Question:
A microbe that follows Monod growth kinetics on a limiting substrate (maximum specific growth rate of 0.5 h\(^{-1}\) and Monod constant of 0.1 mg L\(^{-1}\)) is cultivated in a continuous reactor for microbial growth (chemostat) with sterile feed. Given that the chemostat volume is 5 L and inlet concentration of limiting substrate is 10 g L\(^{-1}\), the minimum inlet feed flow rate for chemostat washout is L h\(^{-1}\). (rounded off to one decimal place)
A microbe that follows Monod growth kinetics on a limiting substrate (maximum specific growth rate of 0.5 h\(^{-1}\) and Monod constant of 0.1 mg L\(^{-1}\)) is cultivated in a continuous reactor for microbial growth (chemostat) with sterile feed. Given that the chemostat volume is 5 L and inlet concentration of limiting substrate is 10 g L\(^{-1}\), the minimum inlet feed flow rate for chemostat washout is L h\(^{-1}\). (rounded off to one decimal place)
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Washout dilution rate is \(D_c = \mu_{max}S_0/(K_s+S_0)\); since \(S_0 \gg K_s\), \(D_c \approx \mu_{max}\). Multiply by reactor volume to get the flow rate.
Updated On: Jul 16, 2026
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Correct Answer: 2.5
Solution and Explanation
Step 1: Understanding the Question.
In a chemostat with sterile feed, "washout" happens when the dilution rate \(D\) (the feed flow rate divided by the reactor volume) becomes so large that cells are removed from the vessel faster than they can grow, so the steady state cell concentration falls to zero. We need the minimum flow rate at which this just starts to happen.
Step 2: Key Formula.
For Monod growth kinetics, the specific growth rate at substrate concentration \(S\) is
\[ \mu = \frac{\mu_{max} S}{K_s + S} \]
At steady state in a chemostat, \(\mu = D\). Washout is approached as \(D \to \mu_{max}\), but since the substrate concentration in the vessel at the point of washout equals the feed (inlet) concentration \(S_0\) (nothing is left to be consumed once cells are washed out), the washout dilution rate is
\[ D_c = \frac{\mu_{max} S_0}{K_s + S_0} \]
Step 3: Substitute the given values (keeping units consistent).
\(\mu_{max} = 0.5\) h\(^{-1}\), \(K_s = 0.1\) mg L\(^{-1}\), and \(S_0 = 10\) g L\(^{-1}\) \(= 10000\) mg L\(^{-1}\). Since \(K_s\) is tiny compared to \(S_0\),
\[ D_c = \frac{0.5 \times 10000}{0.1 + 10000} = \frac{5000}{10000.1} = 0.49999 \ \text{h}^{-1} \]
This is essentially equal to \(\mu_{max}\), because the feed substrate concentration is far above the Monod constant, so the cells are already growing almost at their maximum possible rate.
Step 4: Convert dilution rate to a flow rate.
Dilution rate is defined as \(D = Q/V\), so the flow rate is
\[ Q_c = D_c \times V = 0.49999 \times 5 = 2.4999 \ \text{L h}^{-1} \]
Final Answer:
Rounded to one decimal place, the minimum feed flow rate that washes out the culture is
\[ \boxed{Q_c \approx 2.5 \ \text{L h}^{-1}} \]
In a chemostat with sterile feed, "washout" happens when the dilution rate \(D\) (the feed flow rate divided by the reactor volume) becomes so large that cells are removed from the vessel faster than they can grow, so the steady state cell concentration falls to zero. We need the minimum flow rate at which this just starts to happen.
Step 2: Key Formula.
For Monod growth kinetics, the specific growth rate at substrate concentration \(S\) is
\[ \mu = \frac{\mu_{max} S}{K_s + S} \]
At steady state in a chemostat, \(\mu = D\). Washout is approached as \(D \to \mu_{max}\), but since the substrate concentration in the vessel at the point of washout equals the feed (inlet) concentration \(S_0\) (nothing is left to be consumed once cells are washed out), the washout dilution rate is
\[ D_c = \frac{\mu_{max} S_0}{K_s + S_0} \]
Step 3: Substitute the given values (keeping units consistent).
\(\mu_{max} = 0.5\) h\(^{-1}\), \(K_s = 0.1\) mg L\(^{-1}\), and \(S_0 = 10\) g L\(^{-1}\) \(= 10000\) mg L\(^{-1}\). Since \(K_s\) is tiny compared to \(S_0\),
\[ D_c = \frac{0.5 \times 10000}{0.1 + 10000} = \frac{5000}{10000.1} = 0.49999 \ \text{h}^{-1} \]
This is essentially equal to \(\mu_{max}\), because the feed substrate concentration is far above the Monod constant, so the cells are already growing almost at their maximum possible rate.
Step 4: Convert dilution rate to a flow rate.
Dilution rate is defined as \(D = Q/V\), so the flow rate is
\[ Q_c = D_c \times V = 0.49999 \times 5 = 2.4999 \ \text{L h}^{-1} \]
Final Answer:
Rounded to one decimal place, the minimum feed flow rate that washes out the culture is
\[ \boxed{Q_c \approx 2.5 \ \text{L h}^{-1}} \]
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