Question

A family uses $8kW$ of power.

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

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

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

Answer 1

A.
Step 1: $\text{Total power generated}$

Lets assume the area of the roof be $A{m}^2$.
Given, total power $=200A$.

Step 2: $\text{Net useful energy}$

Out of the total power, only $20\%$ is useful,
$=\frac{20}{100}\times \left(200A\right)=40A=8000W$

Step 3: $\text{Area required}$

Therefore, the required area for, $8000W$ power is,
$A=\frac{8000}{40}=200sq.m$

B.
The area needed is comparable to the roof of a large house of dimension $14m\times 14m$.