(15分)在常压填料逆流吸收塔中,用清水吸收混合气体中的氨,混合气量为2000m^3/h,其中氨的流量为160m^3/h,出口气体中的氨流量为4m^3/h(均系操作温度20℃下的体积),平衡关系为Y=1.5X,传质系数K_(y)=0.45kmol/(m^2.h.Δy),(均按摩尔比表示),试求:(1)吸收率η为多少?若吸收剂用量为最小用量的1.2倍时,求溶液的出口浓度。(2)已知塔径为1.2m,内充25×25×3的乱堆填料拉西环,填料有效比表面积约200m^2/m^3,求填料层高度。(3)若使V、Y、η、X不变,而使吸收剂改为含氨0.1%(mol%)的水溶液时,填料层高度有何变化(K可视为不变)。

答案①Y_{1} =160/(2000-160)=0.08696;Y_{2} =4/(2000-160)=0.00217η=(Y_{1} -Y_{2} )/Y_{1} =(0.08696-0.00217)/0.08696=0.975=97.5%(L_{S} /G_{B} )_{min} =(Y_{1} -Y_{2} )/X_{1} =(Y_{1} -Y_{2} )/(Y_{1} /m)=1.5×0.975=1.4625L_{S} /G_{B} =1.2(L_{S} /G_{B} )_{min} =(Y_{1} -Y_{2} )/X_{1} X_{1} =X_{1} ^{*} /1.2=Y_{1} /m/1.2=0.08696/1.5/1.2=0.0483②S=m/(L/G)=1.5/(1.2×1.4625)=0.8547N_{OG} =1/(1-S)ln[(1-S)Y_{1} /Y_{2} S]=13.07G_{B} =(1840×273/293)/(22.4×0.785×1.2^{2} )=67.7kmol/(m^{2} h)H_{OG} =G_{B} /K_{ya} =67.7/(0.45×200)=0.752mh=N_{OG} ×H_{OG} =13.07×0.752=9.83m③x_{2} =0.001,X_{2} ≈0.001N_{OG} ^{’} =1/(1-S^{’} )ln[(1-S^{’} )(Y_{1} -mX_{2} )/(Y_{2} -mX_{2} )S^{’} ]=18.84∵S=1.5/(L/G),(L/G)^{’} =(Y_{1} -Y_{2} )/(X_{1} -X_{2} ^{’} )=1.793S^{’} =1.5/1.793=0.8368H_{OG} 不变;∴h^{’} =N_{OG} ^{’} ×H_{OG} =18.84×0.752=14.17mΔh=h^{’} -h=14.17-9.83=4.34m。