Question

The inner and outer surfaces of a 25 -cm-thick wall in summer are at \(27^{\circ} \mathrm{C}\) and \(44^{\circ} \mathrm{C}\), respectively. The outer surface of the wall exchanges heat by radiation with surrounding surfaces at \(40^{\circ} \mathrm{C}\) and by convection with ambient air also at $40^{\circ} \mathrm{C}\( with a convection heat transfer coefficient of \)8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Solar radiation is incident on the surface at a rate of \(150 \mathrm{~W} / \mathrm{m}^{2}\). If both the emissivity and the solar absorptivity of the outer surface are \(0.8\), determine the effective thermal conductivity of the wall.

Short answer

Answer: The effective thermal conductivity of the wall is 1.634 W/m·K.

Step-by-step solution

1. Calculate the heat transfer due to solar radiation

First, we need to find the absorbed heat flux of solar radiation (\(Q_{solar}\)): $$ Q_{solar} = (\text{solar absorptivity}) \times (\text{solar radiation incident}) $$ $$ Q_{solar} = 0.8 \times 150 \mathrm{~W} / \mathrm{m}^{2} $$ $$ Q_{solar} = 120 \mathrm{~W} / \mathrm{m}^{2} $$

2. Calculate the heat transfer due to convection

Next, we need to find the heat transfer due to convection (\(Q_{conv}\)) using Newton's law of cooling: $$ Q_{conv} = h \times A \times (T_{surface} - T_{ambient}) $$ To use this equation, we must first recognize that for this problem, the surface area \(A\) and the term \((T_{surface}-T_{ambient})\) will be the same for both the convection and radiation terms. So, let's simply find the heat transfer coefficient (\(h\)) for now, which is given as: $$ h = 8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} $$

3. Calculate the heat transfer due to radiation

Next, we need to find the heat transfer due to radiation (\(Q_{rad}\)) using the Stefan-Boltzmann equation: $$ Q_{rad} = \varepsilon \sigma A(T_{surface}^{4} - T_{surroundings}^{4}) $$ Again, we know that the surface area \(A\) and the term \((T_{surface} - T_{ambient})\) will be the same as before. So, let's find the effective radiation heat transfer coefficient (\(h_r\)) for now: $$ h_{r} = \dfrac{\varepsilon \sigma (T_{surface}^{4} - T_{surroundings}^{4})}{T_{surface} - T_{ambient}} $$ Plugging in the values: $$ h_{r} = \dfrac{ (0.8)(5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}^{4})[(317 \mathrm{K})^{4} - (313 \mathrm{K})^{4}]}{44^{\circ} \mathrm{C} - 40^{\circ} \mathrm{C)} $$ $$ h_{r} = 6.171 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} $$

4. Calculate the overall heat transfer coefficient

Now, we can find the overall heat transfer coefficient \(U\): $$ U = h_{conv} + h_{rad} $$ $$ U = 8 + 6.171 $$ $$ U = 14.171 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} $$

5. Calculate the total heat gained

To calculate the total heat gained by the wall, we must combine the heat transfer due to solar radiation, convection, and radiation: $$ Q_{total} = Q_{solar} + U(T_{surface} - T_{ambient}) $$ $$ Q_{total} = 120 + 14.171(44 - 40) $$ $$ Q_{total} = 175.684 \mathrm{~W} / \mathrm{m}^{2} $$

6. Calculate the effective thermal conductivity

Finally, we calculate the effective thermal conductivity of the wall \(k\) using Fourier's Law of heat conduction: $$ Q = \dfrac{k \times A \times (T_{inner} - T_{outer})}{\text{wall thickness}} $$ We know that in this case, $$ Q = Q_{total} $$ $$ (T_{inner}, T_{outer}, \text{wall thickness}) = (27^{\circ} \mathrm{C}, 44^{\circ} \mathrm{C}, 0.25 \mathrm{m}) $$ Solving for k, we get: $$ k = \dfrac{Q_{total} \times \text{wall thickness}}{A \times (T_{inner} - T_{outer})} $$ $$ k = \dfrac{175.684 \mathrm{~W} / \mathrm{m}^{2} \times 0.25 \mathrm{m}}{(27^{\circ} \mathrm{C} - 44^{\circ} \mathrm{C})} $$ $$ k = 1.634 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} $$ The effective thermal conductivity of the wall is \(1.634 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).