Question

Consider a wall that consists of two layers, \(A\) and \(B\), with the following values: $k_{A}=1.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, L_{A}=8 \mathrm{~cm}\(, \)k_{B}=0.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, L_{B}=5 \mathrm{~cm}\(. If the temperature drop across the wall is \)18^{\circ} \mathrm{C}$, the rate of heat transfer through the wall per unit area of the wall is (a) \(56.8 \mathrm{~W} / \mathrm{m}^{2}\) (b) \(72.1 \mathrm{~W} / \mathrm{m}^{2}\) (c) \(114 \mathrm{~W} / \mathrm{m}^{2}\) (d) \(201 \mathrm{~W} / \mathrm{m}^{2}\) (e) \(270 \mathrm{~W} / \mathrm{m}^{2}\)

Short answer

Answer: (a) 56.8 W/m²

Step-by-step solution

Identify the variables

In this exercise, we have the following values for each layer of the wall: Layer A: \(k_{A} = 1.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (thermal conductivity) \(L_{A} = 0.08 \mathrm{~m}\) (thickness, converted from cm to m) Layer B: \(k_{B} = 0.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (thermal conductivity) \(L_{B} = 0.05 \mathrm{~m}\) (thickness, converted from cm to m) Temperature drop across the wall: \(\Delta T = 18^{\circ} \mathrm{C}\)

Recall the formula for heat transfer through multiple layers

The formula for the rate of heat transfer per unit area through a wall with multiple layers is given by: \(q = \frac{\Delta T}{R_{total}}\) where \(q\) is the rate of heat transfer per unit area and \(R_{total}\) is the total thermal resistance of the wall, which is the sum of the thermal resistances of each layer: \(R_{total} = \sum R_{i}\) For each layer, the thermal resistance \(R_{i}\) can be calculated using: \(R_{i} = \frac{L_{i}}{k_{i}}\) where \(L_{i}\) is the thickness of the layer and \(k_{i}\) is its thermal conductivity.

Calculate the total thermal resistance of the wall

Using the given values, we can calculate the thermal resistance for layers A and B as follows: \(R_{A} = \frac{L_{A}}{k_{A}} = \frac{0.08 \mathrm{~m}}{1.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}} = 0.0667 \mathrm{~K} / \mathrm{W}\) \(R_{B} = \frac{L_{B}}{k_{B}} = \frac{0.05 \mathrm{~m}}{0.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}} = 0.25 \mathrm{~K} / \mathrm{W}\) Now, we can calculate the total thermal resistance: \(R_{total} = R_{A} + R_{B} = 0.0667 \mathrm{~K} / \mathrm{W} + 0.25 \mathrm{~K} / \mathrm{W} = 0.3167 \mathrm{~K} / \mathrm{W}\)

Calculate the rate of heat transfer per unit area

Using the total thermal resistance and the temperature drop across the wall, we can calculate the rate of heat transfer per unit area: \(q = \frac{\Delta T}{R_{total}} = \frac{18^{\circ} \mathrm{C}}{0.3167 \mathrm{~K} / \mathrm{W}} = 56.8 \mathrm{~W} / \mathrm{m}^{2}\)

Choose the correct answer

The calculated rate of heat transfer per unit area is \(56.8 \mathrm{~W} / \mathrm{m}^{2}\). Thus, the correct answer is: (a) \(56.8 \mathrm{~W} / \mathrm{m}^{2}\)