3. A very large distillation column is separating p-xylene (more volatile) from o-xylene. The column has two feeds that are saturated liquids. Feed 1 flows into the column at a rate of 90 kmol/h and contains 42 mol% p-xylene. Feed 2 flows at a rate of 20 kmol/h and contains 9 mol% p-xylene. The bottoms product should be 97 mol% o-xylene, and the distillate product should be 99 mol% p-xylene. Compute the distillate (D) and the bottoms (B) products flow rates

the distillate (D) and the bottoms (B) products flow rates are;

B = 72.1875 kmol/hr

D = 37.8125 kmol/hr

Explanation:

Given the data in the question;

as illustrated in the image below;

F1 stream, ( 90 kmol/hr ) contains 42 mol% p-xylene and Feed 2 ( 20 kmol/hr) contains 9mol% p-xylene, so;

more volatile → p-xylene

less volatile → o-xylene

now, Overall balance;

F1 + F2 = D + B

90 + 20 = D + B

D + B = 110    ------------- let this be equation 1

p-xylene. component balance;

0.42F1 + 0.09F2 = 0.99D +  0.03B

0.42(90) + 0.09(20) = 0.99D +  0.03B

37.8 + 1.8 = 0.99D +  0.03B

0.99D +  0.03B = 39.6 --------------- let this be equation 2

from equation 1 and 2

input (D = 110 - B) into equation 2

0.99(110 - B) +  0.03B = 39.6

108.9 - 0.99B + 0.03B = 39.6

108.9 - 39.6 = 0.99B - 0.03B

69.3 = 0.96B

B = 69.3 / 0.96

B = 72.1875

now, input B = 72.1875 into equation 1

D + B = 110

D + 72.1875 = 110

D = 110 - 72.1875

D = 37.8125

Therefore, the distillate (D) and the bottoms (B) products flow rates are;

B = 72.1875 kmol/hr

D = 37.8125 kmol/hr

where is the following

explanation: