#### Background numbers

I'll do this for the US in detail. To really roughly approximate world usage, just multiply the US results by 5. The relevant number I will use throughout is that the US uses approximately 100x10E15 BTUs per year.

Now to get a feel for how much this is...

**How much coal is this?**

We get roughly 10,000 BTUs/pound of coal and the actual conversion rate to kWh for coal is ~10,000 BTU/kWh. Thus, we get ~1 kWh per pound of coal, or using 1 kg ~= 2.2 pounds, 2.2 kWh of usable energy per pound of coal. This will of course be higher if you're burning the coal in your house for heat, but that's not really a thing and would only be off by about a factor of 2 so I'll ignore it.

Assume we only care about electricity generation for a moment. The US uses ~4100 TWh/year in electricity, so that means the approximate amount of coal needed to have a 100% coal electric system in the US is:

(4100 TWh)*(1E9 kWh/TWh)/(2.2 kWh/kg) ~= 2 trillion kg of coal. At coal's rough density of ~1000 kg/m^3, that means that we need roughly 2 billion cubic meters of coal if we want to power the US electrical grid with coal. A couple of conversions that might make that more relatable:

####

Assume we only care about electricity generation for a moment. The US uses ~4100 TWh/year in electricity, so that means the approximate amount of coal needed to have a 100% coal electric system in the US is:

(4100 TWh)*(1E9 kWh/TWh)/(2.2 kWh/kg) ~= 2 trillion kg of coal. At coal's rough density of ~1000 kg/m^3, that means that we need roughly 2 billion cubic meters of coal if we want to power the US electrical grid with coal. A couple of conversions that might make that more relatable:

- The state of Delaware is ~5,000 km^2, or ~5,000,000,000 m^2. That means this amount of coal spread over the state of Delaware would be ~0.4 m thick which is over 1 foot.
- The Great Pyramid's volume is ~2.5 million cubic meters, which means that this amount of coal is equivalent in size to 800 Great Pyramids.

That's just for our power grid. Going back to the number for total energy usage, 100E15 BTUs for the US means ~4.5 trillion kg or ~4.5 billion m^3 of coal (repeating earlier calculations), so doing those same comparisons and adding in that there are ~300 million Americans and an average bathtub is ~0.17 cubic meters:

- The amount of coal the US would need for all energy in a year is approximately what it would take to cover the state of Delaware with a pile about 3 feet high.
- The amount of coal the US would need for all energy in a year is approximately 2,000 Great Pyramids.
- The average American effectively uses the energy of burning 90 bathtubs full of coal per year.

**How much oil is this?**

Doing the same things except using 140,000 BTUs/gallon of petroleum and the fact that 1 cubic meter ~= 264 gallons, you find that the US would need:

(100E15 BTU/year)*(1 gallon/140,000 BTU)*(1 m^3/264 gallons) = 2.1 billion cubic meters of oil (1 billion for electricity generation alone) Using roughly the same thing as before:

(100E15 BTU/year)*(1 gallon/140,000 BTU)*(1 m^3/264 gallons) = 2.1 billion cubic meters of oil (1 billion for electricity generation alone) Using roughly the same thing as before:

We do not use this much cooking oil each year...

- The amount of oil the US would need for all energy in a year is approximately what it would take to cover the state of Delaware with a pool about 1.5 feet deep.
- The average American effectively uses the energy of burning 40 bathtubs full of oil per year.

#### How much uranium is this?

A given amount of uranium produces ~16,000 times as much usable energy as that same amount of coal. From above, this means we'd need something like (4.5 trillion kg coal)*(1 kg U/16,000 kg coal) ~= 280 million kg of uranium. Uranium is much denser than coal, so taking a value of ~19,000 kg/m^3 as its density, we need (280 million kg)/(19,000 kg/m^3) ~= 15,000 cubic meters of uranium.

A reasonably-sized house might be 1500 square feet and has 8 foot ceilings which yields a volume of ~340 cubic meters. One golf ball has a volume of roughly 4E-5 cubic meters. Using those two numbers with the values above:

A reasonably-sized house might be 1500 square feet and has 8 foot ceilings which yields a volume of ~340 cubic meters. One golf ball has a volume of roughly 4E-5 cubic meters. Using those two numbers with the values above:

- The amount of uranium the US would need for all energy in a year is approximately enough to fill roughly 45 houses.
- The average American effectively uses the energy we'd get from a ball of uranium about the size of a golf ball.

#### How many solar panels is this?

Solar panels make more sense in terms of area than volume, so I'll switch to that and focus on electricity for now. It looks like you need (2.8 acres/1 GWh)*(4047 m^2/acre)*(1000 GWh/1 TWh) ~= 11 million square meters of solar panels per TWh of electricity generation. Recalling from above that the US uses ~4100 TWh of electricity per year, that means we'd need (11 million m^2/TWh)*(4100 TWh) ~= 45 billion square meters of solar panels. The US currently has ~18,000 square miles of paved roads which is (18,000 square miles)*(2.6 million square meters/square mile) ~= 46 billion square meters. Thus,

- To replace the US electric grid with solar panels, you'd need approximately the total area of all paved roads in the US.
- To replace the US electric grid with solar panels, you'd need enough to cover the state of Delaware ~9 times.

####
**Additional reading:**

There is actually a unit created for this sort of thing called 'cubic mile of oil'. If you google that you can find more things like this.

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