Drive is not the same as collision energy. A 15 KeV particle energy will have 60 KeV collision energy. (forgetting about relativity).KitemanSA wrote:That is NOT what got inserted into the FAQ on that question. The lowest "optimum" for DD was ~83keV. Even the DT was at about 50. Why is your "optimum" so different, and why didn't you speak up during the FAQ drafting?TallDave wrote:OK, maybe I should go with 18KV drive, 15K well depth. I believe there's wide agreement that 15KV is the optimum D-D temp.
Debye Length
Engineering is the art of making what you want from what you can get at a profit.
It is center of mass vs laboratory frame. i.e assume one particle is stationary and the other particle is heading for it at 2V (the V you get from accelerating a single particle through the electric field.)rcain wrote:erm... hows that worked out then?
Engineering is the art of making what you want from what you can get at a profit.
Um, I'm confused (again). If you have significant convergence and most fusion reactions are beam- beam, then the collision energy would be twice the energy of each particle. This assumes a Z of one, like with D-D or D-T fusion. If boron or helium 3 is involved, then the kinetic energy goes up as a funtion of the ionization state (+5 for boron).MSimon wrote:Drive is not the same as collision energy. A 15 KeV particle energy will have 60 KeV collision energy. (forgetting about relativity).KitemanSA wrote:That is NOT what got inserted into the FAQ on that question. The lowest "optimum" for DD was ~83keV. Even the DT was at about 50. Why is your "optimum" so different, and why didn't you speak up during the FAQ drafting?TallDave wrote:OK, maybe I should go with 18KV drive, 15K well depth. I believe there's wide agreement that 15KV is the optimum D-D temp.
Even if you are referring to the ~ average kinetic energy of a proton- boron11 collision it should be ~ 6X the accelerating voltage (assuming complete ionization and head on collisions).
The graphs of fusion crossections generally are beam- target, so with beam- beam (which the Polywell hopefully is), the kinetic energy in the collision would be twice the potential well for D-D fuel. In that sence a well depth ~ 7.5 kV should give the optimum results if 15 KeV is the target (where the slope of the crossection curve is the steepest).
But when power density (reaction rate), coumlomb scattering/ thermalization collisions, etc are considered, higher energies may be better.
Remember, while the slope of the crossection curve is greatest at ~ 15 KeV* for D-D, the slope remains greater than two up to ~ 80 KeV (guesstimate), so continued gaines in Q are made, just not as rapidly as at 15 KeV. At least that is my understanding.
* Actually, I'm not sure whether the model which was presented last fall assumed beam- beam or beam- target fusions and whether he was referring to the well depth or the net particle energy, so the potential actual potential well could vary accordingly.
Dan Tibbets
To error is human... and I'm very human.
Ya, point of diminishing returns on temp. Art stated 15KV was optimal at one point, but now I wonder if that may have been tokocentric thinking. I'm not sure Bussard or Nebel said anything on an optimal D-D temp. WB-6 ran at 12/10 KV for its 1E9 fusions/sec run, as did WB-4 for its 1E6 runs. Maybe it makes sense to stick with those for D-D in WB-8 for comparability. I think we're looking at expecting around 1E12, temp being equal.Remember, while the slope of the crossection curve is greatest at ~ 15 KeV* for D-D, the slope remains greater than two up to ~ 80 KeV (guesstimate), so continued gaines in Q are made, just not as rapidly as at 15 KeV. At least that is my understanding.
I was able to dig up Nebel saying WB-7 input was 10MW, which is helpful, as input is basically losses (assuming ions are confined). I am a little unsure if that includes magnet current, which we'd probably want to ignore.
n*kBolt*Te = B**2/(2*mu0) and B^.25 loss scaling? Or not so much? Hopefully we'll know soon...
Nebel did say this, though I still think he meant a demo - breakeven machine of perhaps 70 cm radius. If 10 MW was required for a modest 10KeV well 15 cm machine, breakeven with 10 Tesla magnets would not be reached till a ~ 1500 cm radius (100 times larger liniar dimensions) reduced to perhaps 900-1000 cm radius if drive voltage is optamized (~30 fold gain in fusion rate / 10 fold increase in drive voltage).TallDave wrote:......Remember, while the slope of the crossection curve is greatest at ~ 15 KeV* for D-D, the slope remains greater than two up to ~ 80 KeV (guesstimate), so continued gaines in Q are made, just not as rapidly as at 15 KeV. At least that is my understanding.
