Magnetically Shielded Fusor Grids--Why Won't This Work?
Arguments in this thread are approaching the disorganization of my submissions. Perhaps brousing chapter 3 (starting on page 35) of the below book will give a more consistant appreciation of the importance of things like cross section, mean free path, density, reaction frequency, etc in determining the final fusion power. The book only deals with thermalized plasma, so things like convergence is not considered.
long book url
If anyone is interested about some information about glow discharges-
http://www.glow-discharge.com/
chrisMB feels that the "Star" mode in gridded fusors is due to a subtractive survival mechanism. I believe it is more of a focusing mechanism. As the pressure drops first 'bugles' are seen, then an electron beam, which then fades till "Star" beams are seen through most if not all of the grid openings.
At least in my experiance the beams at the various pressure regions appear as quickly as the voltage is delivered. The appearent delay is presumeably due to the charging time of the capaciters, if present. Also, the active plasma heats the gas in the chamber so that the pressure drops somewhat (presumable because increased speed of the particles allow more rapid pumping in this molecular flow vacuum pumping region and/ or some of the ions and accelerated neutrals are embeded in the walls).
If a few seconds is needed for the development of the visual Star mode due to survival physics only,it would not make any sense as the time frame for the ion survival lifetimes (even those that are preserved longer in the Star beams) are probably much shorter than a millisecond.
Dan Tibbets
long book url
If anyone is interested about some information about glow discharges-
http://www.glow-discharge.com/
chrisMB feels that the "Star" mode in gridded fusors is due to a subtractive survival mechanism. I believe it is more of a focusing mechanism. As the pressure drops first 'bugles' are seen, then an electron beam, which then fades till "Star" beams are seen through most if not all of the grid openings.
At least in my experiance the beams at the various pressure regions appear as quickly as the voltage is delivered. The appearent delay is presumeably due to the charging time of the capaciters, if present. Also, the active plasma heats the gas in the chamber so that the pressure drops somewhat (presumable because increased speed of the particles allow more rapid pumping in this molecular flow vacuum pumping region and/ or some of the ions and accelerated neutrals are embeded in the walls).
If a few seconds is needed for the development of the visual Star mode due to survival physics only,it would not make any sense as the time frame for the ion survival lifetimes (even those that are preserved longer in the Star beams) are probably much shorter than a millisecond.
Dan Tibbets
To error is human... and I'm very human.
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Re: a brief thought experiment
This is essentially a minimalistic Penning trap. It works great for a single-component plasma, but it can't confine a quasi-neutral (i.e. high density) plasma.TDPerk wrote:The simplest grid which could be magnetically shielded would be a single coil, correct?
Re: a brief thought experiment
Um... A Penning trap in the sense that the charged particles are held away from the electrode, but the particles just sit in the center in a thremalized plasma. There is no arrangement to accellerate the ions past the electrodes, as the ions are traped between the two electrodes.Art Carlson wrote:This is essentially a minimalistic Penning trap. It works great for a single-component plasma, but it can't confine a quasi-neutral (i.e. high density) plasma.TDPerk wrote:The simplest grid which could be magnetically shielded would be a single coil, correct?
http://www.aip.org/png/html/penning.htm
I supose perhaps adding an additional electrode in the center at higher potential and/or pulsating electrical field strengths would would have a similar effect, but in a a line, not in a sphere, so any convergence ( if you accept that this is a significant operating perameter of spherical IEC fusors) is much less.
http://iec.neep.wisc.edu/overview.php
Err, upon further research, I guess Penning traps can also be built in three demensions, which would make them very similar (identical?)to a magnetically shielded central cathode grid fusor. The difference from a Polywell design would be in the geometry details. I'm guessing that increasing charged particle densities in the Penning trap (at least one based on the drawing in the link) would tend to force open the cusp, as opposed to the Wiffleball effect pinching them almost closed. Also, recirculation is an issue.
The simplist neutron producing 'grid' that I have heard of is a needle.
Dan Tibbets
To error is human... and I'm very human.
Well, OK, it's erroneous in the sense that soving the grid problem isn't enough (i.e. "fusors can't reach unity because of the grid" is true, "fusors can't reach unity ONLY because of the grid" is false). Even if you somehow solved the thermalization problem, the grid problems would still prevent you from reaching net power, and vice-versa. Necessary, but not sufficient.Gee... this debate isn't about what a fusor could become, it's about the erroneous comment that a fusor doesn't get to break-even BECAUSE of the grid.
Agreed. But even if you somehow fixed ALL the other problems, I think you still wouldn't get to unity, even for a millisecond, with the grid in there. Even if that weren't true, a millisecond fusor has few practical applications. Bussard wanted something practical.Even if the grid lasted a millisecond and then burnt out, you'd still see if net-power was gained. It isn't.
You would also eliminate the largest source of losses.But this isn't the issue. This merely shows that if you fix the grid, you'll gain x2 improvement.
Over-unity means outputs greater than inputs. If any fraction of your inputs is lost without helping make output, obviously that makes unity harder to get to. So grid losses are one reason why fusors can't get to over-unity: besides melting the electrode, confinement is very poor when you have a chunk of metal sitting at the core. If you didn't have collisions with the grid, it would be much easier to get to unity, assuming you could solve the other problems.Now you're confusing 'loss factor' with why the fusor can't get to over-unity.
