GATE 2022 CH – Question 50

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

Consider the process flowsheet in the figure. An irreversible liquid-phase reaction $A \to B$ (reaction rate $-r_A = 164x_A$ kmol m$^{-3}$ h$^{-1}$) occurs in a 1 m$^3$ continuous stirred tank reactor (CSTR), where $x_A$ is the mole fraction of A. A small amount of inert, $I$, is added to the reactor. The reactor effluent is separated in a perfect splitter to recover pure $B$ product down the bottoms and a $B$-free distillate. A fraction of the distillate is purged and the rest is recycled back to the reactor. At a particular steady state, the product rate is 100 kmol h$^{-1}$, the recycle rate is 200 kmol h$^{-1}$ and the purge rate is 10 kmol h$^{-1}$. Given the above information, the inert feed rate into the process is ____________ kmol h$^{-1}$ (rounded off to two decimal places).

fresh A and inert I feeds mix with a recycle of 200 kmol/h and enter the 1 m³ CSTR. The effluent goes to a perfect splitter that gives pure B (100 kmol/h) at the bottom and a B-free distillate. The distillate splits into the purge (10 kmol/h) and the recycle.

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Correct answer: 0.99 to 1.01

Explanation

The B product of 100 kmol/h means that A is consumed at 100 kmol/h in the reactor: $164x_A \times 1 = 100$, so $x_A = 0.6098$. The reactor effluent is $100$ (B) $+ 210$ (the distillate, recycle plus purge) $= 310$ kmol/h, so the A in it is $0.6098 \times 310 = 189.0$ kmol/h, all of which goes to the distillate. The inert in the distillate is $210 - 189.0 = 21.0$ kmol/h. The purge is $\frac{10}{210}$ of the distillate, so it carries $21.0 \times \frac{10}{210} = 1.00$ kmol/h of inert. At steady state, the inert fed equals the inert purged: 1.00 kmol/h.