I was able to dig up Nebel saying WB-7 input was 10MW, which is helpful, as input is basically losses (assuming ions are confined). I am a little unsure if that includes magnet current, which we'd probably want to ignore.
If the input energy for WB6 and 7 was ~ 500,000 watts, as I believe, the breakeven size at 10 Tesla magnetic field strengths would be a little less than 70 cm radius.
I used B^4 r^3 scaling for fusion output, and r^2 for loss scaling. In all calculations 10 Tesla was used as this seems to be a common target. Since this is a constant the breakeven balance derives from r^3/r^2 scaling. Starting at 10 MW instead of 500 KW makes a big difference.
eg: scaled from WB6's 10KV potential well and 0.1 Tesla yielded ~1 milliWatt of fusion power (10^9 fusions per second). Increasing B-field to 10T is a 100 fold increase, or 100,000,000 gain in fusion rate. That times 1 milliWatt original output yields 100 KW of fusion power at the same radius (15 cm). This would be a Q of ~ 0.2. If 10MW was the input/loss figure the Q at 10 T and 15cm would be 0.01. With the isolated radius scaling of r^1.5, it takes a long time to make up the difference.
This reasoning reinforces information from WB6 final report figures / graphs that show electron input at ~ 40 amps X 12 KV while Beta was near 1. If you assume the several thousand amps flowing through the magnets was at the same voltage as the cases of the magrid casings, the input power from that alone would be ~ 20 MW. But, why waste all that energy to drive the amps through ~ 600 meters of ~12-16 gauge(?) copper wire. I'm not sure of the resistance , but I imagine no more than several dozen volts would be needed. Of course the magnetic coil windings would be insulated from the high voltage magrid casings.
Dan Tibbets
To error is human... and I'm very human.
Dan,
Hmmm, let's see... well, I was going to try B^.25*R^2, but then I realized its useless if I don't know the proportion that went into the magnets. But let's see if my math matches yours if we assume all 10MW was e-drive and B is 15x stronger in a reactor 10x larger...
10,000,000 * 15^.25 = 20MW... and then the radius puts it at 2000MW to get 100MW of power? Hmmm, not so good. I agree, that can't be right. So either magnets were a big proportion of the current or he misstated that. 10MW seems like an awful lot anyway, even if it is only for milliseconds.
I think he did say at one point they were looking at 5MW or 10MW for a reactor. The post I was looking at seemed to be talking about WB-7, though. I'll check it tomorrow, it's on my other computer.
I don't know why it took me so long to realize the WB-6 losses were just the current. I'd been thinking for the longest time we didn't have any way of knowing what the losses were because it hadn't been "measured." OTOH, the device isn't holding at beta=1, so it isn't truly the beta=1 input. But it should be close enough for some educated guesswork -- if losses were higher it would take more current to drive to beta=1.
Hmmm, let's see... well, I was going to try B^.25*R^2, but then I realized its useless if I don't know the proportion that went into the magnets. But let's see if my math matches yours if we assume all 10MW was e-drive and B is 15x stronger in a reactor 10x larger...
10,000,000 * 15^.25 = 20MW... and then the radius puts it at 2000MW to get 100MW of power? Hmmm, not so good. I agree, that can't be right. So either magnets were a big proportion of the current or he misstated that. 10MW seems like an awful lot anyway, even if it is only for milliseconds.
I think he did say at one point they were looking at 5MW or 10MW for a reactor. The post I was looking at seemed to be talking about WB-7, though. I'll check it tomorrow, it's on my other computer.
Ah, thank you. 480KW sounds much better than 10MW. And now, we have a basis to propose a range of losses (input) in WB-8! That will be great for the wiki.This reasoning reinforces information from WB6 final report figures / graphs that show electron input at ~ 40 amps X 12 KV
I don't know why it took me so long to realize the WB-6 losses were just the current. I'd been thinking for the longest time we didn't have any way of knowing what the losses were because it hadn't been "measured." OTOH, the device isn't holding at beta=1, so it isn't truly the beta=1 input. But it should be close enough for some educated guesswork -- if losses were higher it would take more current to drive to beta=1.
n*kBolt*Te = B**2/(2*mu0) and B^.25 loss scaling? Or not so much? Hopefully we'll know soon...
You haven't run the numbers. I have and they have been checked over by a physicist.the collision energy would be twice the energy of each particle.
http://iecfusiontech.blogspot.com/2007/ ... r-b11.html
and
http://iecfusiontech.blogspot.com/2007/ ... r-iec.html
Engineering is the art of making what you want from what you can get at a profit.
