Nuclear fusion energy calculations

I'm confused again. How does this work? How does 1kg and 0.666kg result in a 1:1 ratio?In summary, a sci-fi technology called an Aneutronic Triple Alpha Fusion Reactor has been created for a sci-fi story. It works by fusing Deuterium with Helium 3 and mimicking the triple alpha process found in stars to maximize fuel use. The reactor produces a single Helium-4 atom and fuses it with other Helium-4 atoms to create Carbon-12, which is the reactor's waste product. The power output of such a reactor would depend on the amount of fuel used and the energy density of the fuel. The calculations for
  • #1
Deleted member 690984
Me again!

For a sci-fi story I'm working on, I've created a sci-fi technology called an Aneutronic Triple Alpha Fusion Reactor. It works via aneutronic fusion, in this case, fusing Deuterium with Helium 3, but it also mimics the triple alpha process found within stars to maximise fuel use. Fusing deuterium with helium 3 produces a single Helium-4 atom; the reactor then fuses the resultant Helium-4 atoms together to create Carbon-12, which is essentially the reactor's waste product.

As I understand it, fusing deuterium with helium 3 produces 18.3 MeV of energy, while fusing Helium-4 together produces a net energy of 7.275 MeV (as I understand that fusing two Helium-4s together is actually an energy loss, but fusing the resultant Beryllium-8 atom with another Helium-4 is what gets you Carbon-12, releasing energy).

Anyway, I want to calculate what the power output of such a reactor would be. Obviously I can just make up a power output in terms of how much power these things produce per hour, but I want to know how much fuel I would need, so I need to know the energy density of the fuel and how much energy it would produce on a large scale.

Could someone help with the calculations as to how much, say, 1kg of deuterium fused with 1kg of helium-3 would produce? It also needs to be accounted for that 1/3 of that fuel mass will also be fused to carbon-12.

@jbriggs444 - you were a huge help on my engine calculations, would you be able to help here?

I read that 1kg of deuterium contains about 3x10^26 atoms, containing about 845 terajoules of energy. However I can't seem to find similar figures for helium-3.
 
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  • #2
paulthomas said:
Fusing deuterium with helium 3 produces a single Helium-4 atom; the reactor then fuses the resultant Helium-4 atoms together to create Carbon-12, which is essentially the reactor's waste product.

As I understand it, fusing deuterium with helium 3 produces 18.3 MeV of energy, while fusing Helium-4 together produces a net energy of 7.275 MeV (as I understand that fusing two Helium-4s together is actually an energy loss, but fusing the resultant Beryllium-8 atom with another Helium-4 is what gets you Carbon-12, releasing energy).
One does not fuse 4He to 8Be, except under extreme conditions (as in a red giant or red supergiant star). 8Be spontaneously decays to 2 alpha particles. In triple-alpha, it is exactly that, three alpha particles combining under extreme conditions that we cannot replicate on earth.
http://hyperphysics.phy-astr.gsu.edu/hbase/Astro/helfus.html#c2
If the central temperature of a star exceeds 100 million Kelvins, as may happen in the later phase of red giants and red supergiants, then helium can fuse to form beryllium and then carbon.
I recommend calculating the pressures involved. p = nkT, and note there is pressure from nuclei and electrons.

Bear in mind that when fusion reactions occur, the products will interfere with subsequent reactions, as in the 'ash' interferes with the continued process, which is why systems are designed to removed products from the system.

One would have better opportunity with fusing two 6Li nuclei.

Of course, if one suspends reality . . .
 
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  • #3
I think there's confusion as to what I'm actually asking here. Let me rephrase:

How much energy would be released from fusing 1kg of Deuterium to 1kg of Helium-3. How do I go about calculating this?
 
  • #4
paulthomas said:
I think there's confusion as to what I'm actually asking here. Let me rephrase:

How much energy would be released from fusing 1kg of Deuterium to 1kg of Helium-3. How do I go about calculating this?
Firstly, one would want to think in terms of atoms (or atomic density), or stoichiometrically, as in 1 nucleus of D + 1 nucleus of 3He, so given 3He is approximately 1.5 the mass, one would combine 0.666 kg 3He D with 1 kg of D 3He. So compute the number of atoms (nuclei) of 3He or D, and multiply by the energy per reaction. One reaction per 1 nuclei of 3He or D.

Bear in mind that one does not get 100% reaction. There are side reactions, like D+D or 3He + 3He.

Edit: see corrections below.
 
Last edited:
  • #5
paulthomas said:
1/3 of that fuel mass will also be fused to carbon-12.
I don't understand. Isn't all of the He-4 produced being fused to C-12?
 
  • #6
paulthomas said:
I think there's confusion as to what I'm actually asking here. Let me rephrase:

How much energy would be released from fusing 1kg of Deuterium to 1kg of Helium-3. How do I go about calculating this?
There’s a general recipe for answering these questions:
1) take the mass of the input elements, which we can look up online.
2) take the mass of the output elements, which we can also look up online.
3) subtract the one from the other to find out how much mass disappears in the reaction
4) use ##E=mc^2## to calculate the energy released
 
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  • #7
Astronuc said:
Firstly, one would want to think in terms of atoms (or atomic density), or stoichiometrically, as in 1 nucleus of D + 1 nucleus of 3He, so given 3He is approximately 1.5 the mass, one would combine 0.666 kg 3He with 1 kg of D. So compute the number of atoms (nuclei) of 3He or D, and multiply by the energy per reaction. One reaction per 1 nuclei of 3He or D.

Bear in mind that one does not get 100% reaction. There are side reactions, like D+D or 3He + 3He.
Can I ask why there wouldn't be a 1:1 ratio, so 0.666kg of Helium-3 with 0.666kg of Deuterium?
 
