Chemegate

GATE 2024 Chemical Engineering, Q62 · Reaction Engineering

The correct answer is 2.9 to 3.1.

Question

A chemostat with cell recycle is shown in the figure. The feed flow rate and culture volume are F = 75 L/h and V = 200 L, respectively. The glucose concentration in the feed CS0=15C_{S0} = 15 g/L. Assume Monod kinetics with specific cell growth rate μg=(1/CC)(dCC/dt)=(μmCS)/(KS+CS),μ_{g} = (1/C_{C})\cdot (dC_{C}/dt)=(μ_{m}\cdot C_{S})/(K_{S}+C_{S}), where μm=0.25h1μ_{m} = 0.25 h⁻¹ and KS=1K_{S} = 1 g/L. The recycle ratio is α=0.5α = 0.5 and the cell concentration factor is β=2.0.β = 2.0. Assume maintenance and death rates to be zero, input feed to be sterile (CC0=0),(C_{C0}=0), and steady-state operation. The glucose concentration in the recycle stream, CS1,C_{S1}, in g/L, rounded off to 1 decimal place, is _________

Figure for Q62 (Reaction Engineering, Bioreactors): A chemostat with cell recycle is shown in the figure. The feed flow rate and culture volume are F =…

Official answer: 2.9 to 3.1

Why

GATE 2024 official key: 2.9 to 3.1. Cell balance around the chemostat gives μ=F[(1+α)αβ]/V=75×0.5/200=0.1875h1.μ = F[(1+α)−αβ]/V = 75 \times 0.5/200 = 0.1875 h⁻¹. Solving Monod: 0.1875=0.25CS1/(1+CS1)CS1=3.00.1875 = 0.25\cdot C_{S1}/(1+C_{S1}) \Rightarrow C_{S1} = 3.0 g/L.

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