化学工程基础习题答案

1、化学工程基础习题其次章 P69 1.解：P vacP o P绝P oP绝即 13.3 103Pa98.7 103P 绝85.4 103Pa3PaaPPP o 85.4 103Pa98.7 1013.3 103Pa2. 解：de44d 1 24d2d 1d2702d 1d23. 解：对于稳固流态的流体通过圆形管道,有u2u 1d 2d 2 2d 221如将直径减小一半,即d 1u2u 14即液体的流速为原流速的 4 倍. 4. 解：H f11Lu2d2gHfL1u 12d 12gHf22L2u22d22gHf22L2u 2212d22d22gHf11L1u 12u 2fd 12g64Re64du

2、4u L1L2,d64L2u22H2d u2d22gHfu 12164L1Hfd u1d12g64L2u222d u2d22gHf164L11 u 242 gHf2 d21u22 d242116Hf121111Hf2 164216Hf1即产生的能量缺失为原先的 16 倍；6. 解： 1）雷诺数Reud1.0m s1其中1000kg m3,ud25mm25 103m1 cp103Ps s故Reud1000 1.025 10331025000 故为湍流；2）要使管中水层流,就Re320222022即Re100025 10m u103解得u0.08m s17. 解：取高位水槽液面为 1-1 ,A-A

3、 截面为 2-2 截面,由伯努利方程z 1p1u2 1z 2p2u2Hf1g2gg2g其中z 110 , m z 22 ;p 1p u 2 10;Hfh gf就1022u216.15u229.89.8解得1）A-A 截面处流量uu 2100 1032m3hu2.17m s13.142） q vAu其中A1d21447.85 103m2360061.32u2.17m s132.17qv7.85 108. 解：对 1-1 截面和 2-2 截面,由伯努利方程得z 1z 1p12 u 1,z 2p2u2 1g2gg2g其中z 2,p 11 m H O 2gh 2gh 1u 10.5 m s1p 1u

4、22 d 1u 10.2 0.120.52.0 m s1d2 2h0.52220.19m29.829.815. 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z 1p1u2 1Hez 2p2u2 1Hf15.4689800.655kw g2gg2g其中：z 12 , m z 210 ; m u 1u 20p 1pvac100 mmHg13.6 10 39.8 0.113332.2p ap 20213332.2He10 19.69.841000 980gHe12.0819.613332.214.08 1.3889.8980gPHeq V15.468 243 53 1

5、0 210210217. 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z 1p1u2 1Hez 2p2u2 1Hf8.05kw g2gg2g其中：z 10,z 150 ; m p 1p 20Hf209.836 1000He502052.059.8Heq V52.05P1021020.6360019. 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z 1p 1u2 1Hez 2p2u2Hf1g2gg2g其中,z 10,z115 ;p 1f0,33 13.6 109.826656p ap 2200 10H1209.826656 9.8

6、120024.971.632kw He15120 9.8PHeqV24.9720 1200102102360020. 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z 1p 1u2 1Hez 2p2u2 1Hfg2gg2g其中,z 10,z25.0 ;p 10,p 22.5 kgf cm22.5 9.85 2.45 10 p a0.014忽视出水管路水泵至压力表之间的阻力缺失,就：衡算系统的阻力缺失主要为吸入管路的阻力缺失：2）PHf0.29.83.0kw u36003610322.2476He5.02.45 1052.220.210009.82

7、9.89.8P5.0250.250.0230.27Heq V30.273610001021023600Heq V3.04.3kw 1020.73）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z 1p1u2 1z 22p2u2Hf1g2gg2g其中：z 10,z4.8 ;p 10,p 2.忽视进水管路水泵中真空表至水泵之间的阻力缺失,就：衡算系统的阻力缺失为吸入管路的阻力缺失：Hf0.29.80.2 9.8 10009.849600pa p 24.82.2229.8得真空表的读数为Pvac49600 Pa23. 解：1）取低位槽液面为 1-1 截面 高位

8、槽液面为 2-2 截面 由伯努利方程得z 1p 1u2Hez 2p2p 2u2Hf11g2gg2g其中,z 10,z220 ;p 10,0Hf5,He20525w e25 9.8245J/kg 2）在管路 A、B截面间列伯努利方程得：z ApAu2z BpB2 u BHfg66Ag2gg2 gpApBz Bz ALu2ggd2g2 u2gdpApBz Bz ALggp Ap BHgH2OgRH2OpAp BHgH2ORH2O6H2 OgH2OH2O6u2gdHgH2ORLH2 O12 g0.050.5042 .03m s6 0.020.976 kw3）P252.032 0.0541021000

