Magnetically Shielded Fusor Grids--Why Won't This Work?

[MSimon]

Tell me - what percentage of the heat in a grid is due to current and what is due to ion collisions? And then, what percentage of mobile ions hit the grid? To answer, without speculating, would require objective measurements of these factors.

Let me see Chris. You have wires from 2 to 10 mm in diameter with a resistance of under 10 ohms carrying a 10 to 100 mA current.

This is another case Chris where you are arguing without facts.

So let me see, assuming a very high resistance and a very high current (10 ohms and 100 ma) what is the power dissipated from current flow? That would be .1 watt.

Run the numbers Chris.

[D Tibbets]

Dan, I have no contest to the points you, or the others, are making. It's just that they don't relate to mine.

The original posit here was that "fusors with grids won't generate net power because the ions collide with the grid before they transit the center enough times to fuse", which is wrong. It is not the grid that means the fusor won't generate net power.

I'm saying that this statement "A follows from B" is false, and you're saying "but A is true and B is true, so A must follow B". I've said nothing explicitly about A or B, I agree that both are true. I'm saying only that one isn't consequent to the other. This is a classic non-sequitur. I think it's called 'denying the conjunct', or someting like that.

You implied that the grid life before melting was limited by Ohmic heating, not ion bombardment. I suspect that you might be referring to is the occurrence of fusion events not being effected much by the ~ 1 % loss rate of energetic ions per pass, but that the input energy to replace those energetic ions has to go way up to maintain the same fusion rate as a grid less fusor. ie- fusion power will not change much between a Polywell and a gridded fusor, though the Polywell will need less input power (at least that portion of the input power that goes into accelerating the ions). I've wondered about this myself. The answer (if you accept Polywell claims) must rest with the nature of the plasma. In a typical gridded fusor the 'plasma' is dirty- a predominance of neutrals. The ion population is more random/ thermalized. The polywell has better focus, a much smaller percentage of neutrals getting in the way (any surviving neutrals will tend to accumulate in the larger volume outside the magrid and be preferentially pumped away, or crash the machine. There is a more pure ion population with better focus (apparently doesn't have to be great- what that means I don't know, all the ions passing thorough the central 1 cm instead of the central 1 mm?) so the effective density (Nebel mentioned ~ 10e22ions/cubic meter) can be much higher than in a gridded fusor. That central density combined with the claimed nonmaxwellian distribution means that most of the ions are at fusion conditions, as opposed to a small tail, and nearly all of the ions are passing through this small central volume on each of their passes.

Weather my understanding of the claimed physics is possible does not have any bearing on the experimental results (which are presumably real ,if unaviable for public consumption). If one rejects or denies the existence of the data or it's accuracy, one can argue any point based on any perceived physics. If the data is real, then the physics arguments have to be squeezed into reality. This currently open ended situation is what makes for entertaining and educational debates on this site.


Dan Tibbets

[chrismb]

Tell me - what percentage of the heat in a grid is due to current and what is due to ion collisions? And then, what percentage of mobile ions hit the grid? To answer, without speculating, would require objective measurements of these factors.

Let me see Chris. You have wires from 2 to 10 mm in diameter with a resistance of under 10 ohms carrying a 10 to 100 mA current.

This is another case Chris where you are arguing without facts.

So let me see, assuming a very high resistance and a very high current (10 ohms and 100 ma) what is the power dissipated from current flow? That would be .1 watt.

Run the numbers Chris.

As electrons move from the grid into the region around them, they will enter a region of different potential, viz, is there not a resistance at the boundary between the grid and the Townsend discharge plasma? It is not a continuous conductor, like a light bulb, but is serving the purpose of creating a discharge in a plasma. A potentially(!) very non-ohmic situation.

