=1.013times (10)^5Pa,-|||-_(1)=1330Pa , _(2)=7P(a)_(2),-|||-_(m)=1020kg/(h)_(2) , _(e)=0.755X,-|||-dfrac ({q)_(nl)}({q)_(nG)}=2(dfrac ({q)_(nt)}({q)_(n)G})_(min)-|||-therefore (y)_(1)=dfrac ({P)_(1)}(P)=dfrac (1330)(1.013times {10)^5}=0.01313-|||-_(2)=dfrac ({P)_(2)}(P)=dfrac (7)(1.013times {10)^5}=6.910times (10)^-5-|||-_(1)=dfrac ({y)_(1)}(1-{y)_(1)}=0.0133-|||-_(2)=dfrac ({y)_(2)}(1-{y)_(2)}=6.910times (10)^-5-|||-_(n)c=dfrac ({Q)_(m)v}(M)=dfrac ({Q)_(mn)}(Mg{y)_(1)+(M)_(min)}(1-(y)_(1)) _(n)((X)_(1)-(Y)_(2))=(a)_(n)((X)_(1)-(X)_(2))-|||-Itdan+24+100301 _(1)=dfrac ({G)_(mG)}({G)_(m)}((X)_(1)-(X)_(2))+(X)_(2)-|||-Itdan+24+100301-|||-(qnL )min=2·(1-√2)/(n-2)=2 Y1-Y2-,Y1-Y2 =dfrac (0.0133-6.910times {10)^-5}(1.502)+0-|||-m =8.809times (10)^-3-|||-m-|||-=2cdot dfrac (0.0133-6.910times {10)^-5}(0.0133/0.755-0)=1.502-|||-_(1)=(C)_(M)cdot (x)_(1)=dfrac ({P)_(s)}(M,)cdot dfrac ({X)_(1)}(1+{X)_(1)}-|||-_(n)=1.502(a)_(n6)=1.502times G(1-(y)_(1))-|||-=dfrac (998.2)(18)cdot dfrac (8.809times {10)^-3}(1+8.809times {10)^-3}-|||-=51.72kmol/h=930.96kg/h-|||-=0.4842kmol(m)^3_(2)=0, . =1.013times (10)^5Pa,-|||-_(1)=1330Pa , _(2)=7P(a)_(2),-|||-_(m)=1020kg/(h)_(2) , _(e)=0.755X,-|||-dfrac ({q)_(nl)}({q)_(nG)}=2(dfrac ({q)_(nt)}({q)_(n)G})_(min)-|||-therefore (y)_(1)=dfrac ({P)_(1)}(P)=dfrac (1330)(1.013times {10)^5}=0.01313-|||-_(2)=dfrac ({P)_(2)}(P)=dfrac (7)(1.013times {10)^5}=6.910times (10)^-5-|||-_(1)=dfrac ({y)_(1)}(1-{y)_(1)}=0.0133-|||-_(2)=dfrac ({y)_(2)}(1-{y)_(2)}=6.910times (10)^-5-|||-_(n)c=dfrac ({Q)_(m)v}(M)=dfrac ({Q)_(mn)}(Mg{y)_(1)+(M)_(min)}(1-(y)_(1)) _(n)((X)_(1)-(Y)_(2))=(a)_(n)((X)_(1)-(X)_(2))-|||-Itdan+24+100301 _(1)=dfrac ({G)_(mG)}({G)_(m)}((X)_(1)-(X)_(2))+(X)_(2)-|||-Itdan+24+100301-|||-(qnL )min=2·(1-√2)/(n-2)=2 Y1-Y2-,Y1-Y2 =dfrac (0.0133-6.910times {10)^-5}(1.502)+0-|||-m =8.809times (10)^-3-|||-m-|||-=2cdot dfrac (0.0133-6.910times {10)^-5}(0.0133/0.755-0)=1.502-|||-_(1)=(C)_(M)cdot (x)_(1)=dfrac ({P)_(s)}(M,)cdot dfrac ({X)_(1)}(1+{X)_(1)}-|||-_(n)=1.502(a)_(n6)=1.502times G(1-(y)_(1))-|||-=dfrac (998.2)(18)cdot dfrac (8.809times {10)^-3}(1+8.809times {10)^-3}-|||-=51.72kmol/h=930.96kg/h-|||-=0.4842kmol(m)^3

吸收:

例1、在 20℃,1atm下,用清水分离氨-空气的混合气体,混合气体中氨的分压为 1330Pa,经吸收后氨的分压降为 7 Pa,混合气体的处理量为 1020 kg/h,操作条件下平衡关系为 Y = 0.755X。若适宜的吸收剂用量为最小用量的2倍,求所需吸收剂用量及离塔氨水的浓度。