Where's the beef?

[Aero]

The WB-6 final report gives the following summary table:

Summary of last 5 tests
1) 5.0kV, 800A B-field, 1 count
2) 9.8kV, 750A B-field, 2 counts
3) 12.5kV, 700A B-field, 2 counts
4) 12.5kV, 800A B-fiel(1 kG) d, 3 counts
5) 14kV 1000A B-field, 1 count (while leg failing)

I hope this helps.

[93143]

Thanks. Edited my post to comply with the posted data.

[Art Carlson]

I think my calculation of the confinement time can be simplified a bit. What we really care about is energy confinement. The energy inventory is more reliable than the particle inventory at a given pressure because it does not depend on the temperature. The energy input is also easy to calculate as I*V, whether or not the dirve current has anything to do with the particle feed rate. The same number will come out, but on a more solid footing.

I still don't have a complete picture of how the losses from non-neutral cusp confinement fit together. One image where at least some of the pieces fit together is that the ion loss is reduced by the square root of the mass ratio. If I throw in this factor, I "predict" about the right confinement time for WB-6. Bussard's extrapolation to the reactor regime is based on "whiffle-ball confinement", which I take to mean an effective loss area proportional to rho^2. I think he wants to multiply this confinement time by 10,000 to take credit for recirculation, but this point is not clear to me. I think there will not be much recirculation at all, and the the effective loss area will be proportional R*rho (line cusp confinement). The scaling for the ratio of fusion power to drive power is R^3*B^4 for Bussard and R^2*B^3 for me. If we assume Bussard did his arithmetic right and got net power by scaling radius and field each a factor of 10 to a 1 m radius and 1 T field, then my scaling law would require a factor of 25 to a 2.5 m radius and 2.5 T field, and "my reactor" would produce 600 times more fusion power than "his". If nature takes the line cusp root, the size of the net power device might be OK, but the fusion power level will be rather too hot to handle. If the smallest possible Bussard reactor produces 100 MW, the smallest possible Carlson reactor would produce 60 GW. (I am accepting here the argument that the field can be scaled with the radius, and that this does not screw up the physics.)

I have the feeling we have been fighting on the wrong battle field, but I'm not sure where we should be. Maybe the key point is that Bussard thinks the power input to run a polywell at beta = 1 is in the 10 MW range, regardless of the size, and with line-cusp scaling the input power rises with R^2, which may put you into an un-usable regime before the fusion power catches up.

[93143]

Hang on, what? As far as I can recall, Bussard has always said that power out scales with the 7th power of the linear dimension (ie: B^4*R^3) and gain scales with the 5th power of the linear dimension. The required input power does increase as the device gets larger, but not as fast as the fusion power.

I'm somewhat mentally exhausted right now for unrelated reasons, but I'll try to get back into these discussions once I can think properly again...

[Art Carlson]

Hang on, what? As far as I can recall, Bussard has always said that power out scales with the 7th power of the linear dimension (ie: B^4*R^3) and gain scales with the 5th power of the linear dimension. The required input power does increase as the device gets larger, but not as fast as the fusion power.

I'm somewhat mentally exhausted right now for unrelated reasons, but I'll try to get back into these discussions once I can think properly again...

If the fusion power increases as R^7 and the gain increases as R^5, then the drive power increases as R^2. If we think in terms of effective loss area A_loss, the drive power should be nkT*v*A_loss, where nkt scales with R^2 because of beta (and B~R). Therefore - if your reading of Bussard is right - A_loss must be independent of R. This would be the case with line cusps, because the length increases with R but the width decreases with B. It would not be so for point cusps, whose area should decrease with B^2.

On a related point, does anybody know whether and how recirculation figures into the gain according to Bussard?

[TallDave]

This would be the case with line cusps, because the length increases with R but the width decreases with B. It would not be so for point cusps, whose area should decrease with B^2.

The cusps aren't loss mechanisms (due to recirculation). Losses are from cross-field transport and unshielded surfaces (interconnects/mounts).

