Question

Steady heat conduction occurs through a \(0.3\)-m-thick, $9-\mathrm{m} \times 3-\mathrm{m}\( composite wall at a rate of \)1.2 \mathrm{~kW}$. If the inner and outer surface temperatures of the wall are \(15^{\circ} \mathrm{C}\) and \(7^{\circ} \mathrm{C}\), the effective thermal conductivity of the wall is $\begin{array}{ll}\text { (a) } 0.61 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} & \text { (b) } 0.83 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\end{array}$ (c) \(1.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (d) \(2.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (e) \(5.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\)

Short answer

Answer: The effective thermal conductivity of the wall is 0.83 W/m⋅K.

Step-by-step solution

Write down the given information

We are given the following information: - Steady state heat conduction: \(q = 1.2 ~\text{kW}\) - Composite wall thickness: \(L = 0.3 ~\text{m}\) - Wall dimensions: \(9\text{ m} \times 3\text{ m}\) - Inner surface temperature: \(T_1 = 15^{\circ}\text{C}\) - Outer surface temperature: \(T_{2} = 7^{\circ}\text{C}\)

Convert heat conduction to Watts

We must first convert the given heat conduction from kilowatts to watts: \(q = 1.2 ~\text{kW} \times 1000 \frac{\text{W}}{1 \text{kW}} = 1200 ~\text{W}\)

Calculate the temperature difference across the wall

Now, let's calculate the temperature difference between the inner and outer surfaces: \(\Delta T = T_1 - T_2 = 15^{\circ}\text{C} - 7^{\circ}\text{C} = 8^{\circ}\text{C}\)

Calculate the wall's surface area

Next, calculate the surface area of the wall: \(A = 9\text{ m} \times 3\text{ m} = 27 ~\text{m}^2\)

Determine the formula for steady state heat conduction

We know that the steady state heat conduction formula in terms of thermal conductivity (\(k\)) is given by Fourier's Law: \(q = kA \frac{\Delta T}{L}\)

Calculate the effective thermal conductivity

From the formula in step 5, we can solve for \(k\): \(k = \frac{qL}{A\Delta T} = \frac{1200~\text{W} \times 0.3 ~\text{m}}{27 ~\text{m}^2 \times 8^{\circ}\text{C}} = 0.83 \frac{\text{W}}{\text{m} \cdot \text{K}}\) So, the effective thermal conductivity of the wall is \(\boxed{0.83 ~\text{W/m} \cdot \text{K}}\), which corresponds to option (b).