July 02, 2006 09:40 AM
Glenn points out a fellow who notes the potential of solar energy:
"If 2 percent of the continental United States were covered with photovoltaic systems with a net efficiency of 10 percent, we would be able to supply all the U.S. energy needs," said Bulovic, the KDD Associate Professor of Communications and Technology in MIT's Department of Electrical Engineering and Computer Science.
Um, ok. Glenn notes "Two percent is a LOT of land", but that's understating the case a bit I think.
Let's do some math! (I was told there would be no math. Shhh!)
According to this page, the total land area of the U.S. is 3,537,379 square miles. Take away Alaska and Hawaii to get the continental U.S., and you are left with 2,959,005 square miles. Two percent of that is...
Fifty-nine thousand, one hundred and eighty square miles. That's 59,180.
For perspective: Over half of the fifty states are smaller in area than 59,180 square miles. The closest in size to that number are Iowa (55,869), Michigan (56,804), and Georgia (57,906).
So: who's for paving over Georgia?
Because unless I'm missing something, that is what we're talking about: literally paving over that much area, and I have to assume utterly destroying any flora, fauna, or other living things that are unlucky enough to have been previously occupying it. Unless they happen to, you know, not require sunlight.
Now that would be one hell of an Environmental Impact Study.
Professor Bulovic seems to have thrown out this statistic as a positive for solar energy, but he's obviously never negotiated with a local zoning board. If that's the best future we can hope for with solar as our primary energy source, I suspect even the most strident environmentalists will cry out, "Bring on the coal!"
Update: Commenter Eric points out that I shouldn't be subtracting Alaska, as Bulovic said continental U.S., and he's right. I'm not updating as the point is still valid (more so!)
Update: Lots of good discussion (and more math!) in the comments. And reader Ric sends in the following, which was a bit too HTML-heavy for the comments section:
Let us begin with data. The U.S. Energy Information Agency estimates total energy usage for the United States for 2004 at 99.74 quadrillion BTU. Converting to watt-hours using Convert.exe gives 29.23 quadrillion watt-hours. A quadrillion is 1015. To make things simple, round up to 3x1016 watt-hours.
The year is 365.242199 days long. If we divide that 3 x 1016 watt-hours by 365.242199, we get about 8.2 x 1013 watt-hours used per day. That energy must be collected while the sun is shining, of course.
The Solar Constant, the amount of energy reaching a flat surface perpendicular to the sun's rays at Earth orbit, is 1.367 kilowatts per square meter. Atmospheric attenuation of incident solar radiation is from 30% to 90% depending on cloud cover. If we assume that we are in a sunny area with minimum attenuation, the amount of energy reaching the surface is 0.7 x 1.367 = 0.957 kw/m2 or 957 watts per square meter.
It's fairly obvious that an array big enough to collect 8.2 x 1013 watt-hours is going to be too big to turn so that it always faces the sun. The United States is mostly well above the Tropics, so in order to get one meter of area perpendicular to the Sun's rays we must have more than one meter of surface area; to be specific, the effective area is n cos alpha, where alpha is the latitude of the site. Most of the nice clear areas are in the desert Southwest, and a good average for the latitude there is 32 degrees. 957 watts per square meter times cosine of 32 degrees gives 811 watts per square meter.
The Sun rises and sets, and the angle of the Solar radiation varies according to a sine wave. Since we can't move the array, we have to calculate the average energy arriving, which is 0.707 (root mean square of a sine wave) of the peak. 811 times 0.707 gives 574 watts per square meter.
If the solar collector is 10% efficient (we'll get back to that) we thus recover 57.4 watts per square meter.
Checking the average hours of daylight we discover that the worst case is December, with only about 10 hours of daylight at 32 degrees latitude. Our solar collector has to keep up, so it has to collect the 8.2 x 1013 watt-hours in only 10 hours, at an average rate of 57.4 watts per square meter. That makes the area of the array 1.43 x 1011 square meters. That's a bit over 55,000 square miles, close to what the Professor came up with.
And there are kickers.
Only 10/24 of that energy will be consumed as it's generated. We have to store 14/24 of that for use when the Sun isn't shining. 4.8 x 1013 watt-hours is a lot of energy to store.
Lead-acid batteries store about 41 watt-hours per kilogram. So we need about 1.17 x 1012 kilograms, or 1.17 billion tonnes, of lead-acid batteries to store the energy. Around 3 million tonnes of lead are mined each year, so that's only a little less than four hundred years of world production. Lead-acid batteries in deep-discharge service last about five years, so the best way to maintain the array would be to round up to 1.4 gigatonnes of batteries; at any given time, 230 megatonnes of batteries would be undergoing reprocessing.
The EPA's gonna love it, don't you think?
Note that NiCADs are even worse. NiMH is better at 95 watt-hours per kilogram, but if you think lead is a problem, go look up nickel production. Lithium-ion is even better at 128 wh/kg, but if the production figures don't give you pause you should look at the safety data pdf.
Oh, by the way, the power's being produced in the Southwest. Power demand is largely in the Northeast. Add 20% (minimum) for transmission losses. Over that large an area, add another 10% for occasional cloud cover, and a further 20-25% for losses due to dust and dirt on the solar arrays.
And -- those solar cells are 10% efficient, remember? Roughly 20% of the loss is in reflection, light reflected back into space from the array. The rest of it, 70% of the incident light or roughly 5.75 x 1014 watts, is heat. How much power will the fans to carry it away use? Will they have blue LEDs inside? Methinks the site will be popular with hang glider enthusiasts, provided they don't mind being warmed a bit by the reflected light...
Do try to think things through a bit. Wishing isn't physics.
Regards,
Ric