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.

Math that matters (Part II–Atmospheric Changes)

Everyone now knows (or denies) that CO2 has been increasing in the atmosphere for the past couple hundred years and this increase is largely responsible for the climate changes that have occurred over that time. And 400 ppm (parts per million) is a number that more and more people are familiar with as well; it represents the average concentration of CO2 in the atmosphere (up from 280 ppm when the Industrial Revolution commenced, ~1750). Well, there are a few other very interesting/compelling numbers that we can determine based on this information alone, including:

(1) What is the average annual growth rate of the CO2 over this period (1750-2017)?
(2) What is 400 ppm in percentage terms?
(3) How many molecules of CO2 are there in a regular sized balloon?
(4) Considering CO2 changes alone, by what percentage has the composition of the atmosphere changed since Industrialization?

Let’s do some calculations:

(1) What is the average annual growth rate of the CO2 over this period (1750-2017)?

Atmospheric CO2 concentration has risen exponentially so we use an exponential growth equation to model it:

(a) Cf = Ci x e^(rt),
where, Cf = final concentration, Ci = initial concentration, e = 2.718 (Euler’s number), r = annual rate of growth, and t = duration of time (^ is used as an exponential symbol)

The variable we want is r (as we have all the other numbers). Solving for r requires using a little “Algebra II”:

First move Ci over by dividing both sides of (a) by Ci and flipping the equation around; this leaves:
e^rt = Cf/Ci

Now, if we take the log of both sides (here the natural log, ln) and use the fact that ln(e) = 1, we get:
ln(e^rt) = ln(Cf/Ci)  rt x ln(e) = ln(Cf/Ci)  rt = ln(Cf/Ci)

Now, if we divide both sides by t, we get an equation solved for r, the desired variable:
r = ln(Cf/Ci)/t

Inputting the values of the three known variables (Cf = 400 ppm, Ci = 280 ppm, and t = 2017-1750 = 267 years) yields the annual growth rate:
r = ln(400/280)/267 = 0.00134 (or 0.134% per year)

If we just look at the growth of CO2 concentration since 1950 (when industrialization became global; CO2’s concentration in 1950 was ~310 ppm), we can once again use the exponential equation above to determine the growth rate in more recent times:
r = ln(400/310)/67 = 0.0038 (or 0.38% per year)

In either case, notice that humans are increasing CO2’s concentration only very slightly each year. (Over the past 267 years, CO2’s concentration has gone up 43%.) However, what may appear “slight” is definitely not slight in terms of its current and future impacts.

(2) What is 400 ppm in percentage terms?

Answering this merely requires us to understand what ppm (parts per million) means. Unlike some words in English, ppm means literally what it says. If the atmospheric concentration of CO2 is 400 ppm then for every million parts (or molecules) of atmospheric gas, CO2 will be 400 of them. With this understanding we can compute a percentage by expressing the amount of CO2 (relative to everything in the atmosphere) as a ratio:
% of CO2 in atmosphere = ppm of CO2/1,000,000 parts of atm = 400/1,000,000 = 0.0004 (or 0.04%)

Again, this is a very, very small amount but as we know this small amount of CO2 is responsible for a significant amount (~30-40%) of the Natural Greenhouse Effect (NGE) and the bulk (~50-60%) of the Anthropogenic Greenhouse Effect (AGE), associated with “global warming” and “climate change.”

(3) How many molecules of CO2 are there in a regular sized balloon?

While (2) tells us what proportion of gases in the atmosphere are CO2 it doesn’t really tell us how much CO2 is a particular volume. If we take a typical balloon (1 foot in diameter; let’s assume it is a sphere), we can find the volume of the balloon using this formula:
V = 4/3 x pi x r^3,
where V = volume, pi = 3.14159, and r is the radius of the balloon. Plugging in the numbers leads to:

V = 4/3 x 3.14159 x (0.5)^3 = 0.52 cubic feet

Since 1 liter = 0.0353147 cubic feet, the volume of the balloon is:
V = 0.52 cubic feet x (1 liter/0.0353147 cubic feet) = 14.7 liters

This seems quite high when one considers a two liter bottle of soda but rest assured this is the correct value.

