Question

The Campus measures approximately 1.5 km by 1.5 km. If using 15% efficient PV (photovoltaic) panels, what fraction of campus would have to be covered in order to meet the total demand of 100,000 MWh in a year, assuming that each panel would be exposed to 5 hours of 1000 W/m² incident sunlight per day? (hint: first work out the peak power of the PV panels in full sun, which should be several times higher than the average campus demand).

Answer 1

Peak power of PV panels in full sun  

Where is the area covered in the campus.

Energy obtained from the PV panels per day is

Energy obtained from the PV panels per one year  is

The PV panels are 15% efficient,

Useful energy obtained from PV panels  

Energy needed for the campus is

Area of the campus is  

Fraction of campus to be covered is .