化学工程基础习题答案(武汉大学第二版)精品资料

1、化学工程基础习题第二章 P691.解： PvacPoP绝即103Pa 98.7103PoP13.3绝P绝85.410 3 PaPaP Po85.4 10 3 Pa 98.710 3 Pa13.3 10 3 Pa2. 解：d12d22de44d14(d1d2 )70d23. 解：对于稳定流态的流体通过圆形管道，有u2d 12u1d 22若将直径减小一半，即 d 12d 2u24u1即液体的流速为原流速的4 倍.4. 解：H fLu 2d2gHHL 1 u12f 11d12gL 2u22f 22d22gL 2u22H f 22d22gH f 11L 1u12d12g64 64 Re duu24u1

2、 , L 1L2 , d 1 2d264L 2u22H f 2d2u 2d22gH f 164L1u12d1u1d12gHH64L 2u22f 2d2 u 2d22gf 164L 1(1u 2 )242d1u 22d22g2 4H f 2116H f 1111121624H f 216H f 1即产生的能量损失为原来的16 倍。ud6. 解： 1）雷诺数 Re其中1000kg m 3 ， u1.0m s 1d25mm 25 10 3 m1cp10 3 Ps sud故 Re10001.02510 310 325000故为湍流。2）要使管中水层流，则 Re2000100025103 m u即 Re

3、1032000解得 u0.08m s 17. 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p1u12p2u12z12gz2H fgg2g其中 z10m, z 2m;12p1p2 ; u10; H fh fg则102u2216.15u29.89.82解得1）A-A截面处流量 uu2u2.17m s12） q vAu其中 A1d 213.14 (100 10 3 )2447.8510 3 m2u2.17 m s 1q v7.85103 2.17 3600 61.32 m3h8. 解：对 1-1 截面和 2-2截面，由伯努利方程得z1p1u12z2p2u12g2gg2 g其中

4、 z1z2 , p11mH 2Ogh1u1 0.5m s 1 , p1gh2u2d12(0.2 20.5 2.0m s12 u1)d20.1h0.5 2220.19 m29.829.815. 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p1u12p2u12z1H e z2gH fg2g2g其中： z12m, z2 10m; u1u20p1pvac100mmHg13.61039.80.113332.2pap02213332.21019.61000gH e(4)9.8980H e12.0819.613332.214.081.38815.4689.8980gH eqV1

5、5.4682(5310 3)2980P40.655kw10210217. 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p1u12p2u12z1H e z2gH fg2g2g其中： z10,z150m; p1p20Hf209.8H e502052.059.8PH e qV52.053610001021020.68.05kw360019. 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z1p1u12H ez2p2u12H fg2gg2g其中， z10,z 115m;p10,p220010 313.61039.826656paH f1209

6、.8H e151202665624.979.89.81200PH eqV24.972012001.632kw102102360020. 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z1p1u12p2u12gH e z2gH f2g2g其中， z10,z 2 5.0m;p0, p2.5kgf .cm 2122.59.82.45105 p0.01 4a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：H f0.29.836u2.23236004(7610)H e2.451052.220.25.09.829.89.81

7、0005.0250.250.0230.27H eqV30.2736 1000P1021023.0kw3600H e qV3.04.3kw2） P0.71023）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p1u12p2u12z1z2gH fg2 g2g其中： z1 0,z 24.8m;p10, 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：Hf0.29.8p2(4.82.220.2 ) 1000 9.849600pa29.89.849600得真空表的读数为PaPvac23. 解：1）取低位槽液面为 1-

8、1截面高位槽液面为 2-2 截面由伯努利方程得p1u12p2u12z1H e z2gH fg2g2g其中， z10,z220m; p0, p 012H f5,H e20525w e25 9.8245J/kg2）在管路 A、B 截面间列伯努利方程得：zAp AuA2zBpBuB2g2ggH f2gpAp BzBzAL u2ggd 2gu22gd( p ApBzB zA )LggpApB ( HgH 2O )gRH 2Og6pA pB ( HgH2O) RH2O6H 2O gH 2Ou2gd(HgH2O) RH2O6L(6)H 2O2g0.050.5042.03m s 160.023） P252.

