The correct answers are:
**South Texas: 959 W/m ^{2}
northern U.S.: 805 W/m^{2}
northern U.S. (winter): 451 W/m^{2}**

These answers are, respectively: 1,067 W/m^{2} * cos 26°; 1,067 W/m^{2} * cos 41°; 1,067 W/m^{2} * cos 65°;

How much solar energy can you collect at night?

NONE!

Even shortly after sunrise or before sunset, solar collectors have a lower efficiency because of the low sun angle (think back to part 3).

**Question 4.**

Consider a collector sitting on the Earth's surface as the Earth rotates once in 24 hours (relative to the Sun). The total distance travelled by the collector is the circumference of the circular path, but the cross section of sunlight actually intercepted corresponds to the diameter of the circular path. Therefore, by dividing the above answers by pi (3.14159265...), the results will (approximately) be the average power available over a 24-hour period.

One more consideration: the above approximation does not take into account the shorter days during winter (it also would not cover longer days during summer). The result for the northern U.S. in the winter, after dividing by pi, must be multiplied by 0.8 to account for the shorter winter daylight period.

Click **here** to check your answers and continue...

© 2001-2002, 2005 by Wm. Robert Johnston.

Last modified 11 January 2005.

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