Helion Energy to demonstrate net electricity production by 2024

[Skipjack]

along with the 75% of the 10GW D-D reaction that ends up in MeV neutrons, of course... might also take a stab at calculating peak neutronicity for the shielding requirements

Only about 33% of the D-D energy is released in neutrons.
Over two reactions you get:

4.85 MeV (66%) in charged particles + 2.45 MeV in neutrons (33%)

[TallDave]

along with the 75% of the 10GW D-D reaction that ends up in MeV neutrons, of course... might also take a stab at calculating peak neutronicity for the shielding requirements

Only about 33% of the D-D energy is released in neutrons.
Over two reactions you get:
4.85 MeV (66%) in charged particles + 2.45 MeV in neutrons (33%)

yes sorry I forgot about the proton side

and I think we're assuming the 1.01MeV tritium is harvested (both as ash and charged product) instead of reacting so ignoring the "fuel cycle"

that should help a bit

http://hyperphysics.phy-astr.gsu.edu/hb ... on.html#c3

[RERT]

Have Helion said how much D & He3 constitutes the reaction plasma?

I know they have a patent which discussed relative concentrations, especially hypothesizing that it might be possible to create as much He3 from DD reactions as lost in DHe3 reactions. Based on that we can probably work out the total volume of ash - T, He4 and P - from a, say, 10MJ pulse of fusion energy.

But what fraction of the input D and He3 react? It might be that almost everything reacts, or it might be that the mass input is relatively large and only a tiny fraction is reacted. I don't recall seeing that data.

[Skipjack]

and I think we're assuming the 1.01MeV tritium is harvested (both as ash and charged product) instead of reacting so ignoring the "fuel cycle"

The Tritium is too hot and non- collisional on the timescale of the pulse and the chamber is evacuated between pulses. So no significant amounts of it react with the Deuterium.

[Skipjack]

Sam Altman:

where @Helion_Energy will soon start to install polaris:

https://x.com/sama/status/1753250549045018728?s=20

HelionPolarisVault.jpg (115.2 KiB) Viewed 1004 times

[Skipjack]

Making vacuum vessels for our fusion systems!

https://x.com/Helion_Energy/status/1755 ... 68320?s=20

HelionVacuumVessels.jpg (77 KiB) Viewed 816 times

[charliem]

Have Helion said how much D & He3 constitutes the reaction plasma?

I know they have a patent which discussed relative concentrations, especially hypothesizing that it might be possible to create as much He3 from DD reactions as lost in DHe3 reactions. Based on that we can probably work out the total volume of ash - T, He4 and P - from a, say, 10MJ pulse of fusion energy.

But what fraction of the input D and He3 react? It might be that almost everything reacts, or it might be that the mass input is relatively large and only a tiny fraction is reacted. I don't recall seeing that data.

I have not seen it either, but it is at least possible to estimate an order of magnitud from the available info.

Let's see:

A 50 MWe (net) fusion reactor needs to generate at least 50 MJ per second, say 100 MJ to compensate for all possible losses.

100 MJ = 6.24E20 MeV

A D-3He reaction gives 17.6 MeV in charged particles [see correction 1].
A D-D reaction gives 2.4 MeV in charged particles (average of the 2 possible outcomes).
If D-D reactions were 1/3 of the total (what's needed to generate as much 3He as is consumed) [see correction 2], the average would be 12.5 MeV per fusion event [see correction 3].

If we only had D-3He reactions, then 3.4E19 fusions would be needed.
If 1/3 of the reactions were D-D, then 5.0E19 fusions would be needed.

So, between 3.4E19 and 5.0E19 fusions per second.


Now, let's estimate how many D and 3He ions there are at each cycle's beginning.

Say, the 50 MWe machine have 20T magnets surrounding the reaction chamber. FRCs exclude the external magnetic flux, concentrating it outside their separatrix, thus increasing the magnetic pressure. Let's say that external 20T results in 25T compressing the FRC.

25T produces a magnetic pressure of ~2.5E8 P (= 2500 bar).

To equal that with plasma pressure (n.kb.T) the product n.T (number density * temperature) has to be ~1.8E31 K/m³ = 1.55E27 eV/m³

Given that the electron-ion temperature ratio is about 1/10, and electron population is 1.5 times that of ions (1 electron per D, 2 per 3He, and supposing a 50-50 mix), ~15% of that would be taken by electrons, and ~85% by ions. So, n.T for ions would be ~1.3E27 eV/m³.

We don't know the ion temperature they will operate at, let's say 40 keV. Then, ion density would be 3.3E22 1/m³.

So, IF plasma volume is 1 m³, and cycle frequency 1 Hz, then "consumption" of D and 3He would be in the order of 0.1%.

