GATE BT 2025 • Q61 NAT • 2 Marks

Section 2: General Biotechnology

Biomolecules, enzyme kinetics

Subtopic: Fundamentals of Biomolecules, enzyme kinetics

Question Statement:

A hot, freshly-sterilised fermentation medium is cooled in a double-pipe heat- exchanger. The medium enters the inner pipe of the exchanger at 95 C and leaves the exchanger at 40 C. Cooling water, flowing counter-currently to the medium, enters the annulus of the exchanger at 15 C and leaves the exchanger at 45 C. The overall heat transfer coefficient is 1350 W m-2 C-1. The rate of heat transfer per unit area will be _______ W/m2. (Round off to the nearest integer)

Official Answer Key:

48000 to 49000

Step-by-Step Solution & Explanation

### **Step-by-Step Solution & Mathematical Derivation (GATE BT 2025 Q.61)** **1. Governing Concept & Definition**: - **Topic**: Bioprocess Engineering & Heat Exchanger Heat Transfer - Rate of heat transfer per unit area ($q''$) in a counter-current heat exchanger is given by: $q'' = \frac{Q}{A} = U \cdot \Delta T_{\text{LMTD}}$ where $U$ is the overall heat transfer coefficient and $\Delta T_{\text{LMTD}}$ is the log mean temperature difference. **2. Detailed Step-by-Step Calculation**: - **Given Stream Temperatures (Counter-current)**: - Hot fluid (medium): $T_{h,in} = 95^\circ\text{C}$, $T_{h,out} = 40^\circ\text{C}$ - Cold fluid (water): $T_{c,in} = 15^\circ\text{C}$, $T_{c,out} = 45^\circ\text{C}$ - Overall heat transfer coefficient $U = 1350\text{ W m}^{-2}\text{ }^\circ\text{C}^{-1}$ - **Step 1: Calculate Terminal Temperature Differences**: - Hot end: $\Delta T_1 = T_{h,in} - T_{c,out} = 95 - 45 = 50^\circ\text{C}$ - Cold end: $\Delta T_2 = T_{h,out} - T_{c,in} = 40 - 15 = 25^\circ\text{C}$ - **Step 2: Calculate Log Mean Temperature Difference ($\Delta T_{\text{LMTD}}$)**: $\Delta T_{\text{LMTD}} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)} = \frac{50 - 25}{\ln(50 / 25)} = \frac{25}{\ln(2)} \approx \frac{25}{0.69315} \approx 36.067^\circ\text{C}$ - **Step 3: Calculate Heat Transfer Rate Per Unit Area ($q''$)**: $q'' = U \cdot \Delta T_{\text{LMTD}} = 1350 \times 36.067 \approx 48690\text{ W/m}^2$ **3. Final Answer**: - **Calculated Value**: **48690 W/m²** - **Official GATE Accepted Range**: **48000 to 49000**

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