9、0340.05210000.976kw1024）根据静力学基本方程式：pBH 2O g (6H )Hg gR'p0pBHg gR'p0H 2O g(6H )pAH 2O ghpBH 2O g6H 2O g(hR) Hg gRpApBH 2O g6H 2 O gRHg gRpAHg gR'p0H 2O g(6H )H 2O g6H 2 O gRHg gRpAp0Hg gR 'H 2O gH(HgH 2O ) gR13.61.211(13.61)0.049.810001.55105 pa第三章 传热过程p105ex1 解：tqRt1t41231231150300.2

10、0.10.0061.070.14450.18711200.000111201243W m20.7140.901q 3001120R R03.73 C m2 W 1(RR0 )R03.73 0.9012.83 C m2W 1ex2 解：(T1T4 )(T1T4 )iii A2 rm L i2L(T1T4 )2L(T1 T4 )1(ri 1ri )1 l nri 1i ( ri 1ri )ri 1irilnri2(26035)L12515518516 ln 200.2l n 250.07 ln 5523.1422514130.2230.7881.4350.01393.946.214160.20.0

11、071413138.96Wm 110.168Ex4 解：空气的定性温度 T2201802200 C200时空气的物性参数为：0.746Kg/ m33.9110 2Wm 1K 12.610 5 PasCp1. 034KJKg1Cu15m s 1 ，du0. 02541 50. 7464Re2. 651. 091 01 0C p1.034 1032.6105Pr3.93110 20.680. 0 2 3 R(e0 . 80 . 3) Pr ()d3. 931 120(1. 094 0.80 . 30. 0231 0 )0. 680. 02545 3. 8W m2 K4020ex5 解：水的定性温度

12、 T30 C230时水的物性参数为：995.7Kg / m30.6176W m 1 K 180.0710 5 PasCp 4.174KJ Kg 1C0.80.40.023( Re )( Pr )当 u 1m s 1 时，du0.02 1995.72486.68Re80.07105C p4.17410380.0710 5Pr0.61765.410.0230.6176(24868.6) 0.8(5.41) 0. 40.0024583.5W m2 K当 u 0.3 时Redu7460.58 ，此时， 2000<Re<100000.02338.851.9650 .8(161057460.5

13、81.8 )7460.581638W m2C1203075 Cex7 解：甲烷的定性温度： T20条件下：0.717Kg / m30.03Wm 1 K11.0310 5 PasCp1.7KJKg 1K 1由于甲烷是在非圆形管内流动，定性尺寸取 dede流体截面积4润湿周边0.1923740.0192440.02550.19370.019du0.0255100.717Re1.0310517728.7C p1.1031.03105Pr0.030.584由于甲烷被冷却， n0.30.80.30.023(Re )( Pr )0.0230.03(17728.7)0.8(0.584)0 .30.02550

14、.023 1.1762505.770.85157.1W m 2 K 1若甲烷在管内流动：Redu0.015 10 0.71710441 .751.0310 564.2Wm 2 K 1Ex8T1401525C ，T21303397 CTmT2T197 2553.1 C （逆流）T2l n97ln25T1在按照折流校正3315P0.161301513040R533150.97Tm0.97 53.151.5 Cex9(1)qm cp T1T21.251.91038030118750 Js1(2)T1302010C ，T2804040CTmT2T1401021.64 CT2ln 4l nT1(3)R1

15、1d0d0RsiRs0K油 didm010.0250.00250.0250.211030.1761031.71030.851030.020.0250.02450.025l n0.020.00251m2 K W 1K1398.91Wm 2K 1R(4)A118750K tm 395.91 21.64 13.74m2若近似按平面壁计算，则1110RsiRs0RK0油im110311030.00250.2110 30.17610 31.70.8545K453.26A12.11m2ex10(1) d 01610 3 m ， di1310 3 m ，1.510 3 mK1d0d01i didm01161