  • #8
paulthomas said:
Can I ask why there wouldn't be a 1:1 ratio, so 0.666kg of Helium-3 with 0.666kg of Deuterium?
Because the ratio in the reaction is 1 nucleus of D to 1 nucleus of He3, and those two nuclei do not have the same mass. So the mass ratio will be the ratio of the masses of the nuclei.
 
  • #9
PeterDonis said:
Because the ratio in the reaction is 1 nucleus of D to 1 nucleus of He3, and those two nuclei do not have the same mass. So the mass ratio will be the ratio of the masses of the nuclei.
Ah I'm with you, so 1kg of deuterium and 0.666kg of He3 will actually have the same number of atoms?
 
  • #10
paulthomas said:
so 1kg of deuterium and 0.666kg of He3 will actually have the same number of atoms?
No, it's the other way around. Look at the masses of the respective nuclei.
 
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  • #11
Astronuc said:
1 nucleus of D + 1 nucleus of 3He, so given 3He is approximately 1.5 the mass, one would combine 0.666 kg 3He with 1 kg of D.
Wouldn't this be the other way around? It's 1 to 1 in nuclei, so it would be 1 to 1.5 D to He3 in mass. That would mean 1 kg of D to 1.5 kg of He3, or 0.667 kg of D to 1 kg of He3.
 
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  • #13
After some cleanup, the thread is reopened.
 
  • #14
PeterDonis said:
Wouldn't this be the other way around? It's 1 to 1 in nuclei, so it would be 1 to 1.5 D to He3 in mass. That would mean 1 kg of D to 1.5 kg of He3, or 0.667 kg of D to 1 kg of He3.
Yes, the reaction goes as 1 D to 1 3He, on an atomic or nuclear basis, or 2 amu D to 3 amu 3He, on an atomic or nuclear mass basis, or 2 kg D to 3 kg 3He, or dividing by 2 gets 1 kg D to 1.5 kg 3He, or dividing by 3, 0.667:1 kg D:3He.
 
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One must bear in mind the side reactions, e.g., D+D, which would have a greater cross-section at temperature compare to D+3He, and much greater than 3He+3He. One cannot avoid those pesky alternative reactions, unless one is firing beams of particles into a pure target, or opposing beams, but then one has to deal with scattering inefficiencies.
 
  • #16
paulthomas said:
Can I ask why there wouldn't be a 1:1 ratio, so 0.666kg of Helium-3 with 0.666kg of Deuterium?
paulthomas said:
Ah I'm with you, so 1kg of deuterium and 0.666kg of He3 will actually have the same number of atoms?
See my correction.
Thanks to PeterDonis for catching my error.
 
  • #17
Deleted member 690984 said:
Me again!

For a sci-fi story I'm working on, I've created a sci-fi technology called an Aneutronic Triple Alpha Fusion Reactor. It works via aneutronic fusion, in this case, fusing Deuterium with Helium 3, but it also mimics the triple alpha process found within stars to maximise fuel use. Fusing deuterium with helium 3 produces a single Helium-4 atom; the reactor then fuses the resultant Helium-4 atoms together to create Carbon-12, which is essentially the reactor's waste product.

As I understand it, fusing deuterium with helium 3 produces 18.3 MeV of energy, while fusing Helium-4 together produces a net energy of 7.275 MeV (as I understand that fusing two Helium-4s together is actually an energy loss, but fusing the resultant Beryllium-8 atom with another Helium-4 is what gets you Carbon-12, releasing energy).

Anyway, I want to calculate what the power output of such a reactor would be. Obviously I can just make up a power output in terms of how much power these things produce per hour, but I want to know how much fuel I would need, so I need to know the energy density of the fuel and how much energy it would produce on a large scale.

Could someone help with the calculations as to how much, say, 1kg of deuterium fused with 1kg of helium-3 would produce? It also needs to be accounted for that 1/3 of that fuel mass will also be fused to carbon-12.

@jbriggs444 - you were a huge help on my engine calculations, would you be able to help here?

I read that 1kg of deuterium contains about 3x10^26 atoms, containing about 845 terajoules of energy. However I can't seem to find similar figures for helium-3.
I have a calculation for endurance in days using 1 kg total of H2 and H3. It would take approximately 3.9 days to burn through a kilo of fuel at constant 1 GW of power, assuming 100% efficiency. For a kilogram of each H-2 and H-3, the endurance would be twice as long. Sadly, He-3 wasn't in my story: https://www.wolframalpha.com/input?...euterium+++mass+of+tritium)*1+kg)/(1+GW/100%)
 

What is nuclear fusion energy?

Nuclear fusion energy is a process that involves combining the nuclei of atoms to release energy. This is the same process that powers the sun and other stars.

How is nuclear fusion energy calculated?

Nuclear fusion energy calculations involve using mathematical equations and models to predict the amount of energy that can be produced from a fusion reaction. This includes factors such as the type of fuel, temperature, and pressure.

What are the potential benefits of nuclear fusion energy?

Nuclear fusion energy has the potential to provide a nearly limitless source of clean energy. It produces no greenhouse gas emissions and does not produce long-lived radioactive waste.

What are the challenges in calculating nuclear fusion energy?

One of the main challenges in calculating nuclear fusion energy is the complexity of the reactions involved. Fusion reactions produce a wide range of particles and radiation, making it difficult to accurately predict the amount of energy released.

What advancements have been made in nuclear fusion energy calculations?

Over the years, scientists have developed more sophisticated models and simulations to improve the accuracy of nuclear fusion energy calculations. Additionally, advancements in technology have allowed for better control and measurement of fusion reactions, leading to more precise calculations.

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