Dodec Magnetic field?

Discuss the technical details of an "open source" community-driven design of a polywell reactor.

Moderators: tonybarry, MSimon

TallDave
Posts: 3140
Joined: Wed Jul 25, 2007 7:12 pm
Contact:

Post by TallDave »

So, if we toss in a 10x improvement for well depth and keep radius at 1.5M, we get a factor of only 18 for B -- 1.8T.

93143
Posts: 1142
Joined: Fri Oct 19, 2007 7:51 pm

Post by 93143 »

Aero wrote:Using R^2 losses gives B^4 * R, and your numbers, gives net power = 1 mW * 100,000,000 * (2/.15) = 1.3 MW
Why are you scaling the power output at the gain exponent? Power out scales as B^4*R^3. Power in scales as R^2 (loss area) or more likely B^2 (density, given nearly-closed point cusps and thus invariant loss area for a given WB topology), or possibly B^2*R (if line cusps are a problem). You have to calculate them both and subtract. There is no power in the system that scales as B^4*R.

MSimon
Posts: 14334
Joined: Mon Jul 16, 2007 7:37 pm
Location: Rockford, Illinois
Contact:

Post by MSimon »

93143 wrote:
Aero wrote:Using R^2 losses gives B^4 * R, and your numbers, gives net power = 1 mW * 100,000,000 * (2/.15) = 1.3 MW
Why are you scaling the power output at the gain exponent? Power out scales as B^4*R^3. Power in scales as R^2 (loss area) or more likely B^2 (density, given nearly-closed point cusps and thus invariant loss area for a given WB topology), or possibly B^2*R (if line cusps are a problem). You have to calculate them both and subtract. There is no power in the system that scales as B^4*R.
I think 93143 has a point.
Engineering is the art of making what you want from what you can get at a profit.

Aero
Posts: 1200
Joined: Mon Jun 30, 2008 4:36 am
Location: 92111

Post by Aero »

Why are you scaling the power output at the gain exponent? Power out scales as B^4*R^3. Power in scales as R^2 (loss area) or more likely B^2 (density, given nearly-closed point cusps and thus invariant loss area for a given WB topology), or possibly B^2*R (if line cusps are a problem). You have to calculate them both and subtract. There is no power in the system that scales as B^4*R.
I believe Dr. Bussard favored R^2 losses and Art favors a much larger loss term.
Let's look at your options. I believe this says that Net Power scales as
1. Pnet = B^4 * R^3 - R^2
or
2. Pnet = B^4 * R^3 - B^2
or
2. Pnet = B^4 * R^3 - B^2 * R

If it says something else, please clarify it. If not, then please run the numbers using the same example as before.
Aero

93143
Posts: 1142
Joined: Fri Oct 19, 2007 7:51 pm

Post by 93143 »

Aero wrote:please run the numbers using the same example as before.
Okay.

Gross power scales as B^4*R^3, so with a B ratio of 100 and an R ratio of 40/3, we get 1 mW*1e8*2370 = about 240 megawatts.

Loss power scales as B^2, so with a B ratio of 100 we get... what was the loss power of WB-6 again? 10 kW? I can't remember, so I'll use that. So 1e4 W *1e4 = 100 megawatts.

Net power is gross power - loss power, so 240 MW - 100 MW = 140 MW, very roughly.

You can get this by using the gain equation: original gain was 0.001/10000 = 1e-7, and gain is proportional to B^2*R^3, so 1e-7*10000*2370 = 2.4.

If it really is R^2 rather than B^2, loss power is 10 kW *178 = 2 MW, for a net power of 235 MW, again very roughly. Gain in this case is B^4*R -> 1e-7*1e8*40/3 = 133.

With B^2*R, we have 10 kW *1e4*40/3 = 1.3 GW. So this predicts no net power from that scheme, assuming my starting point is correct (dubious). Solution? Bump up the drive voltage, make it bigger, reduce B somewhat. I believe drive voltage is an independent parameter unrelated to the scaling laws, so it's fair game for boosting.

This is all very ballpark...

MSimon
Posts: 14334
Joined: Mon Jul 16, 2007 7:37 pm
Location: Rockford, Illinois
Contact:

Post by MSimon »

Ultimate gain is about 9 for pB11 at peak cross section and about 22 at the resonance peak. (pwr in + pwr out)/pwr in

You can not get a power gain better than that because it assumes 100% efficient drive power and all the energy from the fusion/fission is converted directly to useful energy. Zero power for auxiliaries. All the drive power makes fusions.
Engineering is the art of making what you want from what you can get at a profit.

93143
Posts: 1142
Joined: Fri Oct 19, 2007 7:51 pm

Post by 93143 »

I was neglecting drive power in order to focus on the scaling laws for gross power and loss power. Not the best of ideas, I guess, but I think it got the point across...

To be honest I wasn't thinking about that at all... Naturally, drive power should track the loss scaling at low gains (<<1) and the gross fusion power scaling at high gains (>>1), so after a while things will stop improving. The nice part about that is that once they do stop improving it means you're already well past net power...

D Tibbets
Posts: 2775
Joined: Thu Jun 26, 2008 6:52 am

Post by D Tibbets »

Does using power input scaling of B*2 make sense? In a superconducter, once the field is established, no further current is needed (?). At least, this seems to be the reasoning in the Tokamak community when they claim they are near breakeven with JET and the Japanese machine. I believe they ignor the magnetic power input and only use the power input needed to heat the plasma.


Dan Tibbets
To error is human... and I'm very human.

