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

(a) Power used by the family, P = 8 kW = 8 × 10³ W

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 × 10^3)/ 40`  =  200 m²

(b) The area necessary for a solar plate to produce 8 kW of electricity closely matches the roof area of a building that measures 14 m × 14 m. (≈ `sqrt200`).

Answer 2

(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 = `(40000  "W")/(200  "Wm"^2)` =200 m²

(b) The area needed is camparable to roof area of a large sized house.