RERT wrote: ↑Tue Jan 30, 2024 4:04 pm

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.

"The problem is not what we don't know, but what we do know [that] isn't so" (Mark Twain)