## Making Electricity with the p-B Polywell

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

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93143
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MSimon wrote:Current flow depends on potential and load. It has zero to do with charge canceling.

Any device capable, even in principle, of running in steady state has to obey null summation of currents. No part of it can supply a current without a replacement current coming in from somewhere else.

The decelerator grid is one end of an open circuit. (I hope you haven't forgotten that at 0 Hz a capacitor looks like an open circuit.) The only way a current could flow in it, steady-state, is for it to flow around the grid between two contacts, which has nothing to do with the decelerator grid's function and won't generate any power.

So what's the other end of this open circuit? A voltage supply. Probably closely connected with the magrid power handling system, since they're at roughly the same large-scale floating voltage (as opposed to the collector).

The MaGrids are practically useless as electrostatic collectors. At even 200 KV (high energy density lower Q machines) they will collect about 15% of the available energy. Provided the collector plates are right at the surface of the magnetic bottle. A big no no as that ruins electron oscillation.

That's not what I meant.

Consider the assembly of magrid, decelerator grid, and collector. What is the potential difference between the collector and the decelerator? 2 MV. What is the potential difference between the magrid and the decelerator? 200 kV or less.

Now remove the decelerator, keeping everything else the same. What is the potential difference between the collector and the magrid?

1.8 MV.

Naturally you can't actually run a BFR this way. But it illustrates my point.

The essence of the power cycle is that the alphas in the core (ie: at the magrid) are at the bottom of a massive potential well. Alphas that are fired out from the core (this is free energy for the purposes of the electrostatic analysis) represent a current across a voltage - between the magrid (which is sucking up electrons at the same rate in amps (net, counting the e-guns) as the alphas are streaming outwards) and the wall (which is giving up electrons to the alphas at the same rate in amps as they are arriving at the wall).

Sounds like a closed circuit to me. Furthermore, it behaves as expected if you short it (the potential difference collapses and you have a thermal system) or isolate the hot terminal (massive voltage buildup, at least for the first nanosecond before the alphas stop being able to reach the wall at all).

The sole purpose of the 'decelerator' grid is to flatten the bottom of this megavolt-range potential well so that a recirculating wiffleball machine can happily operate in it.

charliem
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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.

-----------------------------------------------------------------------------------

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.
Last edited by charliem on Tue Jul 08, 2008 12:43 pm, edited 1 time in total.

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

I propose we use these terms from now on:

1-TRAP GRID
2-DECELERATOR
3-COLLECTORS.

OK, just so I have a clear idea of what this structure looks like....

In the center, you have a region where the p and B11 collide, producing α-particles with energies of 2-3MeV each. This is at the center of a virtual cathode formed by...

a "quasi-spherical" plasma of mostly electrons in a magnetically contained "whiffle-ball"/polywell. Passing through this plasma, trapped by the negative electrostatic field of the plasma, are the p and B11 ions. Because they are trapped in the potential well, which is mostly radial, they pass through the center repeatedly until they fuse. But I suppose that the low kinetic energy ions near the top of the electrostatic potential well will form a positively charged "halo" (but still negative overall potential) in the space between the whiffle-ball and the MAgrid.

The magnetic containment of the polywell is formed by the next layer from the center, the MAgrid -- a collection of solenoid coils arranged as a cube or dodecahedron, energized so all magnetic fields are directed inward. These solenoids have conductive casings so any electrons which should hit the solenoids can be extracted. The MAgrid is the first thing that the α-particles escaping from the core are likely to interact with -- by smashing into it, so a good design will minimize the shadow area of the MAgrid as seen from the center. The MAgrid casing also forms the anode of the system, being the collection point for excess electrons. It is electrically grounded.

Outside the MAgrid is your "TRAP" grid, negatively charged, intended to (a) encourage electrons escaping to return to the polywell, and (b) shield the MAgrid from the decelerator grid further out. It seems to me that this grid has to be (relatively) fine to do this job (relative to the MAgrid, at least), meaning it's likely to have an even larger shadow effect on the outgoing α-particles than the MAgrid.

Further out is the "decelerator" grid, which is at a significantly positive potential, forcing the outgoing α-particles to notice it and lose most of their kinetic energy, so they can be caught at low kinetic energy (but high potential energy) by...

the "collector", a solid electrically conductive shell which captures the lower-speed α-particles and acts as the cathode of the system, neutralizing the positive charged α-particles impinging on it.

Is there any reason we shouldn't expect α-particles to be lost by hitting the trap and decelerator grid?

(As an aside, I noticed that &alpha; didn't give me an α. Is there anyway to make HTML-entities work?)

93143
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In my conception the trap grid and the decelerator grid are the same thing.

It's at a high negative potential relative to the collectors further out, and a slightly negative potential with respect to the magrid.

Net result: Electrons can't climb the potential to the trap grid. Alphas accelerate towards it, miss (hopefully it can be in the magrid shadow), and just barely manage to make it across the much larger second potential difference to the collectors.

