dodecahedron

[Torulf2]

From side reaktions as.
11B + α → 14N + n + 157 keV
and
11B + p → 11C + n - 2.8 MeV


http://en.wikipedia.org/wiki/Aneutronic_fusion

[blaisepascal]

There seems to be objections being made to the elliptical toroid cross-section idea of Colonel_Korg.

Why weren't those same objections made in June, in response to viewtopic.php?t=552 where a similar idea (also involving oval toroid cross-sections to minimize alpha-impinging) was discussed?

[blaisepascal]

There seems to be objections being made to the elliptical toroid cross-section idea of Colonel_Korg.

Why weren't those same objections made in June, in response to viewtopic.php?t=552 where a similar idea (also involving oval toroid cross-sections to minimize alpha-impinging) was discussed?

[MSimon]

The idea that a non-thermalized plasma will not produce side reactions is incorrect.

The energy distribution only affects the probability. It may reduce the rate several more orders of magnitude. It will not go to zero.

[MSimon]

The idea that a non-thermalized plasma will not produce side reactions is incorrect.

The energy distribution only affects the probability. It may reduce the rate several more orders of magnitude. It will not go to zero.

I agree Simon. Hopefully careful design will reduce the secondary effects to something that regular neutron absorbing materials can cope with with out inducing any long lived radioactivity or damage to the reactor components. That would fulfill Dr Bussard's vision of clean fusion power and make having one of these near by not something the NIMBY folks will object to.

Paul

Paul,

Even if we reduced neutron generation ten orders of magnitude from D-D (a very tall order) the reactor could not operate unshielded. So the NIMBYs will still find something to complain about.

[rnebel]

We've done the calculations. Neutron yield from a P-B11 Polywell machine (nonthermal) is about 1.0e12/sec. for a 100Mwe reactor. That's about 8 orders of magnitude less than a comparable D-T machine.

[MSimon]

Let us assume a machine 6 m dia, 3 m radius for the reactor.

Area = 4 pi r^2 = 4 * pi * 300 * 300 (cm) =1130973.35 sq cm

with 1 E12 n/sec = 884194.12 n/cm^2 /second. Assume a shield at 5 m

aprox 320k n/sq cm to reduce that to 1 n/sq cm with 6" shield = 1/10th reduction is a shield about 2 3/4 ft thick. Say 3 ft thick to get well under 1 n/ sq cm at the surface.

It is still not a basement reactor.

[MSimon]

Of course that assumes concrete.

If the neutrons get thermalized with a water jacket of 6" or 8" and then absorbed with elemental boron or a borax sludge it could be thinner.

The NRC will still want to keep an eye on you.

[Barry Kirk]

Of course that assumes concrete.

If the neutrons get thermalized with a water jacket of 6" or 8" and then absorbed with elemental boron or a borax sludge it could be thinner.

The NRC will still want to keep an eye on you.

Would that elemental boron or borax sludge by Boron 10 or 11? What does Boron do when it absorbs neutrons? Maybe the shielding could be made of "depleted Boron 10" leftover from enriching the Boron to make the Boron 11 fuel.

[93143]

Would that elemental boron or borax sludge by Boron 10 or 11?

Boron 10.

What does Boron do when it absorbs neutrons?

It, uh, fissions and gives off 0.5 MeV gamma rays... Now for each 1/10 reduction you need a centimetre of lead... or 4" of concrete. If you want the same total order of magnitude reduction, you've only saved between 3 and 5 inches if you go with concrete, although lead would be smaller.

I don't know whether you'd necessarily need that, because I'm not a nuclear engineer.