Question Bank • QB_DFE_HMT_011 NAT • 2 Marks

Section 7: Dairy and Food Engineering

Heat and Mass Transfer

Subtopic: Steady state heat transfer in conduction, convection and radiation

Question Statement:

A refrigerated cold store wall has dimensions $4.0\text{ m} \times 6.0\text{ m}$ (area $A = 24.0\text{ m}^2$) and consists of an inner layer of polyurethane foam ($x_1 = 0.10\text{ m}$, $k_1 = 0.025\text{ W/(m}\cdot\text{K)}$) and an outer layer of brick masonry ($x_2 = 0.15\text{ m}$, $k_2 = 0.60\text{ W/(m}\cdot\text{K)}$). The inside foam surface is at $T_{s,1} = -18.0^\circ\text{C}$ and the outside brick surface is at $T_{s,2} = 32.0^\circ\text{C}$. The rate of heat ingress through the wall in Watts ($\text{W}$) is ________ (round off to one decimal place).

Answer Key:

282.4

Step-by-Step Solution & Explanation

Multi-Approach Solution: **Method 1: Thermal Resistance in Series** 1. Conduction resistance of polyurethane foam: $R_1 = \frac{x_1}{k_1 A} = \frac{0.10\text{ m}}{0.025\text{ W/(m}\cdot\text{K)} \times 24.0\text{ m}^2} = \frac{0.10}{0.60} = 0.16667\text{ K/W}$ 2. Conduction resistance of brick masonry: $R_2 = \frac{x_2}{k_2 A} = \frac{0.15\text{ m}}{0.60\text{ W/(m}\cdot\text{K)} \times 24.0\text{ m}^2} = \frac{0.15}{14.40} = 0.01042\text{ K/W}$ 3. Total resistance: $R_{\text{total}} = R_1 + R_2 = 0.16667 + 0.01042 = 0.17709\text{ K/W}$ 4. Heat ingress rate ($q$): $q = \frac{\Delta T}{R_{\text{total}}} = \frac{32.0 - (-18.0)}{0.17709} = \frac{50.0\text{ K}}{0.17709\text{ K/W}} = 282.34\text{ W} \approx 282.4\text{ W}$ **Method 2: Overall Heat Transfer Coefficient ($U$)** $U = \frac{1}{\frac{x_1}{k_1} + \frac{x_2}{k_2}} = \frac{1}{\frac{0.10}{0.025} + \frac{0.15}{0.60}} = \frac{1}{4.0 + 0.25} = \frac{1}{4.25} = 0.23529\text{ W/(m}^2\cdot\text{K)}$ $q = U A \Delta T = 0.23529 \times 24.0 \times 50.0 = 282.35\text{ W}$
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