Don't get me wrong, I am not speculating that the heat in the grid is only due to resistance heating. Of course not. Actually, I think it's more to do with the central plasma being in contact with the grid. The fast ions loose their energy to the central plasmoid, then the plasmoid looses its energy to the grid. But I refer to my previous post. I am arguing that A doesn't necessarily follow B just because they're both true. There are many other factors.

[MSimon]

Tell me - what percentage of the heat in a grid is due to current and what is due to ion collisions? And then, what percentage of mobile ions hit the grid? To answer, without speculating, would require objective measurements of these factors.

Let me see Chris. You have wires from 2 to 10 mm in diameter with a resistance of under 10 ohms carrying a 10 to 100 mA current.

This is another case Chris where you are arguing without facts.

So let me see, assuming a very high resistance and a very high current (10 ohms and 100 ma) what is the power dissipated from current flow? That would be .1 watt.

Run the numbers Chris.

As electrons move from the grid into the region around them, they will enter a region of different potential, viz, is there not a resistance at the boundary between the grid and the Townsend discharge plasma? It is not a continuous conductor, like a light bulb, but is serving the purpose of creating a discharge in a plasma. A potentially(!) very non-ohmic situation.

Don't get me wrong, I am not speculating that the heat in the grid is only due to resistance heating. Of course not. Actually, I think it's more to do with the central plasma being in contact with the grid. The fast ions loose their energy to the central plasmoid, then the plasmoid looses its energy to the grid. But I refer to my previous post. I am arguing that A doesn't necessarily follow B just because they're both true. There are many other factors.

I don't disagree. But that is not the position you started from if I understand your words.

As to losses. The grids introduce a loss factor of 2% to 5% per pass of the ions. It takes 60 passes to get a fusion (average). So the fusor cannot generate net power due to the loss of high energy ions to the grid (I X E vs IR squared).

The estimate is that the fusor could get to break even with grid losses of under 1%. Thus magnetic shielding of the grids.

However magnetic shielding changes how the device operates. Which has been very productive of discussions.

[chrismb]

As to losses. The grids introduce a loss factor of 2% to 5% per pass of the ions. It takes 60 passes to get a fusion (average).

You're in a dream-world!

It takes 60 billion passes, more like, to get an 'average' fusion! You can see instantly that if there really were 1% losses per pass, then there'd be essentially no chance at all!

[MSimon]

As to losses. The grids introduce a loss factor of 2% to 5% per pass of the ions. It takes 60 passes to get a fusion (average).

You're in a dream-world!

It takes 60 billion passes, more like, to get an 'average' fusion! You can see instantly that if there really were 1% losses per pass, then there'd be essentially no chance at all!

60 or 60 billion? Depends on density.

Which takes us to a whole different set of discussions. When it comes to fusors the number 60 is generally accepted. Fusors operate at relatively high pressures. Polywell operates at low pressure with a high density ratio (according to some) external to internal.

Can such a density difference be maintained? Depends on how leaky you think the balloon is.

[KitemanSA]

I don't think you guys are using the term "pass" in the same way. Sounds like Chris is using it to mean each time one ion passes another while MSimon is using it to mean trip thru the core. No?

[chrismb]

As to losses. The grids introduce a loss factor of 2% to 5% per pass of the ions. It takes 60 passes to get a fusion (average).

You're in a dream-world!

It takes 60 billion passes, more like, to get an 'average' fusion! You can see instantly that if there really were 1% losses per pass, then there'd be essentially no chance at all!

60 or 60 billion? Depends on density.

Which takes us to a whole different set of discussions. When it comes to fusors the number 60 is generally accepted. Fusors operate at relatively high pressures. Polywell operates at low pressure with a high density ratio (according to some) external to internal.

Can such a density difference be maintained? Depends on how leaky you think the balloon is.

We're talking about FUSORS!

A typical fusor running at 40kV and 20 microns will give a healthy output of detectable neutrons, and is around the target settings for an amateur fusor (with these 'supposed grid losses').

