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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Postby 93143 » Fri Jul 11, 2008 6:34 pm

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


The colour map is local electrical potential. The arrows are electric field. It's axisymmetric around the vertical axis (the left side of the coloured area).

Here is my picture.


Great. Turn it 90 degrees counterclockwise and you have the one I modeled (the first picture has that ground arrangement).

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?


There isn't a zero potential line straight down the axis between the source and the target. If there were, no deceleration would occur because there wouldn't be any electric field along that line. As it is now, there's a 'hill' of potential that the alpha climbs (because of its high initial kinetic energy) and then goes back down the far side of (regaining all its kinetic energy in the process).

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


I DID draw the field. Well, the software did. Those arrows. It's not as easy to see as lines would have been, I guess, but all you really have to do is look at the colour gradients, because those are the field (the potential gradient) too.

Yes, there's a funnelling effect. No, it doesn't have the result you describe.

You cannot understand electrostatic accelerators without drawing the eqipotential lines.


Actually, you can see those because the quality of the plots isn't super high. They are the divisions between colours.

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.


That's not what "conservative field" means. Maybe you should Google it...

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Postby MSimon » Fri Jul 11, 2008 10:23 pm

TheRadicalModerate wrote: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?


Yeah. A charged capacitor is an open circuit. Can you draw energy off it?

If I induce a voltage on a capacitor can I draw a current off that capacitor?

The power supples are NOT IMPORTANT for this gedanken experiment. But hey. If you want to draw a supply I have no problem. You slip it in between 1 MV (aprox.) and ground. Want a load? Put it across the power supply. Make the load adjustable so that all the current delivered by the voltage induction is drained off. In fact once the electrode is charged you can throw away the supply since its only purpose is to define the initial charge. Given that we have perfect insulators, perfect vacuum etc. there are no losses in the system.

The electrons of course are exactly what is required to maintain the voltage given the induced charge. Where do the electrons go? Back to the metal they came from. They neutralize some of the positive charge.

At least you get that in the example I provided that the voltage on the + electrode must rise. Small progress. However, once you admit the voltage must rise the game is over. Because that is exactly my point. To maintain the voltage you must draw a current. The current must be between the + voltage and ground.

We also note in this thought experiment that the neutralizing current has nothing to do with the power current. They are two different flows. Or can be. Depending on the arrangement of the electrodes. My arrangement separates the two currents.

The level of confusion here from people supposedly trained in the subject is astounding.

BTW only magnetic fields are conservative WRT particles. Because they change the velocity without changing speed. Electrostatic fields are not conservative. I have been waiting for some of you trained physicists to pop up with that for a couple of days. No such luck. Happy to be of service.

I think this is what Feynman was alluding to when he talked about people who studied physics knowing all the formulas without a clue on how to apply them in circumstances different from the examples taught in class. "Cargo cult science" he called it. Lots of knowledge. No understanding. I thank my lucky stars I had PSSC Physics which emphasized understanding over rote knowledge. They used to teach it to high school kids when I was growing up. Gone are the days.
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Postby MSimon » Fri Jul 11, 2008 11:08 pm

There isn't a zero potential line straight down the axis between the source and the target. If there were, no deceleration would occur because there wouldn't be any electric field along that line. As it is now, there's a 'hill' of potential that the alpha climbs (because of its high initial kinetic energy) and then goes back down the far side of (regaining all its kinetic energy in the process).


The decelerating field is between the 0V and the drift tube. Inside the drift tube along the centerline there is a zero field line near the collector.

What you do is move the grounded electrode far enough back in the drift tube so that it has a negligible effect on the field between the 0V point and the entrance to the drift tube. Then all the deceleration takes place in free space.

BTW since there is a 0 V electrode in the drift tube there has to be a 0V field line in the drift tube. I claim that line is along the centerline.

