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Reading about potential fusion plants it's easy to get the impression that we have nearly infinite access to the fuels required. But looking aside form how much we have available, I am interested in something I have a hard time finding information about; how much fuel would a fusion plant require? I assume this depends on the type of fuel, and it obviously depends on how much energy a particular plant would produce, but do we know approximately how much fuel (over time) can produce how much energy in a (reasonably efficient) fuel power plant?

To go for another approximation, can we (potential problems with heat dissipation and radiation aside) for example put a fusion reactor on a submarine or a spacecraft and have it run for weeks/months/years or would it require very frequent resupply of fuel to keep the fusion reaction going?

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Factually we have next to no fuel for fusion reactors, neither hypothetically nor practically. All of the tritium that would be required will have to be bred either in fission reactors or the fusion reactors themselves. Since a fusion reactor does not have as many excess neutrons, special neutron multiplier enriched blankets will have to be used. Even then realistic tritium breeding ratios will be in the ranges of 1.04 to (optimistically) 1.14. As a result it will take many decades for fusion reactors to even establish their own fuel cycle.

As for the demand... the ITER homepage cites the requirement for 300g tritium/day for an 800MW demonstrator: iter.org/mach/tritiumbreeding. According to en.wikipedia.org/wiki/Tritium current world demand is 400g per year at a cost of $30,000 per gram. The total tritium still available (most of it in thermonuclear weapons?) is said to be around 75kg, so we couldn't even run a single small reactor for much longer than half a year.

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In the course of trying to answer this question, I'll have to make a few assumptions or simplifications but the final numbers should give you an idea on how much fuel is needed.

Let's first assume we have a fusion reactor based on the D+T reaction running as this is the easiest fusion reaction to achieve (ITER is also aiming on that). To calculate the released energy per fusion reaction, we have to look at the masses. Let's first remind ourselves of the fusion reaction: $$ \mathrm{D} + \mathrm{T} \rightarrow\ ^4\mathrm{He} + \mathrm{n}. $$ The respective atomic masses are $m_\mathrm{D} = 2.0141u$, $m_\mathrm{T} = 3.016u$, $m_\mathrm{He} = 4.0026u$, and $m_\mathrm{n} = 1.008665u$ with $u=1.6605\cdot 10^{-27}\,\mathrm{kg}$. The right hand side of the reaction is missing the mass $\Delta m = 0.018835 u$ (this is the mass defect being a results of different binding energies per nucleon). The resulting energy released per fusion reaction can then be estimated to $$ \Delta E = \Delta m c^2 \approx 2.815\cdot10^{-12}\,\mathrm{J}. $$ To power a bright LED of $10\,\mathrm{W}$ would require $10\,\frac{\mathrm{J}}{\mathrm{s}}/(\Delta E)\approx 3.6\cdot10^{12}$ fusion reactions per second. Note that $1\,\mathrm{kg}$ of pure deuterium would contain $1\,\mathrm{kg}/m_{\mathrm{D}}\approx3\cdot10^{26}$ deuterium atoms and $1\,\mathrm{kg}$ of pure tritium would contain $1\,\mathrm{kg}/m_{\mathrm{T}}\approx2\cdot10^{26}$ tritium atoms. That means to power our LED for a full day, we require $$ 3.6\cdot10^{12}\frac{1}{\mathrm{s}}\cdot(3600\cdot24)\,\mathrm{s} \approx 3.1\cdot10^{17} $$ fusion reactions. Since we need an equal amount of tritium and deuterium atoms for the fusion reactions, the fuel mixture has to be 50:50. Therefore, an amount of roughly $\frac{3.1\cdot10^{17}}{3\cdot10^{26}}\approx1\,\mathrm{\mu g}$ of deuterium fuel and $\frac{3.1\cdot10^{17}}{2\cdot10^{26}}\approx1.6\,\mathrm{\mu g}$ of tritium fuel is "burned" per day.

Scaling this up to a 1 GW power plant, results in a fuel demand of $150\,\mathrm{g}$ of tritium and $100\,\mathrm{g}$ of deuterium per day. This corresponds to 60 kg of tritium and 40 kg of deuterium per year.

Note: these are just some rough extrapolations, but you get the idea (and they should be correct to an order of magnitude).

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