化工原理习题第四部分吸收答案

G?100?4.464kmol/h22.4

0.1?1000L?5.556kmol/h18L5.556??1.245 G4.464mG0.526??0.4225 S?L1.245操作液气比

??Y?mXa1ln?(1?S)b?S?1?S?Ya?mXa? 10.026???ln?(1?0.4225)??0.4225??3.1581?0.4225?0.0026?NOG?h?HOGNOG?1.2mHOG 1.2??0.3799m3.1587Kya?GHOG??4.4640.3799??0.22?36004??0.1039kmol/(m3?s)

(2)最小液气比为

Y?Ya0.02?60.0026?L???0.473 ???b 4*GX?X0.026/0.?5260??minba

L/G1.245??2.6 3(L/Gm)in0.4734(3)Ya=0.001,液气比不变,所以S不变

??Y?mXa1ln?(?1Sb)?S?1?S?Ya?mXa?

10.026??ln?(?10.42?25)?1?0.42250.001?NOG??0?.4225??

4.739HOG?0.3799m

h?HOGNOG?0.3799?4.739?1.8m

?1?.2填料层增高 ?h?1.8m0 .6

18、在差压逆流操作的吸收塔内用纯溶剂吸收混合气体中的溶质,入塔气体中溶

质的组成为0.03(摩尔分率,下同),要求溶质的回收率达到95%,已知操作条件下,mG/L=0.89全塔范围内可取为常数),与入塔气体成平衡的液相组成为0.031.

试求(1)操作液气比与最小液气比之比; (2)出塔洗手液的的浓度;

(3)完成上述任务所需的气相总传质单元数NOG。

【解】(1)Yb?yb0.03??0.031 1?yb1?0.03xb*0.031X???0.032

1?xb*1?0.031*bYa?Yb(1??)?0.031?(1?95%)?0.00155

Y?Ya0.031?0.00155?L???0.922 最小液气比为:???b*GX?X0.032?0??minba m?所以

Yb0.031??0.96 9*Xb0.032Lm0.969???1.21 1G0.80.8L?L????1.211/0.922?1.313 G?G?min(2)

Y?YL0.03?10.00155?ba??1.21 1GXb?XaXb求出吸收塔液的浓度 Xb?0.0243xb2?,/L?0. 8(3) S?mG 20.0所以传质单元数

?Y?mXa1?ln?(1?S)b?S?S?Ya?mXa? 10.031???ln?(1?0.8)??0.8??7.8431?0.8?0.00155?NOG?1?

19、用含苯0.02%的没有在内径为1.5m的填料塔中逆流吸收某混合气体中的苯

蒸汽,入塔气体中含苯3%,(摩尔分数),混合气体流率为0.022kmol/s,要求苯回收率不低于99%,已知操作条件下相平衡关系为y*=0.4x,吸收剂用量为最小用量的1.5倍,总传质系数为Kya=0.015kmol/(m3.s) 试求(1)煤油的用量 (2)所需填料层高度

【解】(1) 由题可知,xa=0.02% ,yb=3% ,

Yb?yb0.03??0.0311?yb1?0.03

Ya?Yb(1??)?0.031?(1?99%)?0.0003最小液气比为

Yb?Ya0.031?0.0003?L????0.397 ??*?G?minXb?Xa0.031/0.4?0.0002LBmin?G??0.397?0.022??0.0049kmol/(m2?S)

40.022?(1?0.03)?1.52??0.0121

4mG0.0121?0.4??0.659 (2)S?L0.00735??Y?mXa1ln?(1?S)b?S?1?S?Ya?mXa? 10.031?0.4?0.0002???ln?(1?0.659)??0.659??11.361?0.659?0.0003?0.4?0.0002?NOG??1.52HOG?G0.0121??0.81m Kya?0.015所以填料层高度为

h0?HOGNOG?0.81?11.36?9.2m

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