What you need to do is do a drawing of the equipotential lines. All the lines past the entrance to the drift tube are 999,999 V until you get near the 0 V electrode. A constant field can not affect the speed of a particle. Near the collector the field tends to drive the particle to the 0V electrode since the field is repulsive.
Engineering is the art of making what you want from what you can get at a profit.
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93143
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Postby 93143 » Fri Jul 11, 2008 11:13 pm

MSimon wrote:Yeah. A charged capacitor is an open circuit. Can you draw energy off it?


Only as much as you put in.

If I induce a voltage on a capacitor can I draw a current off that capacitor?


Yes. No problem.

The power supples are NOT IMPORTANT for this gedanken experiment. But hey. If you want to draw a supply I have no problem. You slip it in between 1 MV (aprox.) and ground. Want a load? Put it across the power supply. Make the load adjustable so that all the current delivered by the voltage induction is drained off. In fact once the electrode is charged you can throw away the supply since its only purpose is to define the initial charge. Given that we have perfect insulators, perfect vacuum etc. there are no losses in the system.


Okay so far.

The electrons of course are exactly what is required to maintain the voltage given the induced charge. Where do the electrons go? Back to the metal they came from. They neutralize some of the positive charge.


Still fine.

At least you get that in the example I provided that the voltage on the + electrode must rise. Small progress. However, once you admit the voltage must rise the game is over. Because that is exactly my point. To maintain the voltage you must draw a current. The current must be between the + voltage and ground.


Yes, BUT... The induced voltage only rises for a moment, while the nearby alpha population is rising. It equilibrates in steady state, and then you have no more current from that grid.

We also note in this thought experiment that the neutralizing current has nothing to do with the power current. They are two different flows. Or can be. Depending on the arrangement of the electrodes. My arrangement separates the two currents.


I have shown you repeatedly that your arrangement doesn't actually stop the alphas. There's a momentary equilibration on the decelerator grid, and then it's a thermal machine. If you pulse it, you just wind up having to put in as much power in between pulses as you get out during them.

BTW only magnetic fields are conservative WRT particles. Because they change the velocity without changing speed. Electrostatic fields are not conservative.


You didn't look it up, did you?
Last edited by 93143 on Sat Jul 12, 2008 12:48 am, edited 2 times in total.

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Postby MSimon » Fri Jul 11, 2008 11:19 pm

Yes, BUT... The induced voltage only rises for a moment, while the nearby alpha population is rising. It equilibrates in steady state, and then you have no more current from that grid.


True enough. That is why you need a steady stream of alphas to keep drawing off energy.
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93143
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Postby 93143 » Fri Jul 11, 2008 11:26 pm

MSimon wrote:What you do is move the grounded electrode far enough back in the drift tube so that it has a negligible effect on the field between the 0V point and the entrance to the drift tube. Then all the deceleration takes place in free space.


I did. That big red region in the front of the drift tube is at almost exactly 1 MV. The problem is the rapid shift from orange to yellow to green to blue (0 V) as you approach the collector.

BTW since there is a 0 V electrode in the drift tube there has to be a 0V field line in the drift tube. I claim that line is along the centerline.


I have news for you. Equipotentials are perpendicular to field lines. There's no such thing as a "0V field line". If you mean a 0 V zero-field line, that's not at all necessary, and in fact doesn't occur in this case. If you mean a 0 V equipotential, that's the surface of the collector.

What you need to do is do a drawing of the equipotential lines. All the lines past the entrance to the drift tube are 999,999 V until you get near the 0 V electrode.


Clearly you didn't approve of my advice to look at the divisions between colours. Very well. I will replot with equipotentials and field lines and repost, once I get home.

A constant field can not affect the speed of a particle.


You mean a constant potential.

Near the collector the field tends to drive the particle to the 0V electrode since the field is repulsive.


There are two 0 V electrodes, yes? The collector being one of them, and the other too far away at the moment. The field is DEFINITELY NOT repulsive right near the collector - it is very strong and pointed straight at the collector. This is what re-accelerates the alphas.