5. In these systems electron loss phenomena are solely
to (metal) surfaces of the machine system. Cross-field
losses are well understood and can be controlled. However,
losses to poorly shielded (by fields) or unshielded surfaces
can constitute major loss channels. From WB-5 and WB-6 it
has been proven that that the fractional area of unshielded
surfaces must be kept below 1E-4 to 1E-5 of the total surface
area, if electron losses are to be kept sufficiently small so
that net power can be achieved. And, further, that no B
fields can be allowed to intersect any such internal surfaces
of the machine.

On a related point, does anybody know whether and how recirculation figures into the gain according to Bussard?

It eliminates the cusp losses.

[Art Carlson]

TallDave, do you have any idea how Bussard calculated electron losses and their scaling then? Classical cross field diffusion? I have also heard no satisfactory reason why ions loss through the cusps should not be an important channel.

[MSimon]

TallDave, do you have any idea how Bussard calculated electron losses and their scaling then? Classical cross field diffusion? I have also heard no satisfactory reason why ions loss through the cusps should not be an important channel.

Because they don't have enough energy to get out and up scattering is not significant due to annealing.

I don't know if those are facts. They are the explanation.

[Art Carlson]

TallDave, do you have any idea how Bussard calculated electron losses and their scaling then? Classical cross field diffusion? I have also heard no satisfactory reason why ions loss through the cusps should not be an important channel.

Because they don't have enough energy to get out and up scattering is not significant due to annealing.

I have presented a calculation that says that the electrons punch a hole in the well seen by the ions. That is why I don't call "they don't have enough energy to get out" a satisfactory explanation.

I don't know if those are facts. They are the explanation.

So you just don't know how Bussard arrived at his gain scaling. Does anybody here? If not, I propose we ignore it.

[kcdodd]

I have presented a calculation that says that the electrons punch a hole in the well seen by the ions. That is why I don't call "they don't have enough energy to get out" a satisfactory explanation.

I don't see how you showed that. The polywell is symmetric and electric fields superimpose, even fields generated by electrons all the way on the other side. Even if there is high net charge inside cusps, they are symmetricaly high on the opposing side. So the way I see it the net effect of electrons outside the core should not be great.

If you are assuming these flows are quasi neutral, which I think you said you said you are, that is against the whole idea of the polywel. Of course if you assume quasi neutrality then by definition there will always be nearly equal ion density accompanying electrons. So of course they must be flowing out with the electrons since they must be there by definition! What have you proven if you already started with the answer? If I missed it please point me to where you showed that the machine must be quasi neutral everywhere.

[Art Carlson]

I have presented a calculation that says that the electrons punch a hole in the well seen by the ions. That is why I don't call "they don't have enough energy to get out" a satisfactory explanation.

I don't see how you showed that. The polywell is symmetric and electric fields superimpose, even fields generated by electrons all the way on the other side. Even if there is high net charge inside cusps, they are symmetricaly high on the opposing side. So the way I see it the net effect of electrons outside the core should not be great.

If you are assuming these flows are quasi neutral, which I think you said you said you are, that is against the whole idea of the polywel. Of course if you assume quasi neutrality then by definition there will always be nearly equal ion density accompanying electrons. So of course they must be flowing out with the electrons since they must be there by definition! What have you proven if you already started with the answer? If I missed it please point me to where you showed that the machine must be quasi neutral everywhere.

To make sure we are on the same page, I am referring to this calculation:

In the polywell (and most magnetic confinement devices), the electron gyroradius is close to the Debye length, lambda_D = sqrt(epsilon*kT/(n*e^2)). The electric charge per unit area [Note: corrected error in original] due to a sheet of charge a Debye length thick is on the order of n*e*lambda_D = sqrt(epsilon*n*kT), and the electric field discontinuity across such a sheet is sqrt(n*kT/epsilon) = (kT/e)/lambda_D. In a geometry with a scale length R, the potential of the sheet is on the order of (kT/e)*(R/lambda_D), which is huge compared to the voltages in the system. I conclude that the potential of the fan of escaping electrons is not a minor perturbation, but is an effect of zeroth order. Not "too small" at all.

