I don\'t know how to start this problem. The solar panels on the roof of a house
ID: 1725870 • Letter: I
Question
I don't know how to start this problem.The solar panels on the roof of a house measure 4.0 mby 6.5 m. Assume they convert12% of the incident EM wave's energy to electrical energy. (a) What average power do the panels supply when the incidentintensity is 1.0 kW/m2 and the panels areperpendicular to the incident light?
kW
(b) What average power do the panels supply when the incidentintensity is 0.80 kW/m2 and the light is incidentat an angle of 67.0° from thenormal?
kW
(c) Take the average daytime power requirement of a house tobe 1.7 kW. What do your answersto (a) and (b) imply for the use of solar panels? I don't know how to start this problem.
(a) What average power do the panels supply when the incidentintensity is 1.0 kW/m2 and the panels areperpendicular to the incident light?
kW
(b) What average power do the panels supply when the incidentintensity is 0.80 kW/m2 and the light is incidentat an angle of 67.0° from thenormal?
kW
(c) Take the average daytime power requirement of a house tobe 1.7 kW. What do your answersto (a) and (b) imply for the use of solar panels?
Explanation / Answer
Area = 4.0 * 6.5 = 26 m2 . (a) avg power = incident intensity * area *efficiency = 1.0 * 26 * 0.12 = 3.12 kW . (b) avg power = incident intensity * area *efficiency* cos67 = 0.8 * 26 * 0.12 *cos67 = 0.975 kW . (c) If the light is directly overhead, there is plentyof power available. If the light is lower, at an angle, the solarpanels will nor provide enough energy for the house. . (a) avg power = incident intensity * area *efficiency = 1.0 * 26 * 0.12 = 3.12 kW . (b) avg power = incident intensity * area *efficiency* cos67 = 0.8 * 26 * 0.12 *cos67 = 0.975 kW . (c) If the light is directly overhead, there is plentyof power available. If the light is lower, at an angle, the solarpanels will nor provide enough energy for the house.Related Questions
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