40kV = 20kV collision energy = ~0.5millibarns.
20 microns = ~n=5E20.
mean distance to a fusion event = 1/(0.5E-31[m^2].5E20[/m^3])= 40 billion metres.

Which takes us to a whole different set of discussions. When it comes to fusors the number 60 is generally accepted.

Accepted by whom??

[Art Carlson]

As to losses. The grids introduce a loss factor of 2% to 5% per pass of the ions. It takes 60 passes to get a fusion (average).

You're in a dream-world!

It takes 60 billion passes, more like, to get an 'average' fusion! You can see instantly that if there really were 1% losses per pass, then there'd be essentially no chance at all!

Fact check: The peak of the D-T cross section is about sigma = 8 barns = 8e-28 m^2. At 10 T the magnetic pressure is (10 T)^2/(2*4*pi*10^-7) = and 100 keV, the density would be 4e7 Pa. If the density of one ion species is n, the that of both species together is 2n, and that of the ions plus electrons is 4n, so if T is 100 keV and beta = 1, then (3/2)*(4n)*(1.6e-19 J/eV)*(1e5 eV) = 4e7 Pa, or n = 4e20 m^-3. That gives us 1/(n*sigma) = 3e6 m. If the diameter is 1 m, then this amounts to 3 million passes, somewhere near the geometric mean of 60 and 60 billion.

It's late and I did this quickly, so I hope someone will tell me if I misplaced 4 orders of magnitude. It wouldn't surprise me. Or did chrismb have p-B11 in mind?

[chrismb]

It's late and I did this quickly, so I hope someone will tell me if I misplaced 4 orders of magnitude.

That seems OK Art. I was talking DD at 40kV drive, because the whole thread started with why fusors are believed not to work, so this discussion is about fusors.

[TallDave]

Well, you can't do much if your grid melts, so I think it's fair to say grids are the biggest problem with IEC reactor viability, even if ion collisions with the grid aren't.

That's not to say Chris isn't correct about the other problems. Fixing the grid problem is necessary, but far from sufficient.

You can see instantly that if there really were 1% losses per pass, then there'd be essentially no chance at all!

I'm not sure what model you think would explain how fusors actually get fusion, then, given the amount of physical space the grid occupies. Would you like to calculate an upper limit on intercept area?

[chrismb]

Fixing the grid problem is necessary, but far from sufficient.

Well, maybe. Is the grid a problem? It's a moot point, thermalisation is the 'problem' to a fusor getting net-outputs. Fix that, then we can see if the grid needs replacing.

I'm not sure what model you think would explain how fusors actually get fusion, then, given the amount of physical space the grid occupies. Would you like to calculate an upper limit on intercept area?

Zero. The intercept area for a beam of fusible 'fuel ions' in a fusor is zero, and around 40% of the mobile ions occupy those channels, providing the beam currents are not so high as to cause divergence. The ion channels are self-generating, for the simple reason that if they didn't run along those beam lines, then they'd not exist! It is self-ionisation of the ions and electrons, they 'feed' off each other - a bit like a laser, if you like. Those ions (read 'photons' for the laser) that do not meet the 'vector criterion' early enough to avoid getting wasted don't participate in further stimulation of further ions/electrons along that beam path. Once you start getting a population of ions with a long residence time on particular beam paths, so they get to 'breed' more and more ions until a stable population builds up - but only along those beam lines.

Ions are generated throughout the fusor, this is true. And it is certainly true that many bombard the grid. Of this there is no doubt. Is this why the grid gets hot? Who cares, a good fraction of the fuel ions never encroach on the grid, they are born, and live, in these stable beam paths.

Am I making sense yet? Look, if an ion is not following a beam path, then say it makes a dozen passes before hitting the grid and in that distance it doesn't get to ionise any more background molecules, so it just 'dies'. But an ion that just happens to be running back-and-forth on a 'good' beam path gets to ionise (breed, if you like) more ions, but on that path. Because they are radials, in line with the efield, so those go on to generate more ions, etc. etc, but specifically on that path. It's a case of 'survivial of the [fittest] straightest'. Whichsoever ions remain the longest in their beam paths get to 'father' the most 'off-spring' ions in their beam path.

