Physical Insights

An independent scientist’s observations on society, technology, energy, science and the environment. “Modern science has been a voyage into the unknown, with a lesson in humility waiting at every stop. Many passengers would rather have stayed home.” – Carl Sagan

The footprints of coal and nuclear fuels.

with one comment

In 2007, coal consumption in the United States – just for electric power generation – was 1.046 billion tons (1,046 million tons).

That’s a lot of coal.

Crushed bituminous coal has a bulk density of about 833 kg/m3, so therefore, that coal occupies a total volume of one cubic kilometer.
(1.046 billion tons / 833 kg/m3)1/3 = 1.04 km.)

Combustion of coal produces carbon dioxide at a rate of about 1.83 kg CO2 per kg coal.
Of course, CO2 is just one particular component of the stream of dangerous waste output from coal combustion – there’s the particulate matter, the SO2, the NO2, the fly ash, the polycyclic aromatic hydrocarbons, and so forth.

So, each year, the use of coal for electricity generation in the United States generates 1.91 billion tons of CO2 output to the atmosphere.

Applying the ideal gas equation, we can find the volume that this CO2 occupies, assuming that it’s at a temperature of 300 K and a pressure of one atmosphere.

(T = 300 K is very frequently used as the value of “room temperature” or “ambient temperature” when performing scientific or engineering calculations in the Kelvin scale, since it’s a nice round number.)

(1.91 x 109 tons * R * 300 K / (44 g/mol * 1 atm))(1/3) = 9.90 km.)

The volume of CO2 produced each year from coal-fired generators in the United States corresponds to a cube of CO2 with a dimension of just under 10 kilometers on a side. Over the course of a decade, that adds up to a column of CO2 which is 9.90 km on a side and occupies the entire thickness of Earth’s atmosphere.

But what about nuclear fuels?

Coal yields an output of thermal energy when it is combusted of about 24 MJ/kg – therefore, 1.046 gigatons of coal corresponds to about 2.28 x 1019 J of energy.

One atom of fissile nuclear fuel like uranium-235 generates about 200 MeV of energy in a nuclear fission – and of course, that atom of 235U has a mass of 235 u (atomic mass units).

So, to generate 2.28 x 1019 J of thermal energy, we need to fission about 277.7 tonnes of 235U.

((2.28 x 1019 J / 200 MeV) x 235 u = 277.7 tonnes.)

(By the way, did I ever mention that I really like Google calculator?)

Uranium is quite a dense material – metallic U has a density of 19.1 g/cm3.

Therefore, the amount of uranium required to replace that billion tonnes of coal is only a volume of uranium metal corresponding to a cube measuring about 2.44 meters, eight feet, on a side.

(((2.28 x 1019 J / 200 MeV) x 235 u / 19.1 g/cm3)1/3 = 2.44 m.)

Now, I know what you’re probably thinking. That assumes that we’re using pure 235U as a nuclear fuel, and that it’s consumed in fission with 100% efficiency, and that is not the case with real, practical nuclear fuels in existing nuclear power reactors.

OK, so let’s revise the calculation a little to reflect the characteristics of a typical, existing light-water reactor more realistically.

A light-water reactor typically uses low-enriched uranium dioxide (UO2) fuel, and a modern power reactor typically achieves a burnup of something like 50 GWd (thermal energy) per tonne U.

Therefore, the generation of 2.28 x 1019 J of thermal energy from a conventional, once-through, inefficient LEU fuel cycle in a light-water reactor requires the use of 5987.4 tonnes of low-enriched UO2 fuel.

((2.28 x 1019 J / (50 GW days/tonne)) x (270/238) = 5987.4 tonnes.)

UO2 has a density of 10.97 g/cm3.

Therefore, the amount of LEU uranium oxide fuel required to replace that billion tonnes of coal is a volume corresponding to a cube measuring about 8.17 meters, 27 feet, on a side. The used, irradiated fuel – even before any recycling or recovery of unused uranium and actinides is performed – occupies about the same volume. Yes, you can put it in a garage, or put it in a basement, or something.

(((2.28 x 1019 J / (50 GW days/tonne)) x (270/238) / 10.97 g/cm3)1/3 = 8.17 m.)

Perhaps soon I certainly shall download SketchUp and have a play around with it in order to make some visualisations of these quantities.

Finally, a hat tip to Jason at Pro Nuclear Democrats for an interesting and very educational post which was my inspiration in creating this post.

Written by Luke Weston

December 4, 2008 at 7:27 am

Posted in Uncategorized

One Response

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  1. Luke, thanks again for the U235 energy equivalent calculation to 1 million barrels of oil. I believe visualizations are very educational for people, especially in energy matters where the average person just doesn’t have the time or inclination to study the matter in any depth. They may read an article or two on wind energy or nuclear, but if there is an interesting graphic to support the points being made, they will remember that far easier.

    I might just have to continue this series with some of your calculations done here – nice job!

    Jason Ribeiro

    December 5, 2008 at 6:48 am


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