化工原理课程设计苯甲苯的分离1 - 图文 下载本文

MD?0.949?78.11?0.051?92.13?78.83kgkmol MW?0.035?78.11?0.?92.13?91.64kgkmol (3)物料衡算

5.0?107?73.44kmolh 原料处理量F?330?24?85.96总物料衡算 73.44 =D+W

苯物料衡算 73.44×0.44=0.949D+0.035 W 联立解得 D=32.54 kmol/h W=40.90 kmol/h

式中 F------原料液流量 D------塔顶产品量

W------塔底产品量 3.3 塔板数的确定

3.3.1理论板层数NT的求取

(1)由手册查得苯一甲苯物系的气液平衡数据,绘出x ~y图,见下图

(2)求最小回流比及操作回流比。

采用作图法求最小回流比。在上图中对角线上,自点e(0.44,0.44)作垂线ef即为进料线(q线),该线与平衡线的交点坐标为

yq=0.653, xq=0.44

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x?y0.949?0.653故最小回流比为Rmin?Dqy?0.66?0.44?1.45

q?xq取操作回流比为R?1.7Rmin?2.287 (3)求精馏塔的气、液相负荷

L?R?D?2.287?32.54?74.42kmolh

V?(R?1)D?3.287?32.54?106.96kmolh

V'?(R?1)D?(1?q)F?106.96kmol/h (泡点进料:q=1) L'?RD?qF?2.287?32.54?73.44?147.86kmol/h

(4)求操作线方程

精馏段操作线方程为

yRR?1xxn?1?n?DR?1?0.696xn?0.290 提馏段操作线方程为

yL'Wm?1?V'xm?V'xw?1.38xm?0.013

(5)逐板法求理论板

又根据Rmin?1xD??1[x??(1?xd)] 可解得 F1?xf?=2.475 相平衡方程 y??x2.475x1?(??1)x?1?1.475x

y1?x1D = 0.949 x1?yy??(1?y?y1?0.88311)y1?2.475(1?y1)

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yR2?R?1xxD1?R?1 xy22??0.792 ?0.696x0.290?0.90y2??????(1?y1?2)y3?0.696x2?0.290?0.842 xy33?y??????(1?y?0.682

33)yy44?0.696x3?0.290?0.765 x4?y??????(1?y?0.568

44)y0.696xy55?4?0.290?0.685 x5?y?????(1?y?0.468

5?5)y xy66?0.696x1?0.290?0.616 6?y(1?y?0.3936??????6)因为x6<xf 精馏段理论板N精=5。进料板位置在第6快塔板。x''1?x6?0.393 y2?1.3x81??0130.?0.295 y'x'2?0.312 y'2?y'?????(1?y'3?1.3x8?2?0130.?0.184 2?2)'x'33?yy?0.225 y'''4?1.3x8?3?0130.?0.982 3??????(1?y3)x'y'44?y?0.146 y'''5?1.3x8?4?0130.?0.189 4??????(1?y4)x'y'55?y''?0.086 y'6?1.3x8?5?0130.?0.0615??????(1?y5)'x'?y56y'?0.045 y''7?1.3x8?6?0.0?10.305??????(1?y5)7

5x?'7y5'y5??????(1?y5)''?0.021

则全塔总理论板层数NT=12(不包括再沸器)。 3.3.2确定实际板数 (1)全塔效率的计算

采用“O?connell的精馏塔效率关联图”将全塔效率ET关联成??m的函数,近似式为 ET?0.49?u(mn?0.254)

液体平均粘度为 ?m??xiui

i?1(查表得各组分黏度?1=0.269,?2=0.277)

?m?xF?1?(1?xF)?2?0.44?0.269?(1?0.44)?0.277?0.321

-0.254ET?0.49(?u)?0.519

精馏段实际板层数N精?NT5??9.63?10 ET51.9%NT6??11.56?12 ET51.9%提馏段实际板层数N提=总实际板层数Np?Np精+Np提=10+12=22 进料板在第11块板

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