化學工程基礎習題答案2

1、化學工程基礎習題 其次章P69 1.解：P vacP o P絕P oP絕即 13.3 103Pa98.7 103P 絕85.4 103Pa3PaaPPP o 85.4 103Pa98.7 1013.3 103Pa2. 解：de442 d 1d 4d 22d 1d2702d 13. 解：對于穩(wěn)固流態(tài)的流體通過圓形管道,有u2u 1d 2d 2d 1d 2212如將直徑減小一半,即u2u 14即液體的流速為原流速的 4 倍. 4. 解：H fLu212d22d2gHf11L1u 12d 12gHf22L2u22d22gHf22L2u 22d22gHf11L1u 12d 12g64Re64duu 2

2、4u L1L2,d64L2u22Hf2d u2d22gHf164L1u 12d u1d12g64L2u22Hf2d u2d22gHf164L11 u 242 g2 d21u22 d2416Hf21Hf1211112 164Hf216Hf1即產(chǎn)生的能量缺失為原先的 16 倍；6. 解：1）雷諾數(shù)Reud其中1000kg m ,u31.0m s1d25mm25 103m1 cp103Ps s故Reud1000 1.025 1010325000 故為湍流；2）要使管中水層流,就Re2022即Re100025 103m u2022103解得u0.08 m s17. 解：取高位水槽液面為 1-1 ,A-

3、A 截面為 2-2 截面,由伯努利方程z 1p1u2 1z 2p2u2Hf1g2gg2g其中z 110 , m z 22 ;p 1p u 10;Hfhfg就1022u2 216.15u29.89.8解得1）A-A 截面處流量uu 2u2.17m s12） q vAu其中A1d23123.141001032m3h447.85 10muv2.17m s132.17360061.32q7.85 108. 解：對 1-1 截面和 2-2 截面,由伯努利方程得z 1p12 u 1z 2p2u2 11g2gg2g其中z 1z 2,p 11 m H O 2gh 1u 10.5 m s1,p 1gh 2u 2

4、2 d 1u 10.2 0.120.52.0 m sd2 2h0.522220.19m29.89.815. 解：選取貯槽液面為 1-1 截面,高位槽液面為 2-2 截面,由伯努利方程得z 1p1u2 1Hez 2p2u2 1Hf15.468g2gg2g其中：z 12 , m z 210 ; m u 1u 20p 1pvac100 mmHg3 13.6 109.8 0.113332.2p ap 20213332.2He1019.6 9.841000 980gHe12.0819.613332.214.08 1.3889.8980gPHeq V15.468 243 53 10 29800.655kw

5、 10210217. 解：取水池液面為 1-1 截面,高位截面為 2-2 截面,由伯努利方程得z 1p1u2 1Hez 2p2u2 1Hf8.05kw g2gg2g其中：z 10,z150 ; m p 1p 20Hf209.836 1000He502052.059.8PHeq V52.051021020.6360019. 解：取貯槽液面為 1-1 截面,蒸發(fā)器內(nèi)管路出口為 2-2 截面,由伯努利方程得z 1p 1u2Hez 2p2u2Hf11g2gg2g其中,z 10,z115 ;26656p a1.632kw p 10,p 2200 1033 13.6 109.8Hf1209.8He1512

6、02665624.979.89.8 1200PHeqV24.9720 1200102102360020. 解：1）取貯水池液面為 1-1 截面,出口管路壓力表所在液面為 2-2 截面,由伯努利方程得z 1p 1u2 1He2z 2p2u2Hf1g2gg2g其中,z 10,z5.0 ;p 10,p 22.5 kgf cm22.5 9.85 2.45 10 p a0.014忽視出水管路水泵至壓力表之間的阻力缺失,就：衡算系統(tǒng)的阻力缺失主要為吸入管路的阻力缺失：2）PHf0.29.83.0kw u36003610322.2476He5.02.45 1052.220.210009.82 9.89.85

