反应式(3.2) 消耗甲醇 N D2CH3OH=0.75 kmol
反应式(3.3) 消耗甲醇 N D3CH3OH=0.0652 kmol
反应式(3.4) 消耗甲醇 N D4CH3OH=6.76 kmol
反应式(3.5) 消耗甲醇 N D5CH3OH=36.04 kmol
反应式(3.6) 消耗甲醇 N D6CH3OH=0.38 kmol
由以上数据,得到甲醇消耗量
N DCH3OH=53.33+ 36.04 + 6.76 + 0.75 + 0.38 + 0.0652=97.32 kmol
尾气中带走甲醇:N 10CH3OH = 0.28 kmol
产品中带走甲醇:N 7CH3OH = 1.17 kmol
故,甲醇总消耗量:N CH3OH = 97.32 + 0.28 + 1.17 = 98.77 (kmol)
W 6CH3OH = 3164.59 (kg)
技术单耗e1:3164.59 / 7500 = 0.42 (t/t)
实物单耗e2:3164.59 / (98% × 7500) = 0.43 (t/t)
此数据与上面给定的条件基本一致。
3) 水量衡算
尾气中带出的水:N 10H2O = 3.75 (kmol) W 10H2O = 67.54 (kg)
产品中带出的水:N 7H2O = 185.15 (kmol) W 7H2O = 3334.5 (kg)
空气中带入的水:N 2H2O = 3.33 (kmol) W 2H2O = 59.97 (kg)
原料甲醇中带入的水:W 1H2O = 3164.59 / 98% - 3164.59 = 64.58 (kg)
N 1H2O = 3.58 (kmol)
反应生成的水:N DH2O = 53.33 + 2 × (6.76 + 0.75) + 0.38 + 0.0652
= 68.80 (kmol)
按配料浓度60%计,应加入的配料水蒸气量:
W 3H2O = 3164.59 / 60% - (3164.59+ 59.97 + 64.58) = 1985.18 (kg)
N 3H2O = 110.23 (kmol)
吸收塔加水量:N 11H2O = (3.75 + 185.15) - (3.58 + 3.33 + 68.80 + 110.23)
= 2.96 (kmol)
W 11H2O = 53.31 (kg)
4)吸收系统物料衡算
假设一吸收塔甲醛吸收率为90%,则一吸收塔吸收甲醛量为:
NE HCHO = 89.37 × 90% = 80.43 (kmol)
WE HCHO = 2414.51 (kg)
二吸收塔吸收甲醛量为:
NF HCHO = 89.37 – 80.43 – 0.094 = 8.85 (kmol)
WF HCHO = 265.68 (kg)
又设二吸收塔循环液中甲醛浓度为15%,并假设转化甲醛除尾气带出外,均在二吸收塔吸收。则二吸收塔循环液入一吸收塔量为:
W 8 = 265.68 / 15% = 1771.2 (kg)
其中,甲醇量:W 8CH3OH = 37.5 (kg)
N 8CH3OH = 1.17 (kmol)
水量:W 8H2O = 1771.2 – 37.5 – 265.68 = 1468.02 (kg)
N 8H2O = 81.51 (kmol)
3.2.3 主要设备物料衡算
(1) 蒸发器的物料衡算
图3.2 蒸发器物料衡算图
1)进料组成
甲醇进料: W 1CH3OH=3202.5 kg N 1CH3OH=69.91 kmol
W 1H2O=42547.5 kg N 1H2O=2362.44 kmol
空气进料: W 2O2 = 1215.62 kg N 2O2 = 38.00 kmol
W 2N2 = 3910.76 kg N 2N2 = 139.62 kmol
W 2H2O = 59.97 kg N 2H2O =3.33 kmol
2)计算过程 甲醇进料+空气进料=原料气出料
M4 = M1 + M2 N4 = N1 + N2
W 4CH3OH=W 1CH3OH N 4CH3OH=N 1CH3OH
W 4H2O=W 1H2O+W 2H2O N 4H2O=N 1H2O+N 2H2O
W 4O2=W 2O2 N 4O2=N 2O2
W 4N2= W 2N2 N 4N2= N 2N2
3)出料组成
原料气出料: W 4CH3OH=3202.5 kg N 4CH3OH=69.91 kmol 甲醇铁钼催化法制甲醛的工艺设计(9):http://www.youerw.com/cailiao/lunwen_2443.html