MATERIAL BALANCE AROUND REACTOR

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