化工基礎(chǔ)第四版上冊(cè)課后習(xí)題答案福建師范大學(xué)上海師范大學(xué)吸收

1、1、30C水的飽和蒸氣壓= 4.24kPa,（p268, 表五（一）求濕空氣中水蒸氣的體積分?jǐn)?shù);4 24x0.042101.30.042 18求水蒸氣的質(zhì)量分?jǐn)?shù):Xm0.042 18 0.958 20.0265求濕空氣的密度:J042 18 °.958 29 丿=1.15 竺273 + 3022.4m32、亨利定律為p =Ex 28600 “.0515%100018197000101.3 4 =197MPa5.15 10相平衡常數(shù);101.3= 19453、根據(jù),Exp =Py空U3 = 0.0232%188000(mol%)0.0232% 44質(zhì)量分?jǐn)?shù):0.0232% 4499.9

2、768 18-0.057%0.03 202.6267= 2.28%(mol % )質(zhì)量分?jǐn)?shù)2.28 17x 二2.28 1797.72 18= 2.2%5、進(jìn)入水的最高含氧量 水的最高含氧量是空氣中氧（21%）相平衡的含氧量。p 0.21 101300x 二E94.06 10= 5.24 105.24 105.24 曲 汛9.32 伏 g =9.32 93 m18kgHzO脫氧后的最低含氧量：脫氧后水的最低含氧量是與吹掃地氮?dú)庵泻跸嗥胶獾暮趿?。x=D02 101300 丸.499 忻E0.499 10“ 3294.06 10= 0.88718J 0.12 丫10.1361 - y 1-0.

3、12被吸收到氨量：1000(1-0.12)(0.136-0.0101)=110.8m丫20.011-0.01= 0.0101皿型 17= 77.03kg22.4298出口氣體體積：V=1000-110.8=889.2m 3E = =16500 =550kPa7、0.036、E對(duì)于稀溶液，Ex = mxpMsm10001812、110001.1.01 10-4 k18 550000m3.PaE 550000.43P 101300理=14705.43m =55鳴5.4310130044 1470 = 3.59 kg，（環(huán)氧乙烷分子量=44） 穩(wěn)態(tài)下通過(guò)氣膜的擴(kuò)散系數(shù)（分子擴(kuò)散的斯蒂芬定律）D p（

4、P1 - P2） PbRmolm .S p1 =20 C水的蒸汽壓=2330Pa：1Pb, 因 為40.257 10中 水 蒸 氣 分 壓 小 。A 8.314 293 0.005單位面積（1m2）盤中的水量:(2330 -0)“.00492矍m =1 0.005 J =5kg = 0.2778kmol = 277.8 啤m全部蒸干所需時(shí)間:13、p2=2330X 60%單位面積（1m2）盤中的水量:277856460S = 16h0.004920.257 10 °亍 2330(1 - 0.6) = 0.04928.314 293 0.2 10mol2m2.Sm =1 0.0051

5、=5kg =0.2778kmol =277.8啤m丄 m277.8/ 小t5646S = 1.6hN 0.0492全部蒸干所需時(shí)間：A14、30 C時(shí)的擴(kuò)散系數(shù)（根據(jù)富勒公式）：/,、1.75'273 +30 1 | =1.06 03 +20 丿D30D2030C水的蒸汽壓=4240Pa叢二旦衛(wèi)巾A RPb根據(jù)斯蒂芬定律：N D30 p 1.06 4.24 (1 一0.5)小1.93N D202.33(1-0.5)kl 1. t 30Nt15.5ht1.931.9315、將平衡數(shù)據(jù)換算成比摩爾分率以第一組為例 然后作出Y X圖。16、將溶解度換算成:100；73®10丫C p

6、關(guān)系:C 少OlNH 3以第三組數(shù)據(jù)為例，換算成X Yd的關(guān)系，仍以第三組為例,y =p/P丫叮y1/170.006610.00665 1 - y 1 -0.006613m3溶液(1100)/1000 _0.58km°l/m= 0.8/101.3 =0.0079°.0079=0.007961 -0.0079X=molNH 3 二0106100/18molH 2OXaPaYaP304 0.1517、Pa 二 PaXaPaPa1084= 0.042X1Xa1 - X1皿乙=0.0441-0.042丫10150.17651-0.15對(duì)于完全吸收,V惰氣流量：丫2=0，吸收劑中不含

7、溶質(zhì)，X2 = 01(1 0.15) =0.038kmol/S22.4LmV(¥ -丫2)VY1X； -X20.09X1O.038 O.1765 =0.152kmol/S =0.152 2200.04433.5kg/S0.0991-0.09Y2=Y1(1-0.99)=0.00099，由題知，ymx0.090.0033726.7清水吸收，X2=0丫1 -丫20.099 - 0.00099V m X1 -X20.00337-29.083L/V =1.2(L/V)m =1.2 29.083 = 34.9吸收SQ量:11kg/S 二丄 kmol/S641V(Y1 -丫2)=V (0.099