16、.510 31611310004014.5901 1.23110 30.041410 30.01181.5W m 2 K 1(2)d0d01K 0 1dmi di01161.510 316110004014.518013146.6W m 2 K 1(3)d0d01K i 1dmi di01161.510 316120004014.590131 0.61610 30.041410 30.01185.78Wm 2K 1ex11解：(1)1qmcp T1T2qm'C p' T1'T2'50qmC p25qm'C p'qm'C p'2qm

17、C p.13560T m92.5l n13560A1150qmC pk1Tmk1 Tm(2)2qm cp t1 t2qm'C p't1't2'70qm C pqm'C p't1't2't1't 2'35 ， t2'50tm1353069.81 ， A2170qmC p135k2 tmk2 t mln30又流量及物性不变， k1k 2A270qmC ptm7092.56475k21.855A150qmC p5069.83490.7Tmk1L2A21.855， L21.855mL1A1ex12 解：(1)1qm

18、d cpdT1 T2KA tm19302.88103(90 50) 230 6 tm360061.76 1032306t mtm44.75 C.90t32T m44.75ln90t32t 29.5 C1 50qmC pA1k1 Tmk1 Tm1qmd c pd T1T2qmh cph t1t 261.76 103qmh4.229.5 18qmh1.279 1034.6kg / h吸收p187ex2 解：（1）PA2026 PaCA7.821 10 3(100 10 3) 2.300 mol.m 3341000PAC AHC A2.3001.135 103mol.m31HPA2026Pa（ 2）

19、EP1.135100010 24.895 107PaHM10 31.8PA4.895107 xA（3） mE4.895107483.22P总1.013105（4）总压提高一倍， E、H 值均不变mE4.89510724212.0265P10ex9解：5y 640.0232995 52964y10.02329Y1 0.023291 y10.02329第一解法：qn,C) minY1Y2(X1X 2qn, B又设 Y2qn,CY1 Y2qn,BminX 20Y126.7Y1X126.71.65 Y Y2又YYqn,C121qn,BX1Y126.7Y11.65X126.70.02385X10.000

20、5426.7 1.65又 512 8mol 64qn,C .X18qn,C 14814.8kmol / h266.66m3h第二解法：设吸收率为则，Y2 1Y1进气量设为a.kg / h512a.5%a.95% kmolqn,B3h2910qn,C) minY1 Y2Y1(1 )Y126.7(X1Y1qn, B26.7qn,Cqn ,C512() 1.65() min 1.65 26.7qn, Bqn ,Ba.5%qn ,C2910a.95%3qn,C14814.8kmol / h 266.0 m3hex10（1） Y10.02Y2 Y1(199%)0.0002X 2 0当液气比为 2.2 时

21、，qn ,CY1Y20.020.0002qn , BX1X 22.2X1X10.009Y1Y1Y10.020.0090.011Y2Y2Y2*0.0002Y1Y20.0110.0002Ym0.002695lnY1ln0.011Y20.0002Y1 Y20.02 0.0002NOG7.35Ym0.002695（2）当液气比为 1.25 时，0.0299 %X10.015841 .25 NOG 15.17(3) 当 X 2=0.00011时X11 (Y1 Y2) X21.250.015840.000110.015950.020.0002NOG19.030.001040(4) 当液气比为 0.8，溶质

22、的最大回收率时溶液出口达到气液平衡，qn ,CY1Y2 0.02 Y20.8qn ,BX 1Y1Y2 0.0041Y1 Y20.02 0.001680%Y10.02ex11.0.030.03093Y110.03Y2Y1(198%)0.030930.020.0006X 20qn, B1.67273(13%)65.16mol.s 122.410 3303qn,CYYmin121.281.2898%1.2544qn, BY1X 21.28q1.254465.1681.74mol.s 1n,C minqn,C81.741.5 122.6mol .s 1X1qn, C(Y1Y2) X2qn ,B65.1

