resistive losses in the plasma

[Art Carlson]

Now that I'm starting to understand that the driving power in Bussard's model is independent of the size, I ask myself what else might start to haunt us as we grow. Bussard's model lives by cutting down the size of the holes (rho^2 ~ b^2) just as fast as he increases the pressure (n ~ B^2). This means that he ends up with a huge machine with the losses concentrated at pin prick holes. Is this realistic? Wouldn't we intuitively expect most of the power to be lost across the big surfaces? What would be the mechanism?

One possibility is the ohmic losses in the current sheath. Rick Nebel has claimed that the current must be concentrated in a layer with a thickness comparable to the electron gyroradius rho. This could be true, since any electrons which spread outside such a layer will come into the cusp outside the whiffle ball loss are, so they should see mirror confinement, which has been stated to be much worse. Since joule heating is closely related to particle diffusion, and since joule heating must be lost one way or another, it seems reasonable to treat this as a power loss per unit area.

The perpendicular Spitzer resistivity is
sigma = n*e^2*tau_e/m_e
where the electron collision time is given by
tau_e = (3.44e5 s) * T_e[eV]^1.5 * n_e[cm^-3]^-1 * ln_Lambda^-1
~ (3e-2 s) * (T_e[eV])^1.5 * n[m^-3]^-1
The joule heating power per unit area when the magnetic field jumps by B over a thickness rho is
P/A = (j^2/sigma)*rho = ((B/mu_0/rho)^2/sigma)*rho = B^2 / (mu_0^2 * sigma * rho)
The rest is algebra. To save some calculations, we use the definition and value of the classical electron radius
r_e = e^2/(4pi*epsilon_0*m*c^2) = e^2*mu_0/(4pi*m) = 2.8e-15 m
so we have
P/A = (2pi)^-1 * (B^2/2mu_0) / ( r_e * rho * n * tau_e )
= (2pi)^-1 * (B^2/2mu_0) / ( r_e * rho * n * tau_e )
= (2pi)^-1 * (B^2/2mu_0) / ( r_e * rho * (3e-2 s/m^3) * (T_e[eV])^1.5 )
= (1.9e15 m^2/s) * (B^2/2mu_0) / ( rho * (T_e[eV])^1.5 )
~ (6e7 m^2/s) * (B^2/2mu_0) / rho

where I have taken T_e ~ 100 keV. First note that this term will rise more rapidly than either point cusp losses or line cusp losses (but not as fast as fusion power production). For a numerical value, take B = 1 T, so (B^2/2mu_0) = 400 kPa and rho = 0.75 mm, and then
P/A = 3.2e16 W/m^2
P = (P/A)*4pi*R^2 = 4e17 W
This number is so enormous, that I can't see how to make it reasonable even by scaling back down to WB-6 and throwing out the assumption that the current carrying layer is only an electron gyroradius thick. Most likely I made a mistake. Can somebody check for me? I'm getting groggy. But the basic question is unavoidable: How big is the ohmic heating term in the power balance of a polywell?

[TallDave]

means that he ends up with a huge machine with the losses concentrated at pin prick holes.

The cusps aren't losses (recirculation). They just need to be small enough for a 1,000:1 or better density ratio.

[drmike]

I'm confused. NRL and other sources say Spitzer resistivity is
eta|| = 1.03e-2 * T^-3/2 * Z * ln_lambda [ohm-cm]
but Chen says half that.

Take P/A = j^2/eta*rho = (B/mu_0/rho)^2/eta*rho = B^2/(mu_0^2*eta*rho)
pick T=10^5, Z = 1, ln_lambda ~ 15 so
eta = 4.9e6 ohm-cm

If you take B = 1T and rho = .075 cm then
P/A = 6.3e11/4.9e6/.075 = 1.73e6 W/m^2

That seems big, but not insurmountable.

[Art Carlson]

means that he ends up with a huge machine with the losses concentrated at pin prick holes.

The cusps aren't losses (recirculation). They just need to be small enough for a 1,000:1 or better density ratio.

It is totally unclear both (a) how Bussard calculates losses, and (b) how he connects electron density from recirculation to the neutral density of the Paschen cruve.

