9-20 若将习题 9-17 用清水吸收含氨气体的过程改在有8层理论板的板式塔中进行,-|||-用水量和气体流率不变,试重新计算该题的(1)小题。-|||-9-17 含氨1.5%(体积分数)的气体通过填料塔用清水吸收其中的氨(其余为惰性气-|||-体),平衡关系为 =0.8x 用水量为最小值的1.2倍,气体流率 =0.024kmolcdot (m)^-2-|||-^-1 总传质系数 _(s)a=0.060kmolcdot (s)^-1cdot (m)^-3, 填料层高度6m。(提示:在试算时,可先-|||-取 _(b)-(y)_(a)approx (y)_(b) 试求:(1)出塔气体中的含氨量;(2)可以采取哪些措施使n达到99.5%?-|||-(3)对n达99.5%的措施做出估算,你选择哪一种?说明理由。-|||-xb皆为未知;不便于以平均推动力法求解需用试差法,含超越函数,初值难设定)。以下-|||-采用吸收因数法。已知-|||-_(oG)=G/(K,alpha )=0.024/0.060=0.40m-|||-=(h)_(0)/(H)_(OG)=6/0.4=15-|||-吸收因数 =L/mG=1/s, 已知 =0.8. 但L/G需由物料衡算得出 ((x)_(a)=0);-|||-=((y)_(b)-(y)_(a))/((x)_(b)-(x)_(b))=((y)_(b)-(y)_(b))/(x)_(b) (a)-|||-现ya未知,但知 _(n)lt (y)_(b), 可初步在 _(b)-(y)_(a) 中忽略y,而有 /Gapprox (y)_(b)/(x)_(b); 用此法近似,-|||-以得到A或S的初值A1、S1:-|||-因为 (dfrac (t)(6))=1.2(dfrac (t)(6))=1.2(dfrac ({y)_(6)}({x)_(6)})=1.2(dfrac ({y)_(i)}({y)_(i)}m)=1.2m(=0.96)-|||-所以 _(1)=(dfrac (L)(mG))=1.2times (S)_(1)=(A)_(1)^-1=dfrac (1)(1.2)=0.8333-|||-当将得到的初值y代入式(a),以求较准确的L/G时,与初值(L/G)比,将因计入-|||-_(n)=2.31times (10)^-4 为大。第二次设 _(22)=2.5times (10)^-4, 代入式(a)可求相应的A2:-|||-y0而略减,S因而略增;从图 9-11 可见 _(b)/(y)_(n) 略减,由此可知y,的实际值应较已算出的-|||-_(b)'=(y)_(b)/m=0.015/0.8=0.01875-|||-(dfrac (1)(6))=1.2(dfrac (1)(6))=1.2(dfrac ({x)_(6)-(y)_(2)}({x)_(6)})=1.2times dfrac (0.01475)(0.01875)=0.944-|||-_(2)=((dfrac {1)(G))}_(2)times dfrac (1)(m)=dfrac (0.944)(0.8)=1.18 _(2)=1/1.18-|||-=15 及 _(2)=1.18 求 _(b)/(y)_(2) 时,图 9-11 欠精确,按式 (9-53) 计算:-|||-(C)_(6)=dfrac (1)(1-s)ln [ (1-s)dfrac ({y)_(b)-m(x)_(a)}({y)_(a)-m(x)_(a)}+s] -|||-=dfrac (1.18)(0.18)ln [ dfrac (0.18)(1.18)times dfrac (0.015)({y)_(A)}+dfrac (1)(1.18)] -|||-^2.288=dfrac (2.288times {10)^-3}({y)_(0)}+0.8475-|||-解得 _(n)=2.54times (10)^-4, 与原设 _(n2)=2.5times (10)^-4 相差甚少,不必进一步试差, _(2)=2.54times -|||-^-4 即为最终值。-|||-为使n提高到99.5%,可加大水量至L`或增高填料层至h0,都可用式 (9-53) 计算。-|||-此时-|||-_(0)=(1-0.995)times 0.015=0.75times (10)^-4-|||-dfrac ({y)_(b)-m(x)_(n)}({y)_(b)-m(x)_(n)}=dfrac ({y)_(b)}({y)_(a)}=dfrac (0.015)(0.75times {10)^-4}=200-|||-在应用式 (9-53) 计算L`时,又需对A或S试差,其解法考虑到式中方括号中的值取对数后-|||-变化小,而写成-|||-https:/img.zuoyebang.cc/zyb_eea0a70a398a927ddaf265c2e7ce9857.jpg-S=dfrac (1)(15)ln [ (1-s)times 200+s] (b)-|||-方括号内的S查图 -11 纵、横坐标分别为15及200)得初值为0.735,代人式(b)方括号-|||-中算得等号左边 https:/img.zuoyebang.cc/zyb_eea0a70a398a927ddaf265c2e7ce9857.jpg-S=0.2656, 故 =0.7344, 与原设基本符合,可用于计算L`:-|||-'=mol/s=0.8times 0.024/0.7844=0.0261kmolcdot (m)^-2cdot (8)^-1-|||-而原用水量据-|||-dfrac (1)(6)=1.2times dfrac ({y)_(b)-(y)_(b)}({x)_(b)-(x)_(b)}=1.2times dfrac ((150-2.54)times {10)^-1}(0.015/0.8)=0.044-|||-得 _(1)=0.0446=0.044times 0.024=0.0227kmolcdot (m)^-2cdot (s)^-1 用水量增大的比例为 'L=-|||-.0261/0.0227=1.15, 即增大15%;一般讲,在操作允许范围内。-|||-增高填料层可从总传质单元数的增加直接计算:-|||-=dfrac (L)(G)dfrac (1)(m)=dfrac (0.944)(0.8)=1.180(S=0.848)-|||-'(C)_(6)=dfrac (1.18)(0.18)ln [ dfrac (0.18)(1.18)times 200+dfrac (1)(1.18)] =22.6-|||-故 _(0)'=(H)_(06)N'(x)_(6)=0.4times 22.6=9.04m-|||-比原来 _(0)=6m 增加了约51%。