Making Electricity with the p-B Polywell

[TheRadicalModerate]

Separate topic from the previous post: I had an idea for how to extract the alpha particles from the machine so they don't wind up poisoning the well.

Instead of a solid anode, you create yet another grid anode, charged to a point where almost-but-not-quite all of the energy from the alpha is dumped into it (see previous comment for my confusion for what form the energy gets dumped into the anode--I have faith that somehow it gets there).

Behind the anode, however, you place a thin-walled cathode that's filled with some fluid that's a good He-4 solvent. The cathode is charged up to, say, a couple of hundred volts. (Obviously, the cathode has to be far enough away from the anode to prevent arcing--no idea whether that will be an issue.)

Now, when the alpha gets close to the anode grid, it's got just enough energy to pass through the grid without hitting it (i.e. the grid can deflect the alphas from hitting the wires but can't repel them back towards the center of the well). So the alphas go through the grid and are now attracted to the cathode. When they hit the cathode at fairly low energy, they penetrate the cathode wall, acquire a couple of electrons to become neutral He-4, and wind up dissolved in the solvent.

If you pump the solvent away from the cathode, you can extract the He-4 atoms far away from the polywell and the next time we see them they're in a "get well soon" balloon or allowing somebody to talk funny.

Will this work?

[charliem]

So here's my problem:

We've got a big anode charged up to some high positive voltage. We're firing 8.6 MeV alpha particles at that anode. Now, we know that there are now Coulomb forces between the alpha and the metallic atoms in the anode.

However, unlike electrons, you can't make the metal atoms go anywhere interesting. So, even though we're producing a huge E-field between the alpha and the anode (even bigger than the E-field of the anode all by itself), we can't generate any current. Instead, conservation of energy says that that 8.6 MeV has gotta go somewhere, so, as best I can tell, the metal atoms in the anode are going to move as far as they can in their lattice and then vibrate, i.e. they're going to heat up.

Maybe you'll find it easier to understand if you start from a different point.

Lets say that when the machine is turned on the voltage between the magnets cover (M) and the exterior collecting grid (G) is cero, and that G is isolated from any circuit.

Of course that means that the alphas coming out of the device wont be decelerated and they'd smashed agains G with all their force (BTW, not 8.6 MeV but 2.5 and 3.5 MeV, because the energy of every p-11B fusion is divided into the three alphas that come from it). Lets imagine that G is so sturdy that it can withstand this bombardment with no problem.

Yes, at this initial moment all the energy of those alphas will become heat, but things are going to change fast.

Every alpha has a charge +2 so, no matter if they stick or bounce, they'll take 2 electrons from G.

As this process goes on G gets more and more positively charged, its potential grows, so, given enough time, it will be high enough (that is the E-field between G and M will get strong enough) that every alpha is repelled before getting to G, its initial energy/speed insufficient to reach it.

Of course that'd mean that all those alpha charges would get in the middle trying to get back to the magnets, and that THIS machine would not give any electricity (most probably just melt down like a hot ice cream ).

Lets correct this. Just before the machine reaches the condition where all the alphas are stopped short of G connect a wire to said G and start injecting electrons in it at such rate that the positive charge of G stops building up and hereafter remains stable.

Because this charge (Q) is not yet high enough, alphas would still get to G, and still take two electrons from it, but because Q is very high they'd get to G like a rain drop instead of like a bombardment, and that means very less heating.

So, our grid G is sucking electrons like mad through that wire. Where can we get them from?. Well, there are plenty of them inside the Polywell. If you connect the other side of the wire to M (the magnets cover) then they'll collect the electrons from the device's interior.

Of course that you dont connect G and M directly, G is at a very high voltage compared to M, and we can take advantage of this to charge a battery, run a machine, or use the electricity any other way we like.

You could say: Hey, if you just take electrons from the inside, electrons that you had to put there in the first place, to later on take them away, then there is no gain. Well, there is a BIG difference between the electrons that you inject in a Polywell and the electrons that go from M to G, and that is that you inject them with, lets say, 50 keV of energy, while the ones that go from M to G gain no less than 2.5 MeV. They are the same in number, but 50 times more energetic, and that's where the gain comes from.

I hope to have been of help.

My regards.

Carlos.

[MSimon]

I'm still confused. I tried to get an answer on this in a previous thread and I'm ashamed to say I didn't completely understand the answer, although looking back on it, I sort of asked the wrong question.

So here's my problem:

We've got a big anode charged up to some high positive voltage. We're firing 8.6 MeV alpha particles at that anode. Now, we know that there are now Coulomb forces between the alpha and the metallic atoms in the anode.

However, unlike electrons, you can't make the metal atoms go anywhere interesting. So, even though we're producing a huge E-field between the alpha and the anode (even bigger than the E-field of the anode all by itself), we can't generate any current. Instead, conservation of energy says that that 8.6 MeV has gotta go somewhere, so, as best I can tell, the metal atoms in the anode are going to move as far as they can in their lattice and then vibrate, i.e. they're going to heat up.

Now, you can obviously run a working fluid through the anode and extract that heat and make steam and turn a turbine, but I just don't see how you generate electrical energy directly using an electrostatically coupled anode. Clearly I was asleep during some crucial day in college physics, but I can't figure this out.

Let us start simply:

An electrostatic accelerator. Does the accelerator draw current when accelerating electrons? If not where does the energy of the electrons come from?

Now run it backwards. Fire the electrons into the accelerator at the accelerator KeV. Will the electrons come to a stop (at least momentarily)? Where did the electron energy go? If you absorb the electron exactly when it is stopped how is the total energy conserved?

[TheRadicalModerate]

Carlos, Simon--

I think the thing that's freaking me out is that we're getting power (i.e. a very small current and a gigantic voltage) out of what's essentially an open circuit. So I think I understand my major malfunction: It's not really an open circuit. We charge the anode up to the appropriate voltage, but there's a grounded load (aka the DC-AC conversion stuff that takes power out of the polywell) in parallel with the anode and its associated charging circuit. (Or, if you use Carlos's method, there is no charging circuit; we just use the alphas to charge the anode up until we decide to switch the load into the circuit.) If it weren't for that load, there'd be no way to get current to flow as the alphas came in.

If I understand you guys correctly, it's not that we're pulling a measly two electrons per alpha through some load that generates our power, it's that we're pulling the measly two electrons across a really high voltage, said high voltage being induced by the fact that the alphas are coming at the anode at multiple MeV.

The other thing that you have to take away from this example is that the neutralization of the alphas at the anode is absolutely essential to get current, and hence power, to flow through the system. In fact, the only source of current is that electrons are being slurped up by the alphas. Without that slurping, we'd build up a bigger and bigger E-field between the anode and the load, pulling more and more electrons from the load to the anode, until we hit some sort of unfortunate capacitative limit, at which time all the energy would get dumped into the anode as heat and it's adios muchachos.

Am I getting close here, or am I even more electrically incompetent than I think?

Final thought: the fact that the alpha population isn't energetically homogeneous is a huge pain in the ass. We pretty much have to eat that extra 1 MeV from the 3.5 MeV alphas as heat (requiring lots of cooling). Otherwise, if we tune the anode to bring the 3.5 MeV guys to rest, the 2.5 MeV guys will drop back into the well, recirculating and managing simultaneously to poison the p-B11 reaction and to rip up the magrid.

Given that you're going to be dumping >11% of the generated energy into the anode as heat (and I'm not even including the synchotron radiation from the alpha deceleration), I guess the next question is whether it's not more efficient from a plant complexity standpoint just to have the alphas go slamming into a thermal blanket and use a heat exchanger for everything, since you're certainly going to have to have a heat exchanger anyway.

Kind of a fun problem, isn't it?

[TheRadicalModerate]

Sorry--once I get to thinking about this stuff I get a tad obsessive.

So, since I think I've convinced myself that ion recombination is essential to get power out of direct generation, my grid anode + backing cathode with solvent idea doesn't work.

However, there may be a trick to be played if you can get the electrons flowing from the load to the anode to migrate into an electrically conductive solvent. You could then simultaneously use the solvent to carry off the He-4 and use it as your working fluid for your heat exchanger. Of course, then you'd need pretty thin walls between the surface of the anode and the fluid network. Having a coolant leak into the vacuum chamber would be, uh, exciting.

[MSimon]

You have to neutralize the ions before the braking potential re-accelerates them in the other direction.

[TallDave]

Final thought: the fact that the alpha population isn't energetically homogeneous is a huge pain in the ass.

Yeah, it's messy. But it's a hell of a lot cleaner than a thermal cycle.

(requiring lots of cooling).

There's going to be lots of cooling regardless. In fact, most likely there will be so much cooling we'll probably use the waste heat in a thermal cycle generator. Might be around 10% of total electric production (I'm guessing 40% of the energy ends up as heat, 80% of which is lost in the thermal cycle).

[charliem]

Final thought: the fact that the alpha population isn't energetically homogeneous is a huge pain in the ass. We pretty much have to eat that extra 1 MeV from the 3.5 MeV alphas as heat (requiring lots of cooling)

There's going to be lots of cooling regardless. In fact, most likely there will be so much cooling we'll probably use the waste heat in a thermal cycle generator. Might be around 10% of total electric production (I'm guessing 40% of the energy ends up as heat, 80% of which is lost in the thermal cycle).

Well, the fact that not all the alphas have the same energy should not be that hard a problem to deal with, just re-read the multi-collector grids proposal in the first page of this thread, and there are more possible configurations using electric, magnetic, or combined fields.

More problematic is the magnets heating, it will be much higher. If their magnetic field is not very high (tenths of tesla) they will be hit by a significat portion of the alphas, easily 1/4 to 1/10 of them (and that's a LOT of heat).

Carlos.

