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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

Sagot :

Answer:

The answer is "[tex]\bold{7.18 \times 10^3 \ m^2}[/tex]".

Explanation:

The efficiency system:

[tex]\eta =\frac{P_{req}}{P} \times 10\\\\P =\frac{P_{req}}{\eta} \times 10\\\\[/tex]

   [tex]=(\frac{2.20 \times 10^6 \ W}{30})\times 100\\\\=(\frac{220 \times 10^6 \ W}{30})\\\\=(\frac{22 \times 10^6 \ W}{3})\\\\=7.33 \times 10^6 \ W[/tex]  

Using formula:

[tex]A=\frac{P}{I}[/tex]

Effective area:

[tex]A= \frac{7.33 \times 10^6 \ W}{1020\ \frac{W}{m^2}}\\\\[/tex]

   [tex]=\frac{7.33 \times 10^6 }{1020}\ m^2 \\\\ =0.0071862 \times 10^6 \ m^2 \\\\=7.1862 \times 10^3 \ m^2 \\\\[/tex]  

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