What is the Most Efficient Fuel for Fusion Reactors: T-D, D-D, or D+He-3?

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SUMMARY

The discussion establishes that Deuterium-Tritium (D-T) fusion is the most efficient fuel for fusion reactors, releasing 17.6 MeV per reaction, significantly outperforming Deuterium-Deuterium (D-D) fusion, which yields an average of 14.4 MeV when accounting for its complex reaction pathways. While D-D fusion can produce energy, it requires higher temperatures and is less efficient compared to D-T. Additionally, Helium-3 fusion with Deuterium can yield up to 18.3 MeV but necessitates even higher temperatures, making it less practical for current reactor designs. The use of neutrons from D-T fusion for generating Tritium via lithium targets is also highlighted as a crucial aspect of reactor efficiency.

PREREQUISITES
  • Understanding of nuclear fusion principles
  • Knowledge of energy release calculations in fusion reactions
  • Familiarity with isotopes: Deuterium, Tritium, and Helium-3
  • Basic concepts of fusion reactor design and operation
NEXT STEPS
  • Research the efficiency of Deuterium-Tritium fusion in ITER projects
  • Explore the role of lithium in Tritium breeding for fusion reactors
  • Investigate the potential of fusion-fission hybrid reactors
  • Learn about the challenges of achieving the necessary temperatures for Helium-3 fusion
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Researchers, nuclear physicists, and engineers involved in fusion energy development, as well as students studying advanced nuclear physics and energy systems.

Basil Currie
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Hi,
Does a T-D or a D-D fusion release more energy? Even if D-D needs higher temperatures, would it produce more energy in a fusion reactor?
 
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In one way the deuterium tritium fusion generates 17.6MeV
D+T=2He4+n +17.6MeV

In the other way the D+D fusion is more complex:
D+D->T+p +4.03MeV (a 50% of the reactions)
D+D->He3+n +3.27MeV (a 50% of the reactions)
But tritium fuses very quickly with deuterons as cross section is much larger than D+D fusion and D+T generates 17.6MeV as said before
Also happens with the Helium-3 that fuses again more easily with deuterons giving 18.3MeV, so if the reactor fuses also helium-3 at the end you should have:
D+D+D->He4+n+p+ (4.03+17.6)*50%
D+D+D->n+p+He4+ (3.27+18.3)*50%
So:
D+D+D->He+n+p+21.6MeV

So D+D =2/3*21.6MeV=14.4MeV that is less than 17.6MeV.
But that is not the exact account as long as Tritium does not exists so it is generated using the generated neutron again lithium target to generate tritium releasing more energy that I did not took in account
 
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D+He-3 needs even higher temperatures for a relevant cross section than D+D.
If you don't need tritium as fuel you can use the neutrons for some exothermic reactions - even fission has been proposed (fusion-fission hybrid).
 

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