A silicon solar cell array works at an efficiency of 8.1% in sunlight and is exposed to an average solar insol?

That average insolation value is low enough that it's probably averaged over a 24-hour period already.

So I would do

3600 seconds / hour * 24 hours / day * 365 days / year * 8 years * 230 W/m^2 * 8.1% efficiency = a certain number of W-seconds/m^2.

1 watt-second = 1 J (Joule)

it is needless to say a homework subject so i'm going to easily provide you tricks approximately the thank you to paintings it out: If the photograph voltaic panel have been a hundred% effectual, that's, if each and every watt of photograph voltaic insulation produced one watt of electrical energy, how lots photograph voltaic panel area is needed? A = (six hundred W) / (196 W / m^2) = 3.06 m^2 If the photograph voltaic panel have been 50% effectual, that's, in ordinary terms 0.5 the photograph voltaic insulation falling on the photograph voltaic panel is switched over to electric potential, does not *two times* the panel area be required? I.e., a million / 0.50 circumstances? If the photograph voltaic panel have been 25% effectual, does not 4 circumstances the panel area be necessary to offer that comparable quantity of electrical energy? that's, a million / 0.25 circumstances? Do you spot the place this is going?

Assume the area is 1 M^2.
How many hours of sunshine ? Assume 12, but very few places never have clouds...

Energy gathered = 230 x .081 x 3.15 x 10^7 x 8 Joules/M^2