I started calling it a decelerator grid because I was beginning to think that "trap grid" was at least as appropriate for either of the other structures (assuming the idea about a 'grid' of collector plates works).

Now, of course, when I've made such a fuss about how its function isn't to generate a large potential well but simply to change its shape, "decelerator grid" is starting to sound inappropriate too...
Last edited by 93143 on Tue Jul 08, 2008 5:35 am, edited 2 times in total.

TallDave
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All that deceleration has to be paid for somewhere. Otherwise you could run this in reverse as a free energy machine by putting alphas on the edge and accelerating them in.

Although, now that I think about it, they'd all be stuck at the bottom of the well, so you couldn't really get work this way.

93143
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TallDave wrote:
All that deceleration has to be paid for somewhere. Otherwise you could run this in reverse as a free energy machine by putting alphas on the edge and accelerating them in.

Although, now that I think about it, they'd all be stuck at the bottom of the well, so you couldn't really get work this way.

You really could do that. You'd need to supply about 100 MW to keep the flow of alphas up (pumping -50 amps of electrons down a 2 MV difference from the alpha source to the magrid), and magrid heating would be maybe five times what you'd get running the reactor forwards (there's not really a good way to get p+11B out of 3x4He...), but you could do that.

You could even get energy out. Hook the magrid cooling system to a steam turbine. You might get 30 or 40 MW...

MSimon
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The decelerator grid is one end of an open circuit.

And "ground" is the other end. There is a potential between ground and the grid. If you connect a load current will flow. If the flow is replenished by induced charges a current can be maintained.

The current flow in the decelerator grids has zero to do with neutralizing alpha charges. Zero.

The sole purpose of the 'decelerator' grid is to flatten the bottom of this megavolt-range potential well so that a recirculating wiffleball machine can happily operate in it.

Nope. The machine will operate quite nicely without a decelerator grid. It would then be a thermal machine.
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93143
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MSimon wrote:
The decelerator grid is one end of an open circuit.

And "ground" is the other end. There is a potential between ground and the grid. If you connect a load current will flow. If the flow is replenished by induced charges a current can be maintained.

Okay, I can get traction on this. There's no current flowing into the decelerator grid from the vacuum chamber. Ideally, nothing hits it. Therefore if you just hooked it to ground, through a load or otherwise, it would simply discharge and that would be that. You need to keep it charged up, statically.

The current flow in the decelerator grids has zero to do with neutralizing alpha charges. Zero.

I can agree with that. They don't neutralize any alphas, so any connection would be kind of weird. As a matter of fact, there's no current in them at all.

The sole purpose of the 'decelerator' grid is to flatten the bottom of this megavolt-range potential well so that a recirculating wiffleball machine can happily operate in it.

Nope. The machine will operate quite nicely without a decelerator grid. It would then be a thermal machine.

That ALSO assumes that the massive potential difference between the magrid and the collector is changed to a small positive one to prevent electrons from leaving. This means the charge on the magrid would stay the same, but do power electronics maintain charge or voltage?

The fact that the charge on the decelerator grid is what physically sets up this potential difference in the direct-conversion configuration does not change the fact that it exists, and that at both ends of this potential difference currents are flowing into/out of the vacuum chamber - currents high enough that when multiplied by 2 MV (as one or the other of them will have to be to reach ground) you get 100 MW.

Last edited by 93143 on Tue Jul 08, 2008 6:17 am, edited 2 times in total.

TallDave
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93143 wrote:
TallDave wrote:
All that deceleration has to be paid for somewhere. Otherwise you could run this in reverse as a free energy machine by putting alphas on the edge and accelerating them in.

Although, now that I think about it, they'd all be stuck at the bottom of the well, so you couldn't really get work this way.

You really could do that. You'd need to supply about 100 MW to keep the flow of alphas up (pumping -50 amps of electrons down a 2 MV difference from the alpha source to the magrid), and magrid heating would be maybe five times what you'd get running the reactor forwards (there's not really a good way to get p+11B out of 3x4He...), but you could do that.

You could even get energy out. Hook the magrid cooling system to a steam turbine. You might get 30 or 40 MW...

I was thinking no Magrid, no Polywell, just some alphas running down a well. It would be a bit like the electrostatic equivalent of a hydropower plant: gaining energy by moving "downhill." I think you're claiming the "free" kinetic energy gained would be offset by the need to replenish electrons the alphas give up.

EDIT: The electrons they pick up, I mean.
Last edited by TallDave on Tue Jul 08, 2008 6:17 am, edited 1 time in total.

93143
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TallDave wrote:I was thinking no Magrid, no Polywell, just some alphas running down a well. It would be a bit like the electrostatic equivalent of a hydropower plant: gaining energy by moving "downhill." I think you're claiming the "free" kinetic energy gained would be offset by the need to replenish electrons the alphas give up.

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.

MSimon
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OK. Here is a question:

If the decelerator grid has zero current flow then where is the alpha energy going? Conservation of mass/energy and all that.
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93143
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MSimon wrote:OK. Here is a question:

If the decelerator grid has zero current flow then where is the alpha energy going? Conservation of mass/energy and all that.