9、4）依据静力学基本方程式：p BH2Og6HHggRp 0RHggRpBHggRp0H2 Og6HpAH2Oghp BH2 Og6H2Og hpApBH2Og6H2OgRHggRH2OgRpAHggRp0H2 Og6Hg6H2OgRHggRH2OpAp0HggRH2OgHHg13.61.21 113.610.049.810001.55105p a第三章 传热过程p105 ex1 解：qtt 1t4115030W1R1230 2.01.0 .0060.1871231.070. 1445112011201243Wm20 .7140 .00010 .901q 3001120RR 03 . 73C2

10、mRR 0R 03 . 73.0 901.2 83Cm2W1ex2 解： T 1T 4 T 1T 446.214iiiA2r Li2L T 1T 42L T 1T 41r ir i1 r ir i11lnr i1i1lnr iirr i226035 L1ln251ln5501ln85162002.25.075523 .1422514130 .2230. 7881. 4350. 01393 .94160 .20. 0071413138.96 Wm110. 168Ex4 解：空气的定性温度T2202180200 C200时空气的物性参数为：0.746 Kg/3 m3.912 10 W m1K12.

11、6 105aP sC p1.034 KJ Kg1Cu15m s1,R edu0.0254150.7461.09 102.6 105P rCp1.0341032.6 1050.6883.931 1020.023dR e 0.8P r0.30.0233.931 1024 1.09 10 0.80.3 0.680.025453.8 W m 2Kex5 解：水的定性温度T4022030C30时水的物性参数为：995 7.Kg/m30 . 6176Wm1K180 . 07105PasC p4.174 KJ Kg1C0.023dR e.08P r.04当u1 ms1时,Redu0.021995.72486

12、8.680.07105P rCp4 .17410380.071055. 410.61760 . 0230 . 617624868 6. 0.8 .541 0. 40 .0024583 . 5 W2 mK当u0.3时Redu7460. 58,此时, 2022Re10000 0 . 02338 .851 . 96507460 . 588. 165 107460 .1 581638 W2 mCex7 解：甲烷的定性温度：T12023075C0条件下：0 . 717 Kg/m30.03 Wm1K11K11 . 03105PasCp.1 7 KJKg由于甲烷是在非圆形管内流淌,定性尺寸取de de4流体

13、截面积润湿周边4420 . 19374.02 019.002550 . 19370 . 019Redu0.0255100. 717177287.1 .03105P rCp1.1031. 031050. 5840 .03由于甲烷被冷却,n0 3.0 . 023deR e.08P r.0 30 . 023.0 03 17728 . 7 08. 0 .584 03.0 . 02550 . 023.11762505 .770 . 85157 1. Wm2K1如甲烷在管内流淌：Ex8 T 1Redu0.015100 .717C10441.751. 0310564.2 Wm2K140 1525C ,T 2

14、1303397C53.1（逆流）T 2T 19725T mlnT 2ln9725T 1在依据折流校正P33150.1613015R1304053315 0.97 T m0.97 53.151.5Cex9 1 2 q mcpT 1T 23021 . 64C1 .2519.10380118750Js1,T 1302010CT 2804040C10T mT 2T 140T 2ln4lnT 13 R11d0d0R siR s 00 . 21103.0176103K0油d idmK10 . 0250 . 00250 . 025.1 73 10.0 853 10.0 02450 . 0250 . 02ln

15、0 . 025.0 02.02 00251 mKW11 R398.91 Wm2K14 AKtm118750.64395.912113.74m2如近似按平面壁运算,就R1 .110 .1100R siR s 00. 211030 .176103K0油imK1130.002571038545453.26A12. 11m2ex10 1 d016103m,di13103m,1 .5103mK1d 0d 01id id m01113161 5.1031610004014 5.9011 . 2311030 . 04141030 .01181 5. Wm2K12 K01d0d011id idm0113161

16、 . 51031610004014 . 5180146 6. Wm2K13 Ki1d03d01161id idm01161.51031320224014.59010. 616100 .04141030 .01185. 78 Wm2K1ex11解：1 1q cpT 1T 2 q C T 1 T 2p50 qm Cp25 q m CpqC2 qmCpmp.Tmk 11356092.5ln135A 160m Cp150 qT mk 1T m2 2q cpt1t2 q Ctt70 qm Cpp1270 qmCp q mC t 1tp2t 1 t35, 2t502tm1353069.81,A 2k 21

17、mln135tk2tm301.855又流量及物性不变,k 1k2A 270 q Cpk2t m70 92.56475A 150 q Cpk 1T m50 69.83490.7L2A 2.1 855,L21 . 855 mL 1A 1ex12解：1 1q mdc pdT 1T 2KA t m50230 6tm1930 36003 2.88 10903 61.76 10230 6tmtm44.75C.Tm90t3244.75ln90t32t29.5CA 1k 1150 qm CpT mk 1T m1q mdcpdT 1T 2q cpht 1t261.76 10 3q mh4.229.5 18q m

