Mass Balance and Distillation Column Design for Methanol Production Process

This article discusses the mass balance and distillation column design for the methanol production process. It covers the assumptions made, calculations performed, and the material and energy balances involved in the process.

Mass balance at the reactor
Assume selectivity = 9
Assume the amount of of input basis of methanol is n3 = 10 kmol/h
7 = 1 + 2
0 = 1 - 2
Subtracting the two equations
2 = 0.7 kmol/h
2 = 6.3 kmol/h
nm,9 = 10 – 6.3*0.7 = 3 kmol/h
0 = n02,8 – 0.5 * 1
n02,8 = 0.5 * 6.3 = 3.15 kmol/h
nN2, 8 = nN2, 9 = n02, 8 * (0.79 kmol/h)/(0.21 kmol/h)
= 3.15 * 0.79/0.21 = 11.85 kmol/h
nN2, 9 = 2 = 0.7 kmol/h
nH2, 9 = 1 = 6.3 kmol/h
nF, 9 = 1 + 2 = 6.3 + 0.7 = 7 kmol/h
nm, 8 = 10 kmol/h
n02, 8 = 3.15 kmol/h
nH20, 8 = 0 kmol/h
nN2, 8 = 11.85 kmol/h
Stream 1
∑ I = 25
Xn = 10/25 = 0.4

X02 = 3.15/25 = 0.126
XN =11.85/25 = 0.474
Stream 2
∑ I = 28.85 kmol/h
Yn = 3/28.85 = 0.104
Y02 = 11.85/28.85 = 0.411
YN =0.7/28.85 = 0.0243
YN =6.3/28.85 = 0.2184
YN =7/28.85 = 0.243
Mass balance at the reactor
nF,12 = 0.243 * 28.85 (1-0.99) = 0.0701 kmol/h
Solubility of formaldehyde =
= 0.468 kmol F
= 18.675 kmol H2O/h
Solubility of methanol = = 0.03113 kmol methanol/ L of water
= 10. 4648 k mol H2O/h
All the methane will dissolve in water and no methanol in the off gas
nm,13 = nm, 10 – nm,12 = 3 kmol methanol /h
assuming that all N2, H2
nN2,12 = nN2,10 = 11.85 kmol/h
nh2, 12 = nH2,10 = 0.7 kmol/h
nF,13 = 0.0243 * 28.85 * 0.99 = 0.694 kmol/h

water inlet stream
n11 = 18.675 kmol/h
gas inlet stream
n10 = 28.85 kmol/h
nm10 = 3 kmol/h
n02, 10 = 0 kmol/h
nH20,10 = 6.3 kmol/h
∑ I = 28.85 kmol/h
Yn = 3/28.85 = 0.104
Y02 = 11.85/28.85 = 0.411
YN =0.7/28.85 = 0.0243
YN =6.3/28.85 = 0.2184
YN =7/28.85 = 0.243
Gas outlet stream
n12 = 11.85 + 0.7 + 0.0701 + 24.975 = 37.5951 kmol/h
nm,12 = 0 kmol/h
nH20,12 = 18.675 + 6.3 = 24.975 kmol/h
ym = 0, y02 = 0
yH20 = 24.975/37.5951 = 0.6643 kmol/h
yH2 = 6.3/37.5951 = 0.1662
yF = 0.0701/376.8951 = 0.001865
yN2 = 11.85/37.8951 = 0.3127
liquid outlet
n13 = 3 + 0.694 = 24.975 = 28.669 kmol/h

n13 = 3
nH20,13 = 24.975 kmol/h
nF = 0.695
ym = 0.105
yH,20 = 0.871
yF = 0.0242
mass balance at distillation column
assumptions
Take the light key be methanol
Take the heavy key be water
Non heavy key = formaldehyde
n14 = L1 = D + B
fraction recovery 1 = 99.7%
fraction recovery 2 = 99%
Dx, m = fraction 1 * n14 * Xm,14
= 0.997 * 28.669 * 0.105
= 3 kmol methanol/h
Bx, m = 1 -fraction 1 * n14 * Xm,14
= 1- 0.997 * 28.669 * 0.105
= 0.009031 kmol/h
BX, H20 = fraction 2 * n14 * XH20,14
= 0.99 * 28.669 * 0.807
= 24.721 kmol/h

