[atonalfreerider]

My two takeaways that make this interesting:

1 - ENERGY RECAPTURE: Helion's existing Trenta (Gen 6) machine, and its future Polaris (Gen 7) machine are one-of-a-kind systems when it comes to converting the energy released from fusion into usable electric current. Where all other fusion and fission systems attempt to convert the fusion energy from escaping neutrons with heat exchangers (Beryllium and steam turbines), Helion is taking the magnetic pressure from the fusion reaction that is exerted on their magnetic confinement field, that runs their electromagnets backwards, to flow current to the grid. This is much more efficient compared to a heat system, and does not create radioactive byproducts.

2 - FUEL: These systems are also one-of-a-kind since they are using Helium-3 fuel. They are acknowledging that they need to synthesize this fuel using a separate Deuterium fusion system since He3 is so rare, and since the system for recapturing He3 from the main reaction has a half-life of 12 years.

[joak]

Exactly, but the fuel cycle is actually more subtle than that:

When compressing the D+He3 plasma, inevitabily some D will also fuse together.

And since one in two D+D reaction yields He3+n, if you fine tune the proportion D/He3, the D+He3 reaction will in total (ie taking into account the D+D side reaction) produce more He3 than it consumes.

But D+D also yields T+p, and T eventually decays into He3 (T half life is 12.5 years)

The total fuel cycle has 3 fusion reactions:

D+He3 -> He4 + p (50%)

D+D -> He3 + n (25%)

D+D -> p + T (25%) (T=He3, after decay)

As you can see the only input is deuterium.

The thing is that you want to avoid D+D -> He3 + n because 1. It produces less energy 2. It produces neutrons that damages your machine.

Hence the proposal of splitting the fuel cycle in two reactors, one doing the clean and powerful D+He3 and the other doing the dirty D+D.

So one reactor will consume He3 and the other will produce He3 in excess.

[pfdietz]

D + D -> 3He + n is actually tolerable, since the neutron is of low energy. In particular, it has a much lower cross section for (n,p) and (n,alpha) reactions in the reactor structure. Those are the really serious reactions that ruin the material by filling it with microscopic very high pressure gas bubbles.

What you want to avoid is D + T on the T generated from the other D + D reaction. In Helion's scheme, I believe the ions do not equilibrate during a pulse, so that tritium doesn't slow down before the plasma reexpands and is exhausted to the divertor. The DT fusion cross section peaks around a 60 keV center of mass energy, so keeping the T energetic can reduce its fusion rate somewhat.

This is another reason to have "field reactors" burning 3He generated elsewhere, with a high 3He/D ratio. The lower D concentration reduces DD reactions, but it also reduces DT reactions on any T that are produced.

There's an additional problem of purifying and reusing the D in the exhaust of the machines. You don't want to have to filter out all the tritium from this before every cycle. It's possible the field machines could have many go-arounds of this gas before sending the sufficiently contaminated D back to be filtered (or used in the dirty DD machines).

[moloch-hai]

Seems like you would want to run the power reactor especially 3He-rich, so the 2H can mostly find a 3He. Presumably 3He is hard to make fuse with itself or with waste He4 or tritium. But 2H is much more eager to fuse with 3H than anything else, so it would be important to get the 3H product out quickly. Separating waste 3H and H from the 2H/3He mix cannot be easy.

I have encountered an assertion that 3He eagerly turns into tritium, I guess when struck by a stray hot electron, which will certainly be milling about. Wondering how much of the 3He they breed will be lost to this.

[joak]

You are right, it's a pulsed system. The operation goes like this:

1. Fill the chamber with D and He3

2. Ionize the fuel and compress the plasma

3. The fusion starts and pushes back the magnetic field producing electricity

4. Stop everything

5. Vacuum and start again

So before the plasma is polluted with p, T and He4, the pulse has ended. The side reaction D+T does not get the time to happen much.

I'm using a different notation than yours: 3He, 3H, 2H, H are He3, T, D, p

T is tritium, D is deuterium, and p is protium (a proton)

[moloch-hai]

Thank you.

Terminology is a nuisance here mainly because we don't have a nice name and one- or two-letter abbreviation for ³He. Somebody suggested "hellion", "Eh". Whoever is first to make it work might deserve naming rights.

[joak]

I'm not aware of He3 easily becoming T. I know T spontaneously becomes He3 (emitting a positron and a neutrino) with some probability, a very low one, not relevant at the scale of a pulse

[moloch-hai]

I am wrong. To absorb an electron would need a coincident neutrino.

Instead, a neutron strikes ³He producing T, H and a photon.

[pfdietz]

Yes, and that's not a problem either here, because the neutrons don't stick around. If you had a lithium blanket making T you'd want to clear out any 3He as it was produced from T decay so they don't hoover up too many thermalized neutrons, though.

I don't think that (n,p) reaction on 3He normally produces a photon, btw.

[moloch-hai]

I suppose the 764 keV could be kinetic:

  n + 3He → 3H + 1H + 0.764 MeV

I expect the concentration of ³He in your blanket would be extremely low, relative to the Li, so an unlikely target.

Really, if it takes neutrons, it is not a problem.

[pfdietz]

> I have encountered an assertion that 3He eagerly turns into tritium,

That assertion is nonsense. That would be a weak nuclear interaction, very low cross section.

[wolfram74]

It's somewhat generous to call it a He3 generator though, since the half life of tritium is 12 years, you'd need to make sure you made most of the He3 you need over a decade in advance.

He3 has uses besides as a fusion fuel though, so maybe they can overshoot their estimated consumption rate by an order of magnitude and sell or stockpile the surplus 10 years later?

[DennisP]

Only half of the D-D reactions produce tritium. The other half produce He3 directly.

[wolfram74]

That helps considerably, they can over produce initially and taper down as the tritium backlog decays.

[bilsbie]

How do they plan to deal with the neutrons though? Won’t it damage the equipment?

How much waste heat will there be?

If waste heat is low this could be ideal for space or mars operation.

[vagabund]

Here's my understanding:

The main electricity-generating fusion reaction is aneutronic -- 3He-D fusion yields regular hydrogen and a proton. But a fuel mixture of 3He+D will inevitably produce some D-D fusion, which yields 50% T+p (tritium and a proton) and 50% of He3 + neutron. The neutrons produced are slower than those produced in D-T fusion (2.45MeV vs 14.06MeV), and are similar to those produced by radiation therapy medical devices.

Since they need the products of D-D fusion anyways (tritium decays into 3He in 12.5 years) they can either run D-rich fuel in the main reactor and extract the yield for future fuel inputs or, if the slow neutrons of D-D fusion still prove too damaging to the device, they can lower the Deuterium ratio in the main reactor's fuel, and designate separate reactors to do the D-D fusion exclusively. That way they can concentrate the cost of neutron damage mostly to one fuel producing reactor, and not have to replace every electricity-generating reactor as often.

[DarmokJalad1701]

IIUC, only the De-De reaction generates neutrons. This is the process they use for generating He-3 (and a tiny bit of power). The video mentions that their plan is to separate the De-De reactors from the from the power-generating De-He3 reactors (which do not create neutrons). Therefore, their main power-generating reactor won't have to deal with damage from neutrons.