Question

At the surface of the earth, there is an approximate average solar flux of 0.75 kW/m2. A family wishes to construct a solar energy conversion system to power their home. If the conversion system is 30% efficient and the family needs a maximum of 26 kW, what effective surface area is needed for perfectly absorbing collectors?

Answer 1

Answer:

$S=115.6 [m^{2}]$

Explanation:

If the family needs 26 kW of power, and knowing that the efficient is 30% the relation between the solar flux and the power would be:

$\Phi_{solar} [kW/m^{2}]\cdot S [m^{2}] \cdot e=26 [kW]$ (1)

e is the efficient (in our case 0.3)

Φsolar is the solar flux

S is the effective surface

Solving (1) for S we have:

$S=\frac{26 [kW]}{0.75 [kW/m^{2}]\cdot 0.3}=115.6 [m^{2}]$

I hope it helps!