Given:
The solar radiation incident on the earth is: I = 700 W/m²
The dimension of the roof are: A = 9.02 m × 17.75 m
To find:
The solar power incident on the given surface area.
Explanation:
The area A of the roof is calculated as:
[tex]\begin{gathered} A=9.02\text{ m}\times17.75\text{ m} \\ \\ A=160.105\text{ m}^2 \end{gathered}[/tex]The power P incident on the given surface area can be calculated as:
[tex]P=I\times A[/tex]Substituting the values in the above equation, we get:
[tex]\begin{gathered} P=700\text{ W/m}^2\times160.105\text{ m}^2 \\ \\ P=112.0735\times10^3\text{ W} \\ \\ P=112.0735\text{ KW} \end{gathered}[/tex]Final answer:
The amount of solar power incident is 112.0735 KW.
