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23 December, 03:09

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?

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  1. 23 December, 05:00
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    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.
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