[6-1]解 计算式为=dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x计算结果见下表=dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x苯-甲苯溶液的=dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x计算数据温度80.184889296100104108110.61.01.0 0=dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x 6-1附图1与习题6-1附图2所示。[6-6]解 (1) 80.1℃时 =dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x 110.6℃时 =dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x 从计算结果可知,温度高,相对挥发度小。 平均=dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x(2) 汽液相平衡方程 =dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x (3)计算=dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x平衡数据,与习题6-1的计算结果接近。[6-10]解(1) 已知馏出液流 精馏段下降液体流量(2) 已知,代入式 求得进料流量 釜液采出量 =dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x(3) 提馏段下降液体流量 =dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x 提馏段上升蒸气流量 =dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x或 =dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x(4) 塔顶操作压力计算=dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x 苯=dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x 甲苯 用露点与汽相组成的关系式 计算=dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x,已知 解得操作压力 [6-14]。 解 (1)精馏段操作线方程 已知=dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x =dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x(2) 提馏段操作线方程 已知=dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x,计算塔釜汽相回流比=dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x。=dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x =dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x [6-18]解 先将进料组成由质量分数0.4换算为摩尔分数。苯的摩尔质量为78,甲苯的摩尔质量为=dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x。=dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x 已知=dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x。相平衡方程=dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x 精馏段操作线方程 塔釜汽相回流比=dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x=dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x 提馏段操作线方程 =dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x =dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x两操作线交点的横坐标 =dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x =dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x理论板数计算:先交替使用相平衡方程(1)与精馏段操作线方程(2)计算如下=dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x =dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x 第7板为加料板。以下交替使用提馏段操作线方程(3)与相平衡方程(1)计算如下=dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x =dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x =dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x 总理论板数为11(包括蒸馏釜),精馏段理论板数为6,第7板为加料板。22解 汽液相平衡方程为 =dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x 已知=dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x(1) 冷液进料, 线方程为 由q线方程与相平衡方程解得 =dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x 最小回流比 =dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x(2) 汽液混合物进料,=dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x q线方程为 =dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x =dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x由q线方裎与相平衡方程解得 =dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x 最小回流比 =dfrac ({rho )^2-({rho )_(B)}^0}({{rho )_(A)}^2-({P)_(B)}^0} =dfrac ({P)_(A)^0}(P)x