Magnetically Shielded Fusor Grids--Why Won't This Work?

[TheRadicalModerate]

So, we know that fusors with grids won't generate net power because the ions collide with the grid before they transit the center enough times to fuse, and that those collisions fry the grid pretty quickly, too.

But we've just spent a whole bunch of time engineering a magrid that prevents electrons from colliding with a grid by magnetically shielding the grid.

So what prevents us from plunking a magrid in the center of some machine and simply negatively charging it? As long as the grid is effectively shielded, ions should accelerate into the center and oscillate, just like they do in a regular fusor, but with considerably lower grid collision rates.

In addition to reversing the charge on the magrid, I'd think that you'd want to change the geometry of the B-field from lots of transverse field lines and very little cusp to lots of cusps with very little transverse field. That way, you minimize the number of ions with big perpendicular velocity components to the B-field, which in turn reduces the average gyroradius, which in turn means that the shielding works better.

So why doesn't this work? Is it just that the ions are too heavy to be effectively shielded?

[TallDave]

Not enough density and too many collisions with the grid (ions are heavy) for a reactor. The best you could hope for is to outperform a fusor. That's before we even get into issues like upscattering.

A Polywell combines aspect of a fusor and a magnetic mirror. The former gives us high temps, the latter better density. The WB effect gives us confinement.

[KitemanSA]

Ya know, with the revelation that the fusion product is anticipated to bounce around a number of times before exiting a cusp of a high field large machine, it strikes me too that this might make a MaGrid Hirsh-Farnsworth work, not just a MaGrid Elmore-Tuck-Watson.

Anyone interested in trying?

[D Tibbets]

Such has been proposed before. There was a website that advertised for this but I could not relocate it. A magnetically shielded central cathode would prevent ion collsions with it, but in doing so it would divert the ions at an angle, or trap them in a sheath around the negatively charged magnet structure, significantly impeading the focus of the ions with resultant lower central densities. The electrons that are generated fronm the cathode would interact with the magnetic field and surrounding sheath of ions with what results? Any electrons escaping this immediate area would quickly stream to the vessel wall. Possibly the ion grid collision losses would be improved, but any central foucus would be worse, unless radial ion focusing between the grids ('Star Mode') was enhanced to a greater extent. Electron losses would presumeably still be great.


Dan Tibbets

[chrismb]

So, we know that fusors with grids won't generate net power because the ions collide with the grid before they transit the center enough times to fuse

I'm afraid to point out that in no way do we 'know' this. It was Bussard's error to think so, and it is one that is generally made. Bussard's whole original thesis for conceiving Polywell was based on this false presumption.

To repeat for the nth time; fusors aren't 'overunity' machines because their fast ions thermalise with the generally-cold background medium over many non-fusing, scattering collisions. The ions that 'beam' through the grids are a large fraction of the total ions in motion, and elimintating 'grid' colliding ion events from the process may improve your efficiency by x2, but not by 9 orders of magnitude.

Whether Polywell works in other improved ways is the point in question, but the benefit gained from elimitating 'the grid' is - precious little.

[TallDave]

To repeat for the nth time; fusors aren't 'overunity' machines because their fast ions thermalise with the generally-cold background medium over many non-fusing, scattering collisions.

Well, that's one problem among many. The conventional wisdom on fusors is that grid collisions are also a very big problem, probably the biggest.

http://en.wikipedia.org/wiki/Fusor#The_ ... wer_source

Nonetheless, grid collisions remain the primary energy loss mechanism for Farnsworth-Hirsch fusors.

Although, with a shielded grid, the thermalizing problem might take precedence.

Whether Polywell works in other improved ways is the point in question, but the benefit gained from elimitating 'the grid' is - precious little.

In addition to the above, it's certainly a problem when your cathode melts.

Complicating issues is the challenge in cooling the central electrode; any fusor producing enough power to run a power plant seems destined to also destroy its inner electrode.

Not losing your cathode is a pretty fundamental benefit.

[chrismb]

These are all pure speculations. Along the lines of believing the earth is the centre of the universe merely because that's the easiest explanation. Just 'cos it's in 'wiki' doesn't make it right!

But let me put it another way. Experimentally, a fusor's efficiency seems to top out at around the 1E-8, say, at best. If you turn up the reaction rate, then I agree that grid collisions are then likely to begin to have an effect, because the beams will diverge due to space charge effects and in this case it will start impacting the temperature of the grid.

