How Much PV Area to Equal U.S. Electric Production?

According to the Energy Information Administration of the U.S. Department of Energy, the U.S. produced 3,678 billon kilowatt-hours of electricity in 1999. Let’s round that up and aim for an even 4,000 billion kWh, which is a production figure we’ll hit sometime between the years 2005 and 2010. Note that this is “production,” not “use.” Transmission inefficiencies and other losses are covered. This is the amount of power we need to stuff into the pipeline inlet.

We’ll want our PV modules in a good sunny area to make the best of our investment, so looking at the National Solar Radiation Data Base (NREL document # TP-463-5607) for Tonopah, Nevada, we see that a fiat-plate collector on a fixed-mount facing south at a fixed-tilt equal to the latitude, 38.07 in this case, saw a yearly average of 6.1 hours of “full-sun” per day in the years 1961 through 1990. A “full-sun” is defined as 1,000 watts per square meter.

Nevada Test Site

The area above marked in gray — the Nevada Test Site and Nellis Air Force Base —
could deliver enough energy to meet the entire US. needs using existing photovoltaic technology.

For IN modules we’ll use the large Astropower 120-watt module, which the California Energy Commission,   rates at 107 watts output, based on observed, real-world performance. 107 watts times 6.1 hours equals 652,7 watt-hours or 0.6527 kilowatt-hours per day per module at our Tonopah site. At 26”x 58.l”this module presents 10.5 square feet of surface area. We’ll allow some space between rows of modules for maintenance access, and for sloping wintertime suit, so let’s say that each module will need 15 square feet.

Conversion from PV module DC output to conventional AC power isn’t perfectly efficient. Looking at the real-world performance figures from the California Energy Commission again,  we see that the Trace Engineering 20kW model PV-20208 is rated at 96% efficiency. We’ll probably be using larger inverters, but this is a typical efficiency for large intertie inverters. So our 0.6527 kWh per module per day becomes 0.6266 kWh by the time it hits the AC grid.

A square mile, 5,280 feet times 5,280 feet equals 27,878,400 square feet. Divided by 15 sq.ft. per module, we can fit 1,858,560 modules per square mile. At 0.6266 kilowatt-hours per module per day, our square mile will deliver 1,164,574 kWh per day on average, or 425,069,510 kWh per year. Back to our goal of 4,000,000,000,000 kWh, divided by 425,069,510 kWh per year per square mile, it looks like we need about 9,410 square miles of surface to meet the electrical needs of the U.S. That’s a square area a bit less than 100 miles on a side. This is a bit over half of the approximate 16,000 square miles currently occupied by the Nevada Test Site and the surrounding Nellis Air Force Range.

As a practical measure, we need to point out that PV power production happens during the daytime, and so long as we persist in turning the llghts on at night, there will continue to be substantial power use at night. Also, so long as we’re out in the desert, solar thermal collection might be a more efficient power generation technology But, however you run the energy collection system, large solar-electric farms on what is otherwise fairly useless desert land could add substantially to the electrical independence and security of any countxy The existing infrastructure of coal, nuclear, and hydro power plants could continue to provide reliable power at night, but non-renewable resource use and carbon dioxide production would be greatly reduced.

Source: Doug Pratt, Real Goods, Solar Living Source Book