Growth of an organism on glucose in a chemostat is characterized by Monod model with maximum specific growth rate = 0.45 h⁻¹ and Ks = 0.5 g/L. Biomass from the substrate is generated as Yxs = 0.4 g/g. The chemostat volume is 0.9 L and media is fed at 1 L/h and contains 20 g/L of glucose. At steady state, the concentration of biomass in the chemostat is ________ g/L.
Official Answer Key:
0 to 0
Step-by-Step Solution & Explanation
### **Step-by-Step Solution & Mathematical Derivation (GATE BT 2020 Q.59)**
**1. Governing Concept & Definition**:
- **Topic**: Bioprocess Kinetics & Chemostat Washout Criterion
- In a continuous chemostat with sterile feed, steady-state biomass concentration $X$ is positive only if the dilution rate $D$ is less than the maximum specific growth rate $\mu_{\max}$:
$D = \frac{F}{V}$
- If $D > \mu_{\max}$, cells are washed out faster than they can reproduce, resulting in complete washout ($X = 0$).
**2. Detailed Step-by-Step Calculation**:
- **Given**:
- Working volume $V = 0.9\text{ L}$
- Feed flow rate $F = 1.0\text{ L/h}$
- Maximum specific growth rate $\mu_{\max} = 0.45\text{ h}^{-1}$
- **Step 1: Calculate Dilution Rate ($D$)**:
$D = \frac{F}{V} = \frac{1.0\text{ L/h}}{0.9\text{ L}} = 1.111\text{ h}^{-1}$
- **Step 2: Compare $D$ with $\mu_{\max}$**:
$D = 1.111\text{ h}^{-1} > \mu_{\max} = 0.45\text{ h}^{-1}$
- Since the dilution rate exceeds the maximum possible microbial growth rate, the culture cannot sustain itself in the chemostat.
- Therefore, the steady-state biomass concentration in the reactor is $0\text{ g/L}$ (Washout condition).
**3. Final Answer**:
- **Calculated Value**: **0 g/L**
- **Official GATE Accepted Range**: **0 to 0**
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