气相甲醇脱水法年产20万吨二甲醚设计

2 合成塔的计算

2.1 衡算物料

99.5/32.04=99.11%

99.5/32.04+0.5/18.02

0.5/18.02=0.89?.5/32.+00.45/18.02

99.95/46.07=99.87%

99.95/46.07+0.05/32.040.05/32.04=0.004%

99.95/476.+00.05/32.04

1-99.87%-0.004%=0.126%

200000×103=25000 kg/h=542.65 kmol /h8000

η=542.65×99.87%=99.8%

FxFx=1371.997kmo/h

543.86×2=1359.649kmol/h

80%×10059.649=1371.997kmol/ h0.9911

-5743.8×62=284.277klm/ho 1371.99表2-1 物料衡算表

组分 二甲醚 甲醇 水 合计

进料 F0/(kmol/h)

0 1371.997 12.211 1384.208

进料 qm0/(kg/h)

0 43958.784 220.042 44184.246

出料 F/(kmol/h)

543.860 284.277 556.070 1384.207

出料 qm/(kg/h) 25055.630 9108.230 10020.380 44184.246

2.2 计算催化剂床层体积 VR=qW22.4=1384.208×=6.201m3 Sv5000 - 7 -

2.3 反应器管数

n=VRπ2×d0×L4=6.201=2863.33 20.785×0.022×5.72.4 热量衡算

Cp1=2.495kJ/（kg/℃） CP2=2.25 kJ/（kg/℃） CP3=4.15 kJ/（kg/℃） μ1=1.75×10-5pa μ2=1.63×10-5pa μ3=1.8×10-5pa λ1=0.03/(m2?k) λ2=0.05624 w/(m2?k) λ3=0.5741w/(m2?k) Q1=(43958.784×2.495+220.042×4.15)×(533.15-298)

=2.6×107 kJ/h

Q2=(25055.63×2.25+9108.235×2.459+10020.38×4.15)×(533.15-298) =2.84×107 kJ/h

QR=1543.86×11770=6.4×107 kJ/h QC=Q1+QR-Q2=4.4×106 kJ/h

λdG0.7d at=f=3.5（P）exp（-4.6P）dtμfdtG=44184.246=39744.598kg/(m2h)

π2926××0.02224μf=1.63×10-5×39.29%+1.75×10-5×20.53%+1.8×105×40.18% =1.723×105Pas dPG0.005×39744.598==3203.95μf1.723×10-5×3600λ=0.05642×39.29%+0.03×20.53%+0.5741×40.18% 2=0.2589W/(mk)=0.9320KJ/(m2hk)所以

at=0.93200.0050.7×3.5×（3203.95）×exp（-4.6×）0.0220.022 =14825.4KJ/(m2hk) - 8 -

Kt=11δdtdt1+++Rstαtλdmd0α1=10.00250.0220.0221+++4.78×10-514825.4167.50.02450.0272717=2333.45KJ/(mhk)

Δtm=(533.15510)+(533.15515)=40.65K

2Qc4×106 A需==42.17m2

ktΔtm2333.45×40.65A实=πldtn=3.14×0.022×5.7×2926=1152.13m2

A实>A需，

dpG11Rem()=3203.95×（）=6161.4

mf1-e1-0.48rfu01-eG21-ξΔP=1.75L=1.75L33dsedsPfξ(39744.598/3600)21-0.48=1.75×××5.7 30.005×627.60.48=1812.76kPa

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