Computing the energy density of nuclear fuel

By Nick Touran, Ph.D., P.E., 2020-10-30 , Reading time: 3 minutes

Energy density is like the miles-per-gallon rating of a power plant. It measures how much energy is released (in megajoules) given a certain mass of fuel (in kilograms). Perhaps the most physically unique thing about nuclear power is that the energy density of nuclear fuel is about 2 million times higher than that of any chemical (like fossil fuel, biofuel, or batteries).

See results graphed on a linear-axis bar graph here.

Details of the computation

The easy way to compute energy density of nuclear fuels is to figure out how much fission energy can be released from 1 mole of the fuel. The equation for energy density in MJ/kg is:

$\text{ED} = \frac{\kappa_{fis} \text{[MeV/fission]} N_A \text{[fissions/mol]}}{A \text{[g/mol]}} \times \frac{1.60217\times10^{-19} \text{[Mega Joules/MeV]}}{0.001 \text{kg/g}} = \text{MJ/kg}$

where:

• $$\kappa_{fis}$$ is the energy release per fission for the nuclide of interest. These values are measured by scientists and collected in nuclear data files such as the ones available at the National Nuclear Data Center. Look for the (n,fis.ene.release) Interpreted field for each nuclide here.

• $$N_A$$ is Avogadro’s number, or 6.022e23. This is the number of atoms per mole. Since we’re assuming 100% of atoms fission, this is equal to the number of fissions per mole.

• $$A$$ is the atomic mass of the nuclide of interest. This can be found on any Chart of the Nuclides, like this one.

Some nuclear energy density calculations

Running the equation, here’s how much energy is contained in certain amounts of nuclear fuel.

MaterialEnergy released per fission (MeV) [1]Atomic weight (g/mol) [2]Energy density (MJ/kg)
U-235193.4235.0479,390,000
U-238/Pu-239198.9238.0580,620,000
Th-232/U-233191.0232.0479,420,000

Table 1. Energy densities of nuclear fuels. Energy per fission does not include the energy lost to neutrinos since it is practically unrecoverable.

Influence of burnup

Another factor worth considering is how complete the fuel is consumed. For example, a wood fire may burn out before all the energy is extracted from the wood. In traditional LWR nuclear power plants, usually only 5-7% of the fuel’s energy is extracted. Furthermore, the fuel has already gone through an enrichment process so only about 1% of the energy of the mined resource is used. Advanced nuclear power plants called breeder reactors such as the liquid metal fast breeder reactor (LMFBR) or the molten salt breeder reactor (MSBR) can extract much more of the mined energy. The fraction of the energy extracted from the fuel in a reactor is called the burnup.

So in a LWR, the effective energy density is around 1% of 80 million, or 0.8 million MJ/kg. Note that from a high-level waste perspective, the burnup of the fuel going into the reactor is what matters (rather than the effective burnup of the mined material), so 5% of 80 million is more appropriate.

Comparison with other fuels

A single pellet of fuel contains as much energy as the following:

Material Energy Density
(MJ/kg)
Equivalent to
fuel pellet in LWR
Equivalent to fuel
pellet in breeder
Coal 30 1.3 tons 22 tons
Oil 42 250 gallons 4350 gallons
Natural Gas 53.5 34,000 cubic ft 590,000 cubic ft
Lithium 43 0.9 tons 16 tons

Details of the calculation are found in the pellet comparison calculation.

Other Complications

In a nuclear reactor, fission isn’t the only process that releases energy. The actinides, fission products, and even structural and coolant nuclides often undergo capture reactions that release energy without fissioning. The fraction of energy released by a nuclear reactor by these reactions can be on the order of 10% of the total power of the reactor.