1?2?31.070.1445

?11201120??1243Wm20.187?0.714?0.00010.901q`?300?1120??R?R0?3.73?C?m2?W?1

?(R?R0)R0?3.73?0.901?2.83?C?m2?W?1

Ex4解：空气的定性温度T?220?180?200?C 2200℃时空气的物性参数为：??0.746Kg/m3

??3.91?10?2W?m?1?K?1

?14J?Kg?? C??2.6?10?5Pa?s Cp?1.03Ku?15m?s?1，

Re?du??Cp??0.025?4?15?52.6?100.7464?1.0?91 01.034?103?2.6?10?5??0.68 Pr??3.931?10?2d23.93?1?10?3? ?0.020.0254?53.W8m2?K???0.023Re(0.8P)r(0.3)8(1?.0490.1?0) 0.03.68ex5解：水的定性温度T?40?20?30?C 230℃时水的物性参数为：??995.7Kg/m3

??0.6176W?m?1?K?1

??80.07?10?5Pa?s

Cp?4.174KJ?Kg?1??C ???0.023(Re)0.8(Pr)0.4

d当u?1m?s?1时， Re? Pr?du??Cp??0.02?1?995.7 ?2486.68?580.07?10?4.174?103?80.07?10?5??5.41

0.6176??0.023?0.6176?(24868.6)0.8?(5.41)0.40.002

?4583.5Wm2?K

当u?0.3时 Re?du???7460.58 ，此时，2000

6?105??0.023?38.85?1.965?7460.58?(1?)7460.581.80.8

?1638Wm2??Cex7解：甲烷的定性温度：T?120?30?75?C 20℃条件下：??0.717Kg/m3 ??0.03W?m?1?K?1

??1.03?10?5Pa?s Cp?1.7KJ?Kg?1?K?1

由于甲烷是在非圆形管内流动，定性尺寸取de

de?4流体截面积润湿周边

??4?4

?0.192?37??4?0.0192?0.0255??0.19?37???0.0199

Re?du??Cp??0.0255?10?0.717?17728.7 ?51.03?10Pr??1.?103?1.03?10?5??0.584

0.03由于甲烷被冷却，n?0.3

??0.023?de(Re)0.8(Pr)0.3?0.023?0.03?(17728.7)0.8?(0.584)0.30.0255

?0.023?1.176?2505.77?0.851?57.1W?m?2?K?1若甲烷在管内流动：

Re?du???0.015?10?0.717?10441.75 ?51.03?10??64.2W?m?2?K?1

Ex8 ?T1?40?15?25?C ，

?T2?130?33?97?C

?Tm??T2??T197?25??53.1?C（逆流） ?T297lnln25?T1在按照折流校正

33?15?0.16130?15

130?40R??533?15P?φ＝0.97

?Tm?0.97?53.1?51.5?C

ex9

(1)

??qmcp?T1?T2??1.25?1.9?10??80?30?3

?118750J?s?1(2) ?T1?30?20?10?C ，

?T2?80?40?40?C ?Tm??T2??T140?10??21.64?C ?T2ln4ln?T1(3)

R??11d?d??0?0?Rsi?Rs0K?0?油di?dm10.0250.0025?0.025?3?3 ???0.21?10?0.176?10330.025?0.021.7?100.85?10?0.0245?0.025ln0.02?0.00251m2?K?W?1K?1?398.91W?m?2?K?1 R(4)

A??K?tm?118750395.91?21.64

?13.74m2若近似按平面壁计算，则

R??111????0?Rsi?Rs0K?0?油i?m110.0025?3?3 ???0.21?10?0.176?101.7?1030.85?10345K?453.26

A?12.11m2

ex10

(1) d0?16?10?3m ，di?13?10?3m ，

??1.5?10?3m

?d0?d01??K?1?????d?d?0?m?ii??161.5?10?3?161??1??13?1000?40?14.5?90???? ?11.231?10?3?0.0414?10?3?0.011?81.5W?m?2?K?1??(2)

?d0?d01??K0?1?????d?d??m0??ii?161.5?10?3?161??1??13?1000?40?14.5?180?? ???146.6W?m?2?K?1ex11解：

(1)

''Cp ?1?qmcp?T1?T2??qm?T1'?T2'?

'' 50qmCp?25qmCp''qmCp?2qmCp .?Tm?135?60?92.5 135ln60A1?

?1k1?Tm?50qmCpk1?Tm

(2)

''?2?qmcp?t1?t2??qmCp?t1'?t2'?