So no, I don't think Bussard was mistaken in thinking eliminating the grid was a good idea.
Exactly! And you've just gone from (rounding to a full 10 decimal places!) 100% loss down to 100% loss by the removal of the grid.TallDave wrote: If any fraction of your inputs is lost without helping make output, obviously that makes unity harder to get to.
Well, if you then look at those figures, 100% loss down to an 'improved' 100% loss without the grid, that statment isn't really sustainable.TallDave wrote: So grid losses are one reason why fusors can't get to over-unity
A fusor is a 'freak show', a curious thing arising from the fringes of statistical probabilities, as it were. A tiny slither of a population of fast ions, at a multi-standard-deviation away from a temporal norm, giving a miniscule chance for statisticially insignificant fusion events. Do the sums on the probabilities, and the 1E-8 efficiency is pretty much what you arrive at due to thermalising ionisation and Coulomb collisions - as it will be with Polywell, if its internal mechanisms turn out to do no more than a fusor.
Your statement is rather silly on its face. The losses in any system are eventually 100%. What's relevant is the outputs produced. If you have more losses for the same output, you're going to have more trouble getting to unity, because you'll need more inputs to get to the point where output exceeds input.
By your reasoning, it wouldn't matter if we produced 100MW using 5MW or using 1000MW, since it's all 100% losses anyway.
If removing the grid removed 99% of the losses, for a given fusion level we would need 1/100th as much input.
Here's another source, since you don't like Wikipedia.
By your reasoning, it wouldn't matter if we produced 100MW using 5MW or using 1000MW, since it's all 100% losses anyway.
If removing the grid removed 99% of the losses, for a given fusion level we would need 1/100th as much input.
Here's another source, since you don't like Wikipedia.
http://ssl.mit.edu/publications/theses/ ... Thomas.pdfThe two main loss mechanisms for a gridded IEC are ions impacting the cathode grids and electrons streaming to the anode grids.
Last edited by TallDave on Sat Jul 11, 2009 11:13 pm, edited 1 time in total.
Yes, the logic is right, but the numbers are wrong.TallDave wrote: By your reasoning, it wouldn't matter if we produced 100MW using 5MW or using 1000MW, since it's all 100% losses anyway.
In a fusor of current technology, you get some 1uW for 1kW. Get rid of the grid and you might get 2uW for 1kW. Sure, you've doubled your efficiency, but you've not halved the gap to over-unity! You could improve your efficiency a thousand fold and still be no where near.
Yes, as I've pointed out about a million times now, there are many other factors preventing fusors from reaching unity. This is not news. But the grid is one of the biggest. You need confinement to be a lot better than you can get with a big chunk of unshielded metal in the center.
The numbers aren't wrong, they aren't intended to represent a current fusor, just to demonstrate how irrelevant the 100% loss number is for any fusion device.
The numbers aren't wrong, they aren't intended to represent a current fusor, just to demonstrate how irrelevant the 100% loss number is for any fusion device.
Yes, but only under the condition that the grid is responsible for ALL losses, which it clearly isn't.TallDave wrote:If removing the grid removed 99% of the losses, for a given fusion level we would need 1/100th as much input.
Even if the ions only made just ONE single, lousy pass before crashing into the grid, there are 60 electrons streaming out of the grid for each ion in the chamber volume. So that'd make the grid losses only 1/60, even in that absolutely simplistic argument, not even accounting for ionisation or thermalisation losses.
I have merely tried to raise attention to a fallacy. You wish to believe that fallacy. I guess that's your choice. Science appears to be becoming subjected to the 'rules' of democracy - if enough people want to believe something, then they have a right to declare it 'science'.
(I heard some folks have even tried to get Ohm's Law repealed! There was a lot of resistance to the move, apparently.)
You seem to want the last word on this, so go ahead....
No, if the grid was 100% of the losses then by removing the grid you would have a machine with no losses, which would be nice but very unlikely since it tends to violate the 2LoT. If factor X is responsible for 1% and the grid is 99%, then removing the grid loss means you need 1/100th the input power for the same output (all else being equal, which it isn't; there are nonlinearities but this is a very rough example), because now you only have X to deal with. (This still would not necessarily get you close to breakeven, of course; the losses in a fusor to a grid can be enormous, so the other factors (such as the one you mentioned) could easily still be large enough to prevent this.)If removing the grid removed 99% of the losses, for a given fusion level we would need 1/100th as much input
Yes, but only under the condition that the grid is responsible for ALL losses, which it clearly isn't.
Which is ironic, because instead you merely committed your own. The fact that removing the grid does not enable fusors to reach breakeven does not mean that "fusors cannot reach breakeven because of the grid" is a fallacy. It means "fusors cannot reach breakeven only because of the grid" is a fallacy, and no one (especially not Bussard) believed that. Sorry, you had a fair point that there are other issues besides the grid, but you're overreaching with this grandiose notion that you're doing the world some great public service by pointing something out that career IEC researchers were unaware of.I have merely tried to raise attention to a fallacy.
