Hacker News new | ask | show | jobs
by WorkLobster 677 days ago
> you would need 100% efficiency to be able to breed your own tritium

How so? As I understand it, the breeding reaction is exothermic (Li-6 + n -> He-4 + T) so all you need are neutrons, for which (as someone pointed out to me two years ago[1]) neutron multipliers would be used.

[1] https://news.ycombinator.com/item?id=32224650

1 comments

I was talking about the neutron efficiency, not the thermodynamic efficiency.

As for neutron multipliers, they are currently so far from being realistic that they are essentially science-fiction.

I'm not sure I understand what's 'science-fiction' about lead or beryllium?
Here's some articles about the neutron multipliers. You can go through them and see that they are all talking about feasibility and theoretical considerations. It's not just as simple as shooting some neutrons at a block of lead of beryllium on one side and more neutrons come on the other side.

[1] https://www.sciencedirect.com/science/article/abs/pii/S00223...

[2] https://scipub.euro-fusion.org/wp-content/uploads/eurofusion...

[3] https://publikationen.bibliothek.kit.edu/1000160103

No, I agree, but you specifically mentioned "science-fiction" in the second post, and implicitly denied even their existence in the first post. I think there's a reasonable distance between that, and economic considerations / the fact that test hardware has yet to exist that's presented in those papers.