But, according to Helion patents, [compressed] plasma volume is going to be 10 to 100 times less, and frequency higher, up to 10 Hz.

Considering those two corrections, worse case is on the order of 10% "fuel consumption", best case 0.1%. (see note 4)


From all I've heard and read regarding their machines, if they are able to make one work, I'd say the real number will be closer to the best than to the worst case. We'll see.

Reviews of these calculations will be welcomed.

Corrections and notes (thanks, Skipjack):
1) D-3He fusion gives 18.3 Mev per reaction, not 17.6 (that's for D-T).
2) To generate as much 3He ions as "consumed", 2/3 of the reactions have to be D-D.
3) To generate as much 3He ions as "consumed", the average energy per reaction, released in charged particles, would be 7.7 MeV, not 12.5
With these 3 corrections, the number of fusions to generate 100 MJ goes from 3.4E19 to 8.1E19, instead of 3.4E19 to 5.0E19.
The final conclusion is not affected significantly.
4) That worst case 10% is for an scenario where: plasma volume = 0.01 m³ + frequency = 1 cycle/second. Sounds improbable.

[Skipjack]

Yeah, I think they are still aiming for 10 Hz for the power plant design.
As for the math:

A D-He3 reaction should be 18.3 MeV in charged particles

2He4 (3.6 MeV) + p ( 14.7 MeV )

IF they have enough He3 from T- decay, then D-D reactions would be 1/2 of the total. If they do not, then they would be 2/3 of the total.

So on average, charged particles make a 10.3 MeV or 7.71 MeV per reaction.

Let's say that external 20T results in 25T compressing the FRC.

I am not sure how you get to that. At a Beta = 1.
The equation for fusion scaling is B^4 * Beta^2.

Operating ion temperature would be around 20 keV to maybe 30 keV, but I believe they are aiming for colder and denser rather than hotter and less dense. I believe that 25 keV would be a good middle ground.

Their FRCs are meter scale but that is before compression. Not sure how big/small they are after compression.

[RERT]

So… if I’m hearing right, 1% consumption (give or take 10x).

If that’s correct, it seems serious question as to why the chamber has to be evacuated, since the exhaust is basically the same as the fuel. Seems like inserting the 0.1- 10% back has to be much quicker than emptying it entirely.

Can anybody explain why not do that and instead (say) evacuate every 2nd or third pulse?

[charliem]

So… if I’m hearing right, 1% consumption (give or take 10x).

If that’s correct, it seems serious question as to why the chamber has to be evacuated, since the exhaust is basically the same as the fuel. Seems like inserting the 0.1- 10% back has to be much quicker than emptying it entirely.

Can anybody explain why not do that and instead (say) evacuate every 2nd or third pulse?

I see one possible reason:

If they don't evacuate the chamber every time, D-T reactions, and subsequently 14 MeV neutrons, become a much bigger problem.

[Skipjack]

If they don't evacuate the chamber every time, D-T reactions, and subsequently 14 MeV neutrons, become a much bigger problem.

This! That is also what they talk about in their fuel cycle patent.

[charliem]

A D-He3 reaction should be 18.3 MeV in charged particles

2He4 (3.6 MeV) + p ( 14.7 MeV )

Yep, you are right. I mixed D-3He and D-T energies. Thanks for the correction.

IF they have enough He3 from T- decay, then D-D reactions would be 1/2 of the total. If they do not, then they would be 2/3 of the total.

If they aim to produce as much 3He from D-D reactions, as it is consumed in D-3He fusions, then yes, 1/3 D-3He->4He+p reactions, 1/3 D-D->T+p reactions, and 1/3 D-D->3He+n.

But I don't see the other scenario. If they had enough 3He, operating at 25 keV, with a 1:1 D-3He mix, then for each 2 D-3He reactions they would have 1 D-D. Am I wrong?

By the way, with enough 3He they could go for a higher temperature. If I'm not mistaken, at 25 keV the DD/D3He reaction ratio is 2:1, at 40 keV it is 4:1.

Let's say that external 20T results in 25T compressing the FRC.

I am not sure how you get to that.

I obtained that from Helion's own papers and presentations. For instance, if you review slide number 10 in Dr. Kirtley's Princeton presentation from December 2022, you will see that Bext = Bvac / (1-Xs^2), where Xs=rs/rc (FRC's separatrix radius / coil radius).

My understanding is that Bvac is the nominal strength of their magnets (20T in my example), and Bext is the field right outside the FRC (25T in my example).

[Skipjack]

But I don't see the other scenario. If they had enough 3He, operating at 25 keV, with a 1:1 D-3He mix, then for each 2 D-3He reactions they would have 1 D-D. Am I wrong?

