Plant capacity = 600 Metric Tons/ Day.
= 25253 Kg/hr.
Basis: Unit hour operation
Further we will use the notations as under
Mono-ethylene Glycol MEG
Di-ethylene Glycol DEG
Tri-ethylene Glycol DEG
If we use 1:12 molar ratio of EO: H2O and selectivity of different products is as under:
MEG = 85 %, DEG = 14 %, TEG = 1 % (Patent = US6605192)
Reactions:
C2H4O+H2O OHCH2CH2OH (MEG)
2C2H4O+H2O OHCH2CH2OCH2CH2OH (DEG)
3C2H4O+H2O OHCH2CH2OCH2CH2OCH2CH2OH (TEG)
Suppose 1 % of MEG is lost during evaporation and distillation then
MEG = 1.01*25253 = 25508 Kg
Total Mixture (MEG+DEG+TEG) =25508/0.85 =30009Kg
DEG = 30009*0.14 =4201Kg
TEG = 30009*0.01 =300Kg
Reaction 1: -
EO + H2O MEG
411.4 Kmole 411.4 Kmole 411.4 Kmole
Reaction 2:-
2EO + H2O DEG
80 Kmole 40 Kmole 40 Kmole
Reaction 3:-
3EO + H2O TEG 6Kmole 2 Kmole 2 Kmole
EO required = 411.4 + 80 +6 = 497.4 Kmole
=21885.6 Kg
Total H2O Require = 411.4+40+2 = 453.4 Kmole
= 8161.2 Kg
As
H2O/EO = 12/1
Total H2O Supplied = 12 * EO = 12 * 497.4 = 5968.8 Kmole
= 107438.4 Kg
Excess H2O = 5968.8– 453.4 = 5515.4 Kmole
= 99277.4 Kg
3.1.1:MATERIAL ENTERING THE REACTOR:
| Component | Mass (kg) | Mass % | Mole | Mole % |
| Ethylene Oxide | 21885.6 | 16.9 | 497.4 | 7.7 |
| Water | 107438.4 | 83.1 | 5968.8 | 92.3 |
| Total | 129324 | 100 | 6466.2 | 100 |
3.1.2 MATERIAL LEAVING THE REACTOR:
| Component | Mass (kg) | Mass % | Mole | Mole % |
| Water | 99277 | 77 | 5515.4 | 92.4 |
| MEG | 25508 | 19.7 | 411.4 | 6.9 |
| DEG | 4201 | 3.1 | 40 | 0.67 |
| TEG | 300 | 0.2 | 2 | 0.03 |
| Total | 129286 | 100 | 5968.8 | 100 |
