Question

A family uses 8 kW of power. 

1. Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW? 

2. Compare this area to that of the roof of a typical house.

Answer 1

1. Power used by family, p = 8kW = 8000 W

As only 20% of solar energy can be converted to useful electrical energy, power to be supplied by solar energy 

= (8000w/20%) 

= 40000 w.

Solar energy is incident at a rate of 200 w/m² hence Area needed.

\(\frac {400w}{200Wm^{-2}}\) = 200 m².

2. The area needed is comparable to roof area of large sized house.