大学【化工】- 搜题酱免费题库

催化裂化技术在中国石油炼制工业中被称为()。A. 四大发明B. 五朵金花C. 三大突破D. 六项创新

汽油的燃点温度为（）°C。A. 180-250B. 280-350C. 380-450D. 480-550

防火防爆的技术措施包括（）。A. 点火源的安全控制B. 化工工艺参数的控制C. 火灾爆炸危险物质的控制D. 化工设备的维护

热交换器冷流体进出口温度一定时,采用逆流操作可以节约热(冷)流体用量。()A. 正确B. 错误

危化品应双人双锁管理(）A. 正确B. 错误

[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

4-8 一套管式换热器内流体的对流传热系数 _(1)=200Ncdot (m)^-2cdot (k)^-1, 管外流体的对流传热系数-|||-(alpha )_(2)=350Wcdot (m)^-2cdot (k)^-1 已知两种流体均在湍流情况下进行换热。(1)假设管内流体流动增加一倍;-|||-(2)假设管外流体流速增加二倍。其他条件不变,试问总传热系数是原来的多少倍?管壁热阻及污垢-|||-热阻可不计。

以单位时间为基准进行物料衡算，更便于分析连续生产过程的物料平衡。

化工燃烧的形式：（）、（）、（）、（）和（）

用一单效蒸发器将1500kg/h的水溶液由5%浓缩至25%(均为质量%)。加热蒸汽压力为190kPa,蒸发压力为30kPa(均为绝压)。蒸发器内溶液沸点为78℃,蒸发器的总传热系数为1450W/(m2·℃)。沸点进料,热损失不计。试计算:①完成液量;②加热蒸汽消耗量;③传热面积。

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