''' 70qmCp?qmCpt1'?t2''t1'?t2?35 ，t2?50

??70qmCp?1135?30?tm??69.81 ，A2? ?135k2?tmk2?tmln30又流量及物性不变，k1?k2

k2?tm70?92.56475A2????1.855 50qCA150?69.83490.7mpk1?Tm70qmCpL2A2??1.855 ，L2?1.855m L1A1ex12解：(1)

?1?qmdcpd?T1?T2??KA?tm?1930?2.88?103?(90?50)?230?6??tm3600

?61.76?103?230?6??tm?tm?44.75?C

90?t?32?44.7590?t ln32t?29.5?C?Tm?.A1??1k1?Tm?50qmCpk1?Tm

?1?qmdcpd?T1?T2??qmhcph?t1?t2?61.76?103?qmh?4.2??29.5?18? qmh?1.279?103?4.6kg/h

吸收

p187

ex2解：

（1）

?PA?2026Pa7.821?10?3100?10?3 CA?()?2.300mol.m?3341000CC2.300??3?3?1PA?A?H?A??1.135?10mol.mPa?HPA2026（2）

E?P10007??4.895?10Pa?3?2HM1.135?10?1.8?10

?PA?4.895?107xAE4.895?107（3）m???483.22 5P1.013?10总（4）总压提高一倍，E、H值均不变

E4.895?107m???242 5P2.026?101ex9．解：

5y?64?0.02329955?2964y10.02329 Y??1?y11?0.02329?0.02329第一解法：

Y1?Y2?qn,BX1??X2????qn,C又设Y2??????qn,BX2?0?Y1??X1?26.7?qn,CY1?Y2Y?Y又??1.6512Y1qn,BX126.7Y1?1.65X126.70.02385X1??0.0005426.7?1.65()min?又512?8mol64qn,C.X1?8qn,C?14814.8kmol/h?266.66m3qn,C?Y1?Y2min???Y1?26.7

h第二解法：

设吸收率为?则，Y2??1???Y1 进气量设为 a.kg/h

512a.5%

a.?95%kmolqn,B?h29?10?3?? 15

(qn,Cqn,Bqn,Cqn,B)min?Y1?Y2Y1?(1??)Y1??26.7?Y1X1?26.7qn,Cqn,B)min?1.65?26.7?512a.5%()?1.65(

?qn,C?29?10?3a.95%3qn,C?14814.8kmol/h?266.0mh

ex10．（1）Y1?0.02

Y2?Y1(1?99%)?0.0002 X2?0

当液气比为2.2时，

qn,Cqn,B?Y1?Y20.02?0.0002??2.2 X1?X2X1 X1?0.009

?Y1?Y1?Y1??0.02?0.009?0.011?Y2?Y2?Y2*?0.0002

?Ym??Y1??Y20.011?0.0002??0.002695?Y10.011lnln0.0002?Y2Y1?Y20.02?0.0002??7.35?Ym0.002695NOG?（2）当液气比为1.25时，

X1?NOG0.02?99%?0.015841.25?15.17

(3) 当X2=0.00011时

X1?1(Y1?Y2)?X21.25

?0.01584?0.00011?0.01595NOG?0.02?0.0002?19.03

0.001040(4) 当液气比为0.8，溶质的最大回收率时溶液出口达到气液平衡，

Y1?Y20.02?Y2??0.8Y1qn,BX1?1Y2?0.004?qn,C

??Y1?Y20.02?0.0016??80%Y10.020.03?0.03093 1?0.03ex11.Y1? Y2?Y1(1?98%)?0.03093?0.02?0.0006

X2?0qn,B??qn,C??q?n,B1.67273?1??(1?3%)?65.16mol.s22.4?10?3303?Y1?Y2min??1.28??1.28?98%?1.2544??Y1??X21.28qn,Cmin?1.2544?65.16?81.74mol.s?1qn,C?81.74?1.5?122.6mol.s?1X1?qn,Cqn,B(Y1?Y2)?X2?65.16?(0.03093?0.0006)?0.016122.6

?Y?1?Y1?Y1?0.03093?1.28?0.016?0.01045?Y?2?Y2?Y2?0.0006?Y1??Y2m??Yln?Y1?Y2?0.01045?0.0006?0.ln0.010450034650.0006N?Y1?Y2

OG?Ym?0.03093?0.00060.003465?8.8Hqn,BOG?S?65.16K?0.6503Y?60?1.671.0H?HOG?NOG?8.8?0.6503?5.72mYy11?1?y?0.05?0.052611?0.05Yy22?1?y?0.00263?0.00263?0.00263721 X61.2581?(1000?61.2)?0.0203318X2?0

12.(1)

