What Quality the Losses?
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
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.
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.
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.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.
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 netpowerish tokamak, costs tens of billions of dollars.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.
Of course in theory your power would then be 880MW (from (r*2)^3) or 12.8GW if you're scaling B too.Thus going to R=3 vice 1.5 would be a doubling and the losses would be 40MW; no?
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...
Does this help?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 Edependence of the fusion crosssection (crosssection proportional to E to the s power), minus 3/2. For DD, s = 24, while for DT, s = 36 in useful ranges of drive energy. For pB11, the cross section scales about as s = 34 over the systemuseful range.
That argues for using the pB11 resonance peak to reduce the losses.KitemanSA wrote:Does this help?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 Edependence of the fusion crosssection (crosssection proportional to E to the s power), minus 3/2. For DD, s = 24, while for DT, s = 36 in useful ranges of drive energy. For pB11, the cross section scales about as s = 34 over the systemuseful range.
Engineering is the art of making what you want from what you can get at a profit.
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 WB8 already. B^(1/4) should be fairly obvious, as the WB8 losses would only double relative to WB7 while power increased 4096fold (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.)
Rick might have some notion by now as to whether Bussard was wildly optimistic, if they're running WB8 already. B^(1/4) should be fairly obvious, as the WB8 losses would only double relative to WB7 while power increased 4096fold (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...
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 wrote:Does this help?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 Edependence of the fusion crosssection (crosssection proportional to E to the s power), minus 3/2. For DD, s = 24, while for DT, s = 36 in useful ranges of drive energy. For pB11, the cross section scales about as s = 34 over the systemuseful range.
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?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.
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?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 Edependence of the fusion crosssection (crosssection proportional to E to the s power), minus 3/2. For DD, s = 24, while for DT, s = 36 in useful ranges of drive energy. For pB11, the cross section scales about as s = 34 over the systemuseful range.
and a power of the E drive energy equal to the Edependence of the fusion crosssection (crosssection proportional to E to the s power), minus 3/2. For DD, s = 24, while for DT, s = 36 in useful ranges of drive energy. For pB11, the cross section scales about as s = 34 over the systemuseful range.
Aero

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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?Aero wrote:Who wants to explain what this E stuff means in such a way that I can understand the parts?
Yes.Art Carlson wrote: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?Aero wrote:Who wants to explain what this E stuff means in such a way that I can understand the parts?
There are alot of tradeoffs. Bussard has mentioned several times (I believe) running a DD Polywall at ~ 80120 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
To error is human... and I'm very human.
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.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 ?????
http://fti.neep.wisc.edu/static/TALKS/1 ... elroge.pdf
Dan Tibbets
To error is human... and I'm very human.
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.KitemanSA wrote: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?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.
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.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?
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.Who wants to explain what this E stuff means in such a way that I can understand the parts?
n*kBolt*Te = B**2/(2*mu0) and B^.25 loss scaling? Or not so much? Hopefully we'll know soon...