## Polywell FOIA

Point out news stories, on the net or in mainstream media, related to polywell fusion.

Moderators: tonybarry, MSimon

chrismb
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TallDave wrote:chrismb,

Now calculate it for D-T with a confinement of 100,000 passes
So, sticking with 1m single-flight fusor, you're implying that'd be at 1/(n.1E-20m^2)=100,000m => n=1E15/m^2 (around 1E-7 torr)

Distance to a fusion event (at DT peak drive) = 1/(1E15.5E-28 ) = 2E12m .

Set 2E7 64keV particles running and estimate 1 fusion in first 100,000 passes with loss of 50% of the input 64keV ions.

Energy out = 14MeV, energy in = 64keV x 1E7 = 640MeV. Efficiency = 0.02

bcglorf
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Joined: Mon Jul 23, 2007 2:58 pm

### Grids?

chrismb wrote:
MSimon wrote:
I've gone through the calculation of fusion cross-section before, to show you how you need, on average, millions of passes for a fusion event, even at the supposedly enormous pressures that will be found in a Polywell, let alone at the micron range of fusors.
That is a pretty good proof that fusors have never generated a single neutron ever. None of them gets millions of passes.
Not at all. It just shows that you are out into the multiple-sigma end of the distribution when generating neutrons, hence why it'll never get to break even.

Look at it this way - to get break even with a, say, 2.5MeV neutron energy capture from a cadre of 100keV deuterons, you'd need at least 1 in 25 to fuse to get Q=1.

The mean free path to a fusion event of a 100keV deuteron in 1 micron is 1/(1E19.1E-29)=1E10m. So if you need at least 1 in 25, so that they each then need to cover 1E10m/25=400,000km so that the total 'mileage' they've covered is the 1E10m total.

OK, now let's think of statistics and the distribution: If a fusor is sized to be 1m total flight per pass, we need 1E10 particles to set off to get the 1E10m total flight required, for a fusion event, in the first pass. In the meantime, the mfp for those deuterons to hit a background is around 1/(1E19.1E-20), or 10m. So we can estimate that this first 1m generates one single fusion event and knocks out 10% of the deuterons. 1E9 deuterons @ 100keV = 1E14eV. The fusion event liberates 2.5E6eV. So the efficiency would be (2.5E6/1E14)=2.5E-8.

Now.... let me guess.... what'd'ya think the efficiency of a fusor is????....
I'm wondering if your calculation isn't actually wildly optimistic. In your estimate for mfp to hitting a background, are you including the atoms that make up the grid inside the fusor in the density?

edited for severe typing mistakes.
Last edited by bcglorf on Wed Mar 10, 2010 10:25 pm, edited 1 time in total.

TallDave
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Sorry, I edited. The link suggests mfp to fusion of 2e8M rather than your 1e10M or 2e12M (higher density).

Either way, you can see you don't really need confinement in the millions of passes to get any kind of reaction rate, and it's well-established we're only getting tens of passes in gridded systems.
n*kBolt*Te = B**2/(2*mu0) and B^.25 loss scaling? Or not so much? Hopefully we'll know soon...

chrismb
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TallDave wrote: Either way, you can see you don't really need confinement in the millions of passes to get any kind of reaction rate, and it's well-established we're only getting tens of passes in gridded systems.
Never said you did. I indicated you need confinement in millions of passes to get Q>1. That's what we were talking about, Q>1.

TallDave
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Well, what you actually said before was:
It needs MILLIONS of passes just to summon up a 'reaction rate', let alone before you even begin to measure efficiency!
I agree that in terms of Q>1 your statement is more reasonable. OTOH, if we accept "millions of passes" for a fusion (which may be overly optimistic) as the paper alludes to then a necessary fraction of 1.6% gives us tens of thousands at Q=1 for D-T. So I think Simon's statement is defensible, if optimistic.
Last edited by TallDave on Wed Mar 10, 2010 11:19 pm, edited 1 time in total.
n*kBolt*Te = B**2/(2*mu0) and B^.25 loss scaling? Or not so much? Hopefully we'll know soon...

chrismb
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Joined: Sat Dec 13, 2008 6:00 pm
TallDave wrote:Well, what you actually said before was:
It needs MILLIONS of passes just to summon up a 'reaction rate', let alone before you even begin to measure efficiency!
I agree that in terms of Q>1 your statement is more reasonable. OTOH, if we accept "millions of passes" for a fusion then a necessary fraction of 1.6% gives us tens of thousands at Q=1 for D-T. So I think Simon's statement is defensible, if optimistic.
That's right. As shown in the calc above. It may be one ion on millions of passes, or millions of ions on one pass. Either way it is millions of passes. It's not like you can tell one ion from another, they don't have names, or registration plates, or stuff like that...

One ion on one pass ain't gonna do it.... Nor is 10,000. >Millions, then you might detect it.
Last edited by chrismb on Wed Mar 10, 2010 11:21 pm, edited 2 times in total.

TallDave
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Ah, I believe I see the source of confusion. Simon was talking about number of passes in terms of confinement, you're talking about the number of passes for fusion.

You (maybe) don't need to confine ions for an average of millions of transits to get Q>1, you do need millions of ion transits to get a fusion.
Last edited by TallDave on Wed Mar 10, 2010 11:24 pm, edited 1 time in total.
n*kBolt*Te = B**2/(2*mu0) and B^.25 loss scaling? Or not so much? Hopefully we'll know soon...

chrismb
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Joined: Sat Dec 13, 2008 6:00 pm
Sure. That seems to be the case. I would suggest confinement times of the ~order of a few second are the required boundary to approach net fusion in IEC type devices. 10,000 passes is an ms, which is, I guess, whereabouts Polywell is. My device appears to be around 0.1s in its current trim.
Last edited by chrismb on Wed Mar 10, 2010 11:38 pm, edited 1 time in total.

TallDave
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I would say ions are probably confined for tens of millions of transits in a Polywell, because of the well + magnatic field (the ion pressure is going to be low at the edge, no hotspots on the wall). Most of the losses should be from electrons.
n*kBolt*Te = B**2/(2*mu0) and B^.25 loss scaling? Or not so much? Hopefully we'll know soon...

KitemanSA
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Location: OlyPen WA
SEVEN

One week and counting.

chrismb
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KitemanSA wrote:SEVEN

One week and counting.
I hope you already have a plan for displaying negative/overrun days!!
Last edited by chrismb on Fri Mar 12, 2010 2:28 pm, edited 1 time in total.

KitemanSA
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Location: OlyPen WA
In theory:

SIX DAYS

or fewer.

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Joined: Thu Sep 17, 2009 11:18 pm
Location: North East Coast
The nice lady at NAVAIR Weapons who is my FOIA coordinator has confirmed to me yesterday that I will get a letter by 18 March. Whether it will be postive or negative or something in between remains to be seen. She did not indicate either way, as I am sure she cannot until it is done.

In any event, because Mr. Bradley asked nicely, and Mr. Carlson put in an appearance after a long absence (but did not ask nicely), I have a check set aside to mail out to NAVAIR if required.

We will get an answer in time, whether we will like it or not remains to be seen.

Oh, they have checked out this board, and probably peek at it to see how restless we are getting because they think it is funny. (I would).

BenTC
Posts: 410
Joined: Tue Jun 09, 2009 4:54 am
ladajo wrote:Oh, they have checked out this board.
Thats good. It would be nice if a peek at the size of membership list was sufficient to satisfy the public interest requirement.
In theory there is no difference between theory and practice, but in practice there is.