DX, H20 = 1- fraction 2 * n14 * XH20,14
= 1 - 0.99 * 28.669 * 0.807
= 0.2497 kmol/h
BX, F = 0.0242 * 28.669
= 0.6938 kmol formaldehyde/h
D = ∑ x, Di = 3 + 0.2497 = 3.2497 kmol/h
B = ∑ x, Bi = 0.009031 + 24.721 + 0.6938 = 25.423831 kmol/h
Xm,D = 3/3.2497 = 0.9232
XH20, D = 0.2497/3.2497 = 0.07684
Xm,B = 0.009031/25.423831 = 0.0003552
XH20, B = 24.721/25.423831 = 0.9724
XF, B = 0.6938/25.423831 = 0.02729
Material Ni = yi *ntotal Molecular
weight
Mi = niMW Mass function
methanol 0.009031 32.042 0.2894 0.0006209
Formaldehyde 0.6938 30.026 20.832 0.04469
Water 24.721 18 444.978 0.95468
Total 466.0994
= > 0.04469/0.95468 = 0.04681 formaldehyde to water ratio
4.469% of formaldehyde

CS1.1
8000000 tons per year
1 year = 365 days
= 8000000 *1000/(365 * 24) = 913242.0091 kg/hr
CS1.2
CS1.3
= 36000 *1000/(350 *24)
= 4285.714 kg/hr
CS1.4
= 0.7 conversion of methanol for the process
CS1.5
= 10 kmol/h

CS1.6
No
CS1.7
28.85 kmol/h
CS1.8
1450C
CS1.9
nF,12 = 0.243 * 28.85 (1-0.99) = 0.0701 kmol/h
Solubility of formaldehyde =
= 0.468 kmol F
= 18.675 kmol H2O/h
Solubility of methanol = = 0.03113 kmol methanol/ L of water
= 10. 4648 k mol H2O/h
All the methane will dissolve in water and no methanol in the off gas
nm,13 = nm, 10 – nm,12 = 3 kmol methanol /h
assuming that all N2, H2
nN2,12 = nN2,10 = 11.85 kmol/h
nh2, 12 = nH2,10 = 0.7 kmol/h
nF,13 = 0.0243 * 28.85 * 0.99 = 0.694 kmol/h
CS1.10
n11 = 18.675 kmol/h
CS1.11

No, at saturation point no more methanol can be dissolved on a solvent at given temperature
CS1.12
From the vapor pressure of water , temperature will be derived from energy balance
Qout = Qn10 + Qn11
= ∫ I,13Cpi, 13 dt + ∫ I,12Cpi, 13 dt = 890 C
CS1.13
Raoult law
YH20 = H2O(750 C)/P
76 cm of water = 55.9025 mm of Hg
= 55.9025/760 = 0.07356
dry air = 1 – 0.07356 = 0.2644
Molar composition of gas phase = 0.2644
It is not possible to determine the dew point of gas leaving the reactor
Ptotal = 4.1 bar
Water pressure = 0.2644 * 4100 = 1084.04
Water has a vapor pressure of 1084.04 mbar at 1500
From the calculation the dew point is above 1000C through it follows the behavior of Raoult law
CS1.14
Reactor : electricity and steam
Absorber : cooling water
Distillation : cooling water
CS1.15
H = mC
= 444.978 * 4.2 * 145
= 270.9916 kJ

If efficiency is 40% then it will be lower
= 40% * 270.9916 kJ = 108.397 kJ
Cs 1.16
a steam boiler in which waste heat is used to evaporate water into steam
CS1.17
Standard enthalpy of the reaction [298 K, 1 atm] is
1 = -156 kJ/mol
2 = +85 kJ/mol
1 = -285.8 kJ/mol
CS1.18
25 kmol/h
Converts water to steam from heat produced in the reactor
CS1.19
Fraction = ½
Yes, for all the provided oxygen to fully participate in the reaction
CS1.20 = 28.85 kmol/h
CS1.21
For efficient utilization of all the hydrogen gas produced to be oxidized to water, and ensuring
that there is no back reaction for formation of methanol
CS1.22
444.978 * 4.2 * (145 – 25) = 224.268 kJ
CS1.23
I8.675 kmol/h