Answer:
83.33% of the roof area will be occupied by the PV cells
Explanation:
Given the data in the question;
time-averaged lighting requirement [tex]P_{lighting[/tex] = 10 W/m²
the annual average solar irradiance [tex]q_{solar[/tex] = 150 W/m²
the PV efficiency η[tex]_{pv[/tex] = 10% = 0.1
battery charging/discharging efficiency η[tex]_{battery[/tex] = 80% = 0.8
we know that; Annual average power to the light = [tex]P_{lighting[/tex] × A[tex]_{roof[/tex]
Now, the electrical power delivered by the solar cell battery system will be;
⇒  [tex]q_{solar[/tex] × A[tex]_{pv[/tex] × η[tex]_{pv[/tex] × η[tex]_{battery[/tex]
[tex]P_{lighting[/tex]A[tex]_{roof[/tex] = [tex]q_{solar[/tex] × A[tex]_{pv[/tex] × η[tex]_{pv[/tex] × η[tex]_{battery[/tex]
Such that;
A[tex]_{pv[/tex] = [tex]P_{lighting[/tex]A[tex]_{roof[/tex] / [tex]q_{solar[/tex] × A[tex]_{pv[/tex] × η[tex]_{pv[/tex] × η[tex]_{battery[/tex]
A[tex]_{pv[/tex] / A[tex]_{roof[/tex] = [tex]P_{lighting[/tex] /  [tex]q_{solar[/tex] × η[tex]_{pv[/tex] × η[tex]_{battery[/tex]
so we substitute
A[tex]_{pv[/tex] / A[tex]_{roof[/tex] = 10 W/m² / [ 150 W/m² × 0.1 × 0.8 ]
A[tex]_{pv[/tex] / A[tex]_{roof[/tex] = 10 W/m² / 12 W/m²
A[tex]_{pv[/tex] / A[tex]_{roof[/tex] = 0.8333
A[tex]_{pv[/tex] / A[tex]_{roof[/tex] = (0.8333 × 100)%
A[tex]_{pv[/tex] / A[tex]_{roof[/tex] = 83.33%
Therefore, 83.33% of the roof area will be occupied by the PV cells.