93143
Posts: 1142
Joined: Fri Oct 19, 2007 7:51 pm

Post by 93143 »

B^2 is plasma density. Assuming the cusps stay the same size as you scale up, and assuming constant recirculation fraction per cusp based on geometric similarity, and assuming unchanged drive voltage, loss power is proportional to plasma density.

icarus
Posts: 819
Joined: Mon Jul 07, 2008 12:48 am

Post by icarus »

93143:
loss power is proportional to plasma density.
I think Art C. made some compelling arguments in another thread that loss power is proportional to not only plasma density but also flux rate of losses through the cusps. He also makes the case that the flux rate of losses through the cusps scales as the radius^2 of the wiffle-ball, since the surface area encompassed by the sheath is 4*pi*r^2 and all the flux tubes enclosing the sheath eventually flow out through the cusps.

In any case, power has units of energy density times volume per second, e.g. plasma energy density times some volume flow rate scale would be a power. Unless you had some intrinsic constant volume scale in mind (like electron gyro-radius^3 per second), your statement doesn't make sense.

93143
Posts: 1142
Joined: Fri Oct 19, 2007 7:51 pm

Post by 93143 »

icarus wrote:I think Art C. made some compelling arguments in another thread that loss power is proportional to not only plasma density but also flux rate of losses through the cusps. He also makes the case that the flux rate of losses through the cusps scales as the radius^2 of the wiffle-ball, since the surface area encompassed by the sheath is 4*pi*r^2 and all the flux tubes enclosing the sheath eventually flow out through the cusps.
I'm not compelled. The "wiffleball" idea implies that the high-energy electrons are not confined to what you would normally consider flux tubes, due to the high B-field gradient; they bounce off the edge region and shoot through the core undisturbed (by magnetic fields at least). Only at or very near the cusps, where the magnetic gradient is more shallow, can they get onto flux tubes and follow them out.

The reason why they're injected at high energy is so that the 'sheath' can penetrate into the wiffleball and produce a deep well. If the differential electron buildup were all at the surface of the wiffleball, the fractional well depth would be much worse than observed.

I'm aware Art disputes this picture. He may even be right, but I get the distinct impression that his theories predict the abject failure of WB-6 and WB-7, so maybe not...
In any case, power has units of energy density times volume per second, e.g. plasma energy density times some volume flow rate scale would be a power. Unless you had some intrinsic constant volume scale in mind (like electron gyro-radius^3 per second), your statement doesn't make sense.
Density times average outward flow velocity of electrons (proportional to sqrt(drive voltage)) times effective cusp area (proportional to gyroradius^2) times electron specific energy (proportional to drive voltage) times recirculation loss fraction (constant due to geometric similarity).

Wait a second here... The gyroradius is going to go down as B goes up. This means that there's an extra B^2 in the denominator, which means that for point cusps, the loss is constant...? For line cusps, then, the losses are B*R, which makes sense. Or the device could produce mostly point cusps, as we've already noted it might, and Art is at least partially right about the flux tube losses.

For a narrower cusp on a bigger machine, wouldn't the recirculation loss fraction go down?

I'm just theorizing here, trying to figure out where Bussard's R^2 comes from...

TallDave
Posts: 3140
Joined: Wed Jul 25, 2007 7:12 pm
Contact:

Post by TallDave »

IIRC, Bussard's r^2 was just the fractional intercept area plus cross-field diffusion. He didn't consider cusps a loss mechanism at all (recirculation)

icarus
Posts: 819
Joined: Mon Jul 07, 2008 12:48 am

Post by icarus »

93143:
I'm just theorizing here, trying to figure out where Bussard's R^2 comes from...
Until you can come up with something plausible that has the correct units I think it is a little generous to call it "theorizing" .... it is more commonly known as hand-waving what you're engaged in at the moment, that's okay if it's made clear.

93143
Posts: 1142
Joined: Fri Oct 19, 2007 7:51 pm

Post by 93143 »

icarus wrote:Until you can come up with something plausible that has the correct units
All right, now I'm annoyed. Pay attention.

Density [kg/m^3] times average outward flow velocity of electrons [m/s] times effective cusp area [m^2] times electron specific energy [J/kg] times recirculation loss fraction [-].

Result? [kg/m^3*m/s*m^2*J/kg] = [J/s] = [W].
I think it is a little generous to call it "theorizing" .... it is more commonly known as hand-waving what you're engaged in at the moment, that's okay if it's made clear.
Okay, fine, it's hand-waving. I thought it WAS clear, considering I've presented at least four possible scenarios thus far with no real effort to determine which one (if any) is correct...

Talldave: I thought intercept area WAS a cusp loss, since that's how the electrons get to the unshielded metal. If the point cusps really do have constant leakage at a given drive voltage, that then implies that cross-field diffusion dominates at net power - but that's not plausible because Bussard was very insistent that net power was impossible if the fractional intercept area was above a certain maximum. So line-like cusps and/or flux-tube leakage mechanisms look more plausible. Unless I've missed something...

icarus
Posts: 819
Joined: Mon Jul 07, 2008 12:48 am

Post by icarus »

93143:
Density [kg/m^3] times average outward flow velocity of electrons [m/s] times effective cusp area [m^2] times electron specific energy [J/kg]
No need for annoyance, so far you have managed only to get the units correct, now you need to propose, explain and justify the scaling laws for each of the variables you have put forward:

- plasma density
- average outward flow velocity of electrons
- effective cusp area
- electron specific energy

or else it is not a plausible argument, merely hand-waving. Half an argument and some bluster doesn't work in the real world.

Post Reply