Question

A community plans to build a facility to convert solar radiation to electrical power. The community requires 2.20 MW of power, and the system to be installed has an efficiency of 30.0% (that is, 30.0% of the solar energy incident on the surface is converted to useful energy that can power the community). Assuming sunlight has a constant intensity of 1 020 W/m2, what must be the effective area of a perfectly absorbing surface used in such an installation

Answer 1

The answer is "
`\bold{7.18 * 10^3 \ m^2}`".

Step-by-step explanation:

The efficiency system:

`\eta =(P_(req))/(P) * 10\\\\P =(P_(req))/(\eta) * 10\\\\`

`=((2.20 * 10^6 \ W)/(30))* 100\\\\=((220 * 10^6 \ W)/(30))\\\\=((22 * 10^6 \ W)/(3))\\\\=7.33 * 10^6 \ W`

Using formula:

`A=(P)/(I)`

Effective area:

`A= (7.33 * 10^6 \ W)/(1020\ (W)/(m^2))\\\\`

`=(7.33 * 10^6 )/(1020)\ m^2 \\\\ =0.0071862 * 10^6 \ m^2 \\\\=7.1862 * 10^3 \ m^2 \\\\`