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Another advantage of hydrogen-boron fusion is that all the reaction products are charged. Therefore (theoretically) it would be possible to use very efficient

by bbojan 5y ago

Another advantage of hydrogen-boron fusion is that all the reaction products are charged. Therefore (theoretically) it would be possible to use *very* efficient magnetohydrodynamic methods of converting the reaction energy to electric energy. More mainstream fusion methods have to go through the typical hot water -> steam -> steam turbine route. There is a theoretical limit to the efficiency of this conversion (around 40%). In addition, the cost of electricity has a lower limit based on the capital expense of the steam turbine/generator. H-B11 fusion could be order of magnitudes cheaper.

ginko 5y ago
So if I understand this correctly this essentially creates an electric charge from initially uncharged particles. Wouldn't running a fusion like this lead to us accumulating more and more of a one-sided charge over time? I guess it doesn't matter so much on Earth, but for instance if you ran a fusion reactor like that on a space station. Wouldn't it become electrically charged?
- DennisP 5y ago
  No, there are still plenty of electrons. You start with hydrogen (one proton, one electron) and boron (five protons, five electrons). You end with three heliums (each with two proton, two electrons). The energy isn't extracted because of a build-up of charge. What happens is that each nuclear reaction is a tiny explosion. The helium nuclei shoot out really fast, and they're positively charged. You just need them to push through a magnetic field, and then you're getting electricity, like any generator moving magnets through a coil.
  - pfdietz 5y ago
    That is not HB11's proposed direct conversion scheme. HB11 proposes to have the alpha particles go up a large potential gradient, in effect charging a capacitor. I don't think their scheme works, btw. Having the ions move against a magnetic field doesn't change their energy. What would be needed for that scheme is to subject the ions to a changing magnetic field to produce an electric field that reduces their energy. Helion proposes something like this.
    - DennisP 5y ago
      Interesting, thanks. Would it be possible to use Helion's scheme for HB11 (assuming they manage net power)?
      - pfdietz 5y ago
        Since I also don't think their central idea works (it's off by a factor of 100) it hardly matters. pdf: https://arxiv.org/ftp/arxiv/papers/2012/2012.14533.pdf https://arxiv.org/ftp/arxiv/papers/2012/2012.14533.pdf
        DennisP 5y ago
        Ok fine, that's a separate issue, maybe you're right. But also maybe there's a chance that there's a way to make it work (as your link suggests at the end). It seems to me that the initial impetus, a small explosion of charged particles, is the same for both HB11 and Helion. So it doesn't seem like it matters if HB11's energy extraction doesn't work, as long as Helion's does.
grogers 5y ago
I feel like worrying about the cost/efficiency of the heat engine of a fusion device at this stage is putting the cart before the horse. 40% efficiency is really good when the reactants are practically free. IMO the reason that HB11 may actually succeed is that the reaction keeps most of the output energy in charged particles in the plasma, so it's more feasible to achieve true ignition than when neutrons carry most of the output energy away from the plasma.

Another advantage of hydrogen-boron fusion is that all the reaction products are charged. Therefore (theoretically) it would be possible to use *very* efficient

by bbojan 5y ago

ginko 5y ago

So if I understand this correctly this essentially creates an electric charge from initially uncharged particles. Wouldn't running a fusion like this lead to us accumulating more and more of a one-sided charge over time? I guess it doesn't matter so much on Earth, but for instance if you ran a fusion reactor like that on a space station. Wouldn't it become electrically charged?

DennisP 5y ago
No, there are still plenty of electrons. You start with hydrogen (one proton, one electron) and boron (five protons, five electrons). You end with three heliums (each with two proton, two electrons). The energy isn't extracted because of a build-up of charge. What happens is that each nuclear reaction is a tiny explosion. The helium nuclei shoot out really fast, and they're positively charged. You just need them to push through a magnetic field, and then you're getting electricity, like any generator moving magnets through a coil.
- pfdietz 5y ago
  That is not HB11's proposed direct conversion scheme. HB11 proposes to have the alpha particles go up a large potential gradient, in effect charging a capacitor. I don't think their scheme works, btw. Having the ions move against a magnetic field doesn't change their energy. What would be needed for that scheme is to subject the ions to a changing magnetic field to produce an electric field that reduces their energy. Helion proposes something like this.
  - DennisP 5y ago
    Interesting, thanks. Would it be possible to use Helion's scheme for HB11 (assuming they manage net power)?
    - pfdietz 5y ago
      Since I also don't think their central idea works (it's off by a factor of 100) it hardly matters. pdf: https://arxiv.org/ftp/arxiv/papers/2012/2012.14533.pdf https://arxiv.org/ftp/arxiv/papers/2012/2012.14533.pdf
      - DennisP 5y ago
        Ok fine, that's a separate issue, maybe you're right. But also maybe there's a chance that there's a way to make it work (as your link suggests at the end). It seems to me that the initial impetus, a small explosion of charged particles, is the same for both HB11 and Helion. So it doesn't seem like it matters if HB11's energy extraction doesn't work, as long as Helion's does.

grogers 5y ago

I feel like worrying about the cost/efficiency of the heat engine of a fusion device at this stage is putting the cart before the horse. 40% efficiency is really good when the reactants are practically free. IMO the reason that HB11 may actually succeed is that the reaction keeps most of the output energy in charged particles in the plasma, so it's more feasible to achieve true ignition than when neutrons carry most of the output energy away from the plasma.