大学【化工】- 搜题酱免费题库

[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

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在中低压下，苯-甲苯系统的汽液平衡可用 Raoult 定律描述。已知苯（1）和甲苯（2）的蒸气压数据见习题 5-21 表。习题 5-21 表苯（1）和甲苯（2）的蒸气压数据| t/^circmathrm(C) | p_1^*/mathrm(kPa) | p_2^*/mathrm(kPa) | t/^circmathrm(C) | p_1^*/mathrm(kPa) | p_2^*/mathrm(kPa) ||------------------------|------------------------|------------------------|------------------------|------------------------|------------------------|| 80.1 | 101.3 | 38.9 | 98 | 170.5 | 69.8 || 84 | 114.1 | 44.5 | 100 | 180.1 | 74.2 || 88 | 128.5 | 50.8 | 104 | 200.4 | 83.6 || 90 | 136.1 | 54.2 | 108 | 222.5 | 94.0 || 94 | 152.6 | 61.6 | 110.6 | 237.8 | 101.3 |试求：（1）90^circmathrm(C)、x_1=0.30 时，体系的汽相组成和压力；（2）90^circmathrm(C)、1.013times10^5mathrm(Pa) 体系的汽相和液相的平衡组成；（3）试确定在 x_1=0.55、y_1=0.75 时该体系的 T、p；（4）将含 y_1=0.3 的汽相混合物冷却到 100^circmathrm(C)、1.013times10^5mathrm(Pa)（总压），求该混合物液化分数与液相组成（液化分数即混合物冷却至 t^circmathrm(C) 时，蒸气的冷凝分数）；（5）在上问中，如果混合物的开始组成为 y_1=0.4，结果如何？

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