Y??2.0X?Y1?Y1?Y1??0.0526?2.0?0.02023?0.01214?Y2?Y2?Y2??0.002637?Ym??Y1??Y20.01214?0.002637??Y10.01214lnln0.002637?Y2?0.0095?0.00621.527Y1?Y2?YmNOG??0.0526?0.002637?8.060.00620.556273?0.95?22.4?10?3298 ?21.602mol.s?1q21.602HOG?V??0.7442??0.8KY?SKY??4KY??57.79mol.m?3.s?1qn,B?（2）

qn,Cqn,B??Y1?Y2X1?X20.0526?0.0026369?2.46980.02023

qn,C=2.4698?21.602=53.35mol.s-1XA1?n

53.35?3600n=0.02023×53.35×3600=3885.37mol m=58×3885.37=225.4kg （1）另解：

qn,Cqn,B??Y1?Y2X1?X20.0526?0.0026369?2.46980.02023

气相传质单元数：

NOG?mqn,BY1?mX2mqn,B?1?ln(1?)??mqn,B?qY?mYq?n,C22n,C???1?qn,C?11?22.4698?8.02820.0526?02? ?ln?(1?)?2.46980.0026372.4698???HOG?H6??0.7474m NOG8.028qn,BKY2?SKY2?qn,BHOG?s?21.6020.7474??57.5mol.m3.s1又HOG??d42

（3）若填料层增加3m，则：

'NOG?H'9??12.042m HOG0.7474NOG?11?22.4698Y2'?0.0011?20.05262?ln?(1?)???2.4698Y2'2.4698? ?又液气比一定，则：

Y1?Y2'?2.4698,X1'?4004.45molm?58n'?232.26kg则：

?m?232.26?225.4?6.9kgX1'?0.02085n'?X1'qn,B?0.02085?53.35?3600

13解：（1）

100?22.432Y1??0.07531001000??22.432

Y2?0.0753?(1?0.98)?0.0015X1?0.0196X2?00.278?1000?12.41mol.s?122.4qn,B(Y1?Y2)?qn,C(X1?X2)qn,B?12.41?(0.0753?0.0015)?qn,C(0.0196?0)qn,C?46.72mol.s?1?0.84kg.s?1

（2）

H?HOG?NOG

Y1=0.073 Y2=0.0015 X1=0.0196 X2=0

Y1*=1.15X1=0.0225 Y2*=1.5X2=0

??Ym?(Y1?Y1*)Y1?Y1*lnY2?Y2*

?0.073?0.0025?0.0015?0.0180.073?0.0025ln0.0015NOG?Y1?Y2?Ym

?0.0753?0.0015?4.10.0180.5?1000HOG?22.4?0.845m

26.41H?4.1?0.845?3.46m

14解：（1）HOG?H NOGY1=0.025 Y2=0.0045 X1=0.008 X2=0

Y1*=1.5X1=0.012 Y2*=1.5X2=0

??Ym?(Y1?Y1*)*Y1?Y1lnY2?Y2*

?0.025?0.012?0.0045?0.0080.025?0.012ln0.0045NOG?Y1?Y2?Ym

?0.025?0.0045?2.560.00810?3.91m 2.56HOG?（2）当Y2'?0.003时

qn,Cqn,B?Y1?Y2X1Y1?Y2? ??X1

0.025?0.00450.025?0.003 ??0.008X1?X1?0.0086

?Y1?Y1?Y1

?0.025?1.5?0.0086?0.0121 22

?Y2?Y2?Y2?0.003 ?YY1??Y2m??ln?Y1?Y2

?0.0121?0.003ln0.0121?0.00650.003NY1?Y2OG??Ym

?0.025?0.0030.0065?3.38H??HOG?NOG?3.91?3.38?13.2m H?13.2?10?3.2m

第六章 P244 ex2.(1)

???qn,F?qn,D?qn,W??qn,FxF?qn,Dxd?qn,Wxw 100?qn,D?qn,W100?0.3?q

n,D?095?qn,W?0.05q?1n,D?27.78?kmol?hqn,W?72.22?kmol?h?1

(2)qn,L??qn,L??qn,F

qn,L?R?D?27.78?3.5?97.23kmol?h?1 ??1

精馏

qn,L??97.23?100?197.23kmol.h?1

qn,V?qn,L?qn,D?97.23?27.78?125.01Kkmol.h?1 qn,V??qn,V?125.01kmol.h?1

??qn,F?qn,D?qn,W(3)?

qx?qx?qx?n,DD n,Ww?n,FF

235?qn,D?qn,W235?0.84?0.98qn,D?0.002qn,Wkmol.h?1

qn,D?201.4qn,W?33.6R?1, ??1

qn,L?201.4

qn,L??qn,L?qn,F?436.4?kmol.h?1

qn,V?qn,L?qn,D?402.8

qn,V?qn,V??402.8?kmol.h?1

(4)??(75.3?30)(0.4?2.68?0.6?4.19)?0.4?1055?0.6?232.

