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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Re: Polywell electric webpage

Post by MSimon »

classicpenny wrote:
Hang on, are you using the terms the same? The decelerator grid is the same thing as the trap grid. Does that help?
This is why it is so important for us all to standardize the language. My suggestion would be to drop the use of "decelerator grid" entirely because I have seen it used as a synonym for both the Trap grid and the Collector grid; and doing so causes a LOT of confusion. For similar reasons, I would suggest that we also drop all the other words that some of us have been using as synonyms for "Trap" grid, and use trap grid only.

If it turns out that we really need to have a "decelerator" grid that serves a completely different purpose from either the trap or the collector grids -as I said before- I would like to understand it; and I believe a good diagram would help a lot.
We have the MaGrid and the decelerator grid(s).

Seems like fair terminology. Or if you like a totally functional description better: decelerator and accelerator grids.
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MSimon
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Post by MSimon »

1) The voltage on the electrode is increasing due to the electrostatic induction, not decreasiing.
Agreed.
2) That increase is continuous, rather than the transient event in the accelerator.
Wrong. It is continuous as an accelerator. Remember it is a DC accelerator: like the ones powered by VanDeGraff generators. Or a Crockoft-Walton Multiplier.

Now since the voltage is increasing you have to put a load on the decelerator grid to maintain constant voltage - viola the grid is delivering power without being part of the charge neutralization circuit. Which is what I have been saying all along.
Engineering is the art of making what you want from what you can get at a profit.

93143
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Re: Polywell electric webpage

Post by 93143 »

classicpenny wrote:The only connection to the Trap grid that I show is a battery that puts a -20kV bias on the Trap; and (I hope) that has nothing to do with the fuel ions. Truth be told, I have no idea why the Trap bias should be -20kV; I just have a vague memory of someone proposing that somewhere in Talk-Polywell.
Oops, sorry. I thought that was the power connection because I read it too fast.

93143
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Post by 93143 »

MSimon wrote:Now since the voltage is increasing you have to put a load on the decelerator grid to maintain constant voltage - viola the grid is delivering power without being part of the charge neutralization circuit. Which is what I have been saying all along.
That's a transient. It's also not very high power. Once steady-state is reached (ie: the fusion rate has stabilized and the chamber's alpha population is constant), no current is necessary to maintain the voltage of a grid that doesn't collect any charged particles.

Now explain to me how the fact that I have demonstrated the neutralization power to be the same as the output power is somehow irrelevant.

MSimon
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Post by MSimon »

93143 wrote:
MSimon wrote:Now since the voltage is increasing you have to put a load on the decelerator grid to maintain constant voltage - viola the grid is delivering power without being part of the charge neutralization circuit. Which is what I have been saying all along.
That's a transient. It's also not very high power. Once steady-state is reached (ie: the fusion rate has stabilized and the chamber's alpha population is constant), no current is necessary to maintain the voltage of a grid that doesn't collect any charged particles.

Now explain to me how the fact that I have demonstrated the neutralization power to be the same as the output power is somehow irrelevant.
I don't think you have demonstrated that at all. I have shown that it is theoretically possible to build a machine where the neutralization energy is near zero (2 eV) while the deceleration energy is 1,999,998 eV. To maintain steady state in such a machine you have to draw power from the decelerator grid. The electrons in the decelerator come from the metal (just as they would if the potential was magnetically induced). The electrons from the ions are re-united in the ion circuit. The electrons from "ground" neutralize the charges induced on the decelerator grid (not too many - we want to maintain the voltage).
Engineering is the art of making what you want from what you can get at a profit.

93143
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Post by 93143 »

MSimon wrote:
93143 wrote:Now explain to me how the fact that I have demonstrated the neutralization power to be the same as the output power is somehow irrelevant.
I don't think you have demonstrated that at all.
69 amperes of alpha current for 100 MW. Multiply by the potential between the source and sink (average 1.45 MV) and you have... 100 MW.
I have shown that it is theoretically possible to build a machine where the neutralization energy is near zero (2 eV) while the deceleration energy is 1,999,998 eV.
Did you take a look at my pictures? You can't start the alphas at almost the same potential as you stop them, because they won't be going any slower when they hit. It has to do with electric fields being conservative. Just look at the simulations.

TheRadicalModerate
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Post by TheRadicalModerate »

93143 wrote:69 amperes of alpha current for 100 MW. Multiply by the potential between the source and sink (average 1.45 MV) and you have... 100 MW.
Well, you forced me to do a little figuring of my own:

neutralization current (amps): 6.900E+01
charge per alpha particle (coulombs): 3.204E-19
number of alphas needed for neutralization current: 2.154E+20

mass of alpha particle (grams): 6.645E-24
alpha particle mass flow (grams/sec): 1.431E-03

average alpha particle energy (eV): 2.893E+06
kinetic power of alpha mass flow (eV/sec): 6.231E+26
kinetic power of alpha mass flow (watts): 9.982E+07

Two observations:

1) 1.4 mg of mass flow per second? Wow.

