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 0.8 for solar radiation. If solar radiation is incident on the plate at a rate of \(450 \mathrm{W} / \mathrm{m}^{2}\) and the surrounding air temperature is \(25^{\circ} \mathrm{C}\), determine the surface temperature of the plate when the heat loss by convection equals the solar energy absorbed by the plate. Assume the convection heat transfer coefficient to be \(50 \mathrm{W} / \mathrm{m}^{2} \cdot^{\circ} \mathrm{C}\) and disregard heat loss by radiation.

Short answer

Answer: The surface temperature of the plate is 32.2°C.

Step-by-step solution

Calculate solar energy absorbed

To calculate the solar energy absorbed by the plate, we will need to multiply the solar radiation by the absorptivity of the plate. \(Q_{absorbed} = S_{solar} \cdot \alpha = 450\,W/m^2 \cdot 0.8 = 360\,W/m^2\)

Set up the convection heat loss equation

Now, we will set up the convection heat loss equation. Convection heat transfer can be expressed as: \(Q_{convection} = h \cdot A \cdot (T_{surface} - T_{ambient})\) In this case, we know the heat transfer coefficient \(h\) and the ambient temperature \(T_{ambient}\). We want to find the heat loss when \(Q_{absorbed} = Q_{convection}\). Since the plate's area (\(A\)) is constant, we can omit \(A\) and modify the equation as: \(360\,W/m^2 = 50\,W/m^2\cdot^{\circ} C \cdot (T_{surface} - 25^{\circ} C)\)

Solve for surface temperature

Now, we will solve for the surface temperature \(T_{surface}\): \(360\,W/m^2 = 50\,W/m^2\cdot^{\circ} C \cdot (T_{surface} - 25^{\circ} C)\) Divide both sides by \(50\,W/m^2\cdot^{\circ} C\): \(7.2^{\circ} C = T_{surface} - 25^{\circ} C\) Add \(25^{\circ} C\) to both sides: \(T_{surface} = 32.2^{\circ} C\) The surface temperature of the plate when the heat loss by convection equals the solar energy absorbed by the plate is \(32.2^{\circ} C\).