Solar cells convert the energy of incoming light to electric energy; a good quality cell operates at an efficiency of 15%. Each person in the United States uses energy (for lighting, heating, transportation, etc.) at an average rate of 11 kW. Although sunlight varies with season and time of day, solar energy falls on the United States at an average intensity of 200 W/m². Assuming you live in an average location, what total solar-cell area would you need to provide all of your energy needs with energy from the sun?

Question

Physics

Marlie Valdez

4 November, 21:15

Solar cells convert the energy of incoming light to electric energy; a good quality cell operates at an efficiency of 15%. Each person in the United States uses energy (for lighting, heating, transportation, etc.) at an average rate of 11 kW. Although sunlight varies with season and time of day, solar energy falls on the United States at an average intensity of 200 W/m². Assuming you live in an average location, what total solar-cell area would you need to provide all of your energy needs with energy from the sun?

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Answers (1)

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Mckinley Vasquez · Answer 1

A = 367 m^2, The required coverage area of the solar panels

Explanation:

Given:

- The average rate of energy used in USA P = 11 KW

- Solar cells operate with an efficiency e = 15 %

- Solar energy has the irradiation flux of G = 200 W/m^2

Find:

what total solar-cell area would you need to provide all of your energy needs with energy from the sun?

Solution:

- The total amount of flux converted into useful energy by solar cells and delivered to transmission lines in USA is as follows:

P = e*G*A

Where, A is the coverage area of the solar cells.

- Since, it is required for the solar panels to produce P = 11 KW to meet all of USA energy requirements:

11*10^3 = 0.15*200*A

A = 11*10^3 / 0.15*200

A = 367 m^2