The Intensity of solar radiation at the Earth's orbit is 1370 W/m^2. However, because of the atmosphere, the curvature of the Earth, and rotation (night and day), the actual intensity at the Earth's surface is much lower. At this moment, let us assume the intensity of solar radiation is 800 W/m^2. You have installed solar panels on your roof to convert the sunlight to electricity.

(a) If the area of your solar panels is 4 m^2, how much power is incident on your array?

(b) Unfortunately, solar panels are only about 20% efficient i. e only 1/5 of the light is converted to electricity. This being the case, how much electrical power are you actually producing for your home at this moment?

Question

Allen Liu

Physics

15 October, 17:15

The Intensity of solar radiation at the Earth's orbit is 1370 W/m^2. However, because of the atmosphere, the curvature of the Earth, and rotation (night and day), the actual intensity at the Earth's surface is much lower. At this moment, let us assume the intensity of solar radiation is 800 W/m^2. You have installed solar panels on your roof to convert the sunlight to electricity.

(a) If the area of your solar panels is 4 m^2, how much power is incident on your array?

(b) Unfortunately, solar panels are only about 20% efficient i. e only 1/5 of the light is converted to electricity. This being the case, how much electrical power are you actually producing for your home at this moment?

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Answers (1)

Know the Answer?

Maverick Gamble · Answer 1

a) Pi = 3200 w

b) Pe = 640 w

Explanation:

Solar radiation is 800 w/m²

a) Area of solar panel is 4 m²

Then Power Incident is:

Pi = Solar radiation x Area of solar panel array, then

Pi = 800 w/m² * 4 m²

Pi = 3200 w or Pi = 3.2 Kw

b) Solar panels efficiency is 20%

Electricity production: Pe

Pe = Pi * η where η is efficiency η = 20 % η = 0,2

Pe = 3200 w * 0,2

Pe = 640 w or Pe = 0,640 Kw