I am not sure what you are getting at here.
Are we just assuming a 1:1 ratio of Deuterium to He3?

Generally, Helion NEEDS a ratio of two D-D reactions for every D-He3 reaction unless they have enough He3 from Tritium decay of Tritium from the second D-D reaction branch (which would take decades).
Within a single machine and even between pulses they can control the ratio of density to temperature. D-D fusion reactions are higher at lower temperatures and higher densities. D-He3 is the other way round.
From what I remember is that they are aiming for ion temperatures ranging between 20 and 30 keV. So 25 is probably a good estimate.

[RERT]

If they don't evacuate the chamber every time, D-T reactions, and subsequently 14 MeV neutrons, become a much bigger problem.

Yes, I was aware, but there may be compensating advantages.

Given the T is highly reactive, is it safe to assume that all of the T from one pulse would react with D in a subsequent pulse? That would boost the output of a second pulse by a few MeV of charged particles per pulse.

If so the total inventory of T does not grow. I guess your view on that depends on whether you see it as an He3 source or a regulatory curse.

If 2/3 of all reactions in the original pulse were DD, then there would be 1/3 T ion for every reaction in that pulse.

Overall you would seem to have a reactor with a neutronicity around 1/3 of DT reactor, which could potentially pulse an order of magnitude faster than the pure DHe3 mode.

I could be wrong, but it also seems like in that case the capacitor bank stays the same size.

Seems like capital cost/kWh is drastically reduced.

You can have a similar discussion around partial evacuation or intermittent evacuation and reduced neutronicity.

[Skipjack]

If they don't evacuate the chamber every time, D-T reactions, and subsequently 14 MeV neutrons, become a much bigger problem.

Yes, I was aware, but there may be compensating advantages.

Given the T is highly reactive, is it safe to assume that all of the T from one pulse would react with D in a subsequent pulse? That would boost the output of a second pulse by a few MeV of charged particles per pulse.

If so the total inventory of T does not grow. I guess your view on that depends on whether you see it as an He3 source or a regulatory curse.

If 2/3 of all reactions in the original pulse were DD, then there would be 1/3 T ion for every reaction in that pulse.

Overall you would seem to have a reactor with a neutronicity around 1/3 of DT reactor, which could potentially pulse an order of magnitude faster than the pure DHe3 mode.

I could be wrong, but it also seems like in that case the capacitor bank stays the same size.

Seems like capital cost/kWh is drastically reduced.

You can have a similar discussion around partial evacuation or intermittent evacuation and reduced neutronicity.

1. I would recommend looking at Helion's patent for their fuel cycle. They do a pretty good comparison between concepts, including with Tritium extraction and without Tritium extraction.
https://patentscope.wipo.int/search/en/ ... ESCRIPTION

2. You already have a 2.45 MeV neutron in 1/3 of the reactions (or half if you have enough He3 from Tritium decay but that is not available in this scenario). Leaving the Tritium in would add another 14 MeV neutron to the mix. And no, you can not compare the two. 14 MeV neutrons are much more difficult to handle than 2.45 MeV neutrons.
The wall loads would set the limit for the output of the machine. So for a machine of the same size, you would at best get the same output for the same size.
The capital cost would also rise for the extra shielding that would be needed. Plus a lot of materials would have to be changed in order to have sufficient longevity of the machine. Maintenance would most likely be much more expensive. There might be some savings on compression magnets in return, but I am not sure they can make up for the other costs.
I would actually estimate that a 50 MWe D-D-T machine (forget about the He3 in that scenario since you would likely run it colder and denser) could be bigger than an D-D-He3 machine. In addition to that you would likely want some sort of steam cycle to make use of those 14 MeV neutrons. That adds more capital cost again. That is unless you use the machine for process heat or district heating.

3. Then there is the issue of siting. A D-He3 machine would easier to get a site license for than a D-D-He3 machine and that in turn would be MUCH easier to get a site license for than for a D-T or D-D-T machine.

4. As long as there is enough of a market for Tritium, it would be much more profitable for Helion to just sell it.
In the longer term, I can see them do D-D-T machines as dedicated He3 breeders that supply dedicated D-He3 machines (which would have favorable site restrictions and might have higher customer demand) . Those would likely be stationed at special sites and would have a slightly different design. They would likely have an increased length to protect the expensive equipment at the ends from the neutrons (via inverse square law).
Magnets would have to be the same strength as they would want to get enough D-D reactions to make He3 and T. I am not sure about the diameter. A D-T machine could have weaker magnets and have a smaller diameter, but these machines would need the D-D reactions.
It will be interesting to watch how this plays out, provided Helion can get it all to work and work economically.