Question

A walk-in deep freezer wall is made of 120 mm thick brick layer on the outside followed by 75 mm thick concrete and 50 mm thick cork layers inside. The mean temperatures measured over inside and outside wall surfaces are -18°C and 24°C, respectively. The thermal conductivity of brick, concrete and cork are 0.69, 0.76 and 0.043 W m$^{-1$ K$^{-1}$, respectively. Considering one square meter wall surface area, the heat transfer rate in W is _____.} \textit{[Round off to one decimal place]}

Hint:

The heat transfer rate through a wall is calculated using the thermal resistance of each layer. Higher resistance means lower heat transfer.

Answer 1

Correct Answer: 29

The heat transfer rate \( Q \) through the wall is given by the formula: \[ Q = \frac{A \Delta T}{\sum R} \] where: - \( A = 1 \, \text{m}^2 \) is the surface area, - \( \Delta T = 24 - (-18) = 42^\circ \text{C} \) is the temperature difference, - \( R = \frac{d}{k} \) is the thermal resistance for each layer, - \( d \) is the thickness of each layer, - \( k \) is the thermal conductivity of each material. The total resistance is the sum of resistances for each layer: \[ R_{\text{total}} = \frac{0.12}{0.69} + \frac{0.075}{0.76} + \frac{0.05}{0.043} = 0.1739 + 0.0980 + 1.1628 = 1.4347 \, \text{m}^2 \, \text{K/W}. \] Now, calculate the heat transfer rate: \[ Q = \frac{1 \times 42}{1.4347} \approx 29.3 \, \text{W}. \] Thus, the heat transfer rate is approximately \( \boxed{29.3} \, \text{W} \) (rounded to one decimal place).