Polywell = Navy Advantage

[Aero]

About half as much. That is some less, not a LOT less.

How do you figure that simon?

I'd say the majority of cost and complexity has less to do with direct shielding and more to do with safety, chemistry and control of activation products. Once you can literally turn off the reactor with a switch and kill all the radiation, it'll be an entirely different thing.

No Co-60 = No Nuke craziness.

This is a hypothetical but will give you the general idea.

1. 6" of concrete reduces neutron flux by a factor of 10
2. A nuke produces 1E12 neutrons/sq cm sec
3. A BFR produces 1E6 neutrons/sq cm sec
4. For safety the neutron flux should be no higher than 1 neutron/sq cm sec

To reduce the flux by a factor of 1E12 requires 12 X 6" of concrete
To reduce the flux by a factor of 1E6 requires 6 X 6" of concrete

It seems perverse that reducing the neutron flux by a factor of 1 million only cuts shielding requirements in half. But there you have it.

Interestingly, not all neutrons are the same, according to wikipedia,
http://en.wikipedia.org/wiki/Neutron_temperature

Neutrons from fusion reactions are usually considerably more energetic than 1 MeV; the extreme case is

deuterium-tritium fusion which produces 14.1 MeV neutrons (1400 TJ/kg, moving at 52,000 km/s, 17.3% of the

speed of light) that can easily fission uranium-238 and other non-fissile actinides.

Fast neutrons can be made into thermal neutrons via a process called moderation. This is done with a neutron

moderator. In reactors, typically heavy water, light water, or graphite are used to moderate neutrons.

Does anyone know the energy of the neutrons from the P-B11 reaction? Might there not be a neutron moderator that is more effective than concrete for these neutrons?

[Tom Ligon]

The few neutrons produced by p-B11 fusion are made by an assortment of stray pathways, and it would take a while to track down the energies. I once found a complete list of the possible reactions, and many of them won't occur to any significant degree in a Polywell because they require a plasma in thermal equilibrium, and presume the plasma is heated by the reaction products (which is deliberately minimized in a p-B11 Polywell as that would rob energy from direct conversion).

Absolutely, there are better neutron poisons than plain concrete. One of the best is boron-10. Interestingly, the nuclear industry refines boron to B10 routinely to make control rods, so there's a fair amount of the more common B11 available partially enriched as a byproduct. A B10-doped moderator (polyetheylene or water most commonly) will easily kill stray neutrons. A little boron in the concrete might be a good enhancement as well.

[KitemanSA]

Tom,
It is so true there are better (perhaps I should say "more efficient") shield matterials than concrete (though concrete is dang cheap). However, I think MSimon's point was that no matter WHAT material is needed, there would probably only be a ~50% reduction of it.

Of course, his assessment was predicated on roughly equal energy states. The effect of D-D vs D-T vs pB11 neutron energies would be interesting.

If you can find that sourse, I'd love to reference in the the FAQ-wiki.

[chrismb]

Also, as P-B11 is aneutronic, alot less shielding would be required.

About half as much. That is some less, not a LOT less.

I'm glad this point is beginning to take more prominence and get further recognition. Also, don't forget that the fast alphas from the p11B won't be so forgiving - neutrons will dissipate their energy throughout the bulk of the shield, whereas fast alphas and protons will selectively irradiate the upper most surface of the shield, potentially leading to surface ablations and contamination of the reaction volume.

p-11B is a lovely sales-pitch, just as when tokamak folks claim that DT fusion is 'clean'! The reality is that whatever way man-kind generates energy for its own use, the risks are proportionate to the degree of benefit gained. There is always harmful risk in generating power, and I don't think the 'proportinality' of risk varies that much. An essentially risk-free power source produces essentially tiny amounts of output.

The ease of use, and ease of availability, mean that [IMHO] DD will be the reaction choice of the future, were fusion power to be developed to viability - even surplanting DT due to the problems of sourcing and handling tritium. Deuterium is nature's little gift to us, if we're clever enough to exploit it. A stable and relatively common isotope in one of the easiest substances [water] we can get a hold of that is also quite willing to be 'fused' and has a very wide range of energies over which it has a good cross-section for fusion. And not forgetting the 3He and T products from that reaction, which then go on to have their own reactions with the D fuel.

