Question

(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

Short answer

The area required to supply the needs of the house is \[66.67\;{{\rm{m}}^2}\], and the solar cells would fit on the roof of an average house.

Step-by-step solution

Understanding solar cells

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

Given data

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 \(n = 9\;{\rm{h/day}}\).

Evaluation of the area required to supply the need of the house

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 \(66.67\;{{\rm{m}}^2}\).

Evaluation of the roof area of the house

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.