## What Quality the Losses?

Discuss how polywell fusion works; share theoretical questions and answers.

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Aero
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### What Quality the Losses?

Erroneous interpretation of loss factors could have negative legacy. See below.
Last edited by Aero on Sat Feb 27, 2010 10:45 pm, edited 1 time in total.
Aero

KitemanSA
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I believe you are starting at the wrong "loss point".

Suppose that for a 100MW BFR the gross output was 110MW and the losses were 10MW. The 10MW should be your starting point, no? Thus going to R=3 vice 1.5 would be a doubling and the losses would be 40MW; no?

His Valencia paper (57th IAC (IAC2006)) gives a bit of math on the loss mechanisms.

TallDave
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There is a general hope that a BFR will generate net energy on the order of radius to the fifth power, while it is generally held that gross fusion will scale as radius to the seventh power. Further, it is hoped that net power can be extracted to high efficiency by direct conversion.

Sort of. Power is B^4*r^3, B just scales about the same as R, so r^7 is kinda the same. Gain is really anyone's guess right now, because we don't know how losses will scale.

That is, for a 3 meter BFR generating about 800 MW net, losses are 8 times as much, on the order of 6.4 GW.

No, the gain becomes a higher and higher percentage of total power as r and B increase. This is true for all fusion machines, which is why ITER, as the most net-powerish tokamak, costs tens of billions of dollars.

Thus going to R=3 vice 1.5 would be a doubling and the losses would be 40MW; no?

Of course in theory your power would then be 880MW (from (r*2)^3) or 12.8GW if you're scaling B too.

The truth is no one really knows what loss scaling will look like. Bussard threw out a simplistic r^2 based on surface area being the only loss scaling factor; Art has thrown around a couple less optimistic possibilities; I don't recall Rick hazarding any kind of guesstimate.
n*kBolt*Te = B**2/(2*mu0) and B^.25 loss scaling? Or not so much? Hopefully we'll know soon...

KitemanSA
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In his Valencia paper, Dr. Bussard wrote:Tests made on a large variety of machines, over a wide range of drive and operating parameters have shown that the loss power scales as the square of the drive voltage, the square root of the surface electron density and inversely as the ¾ power of the B fields. At the desirable beta = one condition, this reduces to power loss scaling as the 3/2 power of the drive voltage, the 1/4 power of the B field, and the square of the system size (radius). Since the fusion power scales as the cube of the size, the fourth power of the B field, and a power of the E drive energy equal to the E-dependence of the fusion cross-section (cross-section proportional to E to the s power), minus 3/2. For DD, s = 2-4, while for DT, s = 3-6 in useful ranges of drive energy. For pB11, the cross section scales about as s = 3-4 over the system-useful range.
Does this help?

MSimon
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KitemanSA wrote:
In his Valencia paper, Dr. Bussard wrote:Tests made on a large variety of machines, over a wide range of drive and operating parameters have shown that the loss power scales as the square of the drive voltage, the square root of the surface electron density and inversely as the ¾ power of the B fields. At the desirable beta = one condition, this reduces to power loss scaling as the 3/2 power of the drive voltage, the 1/4 power of the B field, and the square of the system size (radius). Since the fusion power scales as the cube of the size, the fourth power of the B field, and a power of the E drive energy equal to the E-dependence of the fusion cross-section (cross-section proportional to E to the s power), minus 3/2. For DD, s = 2-4, while for DT, s = 3-6 in useful ranges of drive energy. For pB11, the cross section scales about as s = 3-4 over the system-useful range.
Does this help?

That argues for using the pB11 resonance peak to reduce the losses.
Engineering is the art of making what you want from what you can get at a profit.

TallDave
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Thanks Kite, I'd forgotten the other parts of the loss scaling. While r^2 is a useful approximation of Bussard's claims the B (non-) scaling from the WB seems to be the critical factor (i.e. is the B scaling really so small we can ignore it?).

Rick might have some notion by now as to whether Bussard was wildly optimistic, if they're running WB-8 already. B^(1/4) should be fairly obvious, as the WB-8 losses would only double relative to WB-7 while power increased 4096-fold (asuming the same radius), whereas with, say, B^(3/2) the losses would be 20 times higher. Go WB effect!

If we really do get B^4 power vs B^1/4 losses... well. I'm not sure that's really realistic though. Who wants to calculate the largest B loss scaling that lets Polywells be competitive with fission? (I bet Rick knows.)
n*kBolt*Te = B**2/(2*mu0) and B^.25 loss scaling? Or not so much? Hopefully we'll know soon...

hanelyp
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KitemanSA wrote:
In his Valencia paper, Dr. Bussard wrote:Tests made on a large variety of machines, over a wide range of drive and operating parameters have shown that the loss power scales as the square of the drive voltage, the square root of the surface electron density and inversely as the ¾ power of the B fields. At the desirable beta = one condition, this reduces to power loss scaling as the 3/2 power of the drive voltage, the 1/4 power of the B field, and the square of the system size (radius). Since the fusion power scales as the cube of the size, the fourth power of the B field, and a power of the E drive energy equal to the E-dependence of the fusion cross-section (cross-section proportional to E to the s power), minus 3/2. For DD, s = 2-4, while for DT, s = 3-6 in useful ranges of drive energy. For pB11, the cross section scales about as s = 3-4 over the system-useful range.
Does this help?

Given that drive voltage would be selected based on the fuel, that boils down to R^2B^0.25 losses, RB^3.75 net gain. Not quite as favorable as the often quoted R^2 losses, RB^4 gain.