If you keep the amp-turns constant (which you do in a SC due to critical field limits) - the smaller you can make the device the more power you can get out of it.
My target radius for the hole in the donut is 50 cm. And 20 T. i.e. double the power of a 1 m & 10 T device. At roughly 1/8th the volume. Pretty slick no?
My target radius for the hole in the donut is 50 cm. And 20 T. i.e. double the power of a 1 m & 10 T device. At roughly 1/8th the volume. Pretty slick no?
Engineering is the art of making what you want from what you can get at a profit.
That's interesting Simon, I'll have to throw some numbers into that.
Now I'm curious again about how small we could get these, too -- for D-D this was stopped pretty quick with first-wall limits, but if we have charged fusion products... gotta try some numbers.
Now I'm curious again about how small we could get these, too -- for D-D this was stopped pretty quick with first-wall limits, but if we have charged fusion products... gotta try some numbers.
n*kBolt*Te = B**2/(2*mu0) and B^.25 loss scaling? Or not so much? Hopefully we'll know soon...
OK extrapolating from WB-6 to WB-8 I get (.8/.1)^.25*(.30/.15)^2*480,000 = 3.2 MW current for WB-8 using Bussard's loss equation.
Now, I think we expect some improvement on the nub removal. So, assuming they can get the thing fusing, we would probably hope for a beta = 1 current that is less than 3.2 MW (and producing around 33W of fusion).
But wait! There's more!
We can also extrapolate this to WB-D/9/100. Let's say we're looking at 150cm and 10T. That would give us 100MW of power (based on .001W from WB-6) at input of (drumroll please) ...142MW. Not so good. The must be expecting improvement on WB-6 confinement. I'll also have to dig up that paper Dan referred to for more context,
But wait! There's still more! We can also attempt to solve for B such that we get 100MW net power. (I did this the cheap way and just varied T till I got what I wanted for the difference.) Keeping it at 150cm, I get 100MW net at 12.71T (for context: ITER will have 13.5T at the central solenoid, 11.8T at the toroidal magnets... and of course it's a hell of a lot bigger). Unfortunately input is 160MW, so we're wasting a lot of energy if confinment looks like this.
But wait, we're not done yet! Now we can have some real fun and ask: what if we had some newfangled 30T magnets available for a mythical WB-X? How small could it get and produce 100MW? The answer, surprisingly, is a radius of only 36cm, at input of 11MW. That's just a little bigger than WB-8, at 3x the input. Of course, I'm ignoring first-wall and etc; this should really be a p-B11 calc to avoid that. And you'd have trouble making a magnet of that power that small anyway, I think.
(Now how much would you pay for this amazing contraption? Operators are standing by! Act now and we'll include this State of Kansas jello mold, absolutely free!)
And of course, all this is extremely sensitive to the values in WB-6, which aren't known with much precision, and it assumes Bussard's B^.25*r^2 scaling is correct, which may be optimistic. Still fun though.
UPDATE: Fixed some numbers (was using r instead of r^2).
Now, I think we expect some improvement on the nub removal. So, assuming they can get the thing fusing, we would probably hope for a beta = 1 current that is less than 3.2 MW (and producing around 33W of fusion).
But wait! There's more!
We can also extrapolate this to WB-D/9/100. Let's say we're looking at 150cm and 10T. That would give us 100MW of power (based on .001W from WB-6) at input of (drumroll please) ...142MW. Not so good. The must be expecting improvement on WB-6 confinement. I'll also have to dig up that paper Dan referred to for more context,
But wait! There's still more! We can also attempt to solve for B such that we get 100MW net power. (I did this the cheap way and just varied T till I got what I wanted for the difference.) Keeping it at 150cm, I get 100MW net at 12.71T (for context: ITER will have 13.5T at the central solenoid, 11.8T at the toroidal magnets... and of course it's a hell of a lot bigger). Unfortunately input is 160MW, so we're wasting a lot of energy if confinment looks like this.
But wait, we're not done yet! Now we can have some real fun and ask: what if we had some newfangled 30T magnets available for a mythical WB-X? How small could it get and produce 100MW? The answer, surprisingly, is a radius of only 36cm, at input of 11MW. That's just a little bigger than WB-8, at 3x the input. Of course, I'm ignoring first-wall and etc; this should really be a p-B11 calc to avoid that. And you'd have trouble making a magnet of that power that small anyway, I think.
(Now how much would you pay for this amazing contraption? Operators are standing by! Act now and we'll include this State of Kansas jello mold, absolutely free!)
