# Math That Matters (Part IV—Solar Power for the U.S.)

(Part A)

How big a solar array is necessary to provide all the electricity currently used in the United States?
Renewable energy (particularly, solar and wind) use is way up over the last 20 years, both globally and in the United States. Annually, rates of growth for PV (photovoltaic) solar and wind are both above 20% over this time, with solar being closer to +40%. This amazing growth appears ready to continue as more and more areas of the world are looking to install solar due to its many benefits (ref 1). However, some believe that solar would have to cover huge expanses of land in order to make a significant contributor to our energy portfolio. Let’s take a look at this belief by asking a simple question, “How big a solar array is necessary to provide all the electricity currently used in the United States?”

Well, there is some background information and a few assumptions that one needs to make in order to calculate this.
First, how much electricity do we use in the U.S.?

Looking this up, we find:
(1) 12.6 Q = 3,704 PWh (ref 2)
(1 Q = 1 Quad BTU = 294 PWh, where 1 PWh is 10 to the 12th power (or 1,000,000,000,000) Watt-hours)

Next, we need to know how much electricity is produced by a typical solar panel. This requires other information as well:
(a) Power rating for solar panel = 345 W
(b) Size of solar panel (61” x 41”) = 17.37 square feet
Thus, Maximum power output = (a)/(b) = 19.9 W/sq. ft

Note: From these values one can compute an efficiency for the panel (which is typically between 15-20%), but one need not calculate the efficiency for our purposes.

Since no panel produces maximally (due to inverting DC current to AC current, losses in wires, snow/dust on panels, etc.), a “de-rating” of 75% is typically used.
(c) Power output expected = Max. power * de-rating = 19.9 W/sq. ft * 0.75 = 14.9 W/sq. ft

Now we need to consider how many hours of sunlight there will be for this panel. Typically, this is done by computing the “average” number of “full-sun” hours per day a panel would be expected to receive at a location. In the U.S., most locations range from 3.5-6.5 hours. We’ll take 4.5 “full-sun” hours to be conservative (central IL has these types of values).

Thus,

(d) True electricity provided = Power output expected * “full-sun” hours (daily) * days in year
(2) = 14.9 W/sq. ft * 4.5 hrs/day * 365 days/year = 24.5 kWh/sq. ft

Now, we can determine how many square feet we need to provide the electricity for the entire nation of the United States:

Size of solar array = Electricity usage (nation)/Electricity production density
= (1)/(2)
= 3,704 PWh/24.5 kWh per sq. ft
= 151,184,000,000 sq. ft

Wow, 151 billion square feet. That’s huge, isn’t it? Let’s convert this to square miles:

# square feet in a square mile => 1 sq. mile = (5280 ft) * (5280 ft) = 27,900,000 sq. ft

So, 151.2 billion square feet is __X__ square miles; where,

X = 151,200,000,000 sq. ft/(27,900,000 sq. ft/sq. mile) = 5418 sq. miles

But, how much is 5,418 square miles?
Just about a squared area with 74 miles on a side!

The area of the state of Illinois is ~58,000 square miles. So, 5,418 square miles is ~9.3% of the state! It is also only 25-times the combined size of the 10 largest airports in the United States. This area, again, if covered with solar panels, would be produce enough electricity to power the entire nation!

In conclusion, the belief that solar panels sufficient to power the U.S. would have to cover a huge amount of area is just plain wrong! Wondering why this information isn’t widely distributed? Well, are you going to distribute it or not? If not, why not? This might provide you part of the answer as to why it isn’t widely known.

(Part B)

Now, what if we wanted to produce all the energy resources we use, not just electricity, with solar PV power? Understandably, most things that use fossil fuels now are not currently able to use electricity (as in, most of the cars/trucks on the road are not yet electric), but most could be made to use electricity if it was available. So, then, if we need to produce 97.3 Quad (not 12.6 Q which is the current electricity demand alone), we’d need ~7.7 times (or 97.3/12.6) more land than stated above. However, since our fossil-fuel dominant energy economy currently requires 37.5 Q of energy to produce 12.6 Q of electricity (due to the inefficiencies in the use of such sources), we actually wouldn’t need to use this wasted energy (or 24.9 Q (37.5 Q – 12.6 Q)) at all; this is a HUGE understated benefit of moving towards solar energy sources. Thus, we would need only to produce 72.4 Quad which would require 5.7 times more land than calculated in Part A, or ~31,000 square miles. This amounts to about 11% of the land area of the state of Texas, not much land considering the size of the United States. In fact, since the area of the U.S. is 3.8 million square miles, we would need to cover less than 1% of the U.S. land surface with solar PV in order to produce all the energy we would need in a fully “electrified” nation! (A recent study by the National Renewable Energy Lab (NREL) found that we could provide 1,432 TWh of electricity by putting solar panels on suitable buildings in the United States. This would be enough to provide 39% of our national electricity needs and 7% of our national energy needs!; see report.)

Caveats: The above calculations only relate to solar PV. The future renewable energy system, that will become dominate in the 21st Century, displacing almost all fossil fuels, will rely on wind, geothermal and hydropower as well. In addition, the need for the world’s poor to use more energy than they do now (in order to live fully actualized lives) will require greater amounts of energy to be produced. Additionally, as the U.S. is one of the more wasteful energy users, its consumption could easily decline (perhaps by 50%) without any detrimental impacts. These factors are important when looking at the future land needs of the entire energy system.