18、h3 1.279 104.6 kg h吸取p187 ex 解：（1）P A2026P a31001032.300mol m33P a1CA7.821 10341000P ACAHCA2.3001.135 103mol mHP A2026（2）EP10007 4.895 10P aHM1.135 1031.8 102P A7 4.895 10 xA（3）mE4 . 8957 10483 . 22P 总1 . 0135 10（4）总压提高一倍, E、H 值均不变mE.48951072421P2 . 0265 10ex9解：5y64 9550 .02329Y29640 . 02329y 11y 11

19、.0 02329.0 02329第一解法：q n Cq n B min0Y 1Y 22q n CminY 1Y 2X 1X又设Y 2q n BY 126.7X2X1Y 126.7又q n CY 1Y 21.65Y 1Y 1Y 2q n BX 126.7Y 11.65X126.7X10.023850.0005426.7 1.65又5128 mol64q n C. X18q n C14814.8 kmol h3 266. 66mh其次解法：设吸取率为a就,Y 21Y 1进气量设为. kg/h512a .5%q n Ba. 95%kmolh1 Y 126.729 103qn C minY 1XY 2

20、Y 1qn B1Y 126.726.7512qn C1.65q n Cmin1.65qn Bq n Ba.5%q n C29 103a .95%q n C14814.8kmol/h266.03 mhex10（1）Y 10 . 02Y2Y 1199 %0 . 0002X20当液气比为 2.2时,q n Cq n BY 1Y 220.020.00022.2X1XX 1X1.0 009Y 1Y 2Y mY 1Y 10.02 0.0090.011Y 2Y 2 *0.0002Y 1Y 20.011 0.00020.002695lnY 1ln0.011 0.0002NOGY 2Y 1Y 20.020.00

21、027.35Y m0.002695（2）当液气比为 1.25时,X10. 0299%0.01584X3 当 X2=0.00011时X20159511Y 1Y 21. 25N1 .25OG15.17NOG0 .020.000219.030. 015840. 000110 .0 .0010404 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稳,q n CY 1Y 20.02Y 20.80.0006q n BX1Y 11Y 20.004Y 1Y 1Y 20.020.001680%0.02ex11. Y 110 .030.030930 .03Y 2Y 11 98%0.03093 0.02X

22、2011.2544q n B1.67327313%65.16mol s22.4 10303qn CminY 1Y 221.281.2898%qn BY 1X1.28q n Cmin1.254465.1681.74mol s1q n C81.74 1.5122.6mol s1X1qn CY 1Y 2X2qn B65.160.030930.00060.016122.6Y 1Y 1Y 1.0 030931 . 280 . 016.001045Y 2Y 2Y 20 . 0006Y mY 1Y 2Y 1lnY 2.0010450 . 00060 . 003465ln.0010450 . 0006NOGY

23、 1Y 2Y m.0030930 . 00068 . 8.0 003465HOGq n B65.160.6503KYS601.67 1.0HHOGN OG8.80.65035.72m12.1 Y 1 y 1 0.050.05261 y 1 1 0.05Y 2 y 2 0.002630.0026371 y 2 1 0.0026361.2 58X 1 1000 61.2 0.0203318X 2 0Y 2.0 XY 1 Y 1 Y 10.0526 2.0 0.02023 0.01214Y 2 Y 2 Y 2 0.002637Y 1 Y 2 0.01214 0.002637Y mln Y 1 ln

24、0.01214Y 2 0.0026370.00950.00621.527N OG Y 1 Y 2Y m0.0526 0.002637 8.060.0062q n B 0.5563 0.95 2 7322.4 10 2981 21.602 mol sq V 21.602H OG 2 0.744K Y S K Y 0.843 1 K Y 57.79 mol m . s（2）q n CY 1Y 2-1q n BX1X20.05260.00263692.46980.02023q n C=2.469821.602=53.35mol.sX A 153.35n3600n=0.02023 53.35 3600

25、=3885.37mol m=58 3885.37=225.4kg （1）另解：q n CY 1Y 22.4698q n BX 1X20.0526 0.0026369 0.02023气相传质单元数：NOG11n Bln1mqn BY 1mX2mq n Bmqq n CY 2mY 2qn Cqn C112ln12 2.46980.0526020.0026372.46982.46988.028HOGH60 . 7474 m21.602d257.53 1mol m sNOG8 . 028又HOGq n BKY 2qn BKY S 2HOGs0.74744（3）如填料层增加 3m,就：N OGH912

