Talk-Polywell.org

I knew that the 2009 numbers are off, I was more interested in the older contracts. They have also not posted any numbers for 2010. Given that the funding from last year should cover the current testing, one would expect 2010 to be $0.

The history numbers seem to jive with what we know from before. In any event, I thought it was a nice summary of the funding to date. Although it does not cover some of the early years funding before EMC2 became an independant vehicle.

This remains a Navy Contract, and although the funding line source is from the stimulus bill, it is still controlled by the navy. It did not impact the navy budget, and as such was a smart move by ONR. It remains that the navy can terminate the project at will.

This is the data from EMC2 itself, except for the post 2007 which I got from the FBO.

http://www.ohiovr.com/polywell-faq/inde ... results%3F

Probably of more concern is the rate of recombination, neutral losses. At the voltages for PB11 reactions I suspect this number would be quite small, but compared to the fuel ion leak rate, it may be significant.

I don't know. Boron recombination seems even less likely than D recombination, because a boron ion has to find five very cold electrons, and acquire them all without running into a faster electron or ion ... and the combined energy of the boron ion and last electron has to be <8eV! That's a very tall order even if the electron distribution is Maxwellian.

Maybe Joel or someone can do a simulation, but at a glance it doesn't seem likely to be a detectable loss.

Axil wrote:“meaning you're saving hundreds of tons on shielding.”

Concrete and dirt are cheap.

I recall large reductions in radiation resulting in trivial shielding savings anyways. One of the many reasons I'm much more skeptical of p-B11 reactors than of D-D/T. But Rick's got some tricky ideas.

Aneutronic reactions shielding savings are less than might seem resonable on first glance, but still it is considerable. To stop a certain amount of radiation requires a certain amount of material. This amount is proportional to 1/10th. If one inch stops 1 unit of radiation, then 2 inches stops 10 units, 3 inches stops 100 units, etc.
If p-B11 produces 1 unit of neutrons, and D-D produces 100,000,000 units, then the radiation shielding ratio would be 1 : 9 (I think). So, if 10 tons of material was needed to shield a P-B11 reactor, 90 tons would be needed for a D-D reactor. If the gamma rays are comparable to neutrons in the radiation damage effects the sheilding savings may be ~ 1:4. Still a significant weight savings for a mobile application.

Also need to consider the inverse square law. If the initial radiation is smaller, the distance needed to acheive the same net flux is reduced.

Also, need to consider the contribution of conversion efficiency to useful electrical power. Direct conversion may give the same useful power at ~ 1/3 the fusion (radiation producing) rate.

And, if in a space application, the largest advantage may be the direct conversion, with the huge weight savings due to less waste heat that has to be dissipated.

Dan Tibbets

D Tibbets wrote:Aneutronic reactions shielding savings are less than might seem resonable on first glance, but still it is considerable. To stop a certain amount of radiation requires a certain amount of material. This amount is proportional to 1/10th. If one inch stops 1 unit of radiation, then 2 inches stops 10 units, 3 inches stops 100 units, etc.

Simply: NO!
It is linear, not exponential. 10" stops only 10x 1".

The problem is that each type of radiation is stopped differently, the "alpha" of a material is in relation to each decay type.

So, alpha particles are stopped with a few millimeters, but gamma radiation takes a significant amount of lead.

Dang it. I am going to have to throw out the whole idea of "tenth thicknesses".



Oh wait, this is the internet...

I guess I'll go back to thinking about hydrogenous materials and their effectiveness for particle shielding.

D Tibbets wrote:If one inch stops 1 unit of radiation, then 2 inches stops 10 units, 3 inches stops 100 units, etc.

I would say it differently: If one inch reduces the radiation by a factor of 10, then the second inch also reduces it by a factor of ten, so the two together reduce it by a factor of 100. Three inches would reduce it by a factor of 1000.

Or, instead of "stops" just write "reduces to a tolerable dose".

If you're really worried about the amount of radiation absorbed in different parts of the shield (say, if you're looking at the heat load), and 1 unit was absorbed in the first inch, then 1/10 unit will be absorbed in the next inch, etc.

All this neglects activation.

ladajo wrote:Dang it. I am going to have to throw out the whole idea of "tenth thicknesses".



Oh wait, this is the internet...

I guess I'll go back to thinking about hydrogenous materials and their effectiveness for particle shielding.

Well, looking into my copy of "Fundamentals of Nuclear Science and Engineering" I see that the mean penetration distance is actually at a rather precipitous slope from almost complete penetration to almost complete stopping for ionized particles. Gammas are another matter entirely, it seems.

