Chapter 12: Problem 7
Why do skiers get sunburned so easily?
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chapter 12: Problem 7
Why do skiers get sunburned so easily?
These are the key concepts you need to understand to accurately answer the question.
All the tools & learning materials you need for study success - in one app.
Get started for freeDetermine the equilibrium temperature of the absorber surface in Prob. 12-98 if the back side of the absorber is insulated.
Consider a surface at \(-5^{\circ} \mathrm{C}\) in an environment at \(25^{\circ} \mathrm{C}\). The maximum rate of heat that can be emitted from this surface by radiation is (a) \(0 \mathrm{~W} / \mathrm{m}^{2}\) (b) \(155 \mathrm{~W} / \mathrm{m}^{2}\) (c) \(293 \mathrm{~W} / \mathrm{m}^{2}\) (d) \(354 \mathrm{~W} / \mathrm{m}^{2}\) (e) \(567 \mathrm{~W} / \mathrm{m}^{2}\)
The sun can be treated as a blackbody at \(5780 \mathrm{~K}\). Using EES (or other) software, calculate and plot the spectral blackbody emissive power \(E_{b \lambda}\) of the sun versus wavelength in the range of \(0.01 \mu \mathrm{m}\) to \(1000 \mu \mathrm{m}\). Discuss the results.
A semi-transparent plate \(\left(A_{1}=2 \mathrm{~cm}^{2}\right)\) has an irradiation of \(500 \mathrm{~W} / \mathrm{m}^{2}\), where \(30 \%\) of the irradiation is reflected away from the plate and \(50 \%\) of the irradiation is transmitted through the plate. A radiometer is placed \(0.5 \mathrm{~m}\) above the plate normal to the direction of viewing from the plate. If the temperature of the plate is uniform at \(350 \mathrm{~K}\), determine the irradiation that the radiometer would detect.
The absorber surface of a solar collector is made of aluminum coated with black chrome ( \(\alpha_{s}=0.87\) and \(\left.\varepsilon=0.09\right)\). Solar radiation is incident on the surface at a rate of \(600 \mathrm{~W} / \mathrm{m}^{2}\). The air and the effective sky temperatures are \(25^{\circ} \mathrm{C}\) and \(15^{\circ} \mathrm{C}\), respectively, and the convection heat transfer coefficient is \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). For an absorber surface temperature of \(70^{\circ} \mathrm{C}\), determine the net rate of solar energy delivered by the absorber plate to the water circulating behind it.
What do you think about this solution?
We value your feedback to improve our textbook solutions.