Reply To: YouTube Focus Fusion Video

[Brian H]

jamesr wrote: I would expect the B10 to be dissolved in the water sheilding as boric acid. So any neutrons produced are slowed down (or moderated in nuclear terminology) to thermal speeds. They then, while bouncing around in the water get captured by a B10. The resulting lithium will not travel very far in water, as it is charged and slow down within a 10-100 nm. Heating up the water a little in the process (similarly for the helium). The lithium will react with the water producing lithium hydroxide. The boric acid and lithium hydroxide can then react, as any acid and base do leaving a salt, lithium borate in this case. If the concentration of the salt gets too high it can precipitate out – taking some of the boron out of the bulk of the water and reducing the chance of neutron capture slightly.

However, given the low levels of neutrons produced, I wouldn’t expect the shielding water to have to be filtered & topped up with fresh boron for at least 5 years if not the life of a reactor.

Separating Boron into B-10 & B-11 is done on an industrial scale already for conventional nuclear plants. I’m not sure what the cost is but it shouldn’t be significant in the whole scheme of things.

Since LPP would be purchasing decaborane, I assume it would be B11 from the source. The B10 would be acquired separately.

What is the penetration and intensity overall of the gamma radiation? Assuming the B10 to be an external shell around the water, would it be a danger in the surrounding service space? Since the =<12s decay time you give for the B10 is so brief, I would assume that the gamma would be produced only while the core was active, and a short period afterwards (a few minutes to subside to insignificant levels).

P.S. I didn’t mean to imply that the shell in contact with the water was B10; actually, I believe it’s exterior to the water container — a shell around the shell, as it were. But don’t quote me. 😉 :cheese: