In this post, we’ll put a physical, comprehendible scale on the amount of energy typical Americans have used in their lifetimes. No judgment: just the numbers.
The task is to estimate our personal energy volume, so that we can mentally picture cubic tanks or bins corresponding to all the oil, coal, natural gas, etc. we have used in our lives—perhaps plunked down in our backyards to bring the idea home. Go ahead and try to guess/picture how big each cube is.
The resulting analysis is more mathy/quantitative than most of my posts, which might not make for the smoothest reading. Don’t let the math bog you down: the details are there if you want them—but if you just want the answers, they are not too hard to find.
Okay, we can capture almost the entire story of energy use if we consider oil, natural gas, coal, nuclear fission, and hydroelectric. These take up approximately 40%, 23%, 22%, 7%, and 7% of our total energy diet. We have to correct the hydro and nuclear numbers because the 7% figures are adjusted to put their electricity contribution on the same (thermal) footing as the raw inputs from fossil fuels. The electricity that each contributes amounts to about 3% of the total energy pie. (The fossil fuels are counted by thermal energy input, but hydro and nuclear are more easily tabulated by electrical output.)
Let’s set the scale and talk units. The global power budget is about 12 terawatts (TW), one-quarter of which is contributed by the U.S. It is useful to know that one Watt is a Joule per second, and it takes 1055 J to make a Btu, and 4184 J to make a kilocalorie (food Calorie; see the page on useful energy relations for more on units of energy/power).
The U.S. share, 3 TW, divided by 300 million people, turns into 10,000 W per person. Let’s be clear about the meaning this number. It’s not 10,000 Watts per day, per year, or per any time. A Watt is already a rate of energy use, like a speedometer—easily confused with a kWh of energy. The average American is responsible for 10 kW of continuous 24/7 energy consumption. Think about this: 10 kW of thermal power is equivalent to the heat generated by one-hundred 100 W incandescent light bulbs; seven space heaters; or five full-blast hair driers—operating around the clock for each person (though at 35% efficiency, a power plant consuming fuel at a rate of 10 kW would only produce enough electricity to run 35 bulbs, etc.).
You may object: certainly you are not so energy intensive, based on what you see in your own house. But only 22% of energy in the U.S. is consumed in peoples’ homes. The rest is in manufacture, food production, transportation of goods, service industries, etc. Since you presumably partake of these goods and services in a typical way, you’re stuck with responsibility for 10,000 W of power.
10 kW. Such a nice round number! We can therefore quickly say that a typical American uses energy from oil at a rate of 4 kW, natural gas at 2.3 kW, coal at 2.2 kW, 300 W of nuclear, and 300 W of hydroelectric power. In a bit, we will find it convenient to convert to kilocalories (kcal). So 1 W of power run for a year totals 31 MJ, or 7,500 kcal.
The Fossil Fuels
First, we’ll tackle the fossil fuels. The chemical energy in fossil fuels is not very different from that in the food we eat. Natural gas has 13 kilocalories per gram, oil/gasoline has about 10 kcal/g, and coal typically ranges from 4–7 kcal/g, depending on grade. We’ll use 6 kcal/g for coal.
Putting numbers together, we see that 4 kW of oil—run for a year—requires 30 million kcal, which amounts to 3 million grams, or 3,000 kg. Three metric tons of oil each year per person! At 135 kg per barrel, this is 22 barrels per year. Does this make sense? The U.S. consumes about 20 million barrels of oil per day, so that each of us uses 24 barrels per year—close enough for a sanity check!
In my lifetime of about 40 years (the brilliance of estimation is that 40 can be “good enough” for a number of years!), I am responsible for the consumption of 120 metric tons of oil. How big of a box do I need? Water requires a cubic meter per metric ton, and oil is about 85% the density of water. So I need 140 cubic meters. The cube root of 140 is about 5.4 meters. So I need a box about 17 feet cubed! That’s going to be an eyesore in my back yard!
For coal, the numbers work out to 17 million kcal per year, or about 2.8 million grams (much like oil!). At an average of 1.2 times the density of water, 40 years amounts to 92 cubic meters, which corresponds to a box 4.5 m, or 15 feet on a side. Not too different from oil. Note that we get much less of our energy from coal than from oil, but the energy density is much lower, so that the volumes are comparable.
Our 2.3 kW of natural gas, like coal, requires 17 million kcal, or 1.3 million grams per year. You might remember from chemistry classes that gas at standard temperature and pressure (0°C and 1 atm) occupies 22.4 liters per mole. Methane is CH4, with a molecular weight of 16 g/mol. Our annual 1.3 million grams is therefore 83 thousand moles and occupies 1.9 million liters, or 1900 cubic meters. Over 40 years, this becomes 76000 m³ The corresponding cube is 42 meters (140 feet) on a side. This would certainly catch the whole neighborhood’s attention. And you’ll need one for each member of the household (sized according to age).
If we forgot that this is just a visualization exercise, we might agitate for compressing the gas, as anyone in their right minds would do for storage purposes. In this case, at 190 atmospheres of pressure (a practical limit for steel tanks), the volume is reduced to 400 cubic meters, or 7.4 m (24 ft) on a side. Note the trick of using 190 atm instead of a round 200. Made the math easy, and we’re totally allowed to do such things while we’re estimating (not engineering).
What is a relevant volume for the hydroelectric energy you’ve consumed? How about the volume of water that has moved through a dam to produce that energy. Here, the story is simple gravitational potential energy, converted to electrical output at an efficiency of about 90%. The gravitational potential energy of a cubic meter of water is its mass times gravitational acceleration times the height of the water behind the dam. A cubic meter of water has a mass of 1000 kg, turning into 10,000 N of weight when multiplied by g=10 m/s². What is a typical height for a hydroelectric dam? I’ll guess 15 m.
So each cubic meter of water delivers 135,000 J of energy, incorporating the dam efficiency. At a rate of 300 W (300 J/s), this cubic meter provides power for about 450 seconds. (Remember that we can be loose with numbers because our guess of 15 m high dams is not sacred). For reference, this water rate corresponds to 50,000 gallons per day—over two thousand times my household water demand. In 40 years, I need a whopping 3 million cubic meters of water to flow through a dam somewhere on my behalf. Now we have a cube 140 meters on a side. I have no idea where I’m going to put it! This is a small lake’s worth of water, in the ballpark of Thoreau’s famous Walden Pond in Massachusetts. Okay, in California, this would be called a small lake, not a pond. On the other hand, a Massachusetts mountain is a California hill.
This one is less straightforward than the preceding ones. Each fissioning uranium nucleus releases about 200 MeV (million electron-volts) of energy. This computes to 30 picojoules per fission event (3×10−11 J), so that we need 10 trillion fissions per second to supply our 300 Watt allocation of nuclear power. But we’ve jumped the gun a bit. The fission event creates heat, and this thermal energy is converted into electricity in a heat engine achieving 40% efficiency. So we really need to make 25 trillion fission events per second per person in the U.S.
Our fission reactors use the 235U isotope of uranium, which only comprises 0.7% of the natural uranium mined in the world. The other 99.3% is 238U, which indirectly contributes some fission power by first absorbing neutrons and converting to plutonium within the reactor. Ignoring this sub-dominant channel, we need to mine 140 uranium atoms for each one that goes pop.
So we need to mine 3.5 quadrillion (7×1015) uranium atoms per second per person. Over 40 years, this turns into 4×1024 uranium atoms. Sounds like a lot! Uranium is most commonly found in the oxide form U3O8, which has the tasty name “yellowcake.” To get 4×1024 atoms of uranium, we need 1.4×1024 molecules of yellowcake, which is a mere 2.3 moles. At 842 grams per mole, that makes about 2 kg of yellowcake. Not so bad, after all.
As for volume, yellowcake is 9 times more dense than water, and 2 kg of water would occupy 2 liters. So our handful of yellowcake is 0.2 liters, or three quarters of a cup, or a cube about 6 cm (2.4 in) on a side—about like an American softball. Put it in your pocket! On second thought, don’t.
