Question

Fresh feed containing 55 wt% A and 45 wt% B flowing at 100kg/h enters a separation system that removes a portion of pure component A only, as a bottom product. The top product stream (tops) of the separator unit contains 10 wt% of componenet A and the balance is B. A small part of the tops stream is recycled and mixed with the fresh feed stream- the remainder of the tops stream is purged. The separtor is designed to removed exactly two-thirds of component A that is fed into it. Determine the recycle ration (that is the ratio of the flow of the recycle stream to the flow of the fresh feed stream).

Answer 1

100 kg/h of fresh feed

Overall process

Total mass balance

100 = m₆ + m₄

Component balance (A):

Input = Output

0.55 * 100 = m₆ + 0.1 m₄

substituting m₆ from overall balance equation

0.55 * 100 = (100 - m₄) + 0.1 m₄

55 = 100 - 0.9 m₄

0.9 m₄ = 100 - 55

0.9 m₄ = 45

m₄ = 50 kg/h

similarly for m6

100 = m₆ + m₄

substituting the value of m4, we get,

100 = m₆ + 50

m₆ = 50 kg/h

System separator

Total mass balance

m₂ = 50 + m₃

Component balance (A):

Input = Output

x _A,2m2 = 50 + m3 (0.1)

Separator removed 2 / 3 of A fed to it in the bottom stream

from the component balance and the relation

75 = 50 + m₃ (0.1)

m₃(0.1) = 25

m₃ = 250 kg/h

substituting m₃ = 250 kg/h into overall mass balance equation arounf the separator

m2 = 50 + m3

m2 = 50 + 250

m2 = 300 kg/h

overall mass balance in spiltter

m₃ = m₄ + m₅

250 = 50 + m₅

m₅ = 200 kg/h