7、.0250.250.0230.27PHeq V30.273610001021023600Heq V3.04.3kw 1020.73）取貯槽液面為 1-1 截面,水泵吸入管路上真空表處液面為 2-2 截面,由伯努利方程得z 1p1u2z 2.p22u2 1Hf1g2gg2g其中：z 10,z4.8 ;p 10,p 2忽視進水管路水泵中真空表至水泵之間的阻力缺失,就：衡算系統(tǒng)的阻力缺失為吸入管路的阻力缺失：p 2Hf0.29.80.2 10009.849600pa 4.82.2229.89.8得真空表的讀數(shù)為Pvac49600 Pa23. 解：1）取低位槽液面為 1-1 截面高位槽液面為 2-2

8、截面由伯努利方程得z 1p 1u2Hez 2p2u2 1Hf01g2gg2g其中,z 10,z220 ;p 10,p 2Hf5,He20525w e25 9.8245J/kg 2）在管路 A、B截面間列伯努利方程得：z ApAu2z BpB2 u BHfg6Ag2gg2 gpApBz Bz ALu2ggd2g2 u2gdpApBz Bz ALggp Ap BHgH2OgRH2OpAp BHgH2ORH2O6H2 OgH2OH2O66u2gdHgH2ORH2 OL2 g0.050.5042 .03m s10.976 kw6 0.023）P25 2.0342 0.0510001024）依據(jù)靜力學基

9、本方程式：p BH2Og6HHggRp 0RHggRpBHggRp0H2 Og6HpAH2Oghp BH2 Og6H2Og hpApBH2Og6H2OgRHggRH2OgRpAHggRp0H2 Og6Hg6H2OgRHggRH2OpAp0HggRH2OgHHg13.61.21 113.610.049.810001.55105p a第三章 傳熱過程p105 ex1解：qtt 1t4115030W1R1230 2.01.0 .0060.1871231.070. 1445112011201243Wm20 .7140 .00010 .901q 3001120RR 03 . 73C2 mRR 0R 03

10、 . 73.0 901.2 83Cm2W1ex2解： T 1T 41 T 1T 4L T 1T 42i2iiAr LiL T 1T 421r ir i11lnr i1r i1r ilnr ir iiir i226035 1ln85L1ln251ln550162002.25.075523 .1422514130 .2230. 7881. 4350. 01393 .946.214160 .20. 0072202180200 C1413138.96 Wm110. 168Ex4 解：空氣的定性溫度T200時空氣的物性參數(shù)為：3.912 10 W m1K13 0.746 Kg m2.6 105aP sC

11、 p1.034 KJ Kg1C/3 mu15m s ,R edu0.0254150.7461.09 1042.6 105P rCp1.0341032.6 1050.683.931 1020.023dR e0.8 P r0.30.0233.931 1024 1.09 10 0.80.3 0.680.02542 53.8 W mKex5解：水的定性溫度T4022030C30時水的物性參數(shù)為：995 .7 Kg0 . 6176Wm1K180 . 07105PasC p4.174 KJ Kg1C0.023dR e.08P r.04當u1 ms1時,1Redu0.021995.724868.680.07

12、105P rCp4 .17410380.071055. 410.61760 . 0230 . 617624868 6. 0.8 .541 0.40 .0024583 . 5 W2 mK當u0.3時Redu7460. 58,此時,2022Re10000 0 . 02338 .851 . 96507460 . 588. 165 1087460 .1 581638 W2 mCex7解：甲烷的定性溫度：T12023075C0條件下：0 . 717Kg/3 m0 .03 Wm1K1 . 03105PasCp.1 7 KJKg1K1由于甲烷是在非圓形管內(nèi)流淌,定性尺寸取 de de4流體截面積 潤濕周邊4

13、.02 019.002554420 . 19370 . 19370 . 019Redu0.0255100. 717177287.1 .03105P rCp1.1031. 031050. 5840 .03由于甲烷被冷卻,n0 .3584 03.0 . 023deR e.08P r.0 30 . 023.0 03 17728 . 7 08. 0 .0 . 02550 . 023.11762505 .770 . 85157 1. Wm2K1如甲烷在管內(nèi)流淌：Ex8 Redu0.015100 .717C10441.751. 0310564.2 Wm2K1T 140 1525C ,1303397C（逆流

14、）T 2T 2T 2T 1972553.1T mlnln9725T 1在依據(jù)折流校正P33150.1613015R1304053315 0.97 T m0.97 53.151.5Cex9 1 q mcpT 1T 2,C1 .2519.1038030118750Js12 T 1302010C4040C21 . 64T 280T mT 2T 14010T 2ln4lnT 13 R11d0d0R siR s 00 . 21103.0176103K0油d idmK10 . 0250 . 00250 . 025.1 73 10.0 853 10.0 02450 . 0250 . 02ln0 . 025.

