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by yufeng66 2308 days ago
There is a warning sign. Without actually running the number, I believe the energy generated by the proposed HB11 fusion should be several order of magnitude higher compared to the electric energy alpha particle can carry. So extreme hot temperature will be created regardless.

Edit: actually read the paper :) From the paper: H + 11B = 3 x 4He + 8.7 MeV

when alpha particle absorb the electron to become helium, it can carry about 50 eV energy. So vast majority of the energy generated will be kinetic energy or photon energy which translate into very hot temperature at macro level.

6 comments

Actually, converting kinetic energy of fast moving ions to electricity is a very efficient process. See https://en.wikipedia.org/wiki/Direct_energy_conversion.
The whole "electrical energy directly so no steam generators necessary" part of the discussion is fairly irrelevant.

Steam generators might have fairly low efficiency, but if hydrogen fusion works at all it'll use so little fuel and have such a low marginal cost that we can just do more of it to make up for any efficiency losses.

It's not so much about efficiency as capital cost.
Except steam generators are pretty cheap - they're off the shelf, and will be a drop in the ocean compared to the total costs of the first fusion plants.
Well, that depends. ITER is tens of billions. The most expensive part of this reactor would probably be the petawatt laser, which is tens of millions for one-off experimental devices. A turbine and generator is about a million dollars per megawatt, so it could be a significant percentage of total reactor cost, especially in mass production.
The waste heat from the inefficieny is an environmental issue if cooling with sea, lake or river water.

The article talks about the advantage of not needing steam cooling apparatus. The stations could be located in urban areas.

I think the idea is to capture the kinetic energy of the Helium ions not through cooling but slowing them down with electromagnets. The high-speed Helium ions carry a large current, it's not about the ionization energy.
I suppose the idea is that most of the energy will be expelled as kinetic energy of the alpha particles, that then can be converted into electric potential energy?
Yeah, I thought the electrical output seemed fishy too. Why are the electrons stripped from the helium? And is that actually due to the energy of the fusion reaction? And how much of the fusion energy is left after?

These are IMO the fundamental questions.

> Why are the electrons stripped from the helium?

Hydrogen + Boron -> 3×Helium + energy

The energy is mostly 2900keV kinetic energy for each Helium ion i.e. the He ions just fuck off really fast and leave the elections behind.

If a He ions hits H atoms a couple of times, the second H hit has just the right energy to then hit another Boron and create an an avalanche of reactions. https://aip.scitation.org/na101/home/literatum/publisher/aip...

>> If a He ions hits H atoms a couple of times, the second H hit has just the right energy to then hit another Boron and create an an avalanche of reactions. https://aip.scitation.org/na101/home/literatum/publisher/aip....

Well that's frightening.

Only the helium nucleus is formed as part of the fusion reaction, so it starts off being positively charged, and not later stripped of its electrons. That, in principle, creates a current which can be used almost directly.
Well, it creates a region in space with a high charge density that can be accelerated to a plate that has a voltage applied to it which it bumps into which then causes a current.

Importantly, this doesn't have to happen in the reactor vessel. The charged gas can be pumped somewhere else.

It does have to happen in the reactor vessel. It's not just that it's a charged gas. It's that it's a charged gas which is exploding with almost 300 kWh of energy per shot.
Much of the difficulty with fusion is getting the fuel to the temperatures needed to sustain fusion - so that's actually a good thing.