[Art Carlson]

I'm confused. NRL and other sources say Spitzer resistivity is
eta|| = 1.03e-2 * T^-3/2 * Z * ln_lambda [ohm-cm]
but Chen says half that.

I believe the value you quote is the perpendicular Spitzer resistivity. The parallel Spizter resistivity is a nearly clean factor of 2 (actually 1.96) smaller. We need the perpendicular value here. (But what's a factor of two among friends?)

Take P/A = j^2/eta*rho = (B/mu_0/rho)^2/eta*rho = B^2/(mu_0^2*eta*rho)
pick T=10^5, Z = 1, ln_lambda ~ 15 so
eta = 4.9e6 ohm-cm.

eta = 1.03e-2 * T^-3/2 * Z * ln_lambda [ohm-cm]
= 1.03e-2 * (1e5)^-3/2 * 1 * 15 [ohm-cm]
= 1.03e-2 * (3.16e-8 ) * 1 * 15 [ohm-cm]
= (4.9e-9) [ohm-cm]
= (4.9e-11) [ohm-m]
It looks to me like you forgot the minus sign in the exponent of the temperature. On the other hand, I would rather multiple j^2 by the resistivity than divide by it. Also note that I have switched from Gaussian (Ohm-cm) to SI (Ohm-m) units.
P/A = j^2*eta*rho
= (B/mu_0/rho)^2*eta*rho
= B^2*eta/(mu_0^2*rho)

If you take B = 1T and rho = .075 cm then
P/A = 6.3e11/4.9e6/.075 = 1.73e6 W/m^2

That seems big, but not insurmountable.

P/A = B^2*eta/(mu_0^2*rho)
= (1)^2*(4.9e-11)/((4pi*1e-7)^2*(7.5e-4))
= 41.4 kW/m^2
This differs from my previous answer by approximately mu_0^2, which may or may not explain anything. This number seems less unreasonable, but I'm surprized it's so low. Anybody like to check this round?

[TallDave]

means that he ends up with a huge machine with the losses concentrated at pin prick holes.

The cusps aren't losses (recirculation). They just need to be small enough for a 1,000:1 or better density ratio.

It is totally unclear both (a) how Bussard calculates losses, and (b) how he connects electron density from recirculation to the neutral density of the Paschen cruve.

Heh, well, it's not entirely opaque. In some cases, he's at least pretty clear about what isn't a loss:

The need for electron recirculation through all cusps of the machine, so that cusp electron flow is not a loss mechanism.

The consequent elimination of the WB trapping factor as a measure of “losses“ it is simply a measure of density ratios inside and outside the machine.

And he at least mentions they have a calculation for cross-field transport:

The determination of electron transport losses across Polywell B fields, and verification of the electron transport loss phenomena (MG transport coefficient) by extensive experimentation in all Polywell machines.

And this clarifies a bit:

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.

[Art Carlson]

The cusps aren't losses (recirculation). They just need to be small enough for a 1,000:1 or better density ratio.

It is totally unclear both (a) how Bussard calculates losses, and (b) how he connects electron density from recirculation to the neutral density of the Paschen cruve.

Heh, well, it's not entirely opaque. In some cases, he's at least pretty clear about what isn't a loss:

The need for electron recirculation through all cusps of the machine, so that cusp electron flow is not a loss mechanism.

The consequent elimination of the WB trapping factor as a measure of “losses“ it is simply a measure of density ratios inside and outside the machine.

And he at least mentions they have a calculation for cross-field transport:

The determination of electron transport losses across Polywell B fields, and verification of the electron transport loss phenomena (MG transport coefficient) by extensive experimentation in all Polywell machines.

And this clarifies a bit:

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.

Dear Dave,

It's nice that Bussards says these things, but it remains the case that

It is totally unclear both (a) how Bussard calculates losses, and (b) how he connects electron density from recirculation to the neutral density of the Paschen cruve.

By the way, another thing that bugs me about what Bussard says is the way he harps on about "unshielded surfaces".

And, further, that no B fields can be allowed to intersect any such internal surfaces of the machine.

