3 ms·
The reactor would only be about 5MW, and the wall of the reactor core can be however far away it needs to be to not get melted by x-rays. It's not like it's sur
by DennisP 7y ago
The reactor would only be about 5MW, and the wall of the reactor core can be however far away it needs to be to not get melted by x-rays. It's not like it's surrounded by superconductors or breeding blanket, it's just vacuum containment. Generally it's neutron radiation that's considered difficult to deal with, and there'd be little of that.
And the reactor won't produce net power at all unless they're right that bremsstrahlung will be suppressed long enough by the plasma's magnetic field. If it is, then most of the energy will go into the alpha particle beam.
- pfdietz 7y agoThe reactor is described to be 5 MW average output. But if it's very bursty the instantaneous power will be far higher. Damage to the wall will be determined by heating of the surface by these flashes (and possibly mechanical damage from shocks induced by vaporization in that surface layer). Keeping the instantaneous power low enough to avoid damage could make the reactor unacceptably large.
- DennisP 7y agoSo your criticism is that maybe, possibly, the reactor will be too big. I'll take it.
- pfdietz 7y agoThat is a central problem with fusion: the volumetric power density of fusion reactors sucks. It's not a trivial criticism, it's an economic showstopper. LPP's concept has other problems, of course, like the aspirational physics of those putative 1000 T plasma knots, and direct conversion by Handwave Engineering.