0.4?1055?0.6?2320?1.094

(5)精:

RXx?dR?1R?1 30.9?x??0.75x?0.22544x?y?x?f??1??1进:???0.8

?????1.0???1.2?y??y??4x?0.25? ?y?xf?0.5 ??y?6x?2.5

（6）

y?1.5x1?0.5x

y?1.200.2x?.40.2 x??0.42 y??0.52 1RminR?0.95?0.521 min?10.95?0.42 Rmin?4.30 （8）

y?0.75x?0.21y??0.5x?0.66

? ? ??1??0.5 1.5xf?0.66交点x?0.36 (10)

y?0.833x?0.15

y1?xd?0.898y3.0x*11?2.0x*?1 10.898?3.0x*12.0x* 1?1 x*1?0.746

x0?x10.898?x1x?x*??0.6010898?0.746 x1?0.806

?13xf?0.44y?0.48 25

y?0.833?0.806?0.15?0.821

第八章 P340 （1）

?dpA?2.58?10?6p2dtApAh???p2A

???p?1?1A?h

化学反应工程基本原理

pv?nRtnp??cvRt(?dpA)?kpp2Adtd(cARt))?kp(cARt)2dtd(cA)2)?kpc2Rt?kccAAdt(?(?kc?kpRt?2.58?10?6?8.314?450?9.65?10?3m3?kmol?h?1

?3?1kcc2?kmol?m?hA2?3?1 ???(kmol)?kmol?m?h3?m???m3?kmol?1?h?1（2）

C2H6?C2H4?H2yA,0?1.0000yA?0.0900?A?s?r?a1?1?1??1 a1yA,0?yAxA?yA,0(1??AyA)1?0.09?83.5%1?(1?1?0.09)?或

yA,0(1?xA)yA?1??AyA,0xA?xA??yA,0?yAyA,0(1??AyA)

1?0.09?83.5%1?(1?1?0.09)3

4NH3?5O2?4NO?6H2O?Q(主)4NH3?3O2?2N2?6H2O?Q(副)8NH3?8O2?4NO?2N2?12H2O?Q

?A?4?2?12?8?81?84yA,0?yAxA?yA,0(1??AyA)0.1152?0.0022?98.1152?(1??0.0022)4?设n0=1，则nA0=0.1152

yA?nA,0(1?xA)nnA,0(1?xA)yA?n??0.1152(1?0.98)?1.04730.0022?nA?0.1152?0.0023?0.1129

?nO2?0.2304?1.0473?0.087?0.1393设主反应消耗NH3的量为Z，副反应消耗NH3的量为F

4NH3?5O2?4NO?6H2O?Q(主)5Z Z44NH3?3O2?2N2?6H2O?Q(副)３F Ｆ4Z?F＝0.112953Z?F＝0.139344

?Z?0.11F?0.0029??0.110.1152?95.5% ??0.110.1129?97.4%5

t?VRq?2.501?250s V00.对于全混流反应器，

E(t)?1e?t??tF(t)?1?e??

F(250)?1?e?t??1?e?250250?0.632350F(250?350)?e?1?e?250?0.121第九章（1）

均相反应器30

A?R?S对于间歇操作反应器一级反应t?11kln1?xA30?1kln11?0.9

t?1kln11?0.99tln10030?ln10t?60s?t?30s2）

A?B?R对于间歇操作反应器二级反应t?xAkcA0(1?x

A)t0.990?0.3?0.2?0.1?150mint?0.99990.3?0.2?0.1?1650min4）

A?B?R?S对于间歇操作反应器二级反应t?xAkcA0(1?xA)对于全混流操作反应器二级反应??xAkcA0(1?xA)2t1??1?x?1?5A1?0.8??5t?50min8）

（（

（ CH3CHO?CH4?COyA,0?1.0?s?r?a1?1?1A?a?1?1cnV?pA0?RT?15.4mol.l?1??VRq?cxAdxAA0V0?0(?rA)?cxAdxAA0?0k?2?cA0(1?xA)??1?x?A??1xAkc(1?xA)2dxAA0?01?xA?1?0.4?4kc?1?x?xA?4ln(1?x?A)?

A0A?0（15）

C3H8?C2H2?CH3yA,0?1.0??r?a1?A?sa?1?11?1?rA?kpPA?kccA??VRxq?cAdxAA0V0?0(?rA)?cxAdxAA0?0kcA0(1?xA)1??AyA,0xA?1xAk?0(1?xA1?x)dxAA?1k(2ln11?x?xA)A?104?(2ln10.4?0.6)?1.233?104VR?1.233?104?2?10?2L?246.6L

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