93143
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Postby 93143 » Fri Jul 11, 2008 11:34 pm

MSimon wrote:
Yes, BUT... The induced voltage only rises for a moment, while the nearby alpha population is rising. It equilibrates in steady state, and then you have no more current from that grid.


True enough. That is why you need a steady stream of alphas to keep drawing off energy.


No no. Even with a steady stream of alphas, the total charge in the area will stabilize, because the neutralization current will supply negative charge at the same rate the alpha stream supplies positive charge.

This means that the induced voltage will stabilize at some fixed grid charge.

Did you really expect that you would be able to supply electrons to one end of an open circuit at a constant rate for years on end? You'd have to keep building up positive charge from the alphas (NO neutralization current) for that to happen. While it does appear that it would work, at least from a linear analysis, it doesn't sound like good engineering to me. We're talking about 69 coulombs of space charge buildup PER SECOND. For YEARS. Much better to let the electrons hauled in by the alphas combine with them, so you don't have a huge, ever-escalating positive charge right next to an ever-escalating negative charge.

And we're back to the power current being the neutralization current.

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Postby MSimon » Sat Jul 12, 2008 12:19 am

OK I get it.

If a particle travels from 0V to 0V it will have the same energy it started with.
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93143
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Postby 93143 » Sat Jul 12, 2008 12:38 am

OK.

EDIT: As promised, here are the equipotential and field line plots. Someone might find them interesting. Please pardon the weird spacing on the field lines; I didn't write the visualizer...

Equipotentials for grounded collector

Field lines and potential map for grounded collector

Equipotentials for high voltage collector

Field lines and potential map for high voltage collector

Note that in both cases the radial component of the field points inward, so there is, as noted, a funnelling effect on the alphas.

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Postby Aero » Sat Jul 12, 2008 3:02 pm

As you know, the Polywell represents new technology. Using this technology, the BFR generates alpha particles. The direct conversion of alpha particles to electricity is not new technology. I am surprised that no one has mentioned existing art, or known problems. A few days ago I posted a reference on this thread to an existing Alpha Generator study but when I clicked my link just now, I find that it is incorrect. Not broken, it just goes to the wrong document. I'm sorry, my bad. I guess copy and paste is not 100% reliable with humans in the loop. The document I refer to is a NASA sponsored Battelle experimental study final report. 115 pages of fundamental physics and technology of the process. It is scanned into PDF, making it big and slow to down load.
I have posted extracts from the document here: http://aeroaero.bravehost.com/ Included is the correct (I hope) link to the study report.
Do a Google search on "direct conversion of alpha particles to electricity" to see other titles.
Aero

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Postby drmike » Sat Jul 12, 2008 7:49 pm

Agreed: here's proof it's approaching 100 years of use:
The Attainment of High Potentials by the Use of Radium

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Postby MSimon » Sun Jul 13, 2008 12:28 am

I have written an apology (weak as it is) and posted it on my blog:

http://powerandcontrol.blogspot.com/200 ... wrong.html
Last edited by MSimon on Tue Jul 15, 2008 4:36 pm, edited 1 time in total.
Engineering is the art of making what you want from what you can get at a profit.
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Postby TallDave » Sun Jul 13, 2008 12:58 am

To err is human. To figure out why and admit it isn't quite divine, but it's as good as we get here on Earth.

93143
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Postby 93143 » Sun Jul 13, 2008 2:18 am

No problem; apology accepted for my part. Thanks for the argument. Sorry if I was unclear.

I was actually starting to wonder there whether I could possibly be full of it. It's happened before...

Anyway, I think I understand direct conversion about five times better than I did, just from hashing it out like that. Hopefully I can figure out whether my proposed geometry (in the Design section) is worth anything...
Last edited by 93143 on Sun Jul 13, 2008 5:12 am, edited 1 time in total.

charliem
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Postby charliem » Sun Jul 13, 2008 3:15 am

Glad this discussion ended alright.


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