Question

A thin metal plate is insulated on the back and exposed to solar radiation on the front surface. The exposed surface of the plate has an absorptivity of \(0.7\) for solar radiation. If solar radiation is incident on the plate at a rate of \(550 \mathrm{~W} / \mathrm{m}^{2}\) and the surrounding air temperature is \(10^{\circ} \mathrm{C}\), determine the surface temperature of the plate when the heat loss by convection equals the solar energy absorbed by the plate. Take the convection heat transfer coefficient to be $25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, and disregard any heat loss by radiation.

Short answer

Answer: The surface temperature of the plate when the solar energy absorbed equals the heat loss by convection is 25.4°C.

Step-by-step solution

Absorbed Solar Energy

Calculate the absorbed solar energy (\(Q_{abs}\)) by the plate. That can be found by multiplying the solar radiation (\(S\)) by the absorptivity of the surface (\(\alpha\)). So, the absorbed solar energy is: \(Q_{abs} = \alpha \cdot S\)

Heat Loss by Convection

Calculate the heat loss by convection (\(Q_{conv}\)). We can use the equation for the heat transfer by convection: \(Q_{conv} = h \cdot A \cdot (T_{s} - T_{\infty})\) Where \(h\) is the convection heat transfer coefficient, \(A\) is the surface area, \(T_{s}\) is the surface temperature, and \(T_{\infty}\) is the surrounding air temperature. Since the problem asks for the situation when the heat loss by convection equals the absorbed solar energy, we can equate \(Q_{conv}\) and \(Q_{abs}\). The surface area (\(A\)) will cancel out in the equation, so we'll have: \(Q_{conv} = Q_{abs}\) \(h (T_{s} - T_{\infty}) = \alpha \cdot S\)

Solve for Surface Temperature

Now, we can plug in the given values and solve for the surface temperature (\(T_{s}\)): \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \cdot (T_{s} - 10^{\circ} \mathrm{C}) = 0.7 \cdot (550 \mathrm{~W} / \mathrm{m}^{2})\) Divide by \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\): \((T_{s} - 10^{\circ} \mathrm{C}) = \frac{0.7 \cdot (550 \mathrm{~W} / \mathrm{m}^{2})}{25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}}\) Calculate the result: \((T_{s} - 10^{\circ} \mathrm{C}) = 15.4^{\circ} \mathrm{C}\) Add \(10^{\circ} \mathrm{C}\) to find the surface temperature: \(T_{s} = 10^{\circ} \mathrm{C} + 15.4^{\circ} \mathrm{C} = 25.4^{\circ} \mathrm{C}\) Thus, the surface temperature of the plate when the heat loss by convection equals the solar energy absorbed is \(25.4^{\circ} \mathrm{C}\).