Question

A community plans to build a facility to convert solar radiation to electrical power. The community requires 1.00 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 1 000 W/m2, what must be the effective area of a perfectly absorbing surface used in such an installation?

Answer 1

We will find that the effective area that must be used to get the desired power is 3,333.33 m^2

The information we have is:

power needed = 1.00 MW = 1,000,000 watts

Energy of the Sun = 1,000 W/m^2

We know that the system has an efficiency of 30%, then per meter square, the energy converted is:

E = 0.3*1,000 W/m^2 = 300 W/m^2

Now we need to find the area (in meters squared) such that:

A*E  = 1,000,000 watts

A*(300 W/m^2) =  1,000,000 watts

A = (1,000,000 watts)/(300 W/m^2) = 3,333.33 m^2

So the effective area that must be used is A =  3,333.33 m^2

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