化工原理蒸馏部分模拟试题及答案⏺_(n+1)=dfrac (R)(R+1)(x)_(n)+dfrac ({x)_(n)}(R+1)=dfrac (3.6)(4.6)x+dfrac (0.95)(4.6)=0.783(x)_(n)+0.207_(n+1)=dfrac (R)(R+1)(x)_(n)+dfrac ({x)_(n)}(R+1)=dfrac (3.6)(4.6)x+dfrac (0.95)(4.6)=0.783(x)_(n)+0.207_(n+1)=dfrac (R)(R+1)(x)_(n)+dfrac ({x)_(n)}(R+1)=dfrac (3.6)(4.6)x+dfrac (0.95)(4.6)=0.783(x)_(n)+0.207 则 _(n+1)=dfrac (R)(R+1)(x)_(n)+dfrac ({x)_(n)}(R+1)=dfrac (3.6)(4.6)x+dfrac (0.95)(4.6)=0.783(x)_(n)+0.207由 _(n+1)=dfrac (R)(R+1)(x)_(n)+dfrac ({x)_(n)}(R+1)=dfrac (3.6)(4.6)x+dfrac (0.95)(4.6)=0.783(x)_(n)+0.207 得 _(n+1)=dfrac (R)(R+1)(x)_(n)+dfrac ({x)_(n)}(R+1)=dfrac (3.6)(4.6)x+dfrac (0.95)(4.6)=0.783(x)_(n)+0.207_(n+1)=dfrac (R)(R+1)(x)_(n)+dfrac ({x)_(n)}(R+1)=dfrac (3.6)(4.6)x+dfrac (0.95)(4.6)=0.783(x)_(n)+0.207 _(n+1)=dfrac (R)(R+1)(x)_(n)+dfrac ({x)_(n)}(R+1)=dfrac (3.6)(4.6)x+dfrac (0.95)(4.6)=0.783(x)_(n)+0.2077有某平均相对挥发度为3的理想溶液,其中易挥发组分的组成为60%(摩尔百分率,以下同),于泡点下送入精馏塔中,要求馏出液中易挥发组分组成不小于90%,残液中易挥发组分组成不大于2%,试用计算方法求以下各项:(1)每获得馏出液时原料液用量;(2)若回流比_(n+1)=dfrac (R)(R+1)(x)_(n)+dfrac ({x)_(n)}(R+1)=dfrac (3.6)(4.6)x+dfrac (0.95)(4.6)=0.783(x)_(n)+0.207为1.5,它相当于最小回流比的若干倍;(3)回流比_(n+1)=dfrac (R)(R+1)(x)_(n)+dfrac ({x)_(n)}(R+1)=dfrac (3.6)(4.6)x+dfrac (0.95)(4.6)=0.783(x)_(n)+0.207为1.5时,精馏段需若干层理论板;(4)假设料液加到板上后,加料板上溶液的组成不变,仍为0.6,求上升到加料板上蒸汽相的组成。解 (1) 原料液用量依题意知馏出液量,在全塔范围内列总物料及易挥发组分的衡算,得:_(n+1)=dfrac (R)(R+1)(x)_(n)+dfrac ({x)_(n)}(R+1)=dfrac (3.6)(4.6)x+dfrac (0.95)(4.6)=0.783(x)_(n)+0.207 _(n+1)=dfrac (R)(R+1)(x)_(n)+dfrac ({x)_(n)}(R+1)=dfrac (3.6)(4.6)x+dfrac (0.95)(4.6)=0.783(x)_(n)+0.207或 _(n+1)=dfrac (R)(R+1)(x)_(n)+dfrac ({x)_(n)}(R+1)=dfrac (3.6)(4.6)x+dfrac (0.95)(4.6)=0.783(x)_(n)+0.207 由上二式解得,收集_(n+1)=dfrac (R)(R+1)(x)_(n)+dfrac ({x)_(n)}(R+1)=dfrac (3.6)(4.6)x+dfrac (0.95)(4.6)=0.783(x)_(n)+0.207的馏出液需用原料液量为: _(n+1)=dfrac (R)(R+1)(x)_(n)+dfrac ({x)_(n)}(R+1)=dfrac (3.6)(4.6)x+dfrac (0.95)(4.6)=0.783(x)_(n)+0.207(2)回流比为最小回流比的倍数以相对挥发度表示的相平衡关系为: 当_(n+1)=dfrac (R)(R+1)(x)_(n)+dfrac ({x)_(n)}(R+1)=dfrac (3.6)(4.6)x+dfrac (0.95)(4.6)=0.783(x)_(n)+0.207时,与之平衡的气相组成为: _(n+1)=dfrac (R)(R+1)(x)_(n)+dfrac ({x)_(n)}(R+1)=dfrac (3.6)(4.6)x+dfrac (0.95)(4.6)=0.783(x)_(n)+0.207由于是泡点进料,在最小回流比下的操作线斜率为:因此解得 _(n+1)=dfrac (R)(R+1)(x)_(n)+dfrac ({x)_(n)}(R+1)=dfrac (3.6)(4.6)x+dfrac (0.95)(4.6)=0.783(x)_(n)+0.207故操作回流比为最小回流比的_(n+1)=dfrac (R)(R+1)(x)_(n)+dfrac ({x)_(n)}(R+1)=dfrac (3.6)(4.6)x+dfrac (0.95)(4.6)=0.783(x)_(n)+0.207倍(3)精馏段理论板数当回流比,相对挥发度为3时,精馏段的平衡关系为式所示,操作线为:_(n+1)=dfrac (R)(R+1)(x)_(n)+dfrac ({x)_(n)}(R+1)=dfrac (3.6)(4.6)x+dfrac (0.95)(4.6)=0.783(x)_(n)+0.207_(n+1)=dfrac (R)(R+1)(x)_(n)+dfrac ({x)_(n)}(R+1)=dfrac (3.6)(4.6)x+dfrac (0.95)(4.6)=0.783(x)_(n)+0.207 由于采用全凝器,故_(n+1)=dfrac (R)(R+1)(x)_(n)+dfrac ({x)_(n)}(R+1)=dfrac (3.6)(4.6)x+dfrac (0.95)(4.6)=0.783(x)_(n)+0.207,将此值代入式_(n+1)=dfrac (R)(R+1)(x)_(n)+dfrac ({x)_(n)}(R+1)=dfrac (3.6)(4.6)x+dfrac (0.95)(4.6)=0.783(x)_(n)+0.207解得。然后再利用式算出,又利用式算出,直至算出的等于或小于为止。兹将计算结果列于本例附表中。(4)_(n+1)=dfrac (R)(R+1)(x)_(n)+dfrac ({x)_(n)}(R+1)=dfrac (3.6)(4.6)x+dfrac (0.95)(4.6)=0.783(x)_(n)+0.207上升到加料板的蒸汽相组成提馏段操作线方程为: _(n+1)=dfrac (R)(R+1)(x)_(n)+dfrac ({x)_(n)}(R+1)=dfrac (3.6)(4.6)x+dfrac (0.95)(4.6)=0.783(x)_(n)+0.207由于泡点进料,故_(n+1)=dfrac (R)(R+1)(x)_(n)+dfrac ({x)_(n)}(R+1)=dfrac (3.6)(4.6)x+dfrac (0.95)(4.6)=0.783(x)_(n)+0.207。又 _(n+1)=dfrac (R)(R+1)(x)_(n)+dfrac ({x)_(n)}(R+1)=dfrac (3.6)(4.6)x+dfrac (0.95)(4.6)=0.783(x)_(n)+0.207 及 将以上数据代入提馏段操作线方程:由题意知,料液加到板上后,板上液相组成仍为0.6,故上升到加料板的蒸汽相组成为:8用常压精馏塔分离某二元混合物,其平均相对挥发度为=2,原料液量F=10kmol/h,饱和蒸气进料,进料浓度x=0.5(摩尔分率,下同),馏出液浓度x=0.9,易挥发组份的回收率为90%,回流比R=2R,塔顶为全凝器,塔底为间接蒸气加热,求:(1)馏出液量及釜残液组成?(2)从第一块塔板下降的液体组成x为多少?(3)最小回流比?(4)精馏段各板上升的蒸气量为多少kmol/h?(5)提馏段各板上升的蒸气量为多少kmol/h?