Question:
A chemostat with cell recycle is shown in the figure. The feed flow rate and culture volume are \( F = 75 \, \text{L h}^{-1} \) and \( V = 200 \, \text{L} \), respectively. The glucose concentration in the feed \( C_{S0} = 15 \, \text{g L}^{-1} \). Assume Monod kinetics with specific cell growth rate:
\[
\mu_g = \mu_m \frac{C_S}{K_S + C_S},
\]
where \( \mu_m = 0.25 \, \text{h}^{-1} \) and \( K_S = 1 \, \text{g L}^{-1} \). Assume maintenance and death rates to be zero, input feed to be sterile (\( C_{S0} = 0 \)) and steady-state operation. The glucose concentration in the recycle stream, \( C_{S1} \), in \( \text{g L}^{-1} \), rounded off to 1 decimal place, is:
\includegraphics[width=0.5\linewidth]{q62 CE.PNG}
A chemostat with cell recycle is shown in the figure. The feed flow rate and culture volume are \( F = 75 \, \text{L h}^{-1} \) and \( V = 200 \, \text{L} \), respectively. The glucose concentration in the feed \( C_{S0} = 15 \, \text{g L}^{-1} \). Assume Monod kinetics with specific cell growth rate:
\[
\mu_g = \mu_m \frac{C_S}{K_S + C_S},
\]
where \( \mu_m = 0.25 \, \text{h}^{-1} \) and \( K_S = 1 \, \text{g L}^{-1} \). Assume maintenance and death rates to be zero, input feed to be sterile (\( C_{S0} = 0 \)) and steady-state operation. The glucose concentration in the recycle stream, \( C_{S1} \), in \( \text{g L}^{-1} \), rounded off to 1 decimal place, is:
\includegraphics[width=0.5\linewidth]{q62 CE.PNG}
Show Hint
In chemostat problems, calculate the dilution rate and use steady-state Monod kinetics to determine substrate or cell concentrations.
Updated On: Jan 24, 2025
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Solution and Explanation
Step 1: Monod kinetics equation for glucose concentration.
At steady state, the dilution rate \( D \) is equal to the specific growth rate \( \mu_g \):
\[
D = \frac{F}{V} = \mu_g.
\]
Substitute \( F = 75 \, \text{L h}^{-1} \) and \( V = 200 \, \text{L} \):
\[
D = \frac{75}{200} = 0.375 \, \text{h}^{-1}.
\]
Step 2: Solve for \( C_S \) using the Monod equation.
The Monod equation is:
\[
\mu_g = \mu_m \frac{C_S}{K_S + C_S}.
\]
Substitute \( \mu_g = 0.375 \), \( \mu_m = 0.25 \), and \( K_S = 1 \):
\[
0.375 = 0.25 \frac{C_S}{1 + C_S}.
\]
Rearrange:
\[
1.5 = \frac{C_S}{1 + C_S}.
\]
\[
1.5(1 + C_S) = C_S \quad \Rightarrow \quad 1.5 + 1.5C_S = C_S.
\]
Simplify:
\[
1.5 = -0.5C_S \quad \Rightarrow \quad C_S = 3 \, \text{g L}^{-1}.
\]
Step 3: Conclusion.
The glucose concentration in the recycle stream \( C_{S1} \) is \( 3 \, \text{g L}^{-1} \).
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