The correct answer is 1,067 W/m2:

If 22% of the flux is lost in the atmosphere, then 78% reaches the ground.

78% of 1,368 W/m2 = 1,067 W/m2.

When the collector is at an angle to the incoming sunlight, this flux from Question 2 is reduced. If (theta) is the angle between the incoming sunlight and a line perpendicular to the collector surface, the flux collected is equal to 1,067 W/m2 * cos (theta).

Why? Examine the diagram below: "A" is the collector, and theta is the angle between the incoming sunlight and a line perpendicular to the collector. In spite of the fact that the collector has covered area "A", it is only intercepting sunlight coming through area "B". If the collector is face on to the sunlight, then "A" and "B" are the same--when you hold a sheet of paper face on to a light, it intercepts the most possible light and casts the largest possible shadow. On the other hand, if the collector is edge on to the sunlight, then "B" is zero--when you hold a sheet of paper edge on to a light, it intercepts nearly zero light and casts little shadow.

Some of you may wonder: why not just set up the collector perpendicular to the sunlight instead of flat on the Earth's surface? This is in fact how you would aim to build collectors, since it requires less building material. However, notice that such a collector would cast a shadow on collectors behind it. Thus, the land area required to collect solar energy ends up the same.

To sum up: If the Sun is not directly overhead at your location on the Earth, then you can't collect all the power you found in Question 2. Since the angle (theta) we have been using is the same as the angle between the location on the Earth's surface and a point directly below the Sun, we can use the formula
(flux collected) = (incident flux) * cos (theta).

Question 3.

Using the incident flux of 1,067 W/m2 (from Question 2) and the given angles for theta, find the flux available at these locations:

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© 2001-2002, 2005 by Wm. Robert Johnston.
Last modified 11 January 2005.
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