Question Bank • QB_DFE_ADV_001 NAT • 2 Marks

Section 7: Dairy and Food Engineering

Heat and Mass Transfer

Subtopic: Working principles of heat exchangers

Question Statement:

A counter-current double tube heat exchanger pasteurizes whole milk at a flow rate of $1800\text{ kg/h}$ ($C_p = 3.9\text{ kJ/(kg}\cdot\text{K)}$) from $15^\circ\text{C}$ to $72^\circ\text{C}$. Hot water ($C_p = 4.18\text{ kJ/(kg}\cdot\text{K)}$) enters the exchanger at $90^\circ\text{C}$ and exits at $45^\circ\text{C}$. The overall heat transfer coefficient $U = 1200\text{ W/(m}^2\cdot\text{K)}$. The required heat transfer surface area is ________ $\text{m}^2$ (round off to two decimal places).

Answer Key:

3.94

Step-by-Step Solution & Explanation

1. Rate of heat transfer ($Q$): $Q = \dot{m}_m C_{p,m} \Delta T_m = \left( \frac{1800}{3600}\text{ kg/s} \right) \times 3900\text{ J/(kg}\cdot\text{K)} \times (72 - 15) = 0.5 \times 3900 \times 57 = 111150\text{ W} = 111.15\text{ kW}$ 2. Logarithmic Mean Temperature Difference (LMTD): In counter-current flow: $\Delta T_1 = T_{h,in} - T_{c,out} = 90 - 72 = 18^\circ\text{C}$ $\Delta T_2 = T_{h,out} - T_{c,in} = 45 - 15 = 30^\circ\text{C}$ $\text{LMTD} = \frac{\Delta T_2 - \Delta T_1}{\ln(\Delta T_2 / \Delta T_1)} = \frac{30 - 18}{\ln(30 / 18)} = \frac{12}{\ln(1.6667)} = \frac{12}{0.5108} = 23.491^\circ\text{C}$ 3. Required heat transfer area ($A$): $Q = U A \text{LMTD} \implies A = \frac{Q}{U \times \text{LMTD}} = \frac{111150\text{ W}}{1200\text{ W/(m}^2\cdot\text{K)} \times 23.491\text{ K}} = \frac{111150}{28189.2} = 3.943\text{ m}^2 \approx 3.94\text{ m}^2$
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