[djolds1]

The ease of use, and ease of availability, mean that [IMHO] DD will be the reaction choice of the future, were fusion power to be developed to viability - even surplanting DT due to the problems of sourcing and handling tritium. Deuterium is nature's little gift to us, if we're clever enough to exploit it.

How far into the future, however? If QED/ARC gives us LEO, then accessing He3 is relatively "easy," and DHe3 is far more aneutronic than DD (if not to pB11 levels). Conditions for burning DHe3 in a thermal fusion machine are roughly comparable to those for burning DD; does the same apply for Polywell? And of course that leaves the trans-Orgasmatron level of burning He3He3 to aim for.

[chrismb]

DHe3 is far more aneutronic than DD

So, how do you avoid the DD neutron emitting reaction happening, if you're running D3He?

That reaction, D3He, is fine if you're doing a nice little cross-section-measuring experiment with a pure 3He target subjected to a deuteron beam. How on earth can you keep deuteriums apart if you then have to have deuterium and 3He all getting mixed up together, one way or another? 'Aneutronic D3He fusion power' is just fantasy talk.

In all cases, you'll need neutron shielding in any case. It is possibly preferable to have a dominantly neutron output as reaction-facing materials might not exist that can tolerate fast MeV ions/alphas.

So my point is - if you can get DD going, why would you really want to play around with any other fuel types when a) if you can get any other reactions going then you would be able to 'do' DD, b) deuterium is easy to get, c) you'll need neutron shielding anyway.

[djolds1]

DHe3 is far more aneutronic than DD

So, how do you avoid the DD neutron emitting reaction happening, if you're running D3He?

??? You don't. As I said, more aneutronic than DD, not as good as pB11. Plenty of DD side reactions. No such thing as perfection this side of heaven.

In all cases, you'll need neutron shielding in any case. It is possibly preferable to have a dominantly neutron output as reaction-facing materials might not exist that can tolerate fast MeV ions/alphas.

So what about the shielding? Par for the course. And as to materials, we know neutron embrittlement is a problem, whereas "ion toleration" is a bit on the speculative side, and ions can at least be somewhat directed via electromagnetic forces (unlike neutrons).

So my point is - if you can get DD going, why would you really want to play around with any other fuel types when a) if you can get any other reactions going then you would be able to 'do' DD, b) deuterium is easy to get, c) you'll need neutron shielding anyway.

DHe3 allows a large fraction of its energy to be captured via direct conversion. DD still requires the humongous thermal plants.

And regardless, the DHe3 cycle is merely the (I think) preferred engineering fallback if pB11 proves to be "difficult" for polywells. There's plenty of boron on this planet and in the solar system beyond it. Fuel availability will not be a concern for several millennia.

[D Tibbets]

D-D fusion I believe results in ~ 50% of the power being in the form of neutrons, the rest is in the form of charged particles. Concidering the conversion efficiency, just harvesting the charged particle's energy might give you as much as ~70%- 100%(?) of the total energy that could be extracted with a steam plant. So a steam plant could be ignored, and only direct conversion plant with a more robust cooling system that also serves as shielding might be feasable. The only minor problems would be building direct converters that could handle the wider range of charged particle energies while surviving the high neutron fluxes on them as well as the magrids.

If a D-H3 mix at ~1:10 (increases the probability of D- H3 reactions over D-D reactions) gives a 1% neutron yield (aneutronic by definition) then how much less shielding is needed compared to D-D fusion. If a million fold decrease in neutrons from P-B11 only reduces the shielding thickness by half, then I guess it would only be a small inprovement.

