Questions around the copper Anode

[jamesr]

From the report linked to above:

From the total energy estimate and the electron energy, we can get an estimate of 4 mC for the total charge in the beam. If we then assume that the beam is spreading out linearly and covers a 1.4 cm radius circle at a distance of 4 cm

This is quite a big area – so I would think if the it were replaced by a fine mesh, rather than a simple hole, that let most of the electrons pass through to another small chamber behind the anode, they could be slowed down as you describe (extracting their energy in the process).

[Aeronaut]

I’ve never been clear about the mechanism that recycles the electrons back into the plasmoid. But the power of that X-ray blast is a much more serious concern to me than the e-beam erosion. If nothing else, a few anodes with varying dimple dimensions could run some tests to minimize erosion caused by the ebeam.

[Allan Brewer]

So do you have doubts about the capability of the “onion” – I thought it was proven technology?

[Aeronaut]

Allan Brewer wrote: So do you have doubts about the capability of the “onion” – I thought it was proven technology?

The science was proven back in Maxwell’s and Hertz’s era. The challenges are going to be in areas like tuning groups of layers for particular energy bands, and developing the tooling, fixtures, procedures, and all of the other engineering challenges that would get FF into mass production.

But given $1M today, I’d love a crack at positioning FF to clean up cleaning up down in the Gulf of Mexico. 😉 So much for the phrase “due diligence”.

[vansig]

psupine wrote: As I understand it, one of the techniques used in electron tubes (aka valves) to reduce electrode erosion is to a maintain a negative potential on the “target” electrode. This decelerates the free electrons (that had been accelerated by the grid potential) so that most of the energy has been taken out of them and the electrons impact the plate at low energy. This is a bit like a lunar lander game played out on a very small scale.

Electrons travel through the vapourized decaborane, heating up the vapour to plasma on their way from the cathodes to the anode. The anode must be positively charged in order to attract the electrons, but the electrons give lots of their energy to the plasma. If 90% of the energy in a 45 keV electron were to go to the plasma, then the electron would be travelling slowly when it hits the anode (but i dont know the actual percentage). This process begins around the outside edge of the anode, where the cathodes are closest to it. The charge of the anode drops considerably as the electrons hit it and the plasma tendrils climb up.

The tendrils climb up and over the edge and become the plasmoid; so the parts of the electrodes most-exposed to heating vary through this pulse. All this happens on the order of nanoseconds, therefore skin effect will be important: the charges will be confined to the surfaces of the electrodes almost exclusively.

My understanding of the superiority of Beryllium is that it is much more transparent to x-rays, so it wont heat as much as copper. But otherwise its heat capacity and thermal conductivity counteract its higher melting temperature. It seems necessary to use a thin coating of a much higher melting temperature, thermally conductive material. (eg: graphite? single-walled nanotubes? )

But if, after the shot, the anode is turned slightly on its axis, then the next shot will contact a different, perhaps cooler, part of the surface.

[jamesr]

vansig wrote:

As I understand it, one of the techniques used in electron tubes (aka valves) to reduce electrode erosion is to a maintain a negative potential on the “target” electrode. This decelerates the free electrons (that had been accelerated by the grid potential) so that most of the energy has been taken out of them and the electrons impact the plate at low energy. This is a bit like a lunar lander game played out on a very small scale.

Electrons travel through the vapourized decaborane, heating up the vapour to plasma on their way from the cathodes to the anode. The anode must be positively charged in order to attract the electrons, but the electrons give lots of their energy to the plasma. If 90% of the energy in a 45 keV electron were to go to the plasma, then the electron would be travelling slowly when it hits the anode (but i dont know the actual percentage). This process begins around the outside edge of the anode, where the cathodes are closest to it. The charge of the anode drops considerably as the electrons hit it and the plasma tendrils climb up.

The tendrils climb up and over the edge and become the plasmoid; so the parts of the electrodes most-exposed to heating vary through this pulse. All this happens on the order of nanoseconds, therefore skin effect will be important: the charges will be confined to the surfaces of the electrodes almost exclusively.

My understanding of the superiority of Beryllium is that it is much more transparent to x-rays, so it wont heat as much as copper. But otherwise its heat capacity and thermal conductivity counteract its higher melting temperature. It seems necessary to use a thin coating of a much higher melting temperature, thermally conductive material. (eg: graphite? single-walled nanotubes? )

But if, after the shot, the anode is turned slightly on its axis, then the next shot will contact a different, perhaps cooler, part of the surface.

You seem to be mixing a few different things.

