Making Electricity with the p-B Polywell

Discuss how polywell fusion works; share theoretical questions and answers.

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MSimon
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Post by MSimon »

There used to be a tool called resistance paper around when tubes were it.

What you did on the paper was draw lines of copper for the grids and put your various (scaled) voltages to them. Then you could measure the fields with a voltmeter. Very handy. I never used it.
Engineering is the art of making what you want from what you can get at a profit.

blaisepascal
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Post by blaisepascal »

MSimon wrote: In the drift space you have a field going from 1 MV at the edge to 0 V in the center (at least near the collector). If the particle is at the 0V gradient when it enters the drift space it is not going to gain any energy. If you are not cutting across the equipotential lines you don't gain energy. It is not possible. Unless the sky is green in your universe.
I don't see this.

What I see, in terms of fields, is very strong E-fields between the end of the tube and the emitter, no E-field within the main body of the tube (at least, NOT near the collector) and very strong E-fields between the body of the tube and the collector plate. I envision the E-Field lines terminating perpendicular to the surface of the collector and to the inside surface of the tube, since E-Fields are perpendicular to conductive surfaces.

My schooling tells me that the equipotential surfaces are perpendicular to the E-Field, so I see a very strong gradient (tight spacing of equipotential surfaces) near the collector, no gradient within the main body (drift space) of the tube, and another strong gradient between the tube and the emitter.

When you fire the 2MeV alpha particle from the emitter to the collector, I see it crossing lots of equipotential surfaces as it heads towards the tube to have near-0 kinetic energy as it enters the drift space. I see no electrostatic forces acting on it while in the drift space (and my intuition feels the currents aren't large enough to have significant magnetic effects). But when it gets close to the collector, it is going to be accelerated by the potential gradient, crossing lots of equipotential surfaces, near the collector, and slam into the collector with lots of energy. If we assume the emitter and collector have the same voltage, it'll hit the collector with 2MeV of energy.

I'll freely admit I'm not an expert. I've done lots of book-reading of physics, and passed with flying colors a college-level calculus-based EM course in high school -- 20 years ago. I'm old enough that I find FETs easier to understand that bi-polar transistors because FETs work like tubes, but that doesn't mean I understand tubes.

You seem to claim my analysis is wrong. I welcome an opportunity to sharpen my intuition and knowledge. Where is my mental picture failing me?

drmike
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Post by drmike »

Another useful method was a saline bath. You could do a 2D problem really easily by setting up the electrodes in the bath and then measure the potential drop everywhere in between directly with a volt meter.

Computers make it easy keeping your hands dry!

Aero
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Post by Aero »

Hey guys, I know that this is no fun, but the problem has been worked before so I don't think we'll need to discover any new science here. Of course it is fun to figure out how it really works, keeps us sharp, too. And the power regime of BFRs is totally new, but adapting to that regime is a different problem.

http://ntrs.nasa.gov/search.jsp?R=21685 ... 4294967207

Of course, its always possible that this report is not applicable to our problem or that NASA made a mistake.
Aero

TheRadicalModerate
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Post by TheRadicalModerate »

Simon:

I understand what you're trying to do with your 0V electrode plus +1MV cylindrical electrode + -1V collector plate, but let's make the gedanken experiment even simpler:

We've got a 0V point source that's (magically) firing 2 MeV alphas at a +1MV infinite-plate electrode. Just before the alphas reach the plate, they are (again, magically) transformed to neutral He4 and whisked away.

So, a list of things we're not worrying about--yet:

1) We aren't worrying about how the alphas get neutralized and collected, and we're assuming that the neutralization current is trivial.

2) We aren't worrying about the fact that the forces experienced by charged particles in a spherical charged shell are 0.

Now, I think we all agree that the plate electrode (which is presumably connected to a voltage source, which is presumably is connected to ground) is effectively an open circuit. And I also think we all agree that our magically appearing and disappearing alphas start at the 0V source with 2 MeV of kinetic energy and end up at the plate with 0 kinetic energy.

Now (and here's where I, with no degree in physics and absolutely no ego invested in this, can be as stupid as I like and you can make fun of me as much as you want), I think you're saying that that kinetic energy has been converted to some incremental potential difference in the plate, which in turn can be bled off as a current connected to a load parallel to the plate and, voila!--power.

But here's what's freakin' me out: When we bleed off that excess voltage as power, we're physically sucking electrons out of ground (0V in our little experiment), through the load, and into plate electrode. However, last time I checked, Kirchoff's Current Law is still on the books. So I ask the question: Where do the electrons go once they've been sucked into the electrode?

Again, all of this stems from the fact that the decelerator is effectively an open circuit. In normal systems, open circuits don't produce power. So, if you could explain why this one is different, I'd be very grateful.

