Monday, 11 April 2011

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                 +          H
2O                                        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

4 comments:

  1. Thanks a lot for sharing ; all topics are covered in this blog this is Very much useful for engineering students to complete their project works.

    ReplyDelete
  2. can i know how you calculate mole of at reaction 1 equation?

    ReplyDelete
  3. Thanks for support for real i get allot of benefits from this project

    ReplyDelete
  4. Can you just update here the picture of the schematic diagram for which these mass balance is done?

    ReplyDelete