Monday 11 April 2011

MATERIAL BALANCE AROUND EG REFINER




Basis = 29919 Kg of Feed
                                   
F = D+W

D+W = 29919 _____ (1)

MEG Balance:

25408 = 0.992D + 0.034W ____ (2)

Put Value of D from Eq. (1)

25408 = 0.992(29919-W) + 0.034W

W = 4271.648/0.958
               =4645 Kg

D = 29919-4645
               = 25274 Kg

DEG Balance:

4201 = 0.0004*25274 + Xw (DEG) (4645)

Xw (DEG) = 0.9020

TEG Balance:

299.19 = Xw (TEG) (4645)
Xw (TEG) = 0.0646

            Component
Mass (kg)
IN
Mass %
IN
Mass(kg) OUT
Mass %
OUT
Water
10
0.033
0
0
MEG
25408
84.923
155
3.3
DEG
4201
14.041
4190
90.2
TEG
300
1.002
300
6.5
Total
29919
100.0
4645
100.0

Distillate
Component   Mass         Mass %
         MEG      25253        99.92
         Water     10              0.036
                      DEG      11              0.043

                      Total     25273         100.0

Material out                  =4645+25273              =29918(kg)
Material In=Material Out

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