# Energy density calculations of nuclear fuel

Energy density is how much work a certain amount of fuel is capable of exerting. In terms of electricity, it is basically a measure of how much fuel you have to burn in a power plant to power the same size city. It’s like the miles-per-gallon rating of your power plant.

For any fuel, the characteristics of the power generation system affect exactly how much usable energy is extracted. For instance, if a power plant makes heat to be converted to electricity, the thermal efficiency ($$\epsilon_{th})$$ determines how much of the heat gets converted to electricity. These values vary from around 33% for coal plants to 35-40% for nuclear plants, to above 60% for combined cycle natural gas plants.

Another factor 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 nuclear power plants, usually only 5-7% of the energy of the fuel 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.

Since so many factors can change how much electricity comes out of a fuel, it’s best to compare the various fuels based on their energy density alone. That is, how much energy could be extracted with 100% thermal efficiency and 100% burnup. This allows us to focus in precisely on the topic of energy density for comparisons.

## 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

Here’s how much energy is contained in certain amounts of nuclear fuel.

MaterialEnergy released per fission (MeV) Atomic weight (g/mol) 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.

## 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.