A family uses 8 kW of power. (a) 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? (b) Compare this area to that of the roof of a typical house.
Solar energy received per square metre = 200 W
Efficiency of conversion from solar to electricity energy = 20 %
Area required to generate the desired electricity = A
As per the information given in the question, we have:
8 × 103 = 20% × (A × 200)
= (20 /100) × A × 200
∴ A = 8 × 103 / 40 = 200 m2
(b) The area of a solar plate required to generate 8 kW of electricity is almost equivalent to the area of the roof of a building having dimensions 14 m × 14 m. (≈ `sqrt200`).
(a) Power used by family, p = 8 KW = 8000 W
As only 20% of solar energy can be converted to useful electrical energy, hence, power
8000 W to be supplied by solar energy = 8000 W/20 = 40000 W
As solar energy is incident at a rate of 200 Wm-2, hence the area needed
A=4000 W/200 Wm-2 =200 m2
(b) The area needed is camparable to roof area of a large sized house.