Material Balances Reklaitis P2.12

Introduction To Material and Energy Balances by  G.V. Reklaitis

Problem 2.12

The feed to a distillation column contains 36% benzene by weight, the remainder being toluene. The overhead distillate is to contain 52% benzene by weight, while the bottoms are contain 5% benzene by weight. Calculate:
(a) The percentage of the benzene feed which is contained in the distillate.
(b) The percentage of the total feed which leaves as distillate.

Answer

(Ilustration)

We can answer this question without degree of Freedom.

Let we assume W1 = W

Total Material Balance
W1 = W2 + W3
W = W2 +W3 … (1)

Material Balance of Benzene
In = Out
0.36xW = 0.52xW2 + 0.05 W3 (both side we x100)
Then we got
36W = 52W2 + 5W3 … (2)

Material Balance of Toluene
In = Out
0.64xW = 0.48xW2 + 0.95xW3 (both side we x100)
Then we got
64W = 48W2 + 95W3 … (3)

Eliminate … (2) and (3)
36W = 52W2 + 5W3   |x95
64W = 48W2 + 95W3 |x5

Then we got,
3420W = 4940W2 + 475W3
  320W =   240W2 + 475 W3

Then we got,
3100W = 4700W2
W2 = 3100W/4700
W2 = 0.66W

Subs to … (1), and we got
W3 = 0.34W

(a) Benzene in = 0.36W
      Benzene,2  = 0.52x0.66W = 0.3432W
Recall that %yield = [Product/Feed]×100%
Then the percentage of benzene feed which is contained in the distillate is
%Benzene,2 = [(W2,Benzene)/(W1,Benzene)] x100%
%Benzene,2 = [(0.3432W)/(0.36W)] x100%
%Benzene,2 = 95.33%
So, the percentage of benzene feed which is contained in the distillate is 95.33%

(b) Feed = W
Distillate = 0.66W
%Leave the distillate = [(0.66W)/(W)] x100% = 66%

So, the percentage of the total feed which leaves as distillate is 66%