(I) Solar cells (Fig. 15–26) can produce about 40 W of electricity per square meter of surface area if directly facing the Sun. How large an area is required to supply the needs of a house that requires 24 kWh/day? Would this fit on the roof of an average house? (Assume the Sun shines about 9 h/day.).

FIGURE 15-26 Problem 53

Solar cells are devices used to directly convert light energy (sunlight) into electricity. They do not require any heat engine.

The power per meter square is \(P = 40\;{\rm{W/}}{{\rm{m}}^2}\).

The supply given to the house is \(E = 24\;{\rm{kWh/day}}\).

The sun shines about \(n = 9\;{\rm{h/day}}\).

The area required to supply the need of the house is calculated as:

\(\begin{array}{l}A = \left( {24\;{\rm{kWh/day}}} \right)\left( {\frac{{1\;{\rm{day}}}}{{9\;{\rm{h}}}}} \right)\left( {\;\frac{{{\rm{1}}{{\rm{m}}^2}}}{{40\;{\rm{W}}}}} \right)\\A = \left( {24 \times {{10}^3}\;{\rm{Wh/day}}} \right)\left( {\frac{{1\;{\rm{day}}}}{{9\;{\rm{h}}}}} \right)\left( {\;\frac{{{\rm{1}}{{\rm{m}}^2}}}{{40\;{\rm{W}}}}} \right)\\A = 66.67\;{{\rm{m}}^2}\end{array}\)

Thus, the area required to supply the need of the house is \(66.67\;{{\rm{m}}^2}\).

terrace at a \(30^\circ \) angle.

The area of the roof is calculated as:

\(\begin{array}{l}{A_{\rm{R}}} = \left( {1000\;{\rm{f}}{{\rm{t}}^2}} \right)\left( {\frac{1}{{\cos 30^\circ }}} \right){\left( {\;\frac{{{\rm{1m}}}}{{3.28\;{\rm{ft}}}}} \right)^2}\\{A_{\rm{R}}} = 107.33\;{{\rm{m}}^2}\end{array}\)

The area of the roof for a small house is \({A_{\rm{R}}} = 107.33\;{{\rm{m}}^2}\), which is larger than the required area of the house that is \(A = 66.67\;{{\rm{m}}^2}\). Thus, the set of cells will fit on the roof of the house.