j06a10013证：设进口气体中A的摩尔比为Y1，出口气体中A的摩尔比为Y2，而X2=0，则(L/V)min=(Y1-Y2)/(Y1/m)=(Y1-Y2)/Y1∴(L/V)min=m。j06a10103解：Y1=0.01Y2=Y1(1-)=0.01(1-0.8)=0.002m=1LS/GB=1.5(LS/GB)min=1.5(Y1-Y2)/(Y1/m)=1.5m=1.510.8=1.2S=m/(LS/GB)=1/1.2=0.833NOG=1/(1-S)ln[(1-S)Y1/Y2+S]=3.06h=NOGHOG=3.06mj06a10104解：①(L/G)min=(y1-y2)/x1*=(y1-y2)/(y1/m)=mL/G=1.2(L/G)min=(y1-y2)/x1x1=y1/(1.2m)=0.08/(1.2×2.5)=0.0267②S=m/(L/G)=m/(1.2×m×)=1/(1.2)=1/(1.2×0.98)=0.8503y1/y2=y1/[y1(1-)]=1/(1-)=1/0.02=50NOG=1/(1-S)ln[(1-S)y1/y2+S]=14.2h=HOGNOG=0.6×14.2=8.50m>6m，所以不合用。j06a10105①y1=p1/P=10/760=0.01316y2=p2/P=0.051/760=6.7110-5L/G=5(L/G)min=5(y1-y2)/(y1/m)=5(0.01316-6.7110)/(0.01316/0.755)=3.756G=1020/28.8=35.42kmol/hL=3.756G=133kmol/h②S=m/(L/G)=0.755/3.756=0.201NOG=1/(1-S)ln[(1-S)Y1/Y2+S]=6.33j06a10106①y1=0.03，x1*=0.03，所以m=1(L/G)min=(y2-y1)/(y1/m)=m=0.95S=m/(L/G)L/G=m/S=1/0.8=1.25∴(L/G)/(L/G)min=1.25/0.95=1.316②x1=(y2-y1)/(L/G)=(0.03-0.03(1-0.95))/1.25=0.0228③NOG=1/(1-S)ln[(1-S)y2/y1+S]=7.843j06a10107L/G=(y1-y2)/(x1-x2)=(0.02-0.004)/0.008=2y1=y1-y1*=0.02-1.150.008=0.0108y2=0.004ym=(y1-y2)/(ln(y1/y2))=0.00685NOG=(y1-y2)/ym=2.336HOG=h/NOG=3/2.336=1.284mj06a10108ky=PkG=100×3.8410-6=3.84×10-4kmol/(m2s)kx=CkL=55.6×1.83×10-4=1.02×10-2kmol/(m2s)1/Ky=1/ky+m/kx=1/(3.84×10-4)+m/(1.02×10-2)m=E/P，E=p*/xm=p*/Px=6.7/(100×0.05)=1.341/Ky=2604+131.4=2735.4Ky=3.656×10-4kmol/[m2s(y)]j06a10109①y1=0.01y2=y1(1-)=0.01(1-0.9)=0.001L/G=(y1-y2)/x1X1=(y1-y2)/(L/G)=(0.01-0.001)/1.5=0.006②HOG=G/Kya=100/(0.785×0.52×3600)/0.1=1.415mNOG=1/(1-S)ln[(1-S)y1/y2+S]S=m/(L/G)=1/1.5=0.667NOG=4.16h=HOGNOG=1.415×4.16=5.88m进塔气体按混合气计：(上面按纯惰性气体计)由于入塔浓度很低，将摩尔分率当作摩尔比使用。①(1)y1=0.01=Y1②Y2=y2=y1(1–)=0.01(1–0.9)=0.001用纯溶剂吸收,X2=0L/V=(Y1–Y2)/(X1–X2)=(Y1–Y2)/X1X1=(Y1–Y2)/(L/V)=(0.01–0.001)/1.5=0.006(2)V=100(1–y1)=100(1–0.01)=99kmol/hHOG=V/(Kya)=99/(0.7850.523600)/0.1=1.40mX2=0，Y2*=0有：NOG=1/(1–S)ln[(1–S)Y1/Y2+S]S=m/(L/V)=1/1.5=0.667NOG=1/(1–0.667)ln