Sputtering From Alpha Impacts

[TallDave]

FWIW, part of the tokamak solutiom was redeposition, but I'm not sure what the exact mechanism is and it probably doesn't apply in a Polywell.

Too bad we can't put tiny particle accelerators on everything and generate current with them.

[MSimon]

FWIW, part of the tokamak solutiom was redeposition, but I'm not sure what the exact mechanism is and it probably doesn't apply in a Polywell.

Too bad we can't put tiny particle accelerators on everything and generate current with them.

Redeposition does apply. The question is how to maintain the balance between deposition and evaporation.

[TallDave]

Hmm, Art Carlson disagrees, fwiw:

I'm afraid sputtering is too complicated, so I can't do much more than say it seems to me likely that the sputter yield will be much greater than unity. Sputtering is indeed a problem that has received a lot of attention in tokamaks, but they differ from polywells in many respects. For one thing, the alphas in tokamaks are contained by the magnetic field long enough that they cool down to the temperature of the plasma, which is only a few eV next to material surfaces. For another, the erosion is to a large extent countered by redeposition, which would be nearly absent in a polywell.

http://en.wikipedia.org/wiki/Talk:Polywell

But again, I don't know the mechanism so I'm not sure why he thinks it wouldn't apply. Something to look into.

[MSimon]

Redeposition would certainly apply to neutralized B11.

It is a question of balance.

It may be that the reactor has to be shut down as a power producer periodically and B11 deposited to continue operation.

There is so much we don't know.

[TallDave]

Hrm. This does seem to fall into the "no way of knowing" category. Some possibly useful links.

Deuterium retention of DIII-D DiMES sample

long url

Erosion and redeposition of wall material in controlled fusion devices

nother long url

Erosion and Redeposition in ASDEX Upgrade

Erosion and redeposition of boron, carbon and tungsten in the divertor of ASDEX upgrade are examined using marker tiles. In the outer divertor, both carbon and tungsten are eroded; the inner divertor, on the other hand, shows heavy deposition of boron and carbon. The boron originates from the main chamber walls, where it is deposited during regular boronizations for wall conditioning.

http://www.ipp.mpg.de/ippcms/eng/for/be ... a2mig.html

"Boronization." Heh.

Sputtering seems to be a main concern of this group:

http://en.wikipedia.org/wiki/Internatio ... n_Facility

So, do the boron ions get neutralized at some point? Or are we talking about injecting neutrals?

[TallDave]

Hey, a sputtering modeller!

http://www.geocities.com/karolewski/kalypso.htm

Too bad everything seems to deal almost exclusively with neutron sputtering. On the plus side, we probably wouldn't have to wait til 2017 to be able to do materials testing for alpha sputtering.

Maybe they can exhume Litvinenko for us.

[scareduck]

Interesting side note from the Talk:Polywell page linked above, by an anonymous commenter, regarding the bunching effect in the core that is supposed to anneal the ions there and keep them from thermalizing (lightly edited to get rid of some grammatical tics):

Hi Tom, that explanation seems a bit lacking to me. In particular I can't get it to add up in terms of entropy. Collisions will necessarily increase the overall entropy of the ion distribution function, and this remains true no matter if they occur in the high density region or out at the perimeter. Now, if a process, or a set of processes, have as the net effect to restore the non-maxwellian velocity distribution, then it follows directly from the second law of thermodynamics that it must require an amount of work corresponding to the negative change in entropy. Now, unless you use D-D fuel (or another fuel consisting of only one ion species) you will necessarily have collisions which have a neglectable chance to contribute to the fusion process. For the p-B plasma you have p-p collisions as well as B-B collisions as an example. For monoenergetic ions the resulting energies after a collisions would range from 0 to 2E_0, where E_0 is the original energy of the ions. It thus appears to be absolutely impossible that these ions will all reach close to the same potential energy without a large input of work to compensate for the decrease in entropy that this would require. This is of course particularly true for ions which have collided in the core, as their kinetic energies before the collisions will be the highest.

Furthermore, it is obviously impossible for the average kinetic energy of the ions to decrease as they "Maxwellianize" (as you call it ) since conservation of energy would require that the potential energy is increased accordingly. For this to occur the ions would have to spontaneously convert their average kinetic energy into potential energy, and if this is to yield a monoenergetic distribution of potential energies you would most certainly see a large decrease in ion entropy without any corresponding input of work. In summary, it would appear to me that any spontaneous process which tends to counteract thermalisation of the ions, would either have to result in a large heat loss, or require the corresponding input of work.

Unless I'm mistaken, this is a restatement of the major thrust of the Nevins paper: unless the ions fuse in the first pass through the core, most will either thermalize to a lower energy level and won't be able to fuse at all, or they'll get to a higher energy level and be lost from the system, with some fraction staying in the sweet spot. The anonymous commenter's point here seems to be that if the annealing process does work as described, it should require a good bit of energy to maintain, or else it should throw off a lot of waste heat. The real question is, is this a major loss mechanism?

[scareduck]

And, BTW, TallDave, could you repost those links as tinyurl.com links so the page doesn't get ridiculously wide? Thanks.

[93143]

Google "gombosi" "kinetic description of shocks" and check the first result (it should be a Google Books sample). VERY interesting graph.

The upshot is that the second law of thermodynamics doesn't apply to an individual component of a system unless it's in thermodynamic equilibrium. You have to take the system as a whole.

[dch24]

Outside of plasma physics (which I don't know much about) wouldn't the positively charged coils lose energy to the upscattered ions? And the magnets would be losing energy to form the well, also?

[TallDave]

The anonymous commenter's point here seems to be that if the annealing process does work as described, it should require a good bit of energy to maintain, or else it should throw off a lot of waste heat. The real question is, is this a major loss mechanism?

I think that comes from Paul Dietz. But it's very easy to point to situations in which disorder is decreased without work being performed. It's just a question of where the lowest energy lies.

For instance, picture some ball bearings all nicely ordered in a row at the bottom of a V. Now come along and smack some of them. They will become disordered, rolling up and down the sides of the V, but in the end they all come to rest at the bottom again, and you didn't have to do any work to make it happen.

Or consider some iron filings next to a magnet, ordered along its field lines. You push some of them a little bit, and they become disordered, but they move back to where they started without any work being performed because that's their lowest-energy equilibrium point.

Essentially, the gravitic/magnetic field acts as a buffer. Disordering them from their lowest-energy equilibrium takes work, and that work creates energy imbalances, which then force them back to their ordered equilibrium state. So the disordering provides the energy needed to reorder them.

So replace "ions" with bearings/filings and you can see that this:

the ions would have to spontaneously convert their average kinetic energy into potential energy, and if this is to yield a monoenergetic distribution of potential energies you would most certainly see a large decrease in ion entropy without any corresponding input of work.

can easily happen in the right circumstances.