That's why if you watch a fusor start up, the beams form slowly, not straight away. If it were otherwise, the beams should spring up straight away, within the time-of-flight of the ions across the chamber.

Capiche? Has the light dawned yet? The beams are self-organising, if you like, but certainly self-sustaining.

[TDPerk]

The simplest grid which could be magnetically shielded would be a single coil, correct?

Has anyone modeled what happens to a highly (electrostatically) charged single coil in a hydrogen plasma? If the chamber walls were far enough away from the charged surface, they don't hit the walls at any especially fast rate, they are attracted to the ions, not the walls. The ions are presumably hot and moving around the coils, but cannot simply impact it because of the magnetic field.

Who's done this before?

[TallDave]

It's a moot point, thermalisation is the 'problem' to a fusor getting net-outputs.

Shrug. One could just easily say the thermalization problem is moot because no grid could withstand net-output energies. The only logically consistent formulation is that both are necessary and neither is sufficient.

Zero.

Huh? That would mean fusors don't make fusion, since they all have nonzero intercept area.

and around 40% of the mobile ions occupy those channels, providing the beam currents are not so high as to cause divergence.

Sure, but what about the other 60% that ARE hitting the grid? This is only roughly the difference between 2% losses and 1% losses.

Who cares, a good fraction of the fuel ions never encroach on the grid, they are born, and live, in these stable beam paths.

The other fraction banging into the grid is generally believed to be the largest loss factor in a fusor. Given their intercept area, I don't see how your objection to that claim stands up.

[chrismb]

It's a moot point, thermalisation is the 'problem' to a fusor getting net-outputs.

Shrug. One could just easily say the thermalization problem is moot because no grid could withstand net-output energies. The only logically consistent formulation is that both are necessary and neither is sufficient.

But the grid burning out ISN'T WHY a fusor can't reach net-power!!! Gee... this debate isn't about what a fusor could become, it's about the erroneous comment that a fusor doesn't get to break-even BECAUSE of the grid. Even if the grid lasted a millisecond and then burnt out, you'd still see if net-power was gained. It isn't. If you crank up voltage and current, the reaction rate levels off at the ~1E-8 level.

Zero.

Huh? That would mean fusors don't make fusion, since they all have nonzero intercept area.

I thought you meant intercept with the grid. That is what we were talking about.

and around 40% of the mobile ions occupy those channels, providing the beam currents are not so high as to cause divergence.

Sure, but what about the other 60% that ARE hitting the grid? This is only roughly the difference between 2% losses and 1% losses.

But this isn't the issue. This merely shows that if you fix the grid, you'll gain x2 improvement. I agree you probably would. And the remainder ~5E7 improvement comes from....???

Who cares, a good fraction of the fuel ions never encroach on the grid, they are born, and live, in these stable beam paths.

The other fraction banging into the grid is generally believed to be the largest loss factor in a fusor. Given their intercept area, I don't see how your objection to that claim stands up.

Now you're confusing 'loss factor' with why the fusor can't get to over-unity. Quite obviously - QUITE obviously - in one way or another the few 100W going into a fusor comes out as heat, and the only things that can get hot are the shell and the grid. The shell's big and in air, the grid's small and in a vacuum. Clearly, it doesn't take a genius to figure out the grid is gonna get hot one way or the other. Between the heat going into the grid and the shell, that is 100% loss.

I think you're trying to say how much power is lost that aren't going into fusible fuel ions. And to this you will go from, say, 1E-12 to 2E-12 of the energy going in. The idea that you've reduced your 'losses' from 1-[1E-12] down to 1-[2E-12] is a point of pedantery not worth commenting on. The error here comes in not understanding the question.