TheRadicalModerate
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Post by TheRadicalModerate »

OK, I finally realized why the spherical electrode objection is bogus: You've got an E-field from any charge on the the spherical electrode (which is everywhere zero) and an E-field from the potential difference between the outer shell and the center of the wiffle ball, which looks like the E-field from a point charge at the center. Superpose the two and you get null field + point field = point field.

Right?

kcdodd
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Post by kcdodd »

MSimon, dude, for real. This is not "big bath".

KE + PE = E, and PE = 2*e*V for alpha.

alpha starts at 0V with 2MeV: 2MeV + 2*e*0V = 2MeV.

we are assuming that there are no other collisions with particles before it is collected, so total energy of the alpha is conserved dE/dt = 0.

to find the KE at 1MV we already know that E = 2MeV from initial conditions, KE + 2*e*1MV = 2MeV, solving for KE = 2MeV - 2*e*1MV = 0. At 1MeV it is zero kinetic energy, this is where you should collect the alpha.

If the alpha reached any spot at 0v: KE = 2MeV - 2*e*0V = 2MeV. It will crash at full energy it started out with.

I don't think anyone is arguing that things can't decelerate. It obviousy did to reach 1MV, but why on earth then put a 0V collector. You are just re-accelerating it. Maybe you should stop insulting everyone else for a second because honestly you have put your foot so far down your throat its not funny any more.
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TheRadicalModerate
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Post by TheRadicalModerate »

kcdodd wrote:I don't think anyone is arguing that things can't decelerate. It obviousy did to reach 1MV, but why on earth then put a 0V collector. You are just re-accelerating it. Maybe you should stop insulting everyone else for a second because honestly you have put your foot so far down your throat its not funny any more.
Why are we arguing about the details of a thought experiment that obviously wasn't intended to be a workable system? It has nothing to do with whether electrostatic induction will actually produce usable power.

kcdodd
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Post by kcdodd »

Conservation of energy is not a "detail". The laws of physics are the same no matter what.


Btw, I have no idea what the thought experiment was originally supposed to show. Maybe I have my foot in my mouth for saying that it wouldn't work in real life.
Last edited by kcdodd on Wed Jul 09, 2008 3:48 pm, edited 2 times in total.
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ravingdave
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Post by ravingdave »

charliem wrote:It would be much less confusing if we could agree on the names of the structures.

I think that the decelerator grid Simon is talking about is the same that someone else named the trap grid, I the shield grid, and some other people called even more names.

I propose we use these terms from now on:

1-TRAP GRID, not far from the magrid and at a negative potential from it. Its function is to repel back inside the electrons that leak through the cups (to "trap" them), and shield the internal fields from external influences (a Faraday cage).

2-DECELERATOR grids/structures, at a quite high positive potential from the trap grid. Their function is to slow the alphas down, but not to catch them.

3-COLLECTORS. The structures that the alphas hit and where they get neutralized.

I would suggest that we use the term "Repeller Grid" and "Collector Grid". I would also suggest that the "Decelerator grid and the collector grid must be one and the same.

Bear in mind that the repeller grid must serve two functions depending on whether it is dealing with ions or electrons. Electrons are repelled by the repeller grid, but ions simply use it to create a potential gradient between it and the collector grid. (they are not repelled. )


charliem wrote: -----------------------------------------------------------------------------------

About E-fields, acceleration/deceleration of charges, and where the energy goes or comes from.

When a charge gets in the middle of an electric field it feels a force that accelerate it in some direction. That acceleration means that its kinetic energy is altered but the difference in energy is NOT taken from the e-field, so this could very well be an static charge field and even so keep influencing more charges indefinitely without lossing any power.

This looks counter-intuitive only if we concentrate on the e-field and the particles and forget to look at the whole system. Just take a broader point of view and the [apparent] contradiction disapears.

EDIT: Decelerator/s are optional. In some configurations there is no deceleration of alphas at all, or simply get the deceleration from the potential difference between Trap grid and Collector.

Yes, I think that's what I was getting at.


David

ravingdave
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Post by ravingdave »

93143 wrote: Yep. You have to pump them downhill, from the emitter to the target, which for negative charges takes energy. More energy than you could possibly get out with a thermal cycle, and just a little more than you could get with near-perfect direct conversion.

Essentially it's because a charge separation has potential energy, and the potential energy of a static charge separation is rapidly dissipated by the acceleration and subsequent smashup/thermal absorption of the arriving alphas, ending up as heat. To maintain the static charge separation in the face of this continued dissipation, you need to do work to push more charge through the potential gradient - continuously replenishing the potential energy of the static charge separation. The work done is current times voltage, and it ends up in whatever the high-speed alphas do with it.

