第八章气体吸收（Chapter8AbsorptionofGases）习题课习题课例8.3常压下，用煤油从苯蒸汽和空气混合物中吸收苯，吸收率为99%，混合气量为53kmol/h。入塔气中含苯2%（体积%），入塔煤油中含苯0.02%（摩尔分率）。溶剂用量为最小用量的1.5倍，在操作温度50℃下，相平衡关系为y*=0.36x，总传质系数Kya=0.015kmol/(m3s)，塔径为1.1米。试求所需填料层高度。设计型举例设计型举例Example8.4Gas,fromapetroleumcolumn,hasitsconcentrationofH2S(sulfuratedhydrogen)reducedfrom0.03kmolH2S/kmolofinerthydrocarbongasto1percentofthisvaluebyscrubbingwithatriethanolamine(三乙醇胺或三羟乙基胺)-watersolventinacountercurrenttower,operatingat300Kandatatmosphericpressure.TheequilibriumrelationforthesolutionmaybetakenasYe=2X.ThesolvententersthetowerfreeofH2Sandleavescontaining0.013kmolofH2S/kmolofsolvent.Iftheflowofinertgasis0.015kmol/m2•softowercross-section,calculate:(a)theheightoftheabsorbernecessary,and(b)thenumberoftransferunitsNOGrequired.TheoverallcoefficientforabsorptionKYamaybetakenas0.04kmol/m3•s•ΔY(unitmolefractiondrivingforce).SolutionDrivingforceattopofcolumn=Ya-Ya*=0.0003Drivingforceatbottomofcolumn=Yb-Yb*=0.03-2×0.013=0.004Logarithmicmeandrivingforce=(0.004-0.0003)/ln(0.004/0.0003)=0.0037/2.59=0.00143Fromequations8-42and8-44:GB(Yb-Ya)S=KYa(Y-Y*)avShthatis:0.015(0.03-0.0003)=0.04×0.00143h∴h=0.000446/0.0000572=7.79=7.8m(say)HeightoftransferunitHOG=GB/KYa=0.015/0.04=0.375mNumberoftransferunitsNOG=h/HOG=7.79/0.375=20.7=21操作型举例操作型举例操作型举例操作型举例操作型举例操作型举例至于xb：仍需用排除法判定，下略。操作型举例操作型举例操作型举例操作型举例操作型举例至于xb：仍需用排除法判定，下略。练习在逆流操作的填料吸收塔中，对某一低浓气体中的溶质组分进行吸收，现因故(1)气体入塔浓度yb变小，(2)G变小，而其它操作条件均不变，试分析出塔气体、液体浓度如何变化？解法一：快速分析yb变小，ya必变小。关于xb：排除法(2)G变小时作业在逆流操作的填料吸收塔中，对某一低浓气体中的溶质组分进行吸收，现因故(1)气体入塔浓度yb变大，(2)G变大，而其它操作条件均不变，试分析出塔气体、液体浓度如何变化？操作型举例操作型举例操作型举例