KitemanSA
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hanelyp wrote: Given that drive voltage would be selected based on the fuel, that boils down to R^2B^0.25 losses, RB^3.75 net gain. Not quite as favorable as the often quoted R^2 losses, RB^4 gain.
Wouldn't the voltage be more related to the B field? Seems to me that the greater the field, the greater the voltage to keep beta=one. At the higher voltage you get higher density of whatever fuel is chosen. No?

Aero
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Since the fusion power scales as the cube of the size, the fourth power of the B field, and a power of the E drive energy equal to the E-dependence of the fusion cross-section (cross-section proportional to E to the s power), minus 3/2. For DD, s = 2-4, while for DT, s = 3-6 in useful ranges of drive energy. For pB11, the cross section scales about as s = 3-4 over the system-useful range.

R^7 is easy to understand and so is R^3*B^4. Who wants to explain what this E stuff means in such a way that I can understand the parts?
and a power of the E drive energy equal to the E-dependence of the fusion cross-section (cross-section proportional to E to the s power), minus 3/2. For DD, s = 2-4, while for DT, s = 3-6 in useful ranges of drive energy. For pB11, the cross section scales about as s = 3-4 over the system-useful range.
Aero

Art Carlson
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Aero wrote:Who wants to explain what this E stuff means in such a way that I can understand the parts?

I always found this strange. The fusion cross section first increases rapidly and then levels off with the velocity, which should scale like the square root of the voltage. The losses (says Bussard, and he's probably more or less right this time) increase with a modest power of the voltage. Therefore there is a particular voltage where the gain is maximum. Why not go there right away? Instead, Bussard assumes that the "useful range" is on the part of the curve where increasing the voltage continues to increase the gain. Is there some other limit he is not telling us about? Arcing or something?

D Tibbets
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Art Carlson wrote:
Aero wrote:Who wants to explain what this E stuff means in such a way that I can understand the parts?

I always found this strange. The fusion cross section first increases rapidly and then levels off with the velocity, which should scale like the square root of the voltage. The losses (says Bussard, and he's probably more or less right this time) increase with a modest power of the voltage. Therefore there is a particular voltage where the gain is maximum. Why not go there right away? Instead, Bussard assumes that the "useful range" is on the part of the curve where increasing the voltage continues to increase the gain. Is there some other limit he is not telling us about? Arcing or something?

Yes.
There are alot of tradeoffs. Bussard has mentioned several times (I believe) running a D-D Polywall at ~ 80-120 Kv well depths. This would maximize gross fusion power but as demonstrated in a recent simulation ~ 15 KV is the optimal voltage to maximize Q. ( the steepest portion of the crossection curve. Practical concerns about arcing, thermal wall loading, power supply limitations, pumping limitations, limitations of the gas puffers, voltatage / B field / current ratios achievable with the limited resources, etc added up to range in which the experments were run. The WB 7 expermaental resources where perhaps (?) more adaptable over greater ranges, which would expand the data set and perhaps increase the confidence and maby even the understanding of the physics.

From an engeenering/ financial standpoint pushing the machine to higher voltages may make more sense, even though the Q would be droping (assumes you have enough buffer in the proformance to do this). ie: a 500 MW gross power reactor that consumes 100MW of input power may be better than 10 X 50 MW gross output reactors that consume 5 MW input power each. These are made up numbers, I don't know how closely this would come to the actual drive voltage scaling. Another way of looking at it is that a less efficient but smaller machine may be cheaper than a more efficient but nessisarily larger machine that is operating at lower drive voltages where the crossection is correspondingly lower, though the realative Q is higher.

Dan Tibbets
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Aero
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Thanks -- But I meant to ask for a mathematical interpretation. Is all this something like

factor = constant *V *(R^2/B)^1/2 ?????
Aero

D Tibbets
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Aero wrote:Thanks -- But I meant to ask for a mathematical interpretation. Is all this something like

factor = constant *V *(R^2/B)^1/2 ?????

Look at this link. I think the authur describes why he chose 15 KV for his Polywell simulation and the relationship between Q and the drive voltage.

http://fti.neep.wisc.edu/static/TALKS/1 ... elroge.pdf

Dan Tibbets
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TallDave
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KitemanSA wrote:
hanelyp wrote: Given that drive voltage would be selected based on the fuel, that boils down to R^2B^0.25 losses, RB^3.75 net gain. Not quite as favorable as the often quoted R^2 losses, RB^4 gain.
Wouldn't the voltage be more related to the B field? Seems to me that the greater the field, the greater the voltage to keep beta=one. At the higher voltage you get higher density of whatever fuel is chosen. No?

The voltage is for the temperature, B field is for the density. With more B you get more density, hence the B^4 power scaling. To keep a certain density at beta = 1 you need a certain current, which depends on your losses.

Therefore there is a particular voltage where the gain is maximum. Why not go there right away? Instead, Bussard assumes that the "useful range" is on the part of the curve where increasing the voltage continues to increase the gain. Is there some other limit he is not telling us about? Arcing or something?

Unless I misunderstand the relationship of power to voltage within the applicable range, if you're assuming your losses are B^.25*r^2 and your power is r^3*B^4, I'm guessing you may not care all that much about maximizing the relatively small gain from optimized voltage (I'm not saying that's necessarily a realistic set of assumptions). There's a lot of knobs and conditions in this thing -- anode height, thermalization time, etc. -- so it's hard to say what Bussard was trading off exactly. Rick has stated that arcing is more complex than a Paschen curve so that seems possible.

Who wants to explain what this E stuff means in such a way that I can understand the parts?

He's just talking about the relationship between temperature/voltage and fusion rate. Someone could probably throw together an equation for this but I don't think it's a simple one.
n*kBolt*Te = B**2/(2*mu0) and B^.25 loss scaling? Or not so much? Hopefully we'll know soon...