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Question
A family uses 8 kW of power.
- 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?
- Compare this area to that of the roof of a typical house.
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Solution 1
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`).
Solution 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.
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