## Solar Energy

If we take a solar cell a square meter in size, put it on top of the atmosphere, and face it directly toward the sun, it would receive solar radiation energy of 1.366 kW/m2. Take it down to the surface of the earth, and the light will be attenuated by the air’s absorption and scattering. The net result is the convenient figure of 1 kW/m2. Over the whole earth, there is enough sunlight in an hour to supply all the energy use in the world for a year! If you find this hard to believe, as I did, we can do a back-of-the- envelope calculation in Box 3.1.

Box 3.1 How Much Sunlight Does the Earth Get in 1 hour?

The radius of the earth is about 6,400 km (4,000 miles). Replace the earth with a disk 6,400 km in radius, and the disk would get the same amount of sunlight. We do not count the back side of the disk, and that takes care of the fact that there is no sunlight at night (Fig. 3.19).

The area of the disk is nr2, as you well remember. That works out to be some 130,000,000 km2. In square meters, the area of the disk is a million times that, which is 130,000,000,000,000 m2. (Those who are meter-challenged can think of a square meter as a square yard.) Each square meter gets 1 kW, so the total power over the earth is that large number of kilowatts. The number is too long to write, but we can use shorthand and write it as 1.3 x 1014 kilowatts, where the 14 stands for the number of decimal places after the “1.” (This is scientific notation, which was explained in Chap. 2.)

To compare this with our energy consumption, we have to convert kilowatts into Quads per year. We can use Table 2.1 in Chap. 2 to make this conversion. It takes several steps, but 1.3 x 1014 kW is the same as 440 Quads per hour. Also in Chap. 2, we found that our civilization consumes about 500 Quads per year, almost the same number. So indeed, sunlight hitting the earth every hour carries about the same energy as we consume in a year!

Fig. 3.19 The same amount of sunlight falls on the earth as on a flat disk of the same diameter |

Figure 3.20 shows the annual variation of sunlight. This shows that the earth is tipped relative to the plane of its orbit. Consider a location in the northern hemisphere, on the upper red line, say. In the summer, the sun would be on the left, so as the earth rotates, more of that red line is in the sunlit region, and less in the blue night region. Days are longer than nights. In the southern hemisphere, the opposite is true. When the earth moves to the opposite side of its orbit, the sun appears to come from the right. The blue region is then sunlit, and less time is spent in there than in the yellow night region. Days are shorter then nights in the northern hemisphere. Furthermore, the sun never gets high above the horizon at high latitudes. Since we cannot easily store solar energy from summer to winter, solar power is inequitably distributed.

(continued)

Box 3.1 (continued)

N Fig. 3.20 Solar power varies by season because of the tilt of the earth |

So why aren’t we all fried by the sun? First, there’s the factor of 2. Except at the equator, sunlight comes in at an angle, not from overhead, so that the power is spread out over a larger area. To figure out how many kW/m2 there are at any given latitude and longitude is a long exercise in spherical trigonometry, but we can average. The area of a hemisphere is 2pr2, happily just twice that of the disk. So the average insolation over the earth is only 0.5 kW/m2. Since the earth’s axis is tilted with respect to its orbit, there are seasons; and people living at high latitudes have a bigger difference between winter and summer. They also get less sun altogether. Figures 3.21 and 3.22 show this. The number 0.5 kW/m2 is averaged over latitude and seasons. Then there are clouds and storms and smog which prevent the sun from shining. That cuts the average to below 250 W/m2, and it is not available everywhere. Even so, it is a lot of energy, if we could only learn how to capture it efficiently. The average person in the USA uses about 500 W of electricity, averaged over 24 hours. Two square meters of solar cells in a good location could generate this if they were 100% efficient. Right now, it is hard to get 10% except in the laboratory.