And of course, all this is extremely sensitive to the values in WB-6, which aren't known with much precision, and it assumes Bussard's B^.25*r^2 scaling is correct, which may be optimistic. Still fun though.
UPDATE: Fixed some numbers (was using r instead of r^2).
Last edited by TallDave on Thu Jul 01, 2010 8:01 pm, edited 2 times in total.
n*kBolt*Te = B**2/(2*mu0) and B^.25 loss scaling? Or not so much? Hopefully we'll know soon...
Another variable that would improve the energy balance is the drive voltage. If this was increased to ~ 24,000 volts the WB6 baseline input would be ~ 960KW and the fusion output would be ~ 10 milliwatts (reading from the crossection graph shows ~ 10 fold increase in the D-D fusion crossection from 10 to 20 KV well depths). Using this voltage, the 150 cm radius and 10 Tesla machine would consume ~ 280 MW and produce ~ 1 GW of fusion power.TallDave wrote:OK extrapolating from WB-6 to WB-8 I get (.8/.1)^.25*(.30/.15)^2*480,000 = 3.2 MW current for WB-8 using Bussard's loss equation.
Now, I think we expect some improvement on the nub removal. So, assuming they can get the thing fusing, we would probably hope for a beta = 1 current that is less than 3.2 MW (and producing around 33W of fusion).
But wait! There's more!
We can also extrapolate this to WB-D/9/100. Let's say we're looking at 150cm and 10T. That would give us 100MW of power (based on .001W from WB-6) at input of (drumroll please) ...142MW. Not so good. The must be expecting improvement on WB-6 confinement. I'll also have to dig up that paper Dan referred to for more context,
I assume this is why Bussard gave a range of 2-3 meters diameter, the breakeven point would depend on the selected drive voltage. Improving losses by decreasing nub losses or increasing the number of faces is gravy on top of this baseline. Some improvements in these loses may be reflected in the 70 cm radius(1.4 M diameter) model that is suppose to reach breakeven at a well depth of ~ 15 KV.
Speaking of which, if this model (I really must find the link again) reflects the improvement from his truncated cube with separate coil standoffs and no nubs, it should be possible to figure out the improvement this change makes compared to the WB6 losses.(?)*. To much work for me now though.
*Assuming of course that his model incorporates this adjustment.
Dan Tibbets
To error is human... and I'm very human.
D Tibbets wrote:Another variable that would improve the energy balance is the drive voltage. If this was increased to ~ 24,000 volts the WB6 baseline input would be ~ 960KW and the fusion output would be ~ 10 milliwatts (reading from the crossection graph shows ~ 10 fold increase in the D-D fusion crossection from 10 to 20 KV well depths). Using this voltage, the 150 cm radius and 10 Tesla machine would consume ~ 280 MW and produce ~ 1 GW of fusion power.TallDave wrote:OK extrapolating from WB-6 to WB-8 I get (.8/.1)^.25*(.30/.15)^2*480,000 = 3.2 MW current for WB-8 using Bussard's loss equation.
Now, I think we expect some improvement on the nub removal. So, assuming they can get the thing fusing, we would probably hope for a beta = 1 current that is less than 3.2 MW (and producing around 33W of fusion).
But wait! There's more!
We can also extrapolate this to WB-D/9/100. Let's say we're looking at 150cm and 10T. That would give us 100MW of power (based on .001W from WB-6) at input of (drumroll please) ...142MW. Not so good. The must be expecting improvement on WB-6 confinement. I'll also have to dig up that paper Dan referred to for more context,
I assume this is why Bussard gave a range of 2-3 meters diameter, the breakeven point would depend on the selected drive voltage. Improving losses by decreasing nub losses or increasing the number of faces is gravy on top of this baseline. Some improvements in these loses may be reflected in the 70 cm radius(1.4 M diameter) model that is suppose to reach breakeven at a well depth of ~ 15 KV.
[EDIT} After reading the paper again, I think I errored. His model was up to 70 cm radius, but I think his calculated breakeven size with his setup was 1.3 M radius or 2.6 M diameter.
http://fti.neep.wisc.edu/iec2009/talks/ ... elroge.pdf
Speaking of which, if this model (I really must find the link again) reflects the improvement from his truncated cube with separate coil standoffs and no nubs, it should be possible to figure out the improvement this change makes compared to the WB6 losses.(?)*. To much work for me now though.
*Assuming of course that his model incorporates this adjustment.
Dan Tibbets
To error is human... and I'm very human.