How do WB-6 results scale into power plant @ rx10?

[TallDave]

If, as pstudier points out, the drive power was 175,000 watts (drive voltage is 12.5 Kev at a current of 14 amps) and the output of 1E9 neutrons is a gain of .001 watts (1E+9 neutrons x 17.6 MeV per fusion x 1.6E-19 eV per joule = .00282 joules/second / 2 neutrons per fusion = .00141 watts), then even at r^5 scaling how do you get to 100MW at rx10?

Moving the exponent 5 places,. I get 141 watts at rx10, and it takes radiusx38 to get to 100MW.

Was there some other factor involved, or did I make a mistake somewhere? Or was rx10 just an approximation?

Or was it the P-B11 reaction he was talking about? That's probably more energetic.

[TallDave]

Oops, no, I think maybe we have the wrong number there. The 1e9 would be the fusion power, not the gain, wouldn't it? The gain would be the power minus the loss, but obviously no attempt was made to determine the loss. So using the r^7 scaling for power, I get 14MW power @ rx10, and we get to 100MW at about rx14. I'm not sure what the gain is, because of course we have no loss numbers for WB-6, but we could infer from r^7 - r^5.

So he wasn't too far off.

[scareduck]

Wikipedia says D-D fusion generates a neutron in 1/2 the outcomes. Shouldn't each neutron represent two fusion events, not one of two fusion events? If so, there are two fusion types, T+p and He-3+n, generating 3.02 MeV and 2.45 MeV; per neutron, this means we ought to be able to total both and say we got 5.47 MeV. The power calculation should therefore look like

P = 1e9 neutrons/s * 2 fusions/neutron * 5.47 MeV/fusion * 1.6e-19 J/eV = 1.75 mW.

That means Q=1e-8.

[TallDave]

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

Any of the reactions above can in principle be the basis of fusion power production. In addition to the temperature and cross section discussed above, we must consider the total energy of the fusion products Efus, the energy of the charged fusion products Ech, and the atomic number Z of the non-hydrogenic reactant.

Specification of the D-D reaction entails some difficulties, though. To begin with, one must average over the two branches (2) and (3). More difficult is to decide how to treat the T and ³He products. T burns so well in a deuterium plasma that it is almost impossible to extract from the plasma. The D-³He reaction is optimized at a much higher temperature, so the burnup at the optimum D-D temperature may be low, so it seems reasonable to assume the T but not the ³He gets burned up and adds its energy to the net reaction. Thus we will count the DD fusion energy as Efus = (4.03+17.6+3.27)/2 = 12.5 MeV and the energy in charged particles as Ech = (4.03+3.5+0.82)/2 = 4.2 MeV.

It looks like the average is 12.5MeV per fusion, but I'm not sure what ratio they occur in, so I'm not sure how to calculate a per-neutron energy.

(1) D + T ? 4He (3.5 MeV) + n (14.1 MeV)
(2i) D + D ? T (1.01 MeV) + p (3.02 MeV)
(2ii) ? ³He (0.82 MeV) + n (2.45 MeV)
(3) D + ³He ? 4He (3.6 MeV) + p (14.7 MeV)

[TallDave]

Oh wait -- your link tells us:

Though more difficult to facilitate than the deuterium-tritium reaction, fusion can also be achieved through the reaction of deuterium with itself. This reaction has two branches that occur with nearly equal probability:

D + D → T + p
→ ³He + n

[TallDave]

So....

1/2 the time we get a neutron and 3He, half the time we get T. The T presumably fuses (it's a lower energy reaction), giving another neutron per above table.

So... for any 2 neutrons, there was probably a (2i), a (2ii), and a (1). We ignore (3) because the energy for that reaction is much higher than D-D.

So we can add (2i), a (2ii), and a (1), divide by 2, and probably get close.

3.5 + 14.1 +1.01 + 3.02 + .82 + 2.45 = 24.9 / 2 = 12.45 MeV fusion power produced per neutron produced.

[Stefan]

1/2 the time we get a neutron and 3He, half the time we get T. The T presumably fuses (it's a lower energy reaction), giving another neutron per above table.

In WB6 only a very small fraction of the fuel ions fused. A similarily small fraction of the T ions might have fused, but the amount of T ions was tiny and they were at 1.01 MeV (thus not likely to be confined).

So this leaves us with one (2i) and one (2ii) reaction per neutron.

[MSimon]

1/2 the time we get a neutron and 3He, half the time we get T. The T presumably fuses (it's a lower energy reaction), giving another neutron per above table.

In WB6 only a very small fraction of the fuel ions fused. A similarily small fraction of the T ions might have fused, but the amount of T ions was tiny and they were at 1.01 MeV (thus not likely to be confined).

So this leaves us with one (2i) and one (2ii) reaction per neutron.

That would make a D-D reactor a tritium producer. That could be useful.

[scareduck]

That would make a D-D reactor a tritium producer. That could be useful.

I would tend to think, though, that the original quote was correct: in a reactor that's even close to breakeven, you still end up consuming the tritium as quickly as it's created because it's born energetic, and that reaction takes much less energy to create than a D-D reaction. Assuming a Polywell device were workable, it seems to me that using D-D fusion to bombard a lithium blanket would be a better way to create tritium.

Edit: Of course, if you were able to harvest the tritium, that would make Polywell a nuclear proliferation risk.