23、6(0 .030930.0006)0.016122.6Y1Y1Y10.030931.280.0160.01045Y2Y2Y20.0006Y1Y2YmY1lnY20.010450.00060.003465ln 0.010450.0006NOGY1Y2Ym0.030930.00068.80.003465H OGqn ,B65.16KYS0.6503601.671.0H HOGNOG8.8 0.6503 5.72 m12.(1)Y1y10.050.05261y110.05Y2y20.002630.0026371y210.00263X161.2 580.02033(100061.2)18X 2 0Y

24、2.0X Y1 Y1 Y10.05262.00.020230.01214Y2Y2Y20.002637Y1Y20.01214 0.002637YmY1ln 0.01214lnY20.0026370.00950.00621.527Y1Y2NOGYm0.05260.0026378.060.0062qn, B0.5560.95 27322.410 329821.602mol.s 1H OGqV21.6020.744K Y S0.82K Y4KY57.79mol.m 3.s 1（ 2）qn,CY1Y2qn ,BX1X 20.05260.00263692.46980.02023qn ,C =2.4698-

25、1X A1n53.353600n=0.02023×53.35×3600=3885.37molm=58×3885.37=225.4kg（1）另解：qn ,CY1Y2qn ,BX1 X 20.05260.00263692.46980.02023气相传质单元数：N OG1ln(1mqn,BY1mX 2mqn ,Bmqn,Bqn,C)mY2qn,CY21qn, C1ln(12) 0.05260222.46980.0026372.469812.46988.028H OGH60.7474mN OG8.028又 HOGqn , BKYqn, B21.6023 1K Y SH O

26、Gsd 257.5mol.m .s20.747424（3）若填料层增加 3m，则：N OG'H '912.042mH OG0.7474N OG1ln(12)0.05262122.4698Y2 '2.46982.4698Y2 ' 0.0011又 液气比一定，则：Y1 Y2'X1 ' 0.02085X1 '2.4698,n' X1 ' qn, B0.02085 53.35 36004004.45molm58n'232.26kg则：m232.26225.46.9kg13 解：（1）10022.4Y1321000.0753

27、100022.432Y20.0753(10.98)0.0015X10.0196X 20qn, B0.278100012.41mol.s 122.4qn, B (Y1 Y2 )qn,C ( X1 X 2 )12.41(0.07530.0015)qn ,C (0.01960)q46.72mol.s 10.84kg .s 1n,C（ 2）H HOG NOGY 1=0.073Y2=0.0015X 1=0.0196X 2=0Y 1* =1.15X1 =0.0225Y 2*=1.5X2=0(YY* )11YmY1*ln Y1Y2Y2 *0.0730.00250.0015ln0.0730.00250.001

28、50.018Y1Y2NOGYm0.0753 0.00154.10.0180.5 1000H OG22.40.845m26.41H4.1 0.8453.46m14 解：（1） H OGHN OGY 1=0.025Y2=0.0045X 1=0.008X2=0Y 1* =1.5X1=0.012Y 2*=1.5X2=0(Y1Y1* )YmY1*ln Y1Y2Y2 *0.0250.0120.0045ln 0.0250.0080.0120.0045Y1Y2N OGYm0.0250.00452.560.008H OG103.91m2.56（2）当 Y2 ' 0.003时qn ,CY1Y2 Y1Y2qn ,BX 1X10.0250.00450.0250.0030.008X1X 10.0086Y1Y1Y10.0251.50.00860.0121Y2Y2Y20.003Y1Y2YmlnY1Y20.01210.0030.0065ln 0.01210.003Y1Y2N OGYm0.0250.0030.00653.38H HOG NOG3.91 3.3813.2mH13.2 103.2m第六章精馏P244ex2.(1)qn, Fqn ,D qn ,Wqn, F xFq