GATE 2024 CH – Question 54

Process Calculations · Recycle, bypass and purge calculations · 2 marks · Numerical answer

Consider the process in the figure for manufacturing $B$. The feed to the process is 90 mol% $A$ and a close-boiling inert component $I$. At a particular steady-state:
• $B$ product rate is 100 kmol h$^{-1}$
• Single-pass conversion of $A$ in the reactor is 50%
• Recycle-to-purge stream flow ratio is 10
The flow rate of $A$ in the purge stream in kmol h$^{-1}$, rounded off to 1 decimal place, is _____

the feed ($x_A = 0.9$, $x_I = 0.1$) mixes with the recycle and goes to a reactor $2A \to B$. The reactor effluent goes to a separator that gives pure B (100 kmol/h) at the bottom and an A and I stream at the top that splits into the purge and the recycle.

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Correct answer: 18.11 to 18.29

Explanation

For $2A \to B$, making 100 kmol/h of $B$ consumes $200$ kmol/h of $A$. Let $P$ be the purge flow and $x$ the mole fraction of $A$ in the purge and the recycle (they have the same composition). The recycle flow is $10P$. The conversion of $A$ in the reactor is 50% of the reactor feed: $0.5(0.9F + 10Px) = 200$, so $0.9F + 10Px = 400$. The overall balance on $A$ is $0.9F - 200 = Px$, so $0.9F = 200 + Px$. Substituting, $200 + 11Px = 400$ and $Px = \frac{200}{11} = 18.2$ kmol/h, which is the flow of A in the purge.