26、. 042 m0 . 05262HOG.0 7474N OG11ln12 2 . 46982Y 22 . 4698Y 22 . 4698.0 0011又 液气比肯定,就：Y 1Y 22.4698,X10.02085X 1nX1q n B0.02085 53.35 36004004.45 mol就：m58 n 232 . 26 kg.6 9 kgm232 . 26225 . 413 解：（1）Y 1100 22.4321000 100 22.4320.07531Y 20.0753 1 0.980.0015X10.0196X20q n B0.278 100012.41 mol s22.4q n B

27、 Y 1Y 2qn CX 1X212.410.07530.0015q n C0.01960q n C46.72mol s10.84kg s1（2）HHOGN OG0.018Y1=0.073 Y2=0.0015 X1=0.0196 X2=0 Y1*=1.15X1=0.0225 Y2*=1.5X2=0 Y m Y 1Y*lnY 1Y 2* Y 1Y 2*0.0730.00250.0015ln0.0730.00250.0015NOGY 1Y 2Y m0.0753 0.0015 0.0184.10.5 1000HOG22.4 26.410.845 mH4.1 0.8453.46 m14 解：（1）HO

28、GHNOGY1=0.025 Y2=0.0045 X1=0.008 X2=0 Y1*=1.5X1=0.012 Y2*=1.5X2=0 Ym Y 1 Y 1 * *ln Y 1 Y 1Y 2 Y 2 *0 . 025 0 . 012 0 . 00450 . 025 0 . 012 0 . 008ln0 . 0045N OG Y 1 Y 2Y m0 . 025 0 . 0045 2 . 560 . 00810H OG 3 . 91 m2 . 56（2）当 Y 2 0 . 003 时q n C Y 1 Y 2 Y 1 Y 2q n B X 1 X 1.0 025 0 . 0045 0 . 025 0

29、. 0030 . 008 X 1X 1 0 . 0086Y 1 Y 1 Y 1.0 025 .1 5 0 . 0086 0 . 0121Y 2 Y 2 Y 2 0 . 003Ym Yln 1Y 1 Y 2Y 2.0 0121 0 . 003 .0 0065ln 0 . 0121.0 003NOGY 1Y 2第六章精馏Y mH0 .0250.0033.380. 0065HOGNOGH3.91 3.3813.2 m13.2 103.2 mP244 ex2.1q n Fq n Dq n W0.051q n Fx Fq n Dx dq n Wx w100q n Dq n W100 0.3q n D09

30、5q n Wq n D27.78kmol h197.23 kmol hq n W72.22kmol h12q n Lq n Lq n Fq n LR D27.78 3.51q n L97.23 100197.23 kmol h11q n Vqn Lq n D97.2327.78125.01 Kkmol hq n Vq n V125.01 kmol h13qn FxFq n Dn Dqn Wq n Wx wqn FqxD 235q n Dq n W235 0.840.98 q n D0.002 q n Wq n D,1201.4kmol.h14. 190.4105506.232.q n W33.

31、6R1201.4q n Lq n Lq n Lq n F436.4kmol h1q n Vq n Lq n D402.8q n Vq nV402.8kmol h1475.33004.2 .680.60.410550 .623201 .0945精: yR1xXd10 . 2251RR3x.09.075 x44y1xfx进:0 8.1 . 01 . 2y4x0 .25yxf05.y6x25.（6）y11 .5x0 . 5210.5 xy1 .2x0 .40 .20 .2x0 . 42yR min0. 95.0521R min10 . 95.042R min4 . 30（8）y .0 75 x 0

32、. 21y 0 5. x 0 . 660 5. 11 31 . 5 x f 0 . 66 fx 0 . 44交点 x 0 . 36 y 0 . 4810 y.0833 x.015.0 6y 1x d.0898y 13 . 0* x 112 . 0 x 1 *.0898*.3 0 x 112 . 0* x 1* 1x0 . 746x 0 x 1.0 898x 1x 0* x 10898.0 7461x0 . 8060.821y0 .8330 .8060. 15第八章 化学反应工程基本原理P340 （1）dpA2 . 58106p22. 314450dtApAp2hAp1h1ApvnRtnpcvRtdpAkpp2dtAdcARtkpcARt2dtdcAkpc2RtkccAAdtkckpRt2.5810689. 65103m3kmolh1k c2 c Akmolm3h113h1kmol2kmolm3 mm 3kmol1h（2）C2H6C2H4H21%1y1.0000A,0y0. 090011AsraAa1xAyA,0yA83.5yA,0 1AyA10. 091110.09或yAyA,01x AyAA.5 %1AyA,0 xAxAyA,0yA ,01Ay110. 0