An alternate take, though not a “fair” comparison: uranium has an average abundance of 2.7 parts per million by mass in the Earth’s crust. So to get the requisite mass of uranium atoms (1.6 kg), we need just over a half-million kilograms of rock. At a density of about 3000 kg/m³, this means about 200 cubic meters of rock, or about 6 meters on a side. This is amazingly close to our fossil fuel volumes! Of course, uranium mining doesn’t work this way: we go for the concentrations, and any random piece of surface rock is likely to fall far short of the average crustal concentration (so don’t try this in your backyard!).
Personalizing the Numbers
I used 40 years for these figures. You can scale volumes (but not cube edge lengths!) by the ratio of your age to 40 years. You may also wish to account for the fact that energy consumption in the past was not as high as today’s numbers we used for the analysis.
But these wrinkles will not change the story very much. And since the exercise is one of visualization, precision is of no consequence.
Picturing the Impact
We’ve worked out all the numbers. Let’s put it into pictures to get a more visceral feel for what the numbers mean. I use a typical San Diego house and yard (2000 ft², and 7200 ft², respectively) for comparison. Note these cubes apply to just me. My wife requires another set (but I’ve run out of room in the backyard!).
The cubes are color coded: dark brown for crude oil, gray for coal, blue for compressed natural gas, and yellow for the amount of average crustal material containing the requisite amount of uranium (though the actual uranium fits in your pocket). Each of these cubes corresponds to 40 person-years of typical usage. Add up the person-years in your house to figure out how to scale these volumes.
Panning out, we set the scene set on top of the appropriate-sized cube of water (deep blue) for hydroelectric production. This is a big volume for only 3% of the energy. But it is replenished, and so is not a good direct comparison to the other resources (we could compare the volume of dam needed, or note that the daily water use is about the size of the yellow cube). We also now see the volume of natural gas in uncompressed form as the light blue cube. This is the appropriate volume of gas as burned at the stove-top, furnace, or water heater.
Also shown for reference is the size a 15% efficient photovoltaic (PV) array would have to be to supply the entire 10,000 W average power an American demands today. The 18×18 m array produces 48 kW in full sun, but I’m only counting on an average of five hours of full-sun equivalent each day (San Diego is closer to 6, actually). Note that the PV array will last approximately the same 40 year timescale, and is thin enough to require far fewer materials than the other streams—even considering eventual replacement. Sitting in the front yard is the amount of material volume needed to construct the 4000 kg of panels, coming out to a cube 1.15 m on a side. The actual high purity silicon is a cube 0.5 m on a side, seen as a dark blue inset in the solar materials cube.
The installed solar array would cost approximately $200,000 (no storage). Meanwhile, the cube of oil, at $100/bbl would cost $88,000, and the coal and natural gas—if used to produce electricity at 35% efficiency and $0.05/kWh—would cost about $40,000 each (over twice this in California), for a total outlay of at least $170,000. Same ballpark, actually. If I instead use gas for cooking and heating, I would pay $26,000 rather than $40,000 for the gas at $1 per Therm (100 ft³ = 2.8 m³ contains 1.02 Therms of energy).
Yes, I used today’s prices and not the average prices over the last 40 years. But the comparison between fossil fuels and solar, for instance, is of interest for the future and not the past. Anyone want to guess whether the fossil fuel prices over the next 40 years will go up or down?
What of It?
The results of this exercise are intended to do little more than put a visual scale on the materials involved in the energy requirements of a typical American citizen. Some may be amazed at how large the volumes are, while others may be amazed by how small. They are what they are, and we can carry these visualizations around in our heads for whatever purpose.
Personally, I am impressed by the relative similarity of many of the block sizes. I don’t find any of them to be mind-blowingly monstrous, but at the same time I am humbled by the invisible (to me) impact these cubes represent in our wider world. I am impressed by the compactness (especially in comparative volume) of the solar array that accomplishes the same energy yield over 40 years—although it is not my intention to trivialize the practical challenges of transitioning to a fully solar energy system, or to suggest that a fully solar system is the appropriate approach. Nonetheless, solar energy is often characterized as pathetically diffuse and expensive. But when multiplied by decades of sun, the materials and volume required are rather advantageous, and the cost comparison is not frightening.