Make the surfaces small, yes, and put them where the field is high, but before electrons can get to a surface along a field line, they first have to cross onto it. If I drill a hole into my coil, there will be field lines in it that intersect the metal suface inside the hole, but I won't have any increased transport as a result. In tokamaks we worry about exposed edges, not because they increase the losses, but only because the increased thermal load makes a hot spot.
Anyway, if Bussard doesn't tell us what he assumes for cross-field transport coefficients, or how he arrives at his scaling in a way that we can follow his argument (referring to unpublished simulations and experiments isn't helpful), then we are on our own.

[Betruger]

And pretty soon we'll have the company of practical results.

[drmike]

Yup, I also blew a factor of 10^4 not converting from cm^2 to m^2.

But when you say

I believe the value you quote is the perpendicular Spitzer resistivity. The parallel Spizter resistivity is a nearly clean factor of 2 (actually 1.96) smaller. We need the perpendicular value here. (But what's a factor of two among friends?)

I get really confused. I did use the perpendicular form, so there's no factor of 2 missing (or added). I do think the factor of 10^3.5 is a bit bigger though!

40 kW/m^2 looks good to me, I'll take it for now.

[MSimon]

40 kW/m^2 looks good to me, I'll take it for now.

Esp when considering alpha losses (pB11 - 100 MW reactor) of 1 MW/m^2.

I'd love to have alpha losses that low. I could cool the coils with a garden fountain pump.

[Art Carlson]

40 kW/m^2 looks good to me, I'll take it for now.

Since I'm the one insisting up and down the block on taking quantitative calculations seriously unless you have a good reason not to, I guess I have to accept the figure for now, too.
Of course, the resistivity may be anomolous, but at present it looks like resistive losses in the reactor regime will be much smaller than the fusion power. (1%?)
That sends us back to cusp losses. If I take all my models and calculations seriously (Which I do only reluctantly. They are intended more to point out potential trouble spots that need serious attention.), I land at a reactor that is unwieldy, but not obviously impossible. In that spirit, and in full awareness of all the additional problems that can crop up when extrapolating over many, many orders of magnitude, I will be awaiting the new experimental results.

[seedload]

That sends us back to cusp losses. If I take all my models and calculations seriously (Which I do only reluctantly. They are intended more to point out potential trouble spots that need serious attention.), I land at a reactor that is unwieldy, but not obviously impossible. In that spirit, and in full awareness of all the additional problems that can crop up when extrapolating over many, many orders of magnitude, I will be awaiting the new experimental results.

Dr. Carlson,

I am not educated well enough to have followed the conversations so far on any more than a high level. I would like to say that since you became involved in this discussion, I believe the quality of information that has been generated on this board has increased exponentially due to you exceptional clarity in questioning and due to your healthy level of skepticism. You, sir, seem to be a very reasonable man. To someone such as me who is interested and learning, your contributions have been invaluable. Thank you.

Glad to see that you now have a modicum of interest in the results of Dr. Nebel’s experiment.

Best Regards

[TallDave]

Make the surfaces small, yes, and put them where the field is high, but before electrons can get to a surface along a field line, they first have to cross onto it.

Yes. The earlier models had poor recirculation due to the fact the coils were touching and field lines went right through the metal. In WB-6, this is mostly eliminated except for things like the supports, which are intersected by the field lines (something has to hold the Magrid together, after all).

Anyway, if Bussard doesn't tell us what he assumes for cross-field transport coefficients, or how he arrives at his scaling in a way that we can follow his argument (referring to unpublished simulations and experiments isn't helpful), then we are on our own

Yep. Hopefully we'll get to see those calcs at some point.

[Solo]

Ok, this may be of some interest:

http://www.askmar.com/Fusion.html
hosts this paper:
Some Physics Considerations

on page 9 Bussard comes out with 57kW for a 1m-radius machine at 100keV and 1kG field.

[D Tibbets]

I've been watching the theoretical/ mathmatical arguments about the Polywell funtion, and the lack of Bussard's detailed reasonings. This 18 year old paper by Bussard address most or all of the arguments. I cannot confirm or even follow most of the mathmatical descriptions provided, but I note that he mentions such things as the radial vs angular motions, lumpy surface effects, cusp losses ( he claims that there are no line cusps in the Polywell configuration). It should provide some meat for those with the physics/ math skills to digest the claims.