Example: Mass Balance for a distillation column A liquid mixture containing 45 %
ID: 703620 • Letter: E
Question
Example: Mass Balance for a distillation column A liquid mixture containing 45 % benzene (B) and 55 % toluene (T) by mass is fed to a distillation column. A product stream leaving the top of the column contains 95 mole % B and a bottom product stream contains 8 weight % of the benzene fed to the column. The volumetric flow rate of the feed stream is 2000 L/h and the specific gravity of the feed mixture is 0.872. Determine the mass flow rate of the overhead stream and the mass flow rate and composition (mass fraction) of the bottom product streamExplanation / Answer
Ans -
Basis - 100 kmol overhead product with 95 kmol Benzene and 5 kmol Toluene
In the feed stream
Mass flow rate m1 = volumetric flow rate x density
= 2000 L/h x 0.872 x 1000 kg/m3 x 1m3/1000L
= 1744 kg/h
In the bottom
Mass flow rate of benzene mB3 = 0.08 x 0.45 x 1744
= 62.784 kg/h
In the overhead
Mass flow rate of Benzene = moles x molecular weight
= 95 kmol x 78 kg/kmol
= 7410 kg
Mass flow rate of toluene = moles x molecular weight
= 5 kmol x 92 kg/kmol
= 460 kg
Mass fraction of B yB2= 7410/(7410+460) = 0.9415
Mass balance of benzene
Benzene in feed = benzene in overhead + benzene in bottom
0.45 x m1 = m2 x (yB2) + mB3
0.451 x 1744 = m2 x (0.9415) + 62.784
m2 = 766 kg/h
Mass balance of toluene
toluene in feed = toluene in overhead + toluene in bottom
0.55 x m1 = (1-yB2) x m2 + mT3
0.55 x 1744 = (1-0.9415) x 766 + mT3
mT3 = 915 kg/h
Mass flow rate of overhead stream = 766 kg/h
Mass flow rate of bottom stream = mB3 + mT3
= 62.784 + 915 = 977.784 kg/h
Mass fraction of benzene in bottom
= 62.784/977.784 = 0.0642
Mass fraction of toluene in bottom
= 915/977.784 = 0.9358
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