Ionizing radiation stopping power is measured in MeV/cm; but looking at the Bragg curves, it's somewhat odd. For instance, a 2.5MeV beta is stopped by ~.00015 cm H2O, but a 20MeV beta needs ~.035cm H2O.

Generally, it seems, charged particles are considered to give up a portion of their energy per unit length traveled through a medium.

Neutrals, of course, just have to collide with a nucleus of an atom to stop; that's clearly a (statistically) linear interaction. The term used is "half thickness" as each thickness stops half, as one would expect for a probabilistic decay.

Photons (gamma radiation) interact by Compton or Raleigh scattering and photoelectric absorption. Extremely high energy photons (>1.9MeV) can even produce a positron-electron pair. These all combine, and give a rather complex curve, but it's measured in mu/rho (cm^2/g).

Alphas and betas behave non-linearly in shielding materials. I recall beta "stopping power" showing a fairly steady loss of energy until just before the particle stops, then a massive burst.

http://www.kayelaby.npl.co.uk/atomic_an ... 4_5_3.html

Alphas and betas tend to be easy to stop, but should be stopped in low-Z materials to avoid x-ray production. The link above shows stopping power includes a bremsstrahlung component in which high energy charged particles in high Z materials produce way more brem. Give 'em a layer of plastic a few mm thick, and be prepared to replace it. Lead is a terrible shield for these because it will spew x-rays.

Neutrons typically are shielded with a combination of low-Z moderator and a poison, such as the left-over Boron 10 from refining our Boron 11.

For gammas, our shortcut was the "half-value thickness". For a given radiation energy, you found the thickness necessary to cut the dose in half, then added enough layers that the factors of two reduction got you down to what you needed. You could work the problem the same way with factors of ten, but factors of ten are a bit heavy-handed for this purpose.

If you need ten inches of shielding to cut down by a factor of ten, one inch is pretty useless. The example that set this discussion off used an inch, but without any specific idea of the radiation and shielding material really involved the units might as well be kilomousehairs.

The shielding savings were also for a nitrogen reactor, which supposedly has no gamma, though I suppose it'll have a similar amount of neutron radiation. Where do the neutrons come from again?

There was something about the nitrogen reaction that produced a lot less radiation than even boron, which is where I originally got the idea. Spaceships and airplanes will need the weight savings, and can probably deal with the more complicated fuel source and cycle just to get those savings.

Was there supposed to be something happening with EMC2 this month, or was it in the spring?

Actually, WisWom, radiation shielding is exponential.

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

"Shielding reduces the intensity of radiation exponentially depending on the thickness."

Jrvc stated it better than I did.

If 1 inch of shielding stops half of the incident radiation the second inch will be exposed to 1/2 of the initial dose and it will again stop 1/2 of the incident dose (or flux)., so the exiting flux would be 1/4th of the original. A third inch of shielding would be exposed to this 1/4th of the original dose, and would again block 50% of this flux so that the exiting flux would be 1/8th of the original.
Using incremental thicknesses that would stop 90% of the radiation would grow more obviously but it is still the same exponential growth.

Radiation shielding needs to stop the most penetrating dangerous radiation. In nuclear fallout this is gamma rays. Anything that stopped the gamma rays (reduced the dose to tolorable levels) would also stop most of the neutrons (unless they have unusually high energies) and their secondary radiation*. Any less penetrating radiation like X-rays, alpha, and Beta radiation are ignored as they are easily stopped in any system that is required to protect from gamma and neutron radiation. If you are talking about local heat loads, secondary contamination of the enviornment, etc. the picture becomes more complex.

* The secondary radiation (or induced radiation) from neutrons can be a problem- eg:neutron bombs. Of course they are also a problem in conventional nuclear bombs, but because of the conventional bombs greater relative blast effects, you are pulverized anyway if you are close enough for the neutron dose to kill you.

Dan Tibbets

WizWom wrote: .....
, but it's measured in mu/rho (cm^2/g).

PS to my prevous post.
You just countered your earlier 'linier' contention with your formula

cm^2 - that is exponential !

It's like the inverse square law for electromagnetic radiation. As the distance doubles (in this case it is the doubled absorbing shielding material that is standing in for distance) the intensity of the radiation drops 2^2 or to 1/4th of the original. Actual inverse square dissipation of the radiation is ignored in this case as the thicknesses of the shielding material is assumed to be much less than the doubling of the distance from the radiation source.

Dan Tibbets

Polywell power plant.
16 sec - 13 Aug 2010
www.emc2fusion.org www.talk-polywell.org
photobucket.com - Related videos