Now that we have the volume, we’ll need to determine how much CO2 is in such a volume. Here we have to use some chemistry laws; here the most well-known of them all, the Ideal Gas Law:

(c) PV = nRT, which can be rewritten as, n = PV/RT,
where P = pressure, V = volume, n = number of moles of gas, R = the Ideal Gas Constant = 0.082 L-atm/mole-K) and T = temperature in Kelvin degrees.

Assuming that we have a standard atmospheric pressure of 1 atmosphere and we are at room temperature of 27 deg Celsius (equivalent to 81 degrees Fahrenheit or 300 K), we can plug the numbers into equation (c):
n = (1 atm)(14.7 L)/((0.082 L-atm/mole-k)*(300 K)) = 0.60 moles

So what is a mole? Chemists define one mole as 6.02 x 10^23 molecules (or 602,000,000,000,000,000,000,000 molecules).

Notice that our use of the Ideal Gas Law was for all the molecules of gas, not just CO2.

Thus, if the 14.7 L (of a balloon) contains 0.60 moles of gas, we calculate the number of molecules of gas in the balloon as such:

# of molecules of gas (in balloon) = 0.60 moles x (6.02 x 10^23 molecules/mole) = 3.61×10^23 molecules

Now to determine how many of these molecules are CO2 we use the information determined in the previous question thusly:

# molecules of CO2 in a balloon = 3.61 x 10^23 molecules of gas x (0.0004) = 1.4 x 10^20 molecules of CO2

There we have it. There are 140,000,000,000,000,000,000 molecules of CO2 in every balloon (assuming it isn’t filled with helium 😊).

Now, while this balloon’s gas is very light, it certainly is filled with a lot of molecules of CO2. This might lead one to wonder how much does all the CO2 in the atmosphere weigh. Well, this calculation is a bit trickier. According to web references, the entire atmosphere weighs approximately 1.1 x 10^19 pounds (or 5.5 quadrillion tons). We might want to just multiply this by the percentage of CO2 in the atmosphere in order to determine the total weight of CO2 in the atmosphere but we would be wrong to do so; though it wouldn’t be a terribly bad approximation (if we did so we get 4×10^15 pounds). The different molecules of gas have different weights. Thus, we would have to do a calculation that would include all of these different gases and their weights. This is a bit more complicated than I choose to get right now, so suffice it to say that all the CO2 in the atmosphere weighs approximately six thousand times more than all the humans on the Earth combined (which is approximately 7×10^11 pounds).

(4) Considering CO2 changes alone, by what percentage has the composition of the atmosphere changed since Industrialization?

Calculation (1) clarified how small CO2’s concentration is the larger scheme of things, while (2) and (3) suggested how much CO2 there is. Notice, it all depends on how we look at things, as do most things.

The question asked here aims to provide a sense of how much change has occurred in our atmosphere since industrialization.
We know that the atmosphere’s CO2 concentration has gone from 280 ppm to 400 ppm (a 43% increase). And we calculated that CO2 comprises 0.04% of the atmosphere. Thus, to determine how big a change in the atmospheric composition has been caused by Industrialization, we need only to multiply these two percentages together, as such:

Change in Atmospheric Composition since industrialization = % Change in CO2 concentration x % of atm. that is CO2.

Putting numbers in yields,

Change in Atmospheric Composition since industrialization = 43% x 0.04% = 0.017%

Since most don’t work with percentages this small, what does this amount of change mean. Well, if you had 5,814 molecules of atmospheric gas–represented by 5,812 red polka dots (each representing a molecule of gas other than CO2) and 2 blue polka dots (each representing a molecule of CO2)—over the 267 years of Industrialization, one more molecule of CO2 has been added to the atmosphere (or one more blue polka dot). Another way to look at it: If you had 5,814 fans sitting in a gymnasium, 267 years later, one more fan would join the fray. Would you notice a change of this amount? Most definitely not (unless of course the fan was decked out in the rival’s colors 😊). But we can thank modern science for building equipment that can measure such small (yet powerful) changes in the atmosphere. Perhaps the most salient question is: have we built a political/social apparatus to respond to the future/predictions that this information clarifies?