Apparently the difference between "necessary" and "sufficient" just isn't computing for you. Let's try one more, time though. Say you need both a ride to get to a concert, and tickets to get in. You tell your friend "I can't go to the concert because I don't have a ticket." This is entirely accurate, even if it omits the other reason. Having a ticket is a necessary, but not sufficient, condition.
Believe me, I'd like to stop explaining the same thing over and over.You seem to want the last word on this, so go ahead....
I see this kind of confusion with tokamak guys sometimes, apparently because they're used to thinking in terms of a machine that reaches ignition and doesn't need inputs past that point (hence the Lawson obsession). IEC devices don't ignite, they always need inputs. What they need is a sufficiently high output/input ratio. The better your confinement, the less input you need for a given output.
Polywells do this nicely via the Wiffle-Ball effect. No chunk of metal in the middle eating ions, the anode is well-shielded from electrons, densities are high.
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In the meantime tokamak guys have gotten used to the fact that even a working reactor will need hefty inputs, mostly for current drive, but also for profile control and stabilization. The numbers go like this: In order not to take a big hit in the economics, you would like to sell at least 80% of the electricity you produce. If you can turn 1/3 of the fusion power into electricity, and put 1/5 of that back into the plasma, then you need a Q = P_fusion/P_input of at least 15. 20 or 30 would be better, but you don't need infinity. The triple product n*T*tau required compared to the Lawson criterion is Q/(Q+1), so the requirement for a driven machine is only about 6% lower than the requirement for an "ignited" machine. The Lawson obsession is justified.TallDave wrote:I see this kind of confusion with tokamak guys sometimes, apparently because they're used to thinking in terms of a machine that reaches ignition and doesn't need inputs past that point (hence the Lawson obsession). IEC devices don't ignite, they always need inputs. What they need is a sufficiently high output/input ratio. The better your confinement, the less input you need for a given output.
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So, to summarize the last 3 pages of this melee, we have three fundamental sources of loss in a fusor:
1) Fuel ions thermalizing.
2) Grid discharges to the wall.
3) Ion collisions with the grid.
Of these, chrismb is putting up a spirited defense that #3 doesn't swamp the other 2 sources of loss, and therefore magnetically shielding the grid doesn't help that much.
So, two questions (well, rhetorical questions, with argument attached):
First, chrismb, you mentioned that the fueld ions thermalize "with the background". What background? As far as I can tell, the only background consists of monoenergetic ions, discharing electrons, and the grid. So the only source of thermalization ought to be from coulomb scattering near the center of the device. I would further suspect that thermalization would be considerably reduced from a single-species D-D reaction, since you're unlikely to get big momentum transfers when everything's the same mass.
Second, can't we minimize electron discharge by a) pre-ionizing the ions, rather than relying on corona effects, and b) simply making the vacuum chamber bigger?
1) Fuel ions thermalizing.
2) Grid discharges to the wall.
3) Ion collisions with the grid.
Of these, chrismb is putting up a spirited defense that #3 doesn't swamp the other 2 sources of loss, and therefore magnetically shielding the grid doesn't help that much.
So, two questions (well, rhetorical questions, with argument attached):
First, chrismb, you mentioned that the fueld ions thermalize "with the background". What background? As far as I can tell, the only background consists of monoenergetic ions, discharing electrons, and the grid. So the only source of thermalization ought to be from coulomb scattering near the center of the device. I would further suspect that thermalization would be considerably reduced from a single-species D-D reaction, since you're unlikely to get big momentum transfers when everything's the same mass.
Second, can't we minimize electron discharge by a) pre-ionizing the ions, rather than relying on corona effects, and b) simply making the vacuum chamber bigger?
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The ions will isotropize in a few passes by bouncing off the non-spherical electrostatic well. If the level of turbulence is small, then converting a mono-energetic distribution to a Maxwellian will indeed require Coulomb collisions, which are still fast compared to fusion collisions.TheRadicalModerate wrote:First, chrismb, you mentioned that the fueld ions thermalize "with the background". What background? As far as I can tell, the only background consists of monoenergetic ions, discharing electrons, and the grid. So the only source of thermalization ought to be from coulomb scattering near the center of the device. I would further suspect that thermalization would be considerably reduced from a single-species D-D reaction, since you're unlikely to get big momentum transfers when everything's the same mass.
Definitely. I didn't mean to imply otherwise.The Lawson obsession is justified.
I would add "electrons hitting the grid" to #3.
You probably also want a #4: because there's no magnetic confinement, density is limited by space charge considerations. This isn't a loss, but it's a big problem.
In terms of being able to build a net power fusor, he's entirely correct. Fixing the horrible grid confinement problem isn't enough.Of these, chrismb is putting up a spirited defense that #3 doesn't swamp the other 2 sources of loss, and therefore magnetically shielding the grid doesn't help that much.
Someone correct me if I'm wrong here, but I think fusors have a built-in background problem in that they operate at relatively high pressures.As far as I can tell, the only background consists of monoenergetic ions, discharing electrons, and the grid.