1-0.05=00526¥2≈0.00263 Y? 6.12/58 (100-6.1218 =0.02023 Y1=2.0X1=2.0×0.02023=0.04046 A≈(00526-004046)-000263 6.22×10 0.0526-0.0406 Y1-Y20.0526-0.00263 =8.0 8.0=0.75m qn=2.4x298×(1-0026)=77 kmolh Kya= gn. B= 206 kmol-m3h-l 0.75 0.82 回收丙酮=qBn(Y1-Y2)=77.7×0.0526-0.00263)=3.88 kmol h 3-27混合气中含10%CO2,其余为空气,于303K及2×10Pa下用纯水吸收,使CO2 的浓度降到0.5%,溶液出口浓度x=0.06%(以上均为摩尔分数),混合气体处理量为 2240m3h1(标准状态的体积),亨利系数E为2×10kPa,液相体积传质总系数Kd=2780 kmoh1m3。问: (1)每小时用水为多少吨? (2)填料层高度为多少米?(塔径定为1.5m) 解:(1)qn 2240 Y2≈y2=0.005Y1 1-0.1 0.1l1 qn, d (X1-X2)=qn, B(Y1-Y2) qnc=90×0.11-0.005 =1.59×104 koml- h 用水量=159×10-×18=2862th 1000 (2) E 100 P2×10 y_0.1 =0.001 X1≈x1=0.001 x2=20005 1009 解:Y1 = 0.05 1 - 0.05 = 0.0526 Y2 0.00263 Y2 * = 0 X1 = 6.12/58 (100-6.12)/18 = 0.02023 Y1 * = 2.0X1 = 2.0 0.02023 = 0.04046 3 6.22 10 0.00263 0.0526 0.0406 ln (0.0526 0.04046) 0.00263 − = − − − Ym = NOG = Y1 -Y2 Ym = 0.0526 - 0.00263 6.22 10-3 = 8.0 HOG = H NOG = 6 8.0 = 0.75m qn, = 2000273 22.4298 (1- 0.0526) = 77.7kmol·h -1 ∵ HOG = qn.B KYaS ∴ KYa = qn.B HOGS = 77.7 0.75 4 0.82 = 206 kmol·m-3·h -1 回收丙酮 = qB,n ( Y1 - Y2) = 77.7 (0.0526 - 0.00263)= 3.88 kmol·h-1 3-27 混合气中含 10%CO2,其余为空气,于 303K 及 2 103kPa 下用纯水吸收,使 CO2 的浓度降到 0.5%,溶液出口浓度 x1 = 0.06%(以上均为摩尔分数),混合气体处理量为 2240m3·h -1(标准状态的体积),亨利系数 E 为 2 105kPa,液相体积传质总系数 Kxa = 2780 kmol·h-1·m-3。问: (1)每小时用水为多少吨? (2)填料层高度为多少米?(塔径定为 1.5m) 解:(1)qn,B = 2240 22.4 (1 - 0.1) = 90koml·h -1 X2 = 0 X1 x1 = 0.0006 Y2 y2 = 0.005 Y1 = 0.1 1 - 0.1 = 0.111 ∵ qn,c(X1 - X2) = qn,B(Y1 - Y2) ∴ qn,c = 90 0.111 - 0.005 0.0006 - 0 = 1.59 104 koml·h -1 用水量 = 1.59 104 18 1000 = 286.2 t·h -1 (2) m = E P总 = 2 105 2 103 = 100 ∵ x1 * = y1 m = 0.1 100 =0.001 ∴X1 * x1 * = 0.001 同理:X2 * x2 * = y2 m = 0.005 100 = 5 10-5