Determine the power from the visible, UV, and IR components of sunlight striking the surface of a PV panel with a total irradiance of 500 W/m2 if the panel is 0.5 m square. What is the energy in kWh if the specified irradiance is on the panel for 10 hours?
What will be an ideal response?
Incident P = (500 W/m2)(0.5 m2) 250 W
Incident W = Pt = (250 W)(10 h) = 2500 Wh = 2.5 kWh
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