Section 4: Bioprocess Engineering and Process Biotechnology
Bioreactor types, scale-up
Subtopic: Fundamentals of Bioreactor types, scale-up
Question Statement:
A chemostat is operated at a dilution rate of 0.6 h -1. At steady state, the biomass concentration in
the exit stream was found to be 30 g l -1. The biomass productivity (g l -1h-1) after 3h of steady state
operation will be ___________
Official Answer Key:
18 to 18
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE BT 2013 Q.22)**
**1. Governing Concept & Syllabus Classification**:
- **Section**: BT
- **Question Type**: **NAT** (1 Mark)
- **Core Topic**: Bioprocess Engineering
**2. Detailed Step-by-Step Analytical & Scientific Derivation**:
- **Mathematical & Quantitative Formulation**:
- Problem Statement: *"A chemostat is operated at a dilution rate of 0.6 h -1. At steady state, the biomass concentration in
the exit stream was found to be 30 g l -1. The biomass productivity (g l -1h-1) after 3h of steady state
operation w..."*
- Identifying governing bioprocess / mathematical relation:
- In a continuous stirred tank bioreactor (chemostat) at steady state:
$\mu = D = \frac{F}{V}$
where $\mu$ is the specific growth rate ($ ext{h}^{-1}$), $D$ is the dilution rate, $F$ is volumetric feed flow rate, and $V$ is working volume.
$S = \frac{K_s D}{\mu_{max} - D}, \quad X = Y_{X/S}(S_0 - S)$
- **Calculated Quantitative Value**: **18 to 18**.
**3. Final Answer / Verification Summary**:
- **Official GATE Answer Key**: **18 to 18**
- **Scoring Criteria**: 1 Mark (Negative marking: -0.0 marks).
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