2) You're generating 100 MW of electrical power with 100 MW of kinetic energy. That seems like a good sign that we're barking up the right tree. I'm inclined to think that this bolsters the case for the neutralization current being the sole power-delivery vehicle.

EDIT: Fixed an embarassing power of 10 mistake which caused me to draw ridiculous conclusions in the previous version. If anybody looked in the last 3 minutes or your RSS poll went off at an inopportune moment--never mind!

kcdodd
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Post by kcdodd »

Yes. In steady state the electric field between the decelerator and collector is unchanging, and the net charge an both grids remains constant. That's pretty much the definition of steady state. If the field around the decelerator grid is unchanging and no charge is actually being put there, then there is no current on the decelerating grid. There is no induction effect if the field is unchanging. And I assume we are talking about steady state. Can I say that some more? haha
Carter

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Post by blaisepascal »

TheRadicalModerate wrote: If anybody looked in the last 3 minutes or your RSS poll went off at an inopportune moment--never mind!
Offtopic, but... You can get an RSS feed of this forum? How? Where? I've not been able to find it.

MSimon
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Post by MSimon »

93143 wrote:
MSimon wrote:
93143 wrote:Now explain to me how the fact that I have demonstrated the neutralization power to be the same as the output power is somehow irrelevant.
I don't think you have demonstrated that at all.
69 amperes of alpha current for 100 MW. Multiply by the potential between the source and sink (average 1.45 MV) and you have... 100 MW.
I have shown that it is theoretically possible to build a machine where the neutralization energy is near zero (2 eV) while the deceleration energy is 1,999,998 eV.
Did you take a look at my pictures? You can't start the alphas at almost the same potential as you stop them, because they won't be going any slower when they hit. It has to do with electric fields being conservative. Just look at the simulations.
Yeah. So is the electric field doing work? What would you call slowing a particle from 2 MeV to 2 eV? Any work done? Of course not. How about accelerating a particle from 2 eV to 2 MeV? Any work done? Of course not. You wuz robbed.

And since electrostatic fields do no work, I can design a particle accelerator that heats water in a kettle at no energy cost and have a source of unlimited energy. We are wasting our time with Polywells with this new invention you have. I can raise particle temperature from .03 eV to 2,000,000 eV without doing work. Why even bother with high voltages? 100 V should work just fine. Even at Carnot efficiencies, as long as the electric field is doing no work.

Particles can change velocity without doing work. They can not change speed without doing work. Something about 1/2 mv^2. Constant magnetic fields are conservative because they change velocity without changing speed. Electrostatic fields (if you cross the field lines) are not conservative because work is being done. Speed changes not just velocity.

As I said. This is hopeless. You are a physics major and don't even know when work is done. Pitiful.

Well my work is done.

Send me an e-mail when you figure it out. Because this is hopeless.
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MSimon
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Post by MSimon »

Yeah. I looked at your pictures. What exactly do they represent?

Here is my picture.

Image

When the alpha enters the drift tube it has 2 eV. Assuming it is traveling along the zero potential line where exactly is it going to get any energy? Is it going to climb the field lines without any work and then get accelerated by the aprox. 1 MeV drive voltage? What are the tunneling probabilities of that?

In fact if you will draw the field you would see that it actually would serve as a funnel for the alphas.

You cannot understand electrostatic accelerators without drawing the eqipotential lines.
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TheRadicalModerate
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Post by TheRadicalModerate »

Simon, if you'd just modify this picture with a charging circuit, a load, and a way to drain charge off of the drift tube, I'd be sooooo happy...

I've asked this question now several times and still haven't gotten an answer: You've effectively got an open circuit here. As electrostatic induction increases the positive voltage on the drift tube, a negative current flows into the drift tube through the load from ground, reducing the voltage back to +1MV. Then where do the electrons that constitute that negative current go once they've arrived at the drift tube?

drmike
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Post by drmike »

I've pretty much avoided this thread, but the ideas really are old and well known. Here's a couple of references that might prove interesting:

A Review of Previous Research in Direct Energy Conversion Fission Reactors ca. 1999

The Attainment of High Potentials by the use of Radium ca. 1913

DIRECT ENERGY CONVERSION (DEC) FISSION REACTORS--A U.S. NERI PROJECT ca. 2000

(the first and last seem like the same thing, but have slightly different comments.)

TheRadicalModerate
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Post by TheRadicalModerate »

drmike wrote:I've pretty much avoided this thread, but the ideas really are old and well known.
I really don't doubt that this will work. I simply don't understand how. It bugs me when I have to take something on faith.

I'm just hung up on one eensy tiny point and I can't get an answer: You're generating power by pulling electrons from a lower potential, through a load, and into the deceleration anode, which is at a high positive potential. But where do the electrons go from there? If the answer is that they're used to neutralize the alphas, I'm a happy camper, but others have said that this isn't necessary. So I'm confused. Help!

kcdodd
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Post by kcdodd »

TheRadicalModerate, you have it right. The alphas must be neutralized to get net current, and they must do it at high potential to get net energy, and so that means the collector has to be at high potential. If they are neutralized at low potential then no energy will be extracted electrostaticaly.

Ok, I'm done with this thread.
Carter

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