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QB_DFE_HMT_001
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NAT • 2 Marks
Section 7: Dairy and Food Engineering
Heat and Mass Transfer
Subtopic: Working principles of heat exchangers
Question Statement:
In a continuous counter-flow double-tube heat exchanger, whole milk ($C_p = 3.90\text{ kJ/(kg}\cdot\text{K)}$) is heated from $10.0^\circ\text{C}$ to $65.0^\circ\text{C}$ by hot water entering at $85.0^\circ\text{C}$ and discharging at $50.0^\circ\text{C}$. The Logarithmic Mean Temperature Difference (LMTD) of the heat exchanger in $^\circ\text{C}$ is ________ (round off to two decimal places).
Step-by-Step Solution & Explanation
Multi-Approach Solution:
**Method 1: Counter-Flow LMTD Definition**
```
Hot Water: 85.0 °C ====================> 50.0 °C
Milk: 65.0 °C <==================== 10.0 °C
Diff (ΔT): 20.0 °C 40.0 °C
```
1. Temperature differences at terminal ends:
$\Delta T_1 = T_{h,\text{in}} - T_{c,\text{out}} = 85.0 - 65.0 = 20.0^\circ\text{C}$
$\Delta T_2 = T_{h,\text{out}} - T_{c,\text{in}} = 50.0 - 10.0 = 40.0^\circ\text{C}$
2. Logarithmic Mean Temperature Difference:
$\text{LMTD} = \frac{\Delta T_2 - \Delta T_1}{\ln(\Delta T_2 / \Delta T_1)} = \frac{40.0 - 20.0}{\ln(40.0 / 20.0)} = \frac{20.0}{\ln(2)} = \frac{20.0}{0.693147} = 28.8539^\circ\text{C} \approx 28.85^\circ\text{C}$
**Method 2: Chen's Approximation Formula**
$\text{LMTD} \approx \left[ \frac{\Delta T_1 \Delta T_2 (\Delta T_1 + \Delta T_2)}{2} \right]^{1/3} = \left[ \frac{20 \times 40 \times 60}{2} \right]^{1/3} = (24000)^{1/3} \approx 28.845^\circ\text{C}$
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