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At a certain location, the solar power per unit area

Conceptual Physics | 12th Edition | ISBN: 9780321909107 | Authors: Paul G. Hewitt ISBN: 9780321909107 29

Solution for problem 4P Chapter 16

Conceptual Physics | 12th Edition

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Conceptual Physics | 12th Edition | ISBN: 9780321909107 | Authors: Paul G. Hewitt

Conceptual Physics | 12th Edition

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Problem 4P

At a certain location, the solar power per unit area reaching Earth’s surface is 200 W/m2, averaged over a 24-hour day. If the average power requirement in your home is 3 kW and you can convert solar power to electric power with 10% efficiency, how large a collector area will you need to meet all your household energy requirements from solar energy? (Will a collector fit in your yard or on your roof?)

Step-by-Step Solution:
Step 1 of 3

Solution 4P The average requirement of my home is = 3 kW = 3000 W But only 10% of solar power is converted to electric power. Therefore, to obtain 3000 W of power, the incident solar power on the collector has to be = 3000 W × 10 = 30000 W Given that, power is 200 W when the collector area is = 1 m 2 Power is 30000 W when the collector area is = 1 × 30000 m 2 200 2 = 150 m 2 Therefore, the required area of the collector should be 150 m .

Step 2 of 3

Chapter 16, Problem 4P is Solved
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Textbook: Conceptual Physics
Edition: 12
Author: Paul G. Hewitt
ISBN: 9780321909107

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At a certain location, the solar power per unit area