Chemegate

GATE 2024 Chemical Engineering, Q54 · Reaction Engineering

The correct answer is 18.1 to 18.3.

Question

Consider the process shown in the figure for manufacturing B via the reaction 2A B. The feed to the process is 90 mol% A and a close-boiling inert component I, mixed with a recycle stream of unreacted A and I from the separator overhead (recycle-to-purge ratio 10), which is also split off as a purge stream (xB=0(x_{B}=0 in both). At a particular steady-state: B product rate is 100 kmol/h (pure B), single-pass conversion of A in the reactor is 50%, and the recycle-to-purge stream flow ratio is 10. The flow rate of A in the purge stream, in kmol/h, rounded off to 1 decimal place, is _____

Figure for Q54 (Reaction Engineering, Recycle Reactors): Consider the process shown in the figure for manufacturing B via the reaction 2A → B. The feed to th…

Official answer: 18.1 to 18.3

Why

GATE 2024 official key: 18.1 to 18.3. With B=100=0.25×Areactor,inAreactor,in=400B=100=0.25 \times A_{reactor,in} \Rightarrow A_{reactor,in}=400 kmol/h (from 2A→B, 50% single-pass conversion). Overall balances on A and inert (fresh feed F: xA=0.9,xI=0.1)x_{A}=0.9, x_{I}=0.1) combined with the recycle/purge split (ratio 10, i.e., 11×purge=11 \times purge = total light stream leaving separator) give A in purge =(0.5×400)/1118.2= (0.5 \times 400)/11 \approx 18.2 kmol/h.

Explanation cross-checked for consistency. Final answer matches the official IIT answer key.