8、-0.00099)64V =64 0.0981= 0.159L =34.9 0.159 =5.55kmol/S =99.5kg/S吸收所得溶液濃度:19、199.5yx= 3700 x=36.53xP 101.34y0.0395P = Py101.30.03950.0010836.530.03950.04111 -0.03950.001081 -x1-0.001080.°.°°2,43%20.02512730.030.0416kmol0.000375kmol求平均推動(dòng)力：22.42938020、吸收所得溶液濃度:0.000375yi0.00890.0416 0.0

9、00375丫1Sy=空塵=0.008981 -0.00890.005/80y2= 0.00150.04160.005/80丫2二衛(wèi)0.00151-0.00150.0020.0021-0.002將題給平衡關(guān)系數(shù)據(jù)變成（X,X2y2 =以第二組數(shù)據(jù)為例：丫）作圖。0.005/800.0416 0.005/80 一 0.0015亠 二竺生"0015 X 05/80 .0.01261 -y 1 -0.001599.5/200作圖近似為直線。丫 = mX ,以第三組數(shù)據(jù)求 m丫 0.0030.12X 0.025丫1X1X；亠空叫0.075 X2 = mX1m 0.120.03 -0.0052

10、0.002/800.03630.0015 =0.01250.122 0.998200丫1= mX1 =0.12 0.0363 = 0.00436= mX0.12 0.002 =0.00024Y = Y1 Y* )“2 丫；)(0.00898 0.00436)(0.0015 0.00024),丫1-丫1In -丫2 -丫20.00462 -0.00126= 0.00259,0.00898-0.00436 ln0.0015-0.00024,0.00462 ln0.00126x1* -X1 -(x2 -X2)_ 0.075-0.0363 -(0.0125 - 0.002),X1 -X1 In *X2

11、 -x20.0387 -0.01050.0216_ 0.075-0.0363 ln0.0125-0.00221、乙炔:26g/mol，乙炔溶液0.3g/1000g 水 空=0.25 1 - y 1-0.203/262.077 101000/18664平衡分壓：p*=0.135 X 10X=0.135 X 10 X 2.077 X 10- =28.0kPa當(dāng)時(shí)氣相中乙炔分壓：p=Py=101.3 X 20%=20.3kPa/ p*>p ,乙炔由溶液向氣相傳遞分解吸)(部28(2)= 0.382*101.3-28氣相推動(dòng)力 Y : Y -Y =0.25 -0.382 - -0.13222、R

12、eSc4 0.81 =0.035m93°.°35 02 1.215073000.018 100.018 10=Uo / ； = 0.5/0.81 = 0.62m/ S由于氨的濃度很稀， pB P41.25 0.155 10-0.93DP 0.90.5Kg =0.015Re ScRTdePB二 0.0150.155 10* 101.38.314 283 0.035 101.30.915070.930.5=2.82 10，725 0.961= 1.97 101 1+23、( 1)、求 KgKgkgEk|Kg =2.99 10 "(2)、 kG ： Kg , 液相阻力只

13、占總阻力的0.025/32 Y124、1/22.4_6mol2m .S.Pa13.0 10J09弓=3.3466 1090.0015 5 10Y1-Y1YmX1V惰氣量：kmol2 m .S.Pa為氣膜控制，屬于易溶氣體0.4%22.4 0.025 ccl0.01751.33 107"0.4%334.66 1032但遼=0.01781-0.0175X1Y1丫2 =Y(1 - 0.9) = 0.0178 0.1 =0.001780.01780.0155*1.15, X2=0, 丫2 =mX2 =022.4(1 -0.0175) =0.0439S一 _V(Y1 -Y2)0.0439(0.

14、0178 -0.00178) 0.0453X； -X2L =1.3Lm =0.059km010.0155X1 X 2Y _V(Y1 池)X10.0439(0.0178 0.00178) _00120.059Y1* =mX1 =1.15 0.012 =0.0138對(duì)數(shù)平均推動(dòng)力：蟲(chóng)丫Y1丫2 一丫*)_ (0.0178-0.0138)-0.00178 000274lnS丫2 一丫2,0.0178-0.0138 ln0.001780.00274丫一丫2O.°178 -°.00178 =5.85丫0.010.020.030.040.050.06X0.00850.01650.024

