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MATERIAL BALANCE
The two reactions taking place in this process are given below. C2H4 + ½ O2 C2H4 + 3O2 Molecular Weights: Component Ethylene Oxygen Ethylene oxide (EO) Carbon dioxide Water Nitrogen Molecular weights 28 32 44 44 18 28 C2H4O (main reaction) 2CO2 + 2H2O (side reaction

Total ethylene oxide to be produced =

100 TPD.

= 100000 Kgs/ day. = 4166.67 Kgs/ hr.

Basis: 1 Hour ofoperation
REACTOR: Total weight of ethylene oxide is to be produced = 4166.67 Kgs. Assuming 2% of the EO produced is lost during the production. Total EO to be produced = 4250 Kgs/ hr. = 96.59 Kg moles/ hr. Assuming a conversion of 50% of ethylene fed and in that 25 % of the ethylene is converted into ethylene oxide. Ethylene to be supplied= 96.59/ 0.25 = 386.36 Kg moles/ hr. = 386.36 X 28 =10818.1 Kgs/ hr Air to Ethylene ratio is 10:1 Air supplied (95 Mole % oxygen and rest is assumed to be nitrogen) = 10818.1 X10 = 108181 Kgs/ hr Molecular weight of the air supplied = (0.95 X 32 + 0.05 X 28) = 31.8 Air supplied = 108181/ 31.8 = 3401.92 Kg moles/ hr Ethylene converted to ethylene oxide = 386.36 X 0.25 = 96.59 Kg moles/ hr. = 2704.52 kgs/ hr. EO produced = 96.59 Kg moles/ hr = 4250 kgs/hr. Ethylene converted into water and carbon dioxide = 96.59 Kg moles / hr = 2704.52 Kgs/ hr.

Carbon dioxide produced = 96.59 X 2 = 193.18 Kg moles/ hr = 8500 Kgs/ hr. Water produced = 96.59 X 2 = 193.18 Kg moles/ hr. = 3477.24 Kgs/ hr. Ethylene unreacted = 10818.1- (2704.52 X 2) = 5409.06 Kgs/hr
= 193.18 Kg moles / hr

Oxygen required for both the reactions. Oxygen entering = 103148.37Kgs/ hr = 3231.824 Kg moles/ hr. Nitrogen entering = 5032.63 Kgs/ hr = 179.73 Kg moles/ hr. Oxygen needed for the production of EO = 96.59/ 2 = 48.295 Kg moles/ hr. = 1545.44 Kgs/ hr. Oxygen needed for water and carbon dioxide = 96.59 X 3 = 289.77 Kg moles/hr = 9272.64 Kgs/ hr. Oxygen unreacted = 103148.37 - (1545.44 + 9272.64) = 92330.29 Kgs/ hr = 2885.32 Kg moles/hr. Nitrogen leaving the reactor =5032.63 Kgs/ hr.

INPUT TO THE REACTOR: Ethylene Oxygen Nitrogen Total = 10818.1 Kgs/ hr = 103148.37 Kgs/ hr = 5032.63 Kgs/ hr = 118999 Kgs/ hr

OUTPUT FORM THE REACTOR: Ethylene oxide Ethylene unreacted Oxygen unreacted Nitrogen Carbon dioxide Water Total = 4250 Kg/ hr = 5409.06 Kgs /hr = 92330.29 Kgs/ hr = 5032.63 Kgs/ hr = 8500 Kgs/ hr = 3477.24 Kgs/ hr = 118999 Kgs/ hr

Ethylene oxideEthylene Oxygen Nitrogen REACTOR Unreacted ethylene Unreacted oxygen Nitrogen Carbon dioxide Water

WATER ABSORBER: The solubility of EO in water is infinity. So the amount of water required for the absorption of EO is the equal amount of water. Water used for absorption = 4250 X1.10 = 4675 Kgs/hr (10 % of extra water is used to scrub all the EO produced.) Total EO scrubbed = 4250 Kg/ hrSolubility data: Component Nitrogen Carbon dioxide Oxygen Ethylene Solubility (Kgs/ Kgs of water) 1.3462 X 10-5 1.379 X 10-3 2.87 X 10 -5 1.482 X 10-2

During the absorption EO in water oxygen, nitrogen, carbon dioxide and ethylene is also absorbed in smaller quantities. Nitrogen absorbed = 4675 X1.3462 X 10-5 = 0.0629 Kgs/ hr Carbon dioxide absorbed = 4675 X 1.379 X 10-3 = 6.4 Kgs/ hr Ethyleneabsorbed = 4675 X 1.482 X 10-2 = 69.2835 Kgs/ hr Oxygen absorbed = 4675 X 2.87 X 10 –5 = 0.134 Kgs/ hr Recycled and Purged streams calculation Ethylene recycled= 5409.06 - 69.284 = 5340 Kgs/ hr

Oxygen recycled= 92330.29 – 0.134 = 92330.156 Kgs/ hr Carbon dioxide purged = 8500 - 6.4 = 8493.6Kgs/ hr Nitrogen purged = 5032.63 – 0.0629 Kgs/ hr = 5032.56 Kgs/ hr INPUT TO THE ABSORBER: Ethylene oxideEthylene unreacted Oxygen unreacted Nitrogen Carbon dioxide Water Total = 4250 Kg/ hr = 5409.06 Kgs /hr = 92330.29 Kgs/ hr = 5032.63 Kgs/ hr = 8500 Kgs/ hr = 3477.24 Kgs/ hr = 118999 Kgs/ hr

Water used for absorption = 4675 Kgs/ hr. Total = 123674 Kgs/hr

RECYCLE AND PURGE STREAM FORM THE ABSORBER: Ethylene Oxygen Carbon dioxide Nitrogen EO Total = 5340 Kgs/ hr = 92330.156 Kgs/hr = 8493.6 Kgs/...