Question 21.8: A Hot Tin Roof (Solar-Powered Homes) GOAL Calculate some bas...
A Hot Tin Roof (Solar-Powered Homes)
GOAL Calculate some basic properties of light and relate them to thermal radiation.
PROBLEM Assume the Sun delivers an average power per unit area of about 1.00 × 10³ W/m² to Earth’s surface. (a) Calculate the total power incident on a flat tin roof 8.00 m by 20.0 m. Assume the radiation is incident normal (perpendicular) to the roof. (b) The tin roof reflects some light, and convection, conduction, and radiation transport the rest of the thermal energy away until some equilibrium temperature is established. If the roof is a perfect blackbody and rids itself of one-half of the incident radiation through thermal radiation, what’s its equilibrium temperature? Assume the ambient temperature is 298 K.
(a) Calculate the power delivered to the roof.
Multiply the intensity by the area to get the power:
P = IA = (1.00 × 10³ W/m²)(8.00 m × 20.0 m)
=\;1.60\times10^{5}\,\mathrm{W}
(b) Find the equilibrium temperature of the roof.
Substitute into Stefan’s law. Only one-half the incident power should be substituted, and twice the area of the roof (both the top and the underside of the roof count).
P=\sigma e A(T^{4}-\ T_{0}^{4})
T^{4}=\ T_{0}{}^{4}+\frac{P}{\sigma e{A}}
=\,(\mathrm{298\,\,K})^{4}+\frac{(0.500)(1.60~\times~10^{5}\,{\mathrm{W/m^{2}}})}{(5.67~\times~10^{-8}\,{\mathrm{W/m^{2}}}~\cdot~{\mathrm{K}}^{4})(1)(3.20~\times~10^{2}\,{\mathrm{m^{2}}})}
T = 333 K = 6.0 × 10¹ °C
REMARKS If the incident power could all be converted to electric power, it would be more than enough for the average home. Unfortunately, solar energy isn’t easily harnessed, and the prospects for large-scale conversion are not as bright as they may appear from this simple calculation. For example, the conversion efficiency from solar to electrical energy is far less than 100%; 10–20% is typical for photovoltaic cells. Roof systems for using solar energy to raise the temperature of water with efficiencies of around 50% have been built. Other practical problems must be considered, however, such as overcast days, geographic location, and energy storage.