It has to do with ultimate gain. Among other things.
If you operate in the resonance region of pB11 the ultimate gain (power out/power in) is about 20. This is where most power systems are economical.
If you operate at the peak reaction rate the ultimate gain is about 8. This is close to the marginal limit of about 5X for a useful generating system..
The reaction rate with the higher gain is 1/10th that for the lower gain region. The drive voltage is 50KV for the gain of 20 region and 200 KV for the gain of 8 region.
If so you need a larger reaction volume for the 50KV drive vs the 200 KV drive. This is compensated to some extent by a lower dead volume outside the reaction area.
Also a 2 MV direct conversion area will increase the radius.
All in all allowing for current uncertainties I'd say we have to allow for a range of TOTAL reactor radius of between 6 and 20 m for a 100 MW reactor with direct conversion. Once we do more tests that can be narrowed. For one thing POPS could make it smaller.
You can support about 10KV per cm in the dead space theoretically. So 2 MV adds about 2 M to the radius minimum - probably twice that or more operationally.
All this can be tightened up once we have some experience.
I prefer the no surprises approach. It is better for morale and back of the envelope cost estimating.
Once we actually know what we are doing things might get a lot better.
Since reactor radius doesn't increase much for 10X the operating power the initial commercial plants might be designed for 1,000 MW output with the capability of throttling back to 100 MW depending on the power required. Given peak vs minimum load reqmts for the grid a 10:1 ratio should handle anything reality throws at us. We hope.
MIT Talks Plasma Details
But given that the magnetic fields taper off by the 1/R^2 law, wouldn't that mean you have to also square the ampere-turns as you enlarge the radius?
Wouldn't that tend to increase losses just from the increased thickness of the coils? Even with superconductors, you can't pump infinite currents through a wire.
Why do we believe that losses won't scale with the power gains?
Wouldn't that tend to increase losses just from the increased thickness of the coils? Even with superconductors, you can't pump infinite currents through a wire.
Why do we believe that losses won't scale with the power gains?
1) most certainly applied to WB1-5. recirculation appears to be essential to answering this problem.TheRadicalModerate wrote:Since we're circling back to the Rider paper yet again, it seems as if that paper makes three fundamental arguments against IEC:
1) That electron losses eat up too much power. ...
2) The chances of ions in the population gaining enough energy through elastic collisions...
3) ...bremmstrahlung...
2) There will be collisions that transfer most of the energy from 1 ion to another, leaving the first trapped in the center of the well and ejecting the second. I don't know how common they'll be.
3) If it's the up-scattered ions that are the source of most Brem, Brem isn't a problem as the up-scattered ions will be lost from the well before they can cause enough Brem to worry about.
For a constant field ampere turns scale linearly with size.scareduck wrote:But given that the magnetic fields taper off by the 1/R^2 law, wouldn't that mean you have to also square the ampere-turns as you enlarge the radius?
Wouldn't that tend to increase losses just from the increased thickness of the coils? Even with superconductors, you can't pump infinite currents through a wire.
Why do we believe that losses won't scale with the power gains?
I'll see if I can find the Java simulator for that. Check back if you don't see it here.
http://hyperphysics.phy-astr.gsu.edu/hb ... urloo.html
The important number is the field in the center of the coil z=0
That determines the energy of electrons that can be deflected.
And yes. In the real world coils have finite dimensions. That will affect the field near the conductors the most. At the center of the coil it should not have much effect at all.
Engineering is the art of making what you want from what you can get at a profit.
Yet more questions after looking at the animation:
1) Why do we believe that most injected fuel atoms won't just get sucked into the electron clouds surrounding them on all sides?
2) Why would ions that meet but fail to fuse not shoot into the electron cloud (the "wall" of the wiffleball), thus likely giving up irrecoverable brehmsstrahlung? Bussard thought he had an answer to this, but it was never clear how he arrived at it.
1) Why do we believe that most injected fuel atoms won't just get sucked into the electron clouds surrounding them on all sides?
2) Why would ions that meet but fail to fuse not shoot into the electron cloud (the "wall" of the wiffleball), thus likely giving up irrecoverable brehmsstrahlung? Bussard thought he had an answer to this, but it was never clear how he arrived at it.
The positive charge on the ions keeps them in the center region due to the positive charge on the grids. Those ions that do get near the grid will be low energy.scareduck wrote:Yet more questions after looking at the animation:
1) Why do we believe that most injected fuel atoms won't just get sucked into the electron clouds surrounding them on all sides?
2) Why would ions that meet but fail to fuse not shoot into the electron cloud (the "wall" of the wiffleball), thus likely giving up irrecoverable brehmsstrahlung? Bussard thought he had an answer to this, but it was never clear how he arrived at it.
Engineering is the art of making what you want from what you can get at a profit.
I feel like I'm asking really dumb questions here, but ... what grids? The only one I can tell that exists is the magrid, which is the recirculating electrons. That would have a negative charge on it, not a positive one.MSimon wrote:The positive charge on the ions keeps them in the center region due to the positive charge on the grids. Those ions that do get near the grid will be low energy.scareduck wrote:Yet more questions after looking at the animation:
1) Why do we believe that most injected fuel atoms won't just get sucked into the electron clouds surrounding them on all sides?
2) Why would ions that meet but fail to fuse not shoot into the electron cloud (the "wall" of the wiffleball), thus likely giving up irrecoverable brehmsstrahlung? Bussard thought he had an answer to this, but it was never clear how he arrived at it.
You connect the positive terminal of the power supply to it. The shell of the machine is negative. A flip answer to be sure. This is better:
Remember the purpose of the magnetic field is to deflect the electrons attracted to the positive charge on the MAgrid.
Remember the purpose of the magnetic field is to deflect the electrons attracted to the positive charge on the MAgrid.
Engineering is the art of making what you want from what you can get at a profit.
What was Bussard's bremmstrahlung argument, exactly? I remember he said it would be much smaller than Rider had calculated, but I don't recall if he said why.
Rider's claim that the system would collapse to equilibrium before fusion could occur seems to be definitively proven wrong by the WB-6 results, if I understand that correctly. Not sure about the brem, though.
Rider's claim that the system would collapse to equilibrium before fusion could occur seems to be definitively proven wrong by the WB-6 results, if I understand that correctly. Not sure about the brem, though.
Did he ever say that, though? His point was that energy loss would always be higher than energy output unless you could find an efficient way to recirculate bremsstrahlung radiation (or avoid it altogether), NOT that you couldn't achieve some level of fusion.TallDave wrote:Rider's claim that the system would collapse to equilibrium before fusion could occur seems to be definitively proven wrong by the WB-6 results, if I understand that correctly.
Really? Inside the grid there are almost no E fields because the grid forms a Faraday cage. You can not compress ions into a region of empty space with electric fields.MSimon wrote:
The positive charge on the ions keeps them in the center region due to the positive charge on the grids. Those ions that do get near the grid will be low energy.
Fusion is easy, but break even is horrendous.