15、0 02.02 00251 mKW11 R398.91 Wm2K14 AKtm118750.64395.912113.74m2如近似按平面壁運算,就R1 .110 .1100R siR s 00. 211030 .176103K0油imK1130.002571038545453.26A12. 11m2ex10 1 d016103m,d i13103m,1 .5103m1 90K1d 0d 01id id m0113161 5.1031610004014 5.11 . 2311030 . 04141030 .01181 5. Wm2K12 K01d0d011id idm0113161 . 510

16、31610004014 . 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 q mCpp1270 qmCpq m Ct 1 tp2 t 1t35, 2t502tm1353069.81,A 2k 21mln135

17、tk2tm301.855又流量及物性不變,k 1k2A 270 q Cpk2t m70 92.56475A 150 q C mpk 1T m50 69.83490.7L2A 2.1 855,L2.1855 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 mh3 1.2

18、79 104.6 kg h吸取 p187 ex 解：（1）P A2026P a31001032.300mol m33P a1CA7.821 10341000P ACAHCA2.3001.135 103mol mHP A2026（2）EP100027 4.895 10P aHM1.135 1031.8 10P A7 4.895 10 xA（3）mE.48957 10483 . 22P 總.1 0135 10（4）總壓提高一倍, E、H 值均不變mE.48951072421P2 . 0265 10ex9解：5y64 9550 .02329Y29640 . 02329y 11y 11.0 02329

19、.0 02329第一解法：q n Cq n B min0Y 1Y 22q n CminY 1Y 2X 1X又設Y 2q n BY 126.7X2X1Y 126.7又q n CY 1Y 21.65Y 1Y 2q n BX 1Y 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：設吸取率為a就,Y 21Y 1進氣量設為. kg/h512a .5%q n Ba. 95%kmolh29 103qn C minY 1XY 2Y 11 Y 126.7qn B1

20、Y 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）1Y0 . 02Y2Y 1199 %0 . 0002X20當液氣比為 2.2時,q n Cq n BY 1Y 220.020.00022.2X1XX 1X1.0 009Y 1Y 2Y mY 1Y 10.02 0.0090.011Y 2* Y 20.0002Y 1Y 20.011 0.00020.002695lnY 1ln0.011 0.0002NOGY 2Y 1Y 20.020.00027.35Y m

21、0.002695（2）當液氣比為 1.25時,X10. 0299%0.015843 當 X 2=0.00011時1. 25NOG15.17X111Y 1Y 2X2.250 .015840. 000110 .01595NOG0 .020.000219.030 .0010404 當液氣比為 0.8,溶質(zhì)的最大回收率時溶液出口達到氣液平穩(wěn),q n CY 1Y 20.02Y 20.811.2544q n BX1Y 11Y 20.004Y 1Y 1Y 20.020.001680%0.02ex11. 1Y10.030.030930 .03Y 2Y 11 98%0.03093 0.020.0006X20q

22、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 1m.0 030931 . 280 . 016.001045Y 2Y 2Y 20 . 0006Y mY 1Y 2lnY 1Y 2.0010450 . 00060 . 003465ln.0010450 . 0006NOGY 1Y 2Y m.0

23、030930 . 00068 . 8.0 003465HOGq n B65.160.6503KYS601.67 1.0HHOGN OG8.80.65035.7212.1 Y 111y 110.050.0526y 10.05Y 21y210.002630.002637y 20.00263X61.2 58 1000 61.2 180.02033X20Y 2.0 XY 1 Y 1 Y 10.0526 2.0 0.02023 0.01214Y 2 Y 2 Y 2 0.002637Y m Yln 1Y 1 Y 2 0.01214 0.002637ln 0.01214Y 2 0.0026370.0095

