A community plans to build a facility to convert solar radiation to electrical power. The community requires 2.80 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 1 180 W/m2, what must be the effective area of a perfectly absorbing surface used in such an installation?

Given Information:

Output power required = Pout = 2.80 MW

Efficiency = η = 30%

Intensity = I = 1180 W/m²

Required Information:

Effective area = A = ?

Answer:

Effective area = A = 7.907x10³ m²

Step-by-step explanation:

A community plans to build a facility to convert solar power into electrical power and this facility has an efficiency of 30%

As we know efficiency is given by

η = Pout/Pin

Where Pout is the output power and Pin is the input power.

Pin = Pout/η

Pin = 2.80x10⁶/0.30

Pin = 9.33x10⁶ W

The effective area of a perfectly absorbing surface used in such an installation can be found using

A = Pin/I

Where I is the in Intensity of the sunlight in W/m²

A = 9.33x10⁶/1180

A = 7.907x10³ m²

Therefore, the effective area of the absorbing surface would be 7.907x10³ m².