That is to say, if you can avoid having a grid, then maybe you can keep up the fusor's 1E-8 efficiency at higher reaction rates/beam currents, otherwise it drops even further.

[MSimon]

These are all pure speculations. Along the lines of believing the earth is the centre of the universe merely because that's the easiest explanation. Just 'cos it's in 'wiki' doesn't make it right!

But let me put it another way. Experimentally, a fusor's efficiency seems to top out at around the 1E-8, say, at best. If you turn up the reaction rate, then I agree that grid collisions are then likely to begin to have an effect, because the beams will diverge due to space charge effects and in this case it will start impacting the temperature of the grid.

That is to say, if you can avoid having a grid, then maybe you can keep up the fusor's 1E-8 efficiency at higher reaction rates/beam currents, otherwise it drops even further.

Chris,

It is not speculation. There are something like 15 or 20 fusors out there made by experimenters and grid thermal limitations are well known. In fact some in the amateur fusion community have built fusors with liquid cooled grids to handle that very problem.

[chrismb]

It is not speculation. There are something like 15 or 20 fusors out there made by experimenters and grid thermal limitations are well known. In fact some in the amateur fusion community have built fusors with liquid cooled grids to handle that very problem.

If you're trying to argue that grid collisions are 'obvious' because the grid gets hot, then think again! The grid is carrying a current as electrons flow through the device - of course it's going to get hot! Tell me - what percentage of the heat in a grid is due to current and what is due to ion collisions? And then, what percentage of mobile ions hit the grid? To answer, without speculating, would require objective measurements of these factors. - Show me these measurements!....

[KitemanSA]

If you're trying to argue that grid collisions are 'obvious' because the grid gets hot, then think again! The grid is carrying a current as electrons flow through the device - of course it's going to get hot! Tell me - what percentage of the heat in a grid is due to current and what is due to ion collisions? And then, what percentage of mobile ions hit the grid? To answer, without speculating, would require objective measurements of these factors. - Show me these measurements!....

But there would be no current (it is supposed to be an electro-STATICly charged grid) without the impacts to remove the electrons (unless you are positing a large discharge current). No impact, no current, NO NEUTRALS. Ok, to be pure about it, my statement shouldn't be so absolutist, but without high impact rates, the problems you describe should also reduce enormously. A MaGrid Farnsworth sounds interesting!

And wouldn't annealing also occur in a MaGrid Farnsorth to help keep the mono-energetic nature?

[chrismb]

But there would be no current (it is supposed to be an electro-STATICly charged grid) without the impacts to remove the electrons (unless you are positing a large discharge current). No impact, no current

You're convoluting two discussions and confusing them. I have not suggested the imposition of a suitably engineered magnetic field won't reduce electron emission currents. Not at all - I've proposed such solutions myself. The issue was to do with the existing hollow-cathode fusor and, yes indeed, it is just a glorified discharge device. The inefficiencies due to the central grid are certainly there, I've not suggested otherwise, but it is a trifling detail compared with the 1E-8 efficiency due to the fact that it IS just a discharge device. The axiom at the start of the thread was that it is inefficient DUE TO the grid. This is not so. There IS an inefficiency due to the grid, but it ain't much compared with the other lossy mechanisms!

[KitemanSA]

You seem to be saying that the grid which is supposed to be electro-statically chargedd in fact has, and fundamentally MUST have a very large discharge current. This I don't understand.

What about the simple system pre-determines so large a discharge that it will always result is something like a magnitude 8 inefficiency?

[chrismb]

You seem to be saying that the grid which is supposed to be electro-statically chargedd in fact has, and fundamentally MUST have a very large discharge current. This I don't understand.

What about the simple system pre-determines so large a discharge that it will always result is something like a magnitude 8 inefficiency?

I'm sorry, I don't understand what you are saying. It is called 'electrostatic' because it isn't 'magnetic' or 'electro-magnetic'. Perhaps it is a false description, it is 'electric' confinement.

A fusor will pull a current of a few 10's of mA. Some of the heat in the grid is therefore due to that current. Some of the heat is due to ions being sputtered off the outer shell, and fully accelerating back into the grid, by electrons accelerating away from the grid and hitting the shell. Some of the heat is due to it being in contact with the thermalised ball of plasma in the centre of the hollow cathode, and some of it is, indeed, due to fast 'fuel' ions and fast charged molecules hitting the grid. You can speculate a) which is most significant in heating the grid and b) which is most degrading of the fusion rate. But they would be speculations.