24、0.00621.527N OG Y 1 Y 2Y m0.0526 0.0026378.060.0062q n B 0.5563 0.95 2 7322.4 10 298121.602 mol sH OG q V 21.6022 0.744K Y S K Y 0.843 1K 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=3885.37mol m=58 3885.37=225

25、.4kg （1）另解：q n CY 1Y 22.4698q n BX 1X20.0526 0.0026369 0.02023氣相傳質(zhì)單元數(shù)：NOG11n Bln1mqn BY 1mX2mq n Bmqq n CY 2mY 2qn Cqn C112ln12 2.46980.0526020.0026372.46982.46988.028HOGOGHn B6Y 20 . 7474 m21.602d23 157.5 mol m sNOG8 . 028又HqKqn BKY S 2HOGs0.74744（3）如填料層增加 3m,就：N OGH912 . 042 m0 . 05262HOG.0 7474N

26、OG11ln12 2 . 46982Y 22 . 4698Y 22 . 4698.0 0011又 液氣比肯定,就：Y 1Y 22.4698,X10.02085X 1nX1q n B0.02085 53.35 36004004.45 mol就：m58 n 232 . 26 kg.6 9kgm232 . 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 Y 1Y 2qn CX 1X212.410.07530.0

27、015q n C0.01960q n C46.72mol s10.84kg s1（2）H H OG N OGY 1=0.073 Y 2=0.0015 X 1=0.0196 X2=0 Y 1 *=1.15X1=0.0225 Y 2 *=1.5X2=0 Y m Y 1 Y* *ln Y 1 Y 1Y 2 Y 2 *N0.0730.00250.00150.018ln0.0730.00250.0015OGY 1Y 2Y m0.0753 0.0015 0.0184.10.5 1000HOG22.40.845 m26.41H4.1 0.8453.46 m14 解：（1）H OG HN OGY 1=0.02

28、5 Y 2=0.0045 X 1=0.008 X 2=0 Y 1 *=1.5X1=0.012 Y 2 *=1.5X2=0 Ym Y 1 Y 1 * *ln Y 1 Y 1Y 2 Y 2 *0 . 025 0 . 012 0 . 00450 . 0080 . 025 0 . 012ln0 . 0045N OG Y 1 Y 2Y m0 . 025 0 . 00452 . 560 . 00810H OG 3 . 91 m2 . 56（2）當 Y 2 0 . 003 時q n C Y 1 Y 2 Y 1 Y 2q n B X 1 X 1.0 025 0 . 0045 0 . 025 0 . 0030

29、. 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 . 003YmY 1Y 2第六章精餾lnY 1Y 2.001210 . 003.0 0065ln0 . 0121.0 003NOGY 1Y 2Y m0 .0250.0033.380. 0065HHOGNOG3.91 3.3813.2 mH13.2 103.2 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 D095q n Wq n D27

30、.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 h14 .190 .410550.6232.3qn FxFqn Dq n Wqn Wxwqn Fq n Dx D 235q n Dq n W235 0.840.98 q n D0.002 q n Wq n D201.4kmol.h1q n W33.6R,11q n L201.

31、4q n Lq n Lq n F436.4kmol h1q n Vq n Lq n D402.8q n Vq nV402.8kmol h1475.3300 .42. 6806.04.105506.23201 .0945精: yR1xXd0 . 225RR13x.09.075 x44y1xfx1進:.0 8.10.12y4x0 .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 x0 . 2101y0 5.x0 . 66

32、10 5.31 . 5 xf0 . 66fx0 . 44交點x0 .36y. 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 化學反應工程基本原理P340 （1）dpA12 . 58106p2dtApAp2 Ahph1ApvnRtcARt22. 314450npcvRtdpAkpp2AdtdcARtkpdtdcAkpc2RtkccdtAAkckpRt2.5810689. 65103m3kmolh1k c2 c Akmolm3h1kmol2kmol1m3h13 m3 mkmol1h（2）C2H6C2H4H21%1y1.0000A,0y0. 090011AsraxAa1yAyA,0A.5yA,0 1AyA8310. 091110.09或yAyA,01x AyAA.5 %1AyA,0 xAxAyA,0yA ,01Ay110. 09