*
http://www.docstoc.com/docs/23853167/Fi ... ER-to-DEMO
*
Just a few minor problems.
ITER Problem Chart
ITER Problem Chart
Engineering is the art of making what you want from what you can get at a profit.
-
Heath_h49008
- Posts: 71
- Joined: Wed Jan 27, 2010 9:12 pm
- Location: Michigan
I'm betting (just a feeling - an intuition - based on nothing more than a few very tenuous clues) that Polywell harnesses natural instabilities instead of fighting them.chrismb wrote:You seem to be thinking that Polywell will not suffer any such issues if it might ever get to GW outputs, rather than nW outputs. Best go do some calculations... it's just that tokamaks are at the point of engineering for it and Polywell hasn't got that far yet.
All the simulations I have seen and many of the fusor experiments go into a star mode. In simulations the arms of the star oscillate. If those oscillating beams dominate the internal energetics I expect the other modes woll be suppressed - not enough free energy to support them.
But as I said. The clues are sparse. And Murphy hates a wise ass.
Engineering is the art of making what you want from what you can get at a profit.
Not sure I understand what you're trying to say,MSimon wrote: I'm betting (just a feeling - an intuition - based on nothing more than a few very tenuous clues) that Polywell harnesses natural instabilities instead of fighting them.
These slides are about ITER/DEMO when it's working *as intended*, not when it's working badly.
I think there is some misunderstanding about how much material (neutrons and high energy charged particles) will get released in a GW reactor. The amount is truly enormous and if you do a few calcs you'll begin to understand the issues. I've tried to explain these in the critiques regarding vacuum pumping and the global disruption due to 100's MA high current alphas that will be leaving that'll generate huge mag fields.
It's all fine and dandy when running *experimental* reactors of a few billion neutrons per sec. Once you get into GW, it'll be a whole different ball-game.
I have actually done some vacuum pumping calcs and came in two or three orders of magnitude below your numbers i.e. from impossible to reasonable.chrismb wrote:Not sure I understand what you're trying to say,MSimon wrote: I'm betting (just a feeling - an intuition - based on nothing more than a few very tenuous clues) that Polywell harnesses natural instabilities instead of fighting them.
These slides are about ITER/DEMO when it's working *as intended*, not when it's working badly.
I think there is some misunderstanding about how much material (neutrons and high energy charged particles) will get released in a GW reactor. The amount is truly enormous and if you do a few calcs you'll begin to understand the issues. I've tried to explain these in the critiques regarding vacuum pumping and the global disruption due to 100's MA high current alphas that will be leaving that'll generate huge mag fields.
It's all fine and dandy when running *experimental* reactors of a few billion neutrons per sec. Once you get into GW, it'll be a whole different ball-game.
There will only be 100's of MAs of alphas if they are circulating. The actual number if they are streaming will run about 50 to 100 Amps. For a 100 MW machine.
Here is a simple calculation that will give you a ROM.
2 MV X 50 Amps = 100 MW.
Engineering is the art of making what you want from what you can get at a profit.
The nice thing about Polywell is you probably don't have to build GW reactors.You seem to be thinking that Polywell will not suffer any such issues if it might ever get to GW outputs, rather than nW outputs.
...
I think there is some misunderstanding about how much material (neutrons and high energy charged particles) will get released in a GW reactor.
I think Simon calculated a while back the sweet spot of cost vs. first wall load was around 100MW, and heat load could be handled with existing materials. And if turns out to be too high, you can probably trade R for B and reduce the load, though this costs more.
That's for D-D/D-T, of course. If your fusion products are charged, the first wall problem is significantly ameliorated by those 1-10T B fields.
It will be fascinating next year to see if the alphas swirl out the cusps of WB-8.1 as predicted (I'm assuming .8T is enough to see this effect, but I haven't done the math on that).
(And isn't it a nice coincidence (for p-B11 operation) that open-cusp machines turn out to work better than closed-box?)
n*kBolt*Te = B**2/(2*mu0) and B^.25 loss scaling? Or not so much? Hopefully we'll know soon...
I thought it was 3 alphas average energy just short of 9 MeV total.KitemanSA wrote:Correct. With pB11, there will be one alpha at ~1.85MV (3.76MeV, 2 charges) and 2 alphas at ~1.25MV for each reaction.
His 50Amps is order of magnitude, almost single digit, but not exact.
Say 3 MeV/alpha. So figure roughly 3 MV * 33 Amps.
My point was not an exact figure. It was just to show that the currents chris saw were available only in his imagination.
Engineering is the art of making what you want from what you can get at a profit.