Now reverse it. The high-speed alphas, instead of crashing into stuff and dissipating their energy as heat, are generated from thin air by fusion. They then fly energetically uphill and are discharged at the collector. In order to maintain charge balance, you have to move electrons uphill along with the alphas. But for a negative charge, uphill is good. So they do that naturally, and you have to stick a load in between (current times voltage) to prevent them from equalizing the source and sink too fast and eliminating the potential difference that stops the alphas.

Or, to put it another way, the alpha flow causes a charge imbalance between the magrid and the collector. This pulls electrons from one to the other, like particles in a linear accelerator (except that they're going through ground and load and transmission wires, so they do work along the way and arrive with very little kinetic energy). The power necessary to maintain this charge imbalance is the power that continually pushes more alphas up the hill - the fusion power.

Oooohhhh This stuff makes my head hurt sometimes !

I might as well interject.


Fuel ions are ions because electrons got stripped off of them. I know that you get one electon off the hydrogen atom, and I believe you get 5 electrons off the boron atom. That gives us 6 electrons to work with.


Hokay, the electronless boron and hydrogen fuse to become 3 atoms of helium (without electrons, thats why we call them ions) each with a charge of plus 2, and a velocity of +1-4 Mev.

By the process of reverse acceleration the veolocity is converted to Voltage, or potential. The current is the electrons flowing from where we stripped them off the ions in the first place to the collector where they recombine with the helium ions.

Now it looks to me like a lot of people are confusing the difference between voltage and current.

In this case, Voltage is the attractive force between the ions and THE electrons flowing inward to the collector grid.

Current is the result of the electrons recombining with their formely intimately connected ions.


This stuff is a lot simpler than people are making it out to be. Haven't you all seen the experiment where you take two plates of metal separated by a small distance, charge them up to some voltage, then move them further apart and notice that the voltage has gone up ?

It requires energy to seperate the plates. The energy required to seperate the two plates is converted to voltage. The numbers of ions and electrons hasn't changed, Just the distance between them. Pulling electons and protons apart REQUIRES energy. This energy is like a spring.

You can get a volage increase without particles recombining !
This just makes the pull between them stronger. It means that they want to recombine more forcefully. The current is caused by recombination.


The decellerator grid thing creates a higher voltage because as they approach the collector grid, the pull on the electrons in the system gets stronger and stronger, even with no current flowing.


I would like to go on, but I have found that sometimes more words has the inverse effect of improving understanding.


David

93143
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Post by 93143 »

With all this talk about visualizing field lines and potentials re: MSimon's thought experiment, no one seems to have noticed that I posted simulation results...

I'm starting to think MSimon is under the impression that I've denied the feasibility of direct conversion.

All I'm arguing is that the current flow for the output power is between the magrid and the collector, not between the magrid and the decelerator or the collector and the decelerator.

kcdodd
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Post by kcdodd »

Well, between whatever ionized the the original ions and the collector. The magrid itself may or may not be the source.
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ravingdave
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Post by ravingdave »

93143 wrote:With all this talk about visualizing field lines and potentials re: MSimon's thought experiment, no one seems to have noticed that I posted simulation results...

I'm starting to think MSimon is under the impression that I've denied the feasibility of direct conversion.

All I'm arguing is that the current flow for the output power is between the magrid and the collector, not between the magrid and the decelerator or the collector and the decelerator.

Perhaps I am mistaken, but what becomes of the electrons originally stripped off the fuel when it is ionized ?

I think the current is between the collector grid and the ionizing guns for the fuel, which I think are at MagGrid potential, but the current flow is from the point at which the electrons are stripped to where they are recombined at the collector.


David

P.S. I like the term "Collector" because it sounds better than "Plate" which would be the tube analogy. Collector, Emiter, and Base are BJT terms.
I suppose we could even use Source, Gate, and Drain, but I like "Collector."

93143
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Post by 93143 »

ravingdave wrote:Perhaps I am mistaken, but what becomes of the electrons originally stripped off the fuel when it is ionized ?

I think the current is between the collector grid and the ionizing guns for the fuel, which I think are at MagGrid potential, but the current flow is from the point at which the electrons are stripped to where they are recombined at the collector.
I'm just lumping all that in as the "magrid". The potential differences within the magrid/fueling assembly system should be small. The assumption was that the fuel was ionized by ECR slightly inside the magrid, but there can be other arrangements.

The upshot is that that system - however it is configured - has to lose electrons through a wire at exactly twice the rate at which it loses alphas, or else fueling (ions+electrons in, vs. just ions out) will cause a negative charge buildup.

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