If you have direct conversion of alpha particles at 80% efficiency then the flux of high energy alphas reaching the walls would would be decreased by a similar amount, reducing the effects on the wall surface. Concidering the concentration effect at the surface of the alpha's compared to neutrons that would deposit thier energy over a thicker area, what would be the balance, what material would survive best under either condition? Since the high energy charged particles are comming out of cusps, the vunerable wall area would presumably only be a small percentage of the total. Special shielding in these limited areas may significantly reduce the operational and maintainance costs. Having most of the beams directed into vacuum ports with decelerater grids lining the walls might kill two birds with one stone.
Also considering the difference in surface bombardment by fusion products, rember that in addition to neutrons, D-D fusion machines would also have a lot of charged particles (P,T,He3) exiting the cusps and potentialy hitting the vessel wall, so you have a double wammy on the material of the wall, neutrons and charged particles.


Dan Tibbets

[MSimon]

Tom,
It is so true there are better (perhaps I should say "more efficient") shield matterials than concrete (though concrete is dang cheap). However, I think MSimon's point was that no matter WHAT material is needed, there would probably only be a ~50% reduction of it.

Of course, his assessment was predicated on roughly equal energy states. The effect of D-D vs D-T vs pB11 neutron energies would be interesting.

If you can find that sourse, I'd love to reference in the the FAQ-wiki.

It takes roughly 22 collisions with H2O (an important ingredient of concrete) to thermalize a fission neutron (2 MeV). It only takes 1 or 2 collisions to get a 14 MeV neutron down to 2 MeV.

So it will not change the shield size much.

In any case I did not plan to give a complete dissertation on the subject. Just an illustration of why my statement was essentially correct.

Note: B10 is only useful for thermal neutrons. So to make use of it in a shield you have to first slow the neutrons down.

So a shield might look like:

Steel plate to absorb generated X-Rays. About 6" to 8" of water to thermalize neutrons. Concrete to absorb X-Rays generated by n-B10 reactions and stray neutrons.

There is a way of putting all this together to minimize shield volume and mass. The Navy does this. Fortunately I forget most of the details - which were classified when I was doing Naval Nuke work.

For a terrestrial plant borated concrete may be the low cost way of shielding ($100 a cu yard). For mobile applications a higher cost to get lower mass/volume is in order.

[TallDave]

Also, as P-B11 is aneutronic, alot less shielding would be required.

About half as much. That is some less, not a LOT less.

Simon, are you basing this on Nebel's p-B11 reactor radiation estimates that he shared here a while back?

I was going to say that seemed incongruous, but now I see you already explained. A million times less neutrons, half as much shielding.

Now it just seems perverse.

[djolds1]

So a shield might look like:

Steel plate to absorb generated X-Rays. About 6" to 8" of water to thermalize neutrons. Concrete to absorb X-Rays generated by n-B10 reactions and stray neutrons.

There is a way of putting all this together to minimize shield volume and mass. The Navy does this. Fortunately I forget most of the details - which were classified when I was doing Naval Nuke work.

IIRC, tungsten polymer does well for absorbing x/gamma, and NASA is looking at Hydrogenated & Borated HDPE as a stand in for LH or H2O rad shielding.

[MSimon]

So a shield might look like:

Steel plate to absorb generated X-Rays. About 6" to 8" of water to thermalize neutrons. Concrete to absorb X-Rays generated by n-B10 reactions and stray neutrons.

There is a way of putting all this together to minimize shield volume and mass. The Navy does this. Fortunately I forget most of the details - which were classified when I was doing Naval Nuke work.

IIRC, tungsten polymer does well for absorbing x/gamma, and NASA is looking at Hydrogenated & Borated HDPE as a stand in for LH or H2O rad shielding.

Hydrogenated & Borated HDPE as a stand in for LH or H2O rad shielding.

This has already been done in some shielding applications. (A nod is as good as a wink)

[Roger]

(A nod is as good as a wink)

to a blind horse.

Rod Stewart.

[MSimon]

Also, as P-B11 is aneutronic, alot less shielding would be required.

About half as much. That is some less, not a LOT less.

Simon, are you basing this on Nebel's p-B11 reactor radiation estimates that he shared here a while back?

I was going to say that seemed incongruous, but now I see you already explained. A million times less neutrons, half as much shielding.

Now it just seems perverse.

The alternative is a million times as much shielding for a million times as much radiation (linear relation vs log). If it was linear nuclear power of any kind would be out of the question as a practical matter.