A fusor is just, plain and simple, a discharge lamp. Like a neon tube. Just the same. It just so happens that if you stick in some 'fusible' deuterium and crank up the volts, then you'll happen to get some fusion going.

[D Tibbets]

You seem to be saying that the grid which is supposed to be electro-statically chargedd in fact has, and fundamentally MUST have a very large discharge current. This I don't understand.

What about the simple system pre-determines so large a discharge that it will always result is something like a magnitude 8 inefficiency?

I'm sorry, I don't understand what you are saying. It is called 'electrostatic' because it isn't 'magnetic' or 'electro-magnetic'. Perhaps it is a false description, it is 'electric' confinement.

A fusor will pull a current of a few 10's of mA. Some of the heat in the grid is therefore due to that current. Some of the heat is due to ions being sputtered off the outer shell, and fully accelerating back into the grid, by electrons accelerating away from the grid and hitting the shell. Some of the heat is due to it being in contact with the thermalised ball of plasma in the centre of the hollow cathode, and some of it is, indeed, due to fast 'fuel' ions and fast charged molecules hitting the grid. You can speculate a) which is most significant in heating the grid and b) which is most degrading of the fusion rate. But they would be speculations.

A fusor is just, plain and simple, a discharge lamp. Like a neon tube. Just the same. It just so happens that if you stick in some 'fusible' deuterium and crank up the volts, then you'll happen to get some fusion going.

The thought that the current flowing through the wire grid is what heats it up is contrary to most of descriptions that I see on Fusor. net where the amateur fusor hang out. And, while stainless steel wire or especially tungsten wire has more resistance than copper, the general size of wire is in the range of 20-24 gauge and this can generally handle 1-2 amps of constant current. Also, keep in mind that the grid is generally 2-4 loops or more of this wire which would increase the effective gage size by up to several numbers. Even with a good vacuum insulation as opposed to plastic insulation on a general household use wire, the grid wires are not long and will conduct the ohmic heat into the metal feed through, so at a current load of only ~ 1-10 percent of the wires rated capacity I would not expect much heating beyond what can be carried away. Meanwhile the mobile charge carriers (ions and electrons) have a kinetic energy up to the applied voltage (say 30,000 eV). When an ion at that velocity hits the wire it will generate a lot of heat . I'll leave it to someone else to calculate the joules of heat generated per ion X estimated number of particles/sec. hitting the grid. My impression is that the increases in current through the wire from the power supply provides a cascading increase in charge carrying ions and secondary electrons, and these charged particles (ions primarily) bombard the grid in logrhythmically increasing numbers.
Also, I have made grids out of heavy copper wire (16 or 18 gauge) and at 20-30 mA at a few thousand volts (and probably ~ 100-200 Microns pressure) it will heat to a cherry red quickly. In that size copper wire the Ohmic heating from that amount of current should be negligible.

ps: You may be confusing the heating of the wire by conducted amps and the Watts that are being produced (A X V). A 1000 watt toa.ster oven heats by Ohmic heating of a wire * . But a 1000 watt microwave generates heat in the target but not through Ohmic heating. In a similar sense the gridded Fusor produces most of it's heat through accelerating particles which then hit the grid or walls (or each other) to generate heat. Or to push it to rediculus extreems, particle acelleraters with superconductors produce thrmendous watts of power, but very little of it goes into heating the wires.

* Ohmic heating of a 1000 watt toaster oven in the US has a heating current of ~ 8 amps. But in Europe with 240V mains, the same 1000 watts would only provide ~4 amps of heating current. Do European toasters have to operate at twice the current or are they generally provided with voltage step down electronics?

Dan Tibbets

[chrismb]

Dan, I have no contest to the points you, or the others, are making. It's just that they don't relate to mine.

The original posit here was that "fusors with grids won't generate net power because the ions collide with the grid before they transit the center enough times to fuse", which is wrong. It is not the grid that means the fusor won't generate net power.

I'm saying that this statement "A follows from B" is false, and you're saying "but A is true and B is true, so A must follow B". I've said nothing explicitly about A or B, I agree that both are true. I'm saying only that one isn't consequent to the other. This is a classic non-sequitur. I think it's called 'denying the conjunct', or someting like that.