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)

X_w (DEG) = 0.9020

TEG Balance:

299.19 = Xw (TEG) (4645)

X_w (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