[classicpenny]

I count three other threads in addition to this one on this particular issue:
1. Where do the electons go?
2. Energy extraction alternative
and -the hot one right now-
3. Electron recirculation.
Each of them has some pretty good stuff, and of course there's a huge amount of overlap. I tried to extract the essential details from all four threads and sum up the basic details of How the pB11 polywell makes electricity at

http://www.polywellnuclearfusion.com/Cl ... ctric.html

I am sure that it is full of errors, but I would like to correct as many as possible, so any comments would be welcome.

[MSimon]

TallDave,

Steam plants are very high maintenance. They are also very expensive.

District heating and/or process steam/hot water is a better way to go.

[TallDave]

Hrm, district heating would seem to require an urban setting. That might be hard to do in America, with the word "nuclear" still a boogeyman to the left-dominated urban areas.

http://en.wikipedia.org/wiki/District_heating

http://en.wikipedia.org/wiki/Beznau_Nuclear_Power_Plant

It certainly is more efficient though. Huh, they can even power air conditioning with it.

http://en.wikipedia.org/wiki/Absorption_chiller

[rcain]

I count three other threads in addition to this one on this particular issue:
1. Where do the electons go?
2. Energy extraction alternative
and -the hot one right now-
3. Electron recirculation.
Each of them has some pretty good stuff, and of course there's a huge amount of overlap. I tried to extract the essential details from all four threads and sum up the basic details of How the pB11 polywell makes electricity at

http://www.polywellnuclearfusion.com/Cl ... ctric.html

I am sure that it is full of errors, but I would like to correct as many as possible, so any comments would be welcome.

what a nice concise description - well done classicpenny.

i don't know if its right (someone else can answer that), but i understand it

[TheRadicalModerate]

classicpenny, I've got the following objections to the deceleration scheme on your page:

1) The force on a charged particle inside a charged conducting sphere is everywhere 0, so deceleration can't happen. (Note that this is not a faraday-cage effect, it's a geometrical effect.)

2) The two-grid decelerator won't work because it's a grid. If you fire an alpha particle at the grid and it doesn't strike the grid (i.e., it goes through empty space in the grid), then it won't recombine and will eventually drop back into the well. I think you need solid conductors, which means you can only have one grid. Also, since it's essential to capture the alphas reliably, you'd need to place the collector so that all particles have enough residual energy to bury themselves in it.

3) A vacuum pump ain't gonna cut it as a means of collecting the He4. The whole machine is already a pretty hard vacuum. The only removal scheme I've been able to think of is to give the alphas enough energy to penetrate the collector and wind up in coolant fluid. The fluid can then be pumped off and the He4 removed.

Of course, objections 2 and 3 are moot in the face of objection 1. This has come up several times on the various threads but we don't seem to have reached consensus on whether it's a problem. Can somebody who can actually do physics properly comment on this objection?

[93143]

1) The force on a charged particle inside a charged conducting sphere is everywhere 0, so deceleration can't happen. (Note that this is not a faraday-cage effect, it's a geometrical effect.)

The force on a charged particle inside an EMPTY charged conducting sphere (or any other closed shape) is everywhere 0. This is true for the same reason that a Faraday cage works. Once the alphas pass the trap grid, they can be slowed down by it until they hit the wall (or collector plates, in a parallel repeller+plate geometry or something).

One other thing that bugs me about classicpenny's explanation is that the "control grid", or what I would call the trap grid, looks too close to the magrid for effective electron recirculation, taking the magnetic field into account. Could be just me.

3) A vacuum pump ain't gonna cut it as a means of collecting the He4. The whole machine is already a pretty hard vacuum. The only removal scheme I've been able to think of is to give the alphas enough energy to penetrate the collector and wind up in coolant fluid. The fluid can then be pumped off and the He4 removed.

This has been bothering me for a while. The helium atoms are essentially going to be in the free molecular flow regime, which means you can't 'suck' them with anything; they just have to bounce around long enough to hit the pump intake. Your idea probably won't work either; alpha particles don't have much penetrating power even at MeV energies.

I don't suppose you could add an extra charge to the alphas and make them negative helium ions, could you? This might work with my repeller+plate idea; there could be a positive (relative to the collector) vacuum pump at the base of the repeller where no alphas go. Negative ions would love it there. Unfortunately I very much doubt that you'd be able to guarantee nearly 100% negative ionization of helium under those circumstances (especially in alpha crossflow...).

[TheRadicalModerate]

The force on a charged particle inside an EMPTY charged conducting sphere (or any other closed shape) is everywhere 0. This is true for the same reason that a Faraday cage works. Once the alphas pass the trap grid, they can be slowed down by it until they hit the wall (or collector plates, in a parallel repeller+plate geometry or something).

Just to be clear, you're back to discussing a trap grid where all the energy is extracted by a negatively charged grid as the alphas fly away from it, rather than classicpenny's positively charged collector.

This one always makes my head hurt. I can't make the charges balance to do useful work. As the alpha passes the grid, it pulls more electrons towards the grid, but they flow back out of the grid as the alpha gets further away. Where's the net current? As usual, I'm sure I'm missing something...