The alpha kinetic energy is transformed into electrical potential energy by the potential difference it has to climb to get from the magrid to the collector. (The fact that the first bit is slightly downhill is irrelevant.) Since the alpha kinetic energy was from fusion, this is essentially free electrical potential energy, and it can (indeed must, if the reactor is to continue to operate) be used to generate a flow of electrons through a load.

The electrons leaving the magrid (at either ground or -1.8 MV) don't need to be the same electrons that ultimately arrive at the collector (at either +1.8 MV or ground), but the principle is the same: somewhere in the device, -50 amps of electrons (the neutralization current) has to traverse a +1.8 MV potential difference. The alpha current is like the internal current in a voltaic cell - the fusion reaction drives the alphas uphill, and this gives rise to the external current at its characteristic voltage, which can then be used to power stuff.

MSimon
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93143 wrote:
MSimon wrote:OK. Here is a question:

If the decelerator grid has zero current flow then where is the alpha energy going? Conservation of mass/energy and all that.

The alpha kinetic energy is transformed into electrical potential energy by the potential difference it has to climb to get from the magrid to the collector. (The fact that the first bit is slightly downhill is irrelevant.) Since the alpha kinetic energy was from fusion, this is essentially free electrical potential energy, and it can (indeed must, if the reactor is to continue to operate) be used to generate a flow of electrons through a load.

The electrons leaving the magrid (at either ground or -1.8 MV) don't need to be the same electrons that ultimately arrive at the collector (at either +1.8 MV or ground), but the principle is the same: somewhere in the device, -50 amps of electrons (the neutralization current) has to traverse a +1.8 MV potential difference. The alpha current is like the internal current in a voltaic cell - the fusion reaction drives the alphas uphill, and this gives rise to the external current at its characteristic voltage, which can then be used to power stuff.

I think you are still confused.

Let us make this a free space problem. You have a 0V electrode and a 1 MV tubular electrode. And inside the tubular electrode a -1 V collector plate. (forget practical difficulties like arcing). You shoot 2 MeV (free energy) alphas into the collector plate with great accuracy from the 0 V electrode. Nothing hits the 1 MV electrode.

Describe the voltages and current flows in the system - assuming that there is enough source/sink such that the system is in the same condition (voltage wise) as before the alpha(s) entered the system.
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93143
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MSimon wrote:Let us make this a free space problem. You have a 0V electrode and a 1 MV tubular electrode. And inside the tubular electrode a -1 V collector plate. (forget practical difficulties like arcing). You shoot 2 MeV (free energy) alphas into the collector plate with great accuracy from the 0 V electrode. Nothing hits the 1 MV electrode.

Describe the voltages and current flows in the system - assuming that there is enough source/sink such that the system is in the same condition (voltage wise) as before the alpha(s) entered the system.

Not sure I completely understand the geometry, but...

The alphas follow a 'valley' of potential from the 0 V electrode to the -1 V collector plate, striking it at 2.000002 MeV. Much heating ensues. A -50 amp (for the sake of argument) electron replacement current flows from source to ground to power supply to collector, requiring -50*-1=50 watts of pumping power (ie: negative useful work).

Remember that charge and potential are not the same thing.

I think a better analogy would be to have both the source electrode and the tubular electrode at -1 MV, and the collector at -1 V. That way, the alphas would be at 2 eV by the time they reach the collector (assuming that the geometry in this configuration allows them to hit the collector at all).

This also removes the otherwise large gradient between the source electrode and the tubular electrode, so that if you wanted to, you could further depress the tubular electrode voltage slightly and force electrons to prefer the source electrode. Note that this would not change the alpha impingement energy, because the potential difference between the source and the collector remains the same.

To return to the direct-conversion Polywell, it is essential to grasp that you can't ground both the collector and the magrid. One of them has to be at high voltage, because the magrid is inside a (reasonable facsimile of a) closed spherical shell possessing a massive negative charge. Thus the magrid MUST be about 2 MV lower than the collector, no matter which one you ground.

If you did ground them both, but kept the decelerator at -1.8 MV, the alphas would gain 3.6 MeV passing between the (now very very positively charged) magrid and the (very very negatively charged, but invisible from inside) decelerator, and lose it again on the way to the (neutral) collector. Impingement kinetic energy = birth kinetic energy, because the potentials at both locations are the same.

MSimon
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The alphas follow a 'valley' of potential from the 0 V electrode to the -1 V collector plate, striking it at 2.000002 MeV. Much heating ensues. A -50 amp (for the sake of argument) electron replacement current flows from source to ground to power supply to collector, requiring -50*-1=50 watts of pumping power (ie: negative useful work).

Nope. By the time the alphas hit the -1V collector inside the 1 MV cylindrical grid they have -2eV of energy.

They lost 2 MeV getting to the -1V collector. According to the potential gradient.

Any way I see why you are confused.
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THE OPEN POLYWELL FUSION CONSORTIUM