15、00.03140.03840.0452Y=0.05 , Y2=Yi(1-0.9)=0.005氣相傳質(zhì)單元數(shù)：25、上列平衡關(guān)系換算得八舟(1一°.0476 八 °.°425 罟X2=0， X1 =00384 （由平衡線上丫1查得）LmV" -丫2)0.0425(0.05 0.005) ° 帖 kmol-.SX-X20.03841.5LmYi丫2X1-X2X1V(Yi -丫2)1.5Lm丿0425©05 -0.005)“ms1.5 0.05作圖,作線。1)、根據(jù)題給Y、X數(shù)據(jù)作出平衡線。（2）、根據(jù)（Xi、Yi）、（X2、丫2）作出操0

16、.0138(0.015 -0.0003)VV0.08(3)、從平衡線和操作線間在丫1=0.05，Y2=0.005范圍內(nèi)求得各Y及其對(duì)應(yīng)當(dāng)垂直距離、Y對(duì)Y -丫作圖，曲線下的面積,y-y*。(4)dY* =2.940.4060.3060.2060.12250.10.44 = 4.5205丫 -丫氣相傳質(zhì)單元數(shù)為 4.526、用解析法1.50.015101.3 -1.5，丫2=0.015(1-0.98)=0.0003，X2=0S.0138 鯉'29SLmV" -丫2)0.0138(0.015 - 0.0003) .kmol0.0103SX； -X20.015/0.76L=1.5

17、x 0.0103=0.0155kmol/S=0.28kg/S溶液出塔濃度:X1V(Y -丫2)= 0.0138(0.015 -0.0003) =0 01310.0155Y1 =0.76X0.010 Y2 =0.76X2 =0J 3 二 0.00167*,丫1 -丫1In *丫2 -丫2,0.015-0.010 In0.0003K2A0 %27、hGKYaA0hGaA01.6 146(0.785 1.52)= 1.94 10，kmo|.'Ym2SK V" -丫2)K Y 二aA0 H.YmYi2000(1 -0.05)273_0 0216kmol一 22.4 3600 273

18、25 一 . S0.050.00250.0526 丫20.00251 -0.051 -0.0025X1*Yi065/58 =0.02021/18( 58為丙酮的mol質(zhì)量)= 2X“ = 2 漢 0.0202= 0.0404 丫* = 0(丫 _¥* )丫2 -Y* ) (0.0526-0.0404)-0.00251 1 2 2 0.00612*丫1KyIn丫2 -丫20.0216(0.0526 -0.0025)ln。.0526-。.04040.002529、4 kmoI,Y-2=7.010 2(0.9 93)(0.785 0.8 ) 6 0.00612m S(1)、填料層增高3米后

19、，因V" -丫2)0.0216(丫1 -丫2)9 =2KYaAcAY：0.0007(0.9 73)(0.785 x 0.82 工丫整理得，：Ym =0.0815(丫1 -丫2)V(£ -丫2)0.0216(0.0526 -0.0025)L0.05360.0202(2)、求 丫2L 0.0536“2.48V 0.0216平衡關(guān)系：Y=2.0X* 2Y1 =2X1(0.0526 -丫2) =0.0424 -0.8072.48丫1 一* 一丫2 =0.0815" -丫2)Xi%1=乞(丫1-丫2)1 L 122.48(0.0526 -丫2)，X2=0= 0.0815&q

20、uot; -%)丫20.0526 - (0.0424 - 0.807丫2) -丫2|n 0.0526 -(0.0424 - 0.807%)= 0.0815(0.0526-%)丫20.0102-0.19362 =0.00429 - 0.0815丫2 ,0.012 0.807Y22In丫2CD(3)、求丙酮回收率丫 = °.0526 一0.001 =98 1%0.0526，用試探法求丫2出口氣體含丙酮0.1 %¥與原條件相比:（丫1 -丫2）/丫（丫1 一丫2）/丫0.0526 -0.0025塔增高后回收率增加 3%00526 弋001=1.0330、氣相傳質(zhì)單元數(shù)為,丫 -丫

21、2HnG6/2=3hG3也Ym，Y1 =0.03/(1-0.03)=0.0309物料衡算關(guān)系得，V(Y1-Y2)=L(X 1-X2)=Lh( X1 -X2)/ Y=mX 二 X=Y/m=Y/18Lm(丫嚴(yán)2円.4皿1"2)X2=0空嘰 0.0012318 1.4X10.0309/18=1.4X 1 ,Y1* =18x0.00123=0.0221 Y2 =mX2 =0z(丫 -Y* )-丫2 -丫2*)(0.0309-0.021)-丫2：Ym =*,丫1十 InY2-Y2用試算法求算，求得出塔氣體含V 丫 一丫2I 0.03090.021=3InSO為 0.71%HK 丫 aA0 包 丫 m31、Y=0.03，丫2=丫1 (1