Subtopic: Fundamentals of Biomolecules, enzyme kinetics
Question Statement:
A CSTR with a volume of 100 m3 is operated in cell recycle mode. At a volumetric
flow rate of 10 m3 day-1 and effluent biomass of 20 mg L-1, the steady state biomass
concentration is 200 mg L-1. The mean cell retention time in the reactor is ____ days.
(answer in integer)
Official Answer Key:
100 to 100
Step-by-Step Solution & Explanation
### **Step-by-Step Solution & Mathematical Derivation (GATE BT 2026 Q.56)**
**1. Governing Concept & Definition**:
- **Topic**: Bioprocess Engineering & Mean Cell Residence Time ($\theta_c$)
- Mean cell residence time (also known as solids retention time or sludge age $\theta_c$) is:
$\theta_c = \frac{\text{Total biomass in reactor}}{\text{Rate of biomass loss in effluent}} = \frac{V \cdot X}{F \cdot X_e}$
**2. Detailed Step-by-Step Calculation**:
- **Given**:
- Reactor Volume $V = 100\text{ m}^3$
- Volumetric Flow Rate $F = 10\text{ m}^3\text{ day}^{-1}$
- Reactor Biomass Concentration $X = 200\text{ mg L}^{-1}$
- Effluent Biomass Concentration $X_e = 20\text{ mg L}^{-1}$
- **Substitution**:
$\theta_c = \frac{100\text{ m}^3 \times 200\text{ mg L}^{-1}}{10\text{ m}^3\text{ day}^{-1} \times 20\text{ mg L}^{-1}} = \frac{20000}{200} = 100\text{ days}$
**3. Final Answer**:
- **Calculated Value**: **100 days**
- **Official GATE Accepted Range**: **100 to 100**
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