Section 4: Bioprocess Engineering and Process Biotechnology
Bioreactor types, scale-up
Subtopic: Fundamentals of Bioreactor types, scale-up
Question Statement:
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)
Official Answer Key:
2.4 to 2.5
Step-by-Step Solution & Explanation
### **Step-by-Step Solution & Mathematical Derivation (GATE BT 2026 Q.61)**
**1. Governing Concept & Definition**:
- **Topic**: Continuous Culture Kinetics & Washout Dilution Rate
- In a continuous chemostat with sterile feed, washout occurs when the dilution rate exceeds the maximum possible specific growth rate in the reactor:
$D_{\text{washout}} = \frac{\mu_{\max} S_0}{K_s + S_0}$
Since $S_0 \gg K_s$ ($10\text{ g L}^{-1} \gg 0.1\text{ mg L}^{-1} = 10^{-4}\text{ g L}^{-1}$), $D_{\text{washout}} \approx \mu_{\max}$.
- The critical washout feed flow rate is:
$F_{\text{washout}} = D_{\text{washout}} \times V$
**2. Detailed Step-by-Step Calculation**:
- **Given**:
- Chemostat volume $V = 5\text{ L}$
- $\mu_{\max} = 0.5\text{ h}^{-1}$
- $D_{\text{washout}} \approx 0.5\text{ h}^{-1}$
- **Calculation of Washout Flow Rate**:
$F_{\text{washout}} = 0.5\text{ h}^{-1} \times 5\text{ L} = 2.5\text{ L h}^{-1}$
**3. Final Answer**:
- **Calculated Value**: **2.5 L/h**
- **Official GATE Accepted Range**: **2.4 to 2.5**
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