Nuclear fusion energy calculations

[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.

 

[Astronuc]

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 . . .

 

[Deleted member 690984]

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?

 

[Astronuc]

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.

 

[PeterDonis]

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?

 

[Nugatory]

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

 

[Deleted member 690984]

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?

 

[PeterDonis]

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.

 

[Deleted member 690984]

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?

 

[PeterDonis]

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.

 

[PeterDonis]

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.

 

[PeterDonis]

Thread closed for moderation.

 

[berkeman]

After some cleanup, the thread is reopened.

 

[Astronuc]

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.

 

[Astronuc]

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.

 

[Astronuc]

Can I ask why there wouldn't be a 1:1 ratio, so 0.666kg of Helium-3 with 0.666kg of Deuterium?

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.

 

[AlexB23]

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%)