At the start the high voltage in the capacitors is switched and the potential on the anode jumps to this 20-45kV. after the initial breakdown electrons in the plasma arc are accelerated by the E-field, but due to the relatively high pressure will undergo many collisions and quickly a drift velocity as the short accelerations between collisions average out. This movement of electrons forms the growing current (as more electrons are involved in it – not that they are going much faster) in the run-down and axial phase over microseconds, not nanoseconds. The energy (temperature) of the electrons reaching the anode in these phases is still fairly small. As you correctly say most of the heating of the bulk of the anode is due to the resistive heating in the thin skin where the MA size current flows. But this only would raise the surface temp by a degree or so each shot. So assuming in the milliseconds between shots, this heat has enough time to conduct down through the bulk of the anode where the cooling is provided, then over many shots the temperature of the surface will not get too high.

The plasmoid which forms at the focus lasts the tens of nanoseconds and heats the electrons and spits them out as a high energy beam of upto the 45keV quoted. It is this small population of very high energy electrons doing damage that we are talking about.

[nemmart]

psupine wrote: I’m still trying to understand some of the subtleties too (I bet we all are!), so please excuse me if this question is so bad that it isn’t even wrong …

Doesn’t conservation of momentum require a significant e-beam to balance the ion-beam from which we hope to extract significant energy? So if optimality has the e-beam energy reduced as much as possible, I’d have thought that the ion-beam is similarly down to nothing or else the plasmoid shoots sideways (or rather in FuFu, up).

I’m sure someone can set me straight. I hope to get there eventually.
Thanks

Lerner wrote: It is very difficult for the electrons to balance the momentum of the ions which are thousands of times heavier. Two things can absorb the momentum of the ion beam–the motion of the plasmoid, or the magnetic field that the plasmoid is tied to. It is probably mostly the latter, since otherwise the plasmoid would smash into the anode before the pulse ended.

I think this is a really interesting question. I don’t understand why one should expect the alpha particles to all go in the right direction, down the axis of the anode. When the excited C12 fissions, I would expect the alpha particles to go shooting out in random directions, with piles of energy. More energy than is in the plasma. Is there a good reason to believe the plasma can capture this energy as opposed to the alpha particle punching through it?

When you run a DPF with deutrium what what percentage of the fusion energy gets captured by the plasma? Does the amount of neutron radiation change with direction?

[jamesr]

nemmart wrote:

I think this is a really interesting question. I don’t understand why one should expect the alpha particles to all go in the right direction, down the axis of the anode. When the excited C12 fissions, I would expect the alpha particles to go shooting out in random directions, with piles of energy. More energy than is in the plasma. Is there a good reason to believe the plasma can capture this energy as opposed to the alpha particle punching through it?

When you run a DPF with deutrium what what percentage of the fusion energy gets captured by the plasma? Does the amount of neutron radiation change with direction?

The alphas for p+B11, or the neutrons in the case of D+D->He3 + n, will of course be emitted in random directions. But the very high magnetic field will cause any alphas to spiral round the field lines with a radius of:

r=mv/qB

where v is the velocity perpendicular to the magnetic field. For a 3MeV alpha emitted perpendicular to a 1GG field the radius is r=2.5E-6m. However the denisty of the plasma where the fusion occurs will mean it will have many collisions before it gets that far and slow quickly to the thermal ion temperature. At 100keV the radius drops to only 4.5E-7m, so the ions are confined easily by the magnetic field in the plasmoid.

Only when the magnetic field collapses creating a large electric field along the axis of the plasmoid are the ions accelerated out in the narrow beam

[vansig]

jamesr wrote:
Only when the magnetic field collapses creating a large electric field along the axis of the plasmoid are the ions accelerated out in the narrow beam

what is the exit velocity?

[jamesr]

vansig wrote:
what is the exit velocity?

I would think there would be quite a spread of speeds, partly from the thermal Maxwellian speed distribution of the ions, but also their orientation to the E & B fields. But the peak would still be around the peak of the thermal distribution. So at 100keV the speed is ~2000km/s or 0.007% of c. Rising to 5400km/s at 600keV

[Francisl]

Beryllium is a fairly reactive metal. Will there be a problem with beryllium electrodes forming beryllium borides? Would a copper surface take care of the problem?

[annodomini2]

Maybe I missed something in an earlier post, why is everyone avoiding the hole in the anode questions?

[Francisl]

annodomini2 wrote: Maybe I missed something in an earlier post, why is everyone avoiding the hole in the anode questions?

The important reactions take place in and around the hole. Please look at these videos: DPF Animation and Focus Fusion movie

[Tulse]

This may have been discussed elsewhere, but why is the anode a single cylinder, rather than a set of rods like the cathodes?

[Henning]

Tulse wrote: This may have been discussed elsewhere, but why is the anode a single cylinder, rather than a set of rods like the cathodes?

This hasn’t been discussed in the forum yet, only the other way round: Why is the outer electrode made out of rods? Old-day DPFs had a hollow cylinder as outer electrode, modern DPFs have those rods. It helps forming of the filaments (if I remember correctly).

Maybe a sawtooth shaped inner electrode would help guiding those filaments? Maybe not. Maybe they break if guided too tightly.

[Author]

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