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GATE 2024 CH – Question 62

Chemical Reaction Engineering · Enzyme kinetics: Michaelis-Menten and Monod models · 2 marks · Numerical answer

A chemostat with cell recycle is shown in the figure. The feed flow rate and culture volume are $F = 75$ L h$^{-1}$ and $V = 200$ L, respectively. The glucose concentration in the feed $C_{S0} = 15$ g L$^{-1}$. Assume Monod kinetics with specific cell growth rate $\mu_g = \frac{1}{C_C}\frac{dC_C}{dt} = \frac{\mu_mC_S}{K_S + C_S}$, where $\mu_m = 0.25$ h$^{-1}$ and $K_s = 1$ g L$^{-1}$. Assume maintenance and death rates to be zero, input feed to be sterile ($C_{C0} = 0$) and steady-state operation. The glucose concentration in the recycle stream, $C_{S1}$, in g L$^{-1}$, rounded off to 1 decimal place, is _________

a chemostat of volume V with the fresh feed F ($C_{C0}$, $C_{S0}$) and a recycle stream of $\alpha F$ coming back from a cell concentrator. The outlet from the chemostat, $(1 + \alpha)F$ with $C_{C1}$ and $C_{S1}$, goes to the cell concentrator, which returns the concentrated cells at $\alpha F$ with the concentration $\beta C_{C1}$ and sends the rest out. $\alpha = 0.5$ and $\beta = 2.0$.

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Correct answer: 2.99 to 3.01

Explanation

Let $D = \frac{F}{V} = \frac{75}{200} = 0.375$ h⁻¹. A steady-state cell balance on the chemostat (the feed has no cells) is $V\mu_gC_{C1} + \alpha F\beta C_{C1} - (1 + \alpha)FC_{C1} = 0$, so $\mu_g = D(1 + \alpha - \alpha\beta) = 0.375 \times (1 + 0.5 - 1.0) = 0.1875$ h⁻¹. From the Monod kinetics, $\frac{0.25C_S}{1 + C_S} = 0.1875$, so $C_S = \frac{0.1875}{0.0625} = 3.0$ g/L. The substrate concentration in the chemostat is the same as in the stream leaving it, so $C_{S1} = 3.0$ g/L.