(More generally, if the sheet has a thickness delta, the potential is on the order of (kT/e)*delta*R/lambda_D^2.)
I'm not sure where I lost you. I don't assume quasineutrality. On the contrary, the method of proof is to assume that the cusp sheets are composed primarily of electrons, and to show that that charge configuration would lead to very high negative potentials in the sheets. (I am also using the fact that the potentials and the electron temperature are comparable.) I then conclude that this potential would be negated by the ions it attracts, i.e. that quasineutrality must then result.
It is true that an un-neutralized sheet on the opposite side of the sphere would make similar contributions to the electric potentials and the electric fields. Perhaps it helps to think of a two-stage process. Initially, i.e. with spherically symmetrical ions but electrons partly in a sphere and partly in sheets, there is a large tangential electric field near the sheets and pointing toward them. (The distant sheets will make only a small net contribution to the tangential electric field.) This will pull the ions to the base of the sheets. Once an ion clump forms, it will largely cancel the electric field at distant points due to its own electron sheet. Then the ions will see primarily a radial electric field, each clump due to its local sheet, which will pull them out of the sphere.

[drmike]

I guess I don't see how the cusp sheet can form. If the ions are radial they will form the "spiky" shape and extend farther out on the cusp lines. Since they have more momentum than electrons they'll tend to hold the electrons in. You still have cyclotron motion along the line cusps so any transverse field helps mirror action.

I do think the bleed rate from the cusps has to be limited. Even a loss of 10^-4 will make net power impossible. That is why there are electron current injectors outside the MaGrid. But I think that loss has more to do with maintaining the virtual cathode which helps sustain fusion.

Or maybe we are saying the same thing with different words.

[TallDave]

TallDave, do you have any idea how Bussard calculated electron losses and their scaling then? Classical cross field diffusion?

In the Valencia paper, there's a reference to equations or programs developed at EMC for cross-field diffusion. The rest would come from metal surfaces not shielded. Bussard threw that together and estimated r^2.

I have presented a calculation that says that the electrons punch a hole in the well seen by the ions. That is why I don't call "they don't have enough energy to get out" a satisfactory explanation.

Hmm, sorry, not buying that. The hole is so tiny compared to the well, and the Magrid is pushing back on any ions that far out. I don't see how they can pull any ions out. With a 1000:1 or greater overall density difference, it seems extremely likely the gradient goes the wrong way, and in any case the ions probably don't get far enough out to be affected by the tiny holes in the cusps.

[kcdodd]

I guess I just don't get your math. To start, what is debye length even supposed to mean when talking about a pure electron sheet. That would mean the thickness of your pure electron sheet would *decrease* with higher densities. That makes no sense to me.

[Art Carlson]

TallDave, do you have any idea how Bussard calculated electron losses and their scaling then? Classical cross field diffusion?

In the Valencia paper, there's a reference to equations or programs developed at EMC for cross-field diffusion. The rest would come from metal surfaces not shielded. Bussard threw that together and estimated r^2.

I take that to mean "no".

I have presented a calculation that says that the electrons punch a hole in the well seen by the ions. That is why I don't call "they don't have enough energy to get out" a satisfactory explanation.

Hmm, sorry, not buying that. The hole is so tiny compared to the well, and the Magrid is pushing back on any ions that far out. I don't see how they can pull any ions out. With a 1000:1 or greater overall density difference, it seems extremely likely the gradient goes the wrong way, and in any case the ions probably don't get far enough out to be affected by the tiny holes in the cusps.

Buy it or not, I would appreciate an answer that makes some sense.

"The hole is so tiny compared to the well": Is that talking about the thickness of the sheet (lambda_D) compared to the radius of the potential well (R)? Yeees, that's already in the calculation.

"the Magrid is pushing back on any ions that far out": If "that far out" is within the magrid radius, then there is no field from the magrid doing anything to the ions. (Faraday cage and all that.) If "that far out" is outside the magrid, then the electric field is directed outward and "pushing back" is the opposite of what is happening.

"With a 1000:1 or greater overall density difference": Why do you think the density ratio will have this value, and what difference does it make? I don't remember in what detail I presented my argument, but I contend that the electron density in the sheet will be comparable to that in the interior for a significant distance, roughly until the electric potential due to the magrid becomes large enough to overcome the thermal energy of the electrons. Would you like me to spell this out in more detail?

"the ions probably don't get far enough out to be affected by the tiny holes in the cusps": I have a calculation that says they do. You can pick a hole in my calculation, or you can offer a calculation that leads to a different conclusion. In the face of a calculation that contradicts your intuition, the way things "seem" to you doesn't carry much weight.