It Is A Different Machine
It Is A Different Machine
Let me start with gyroradius.
r = mv/qB
where m is mass in kg
v is velocity in m/S
q is charge in coulombs
B is field in Teslas
So let us talk fusion alpha energy. I'm going to pick a number that is twice the expected energy based on equal energy division from the Be fission part of the reaction. That would be roughly 6,000 KeV. That gives a gyroradius of about 3.5 cm with a 10T field. About 1/30th the radius of the coils in a proposed 100 MW machine (1 m radius). A 1 T machine would be 1/3rd the coil radius. And a .1T machine about 3.3 m gyroradius.
At 10 T it is a different machine. The transition is at about .35 T where the gyroradius = coil radius.
Some helpful urls:
http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=33
Tokamak oriented
r = mv/qB
where m is mass in kg
v is velocity in m/S
q is charge in coulombs
B is field in Teslas
So let us talk fusion alpha energy. I'm going to pick a number that is twice the expected energy based on equal energy division from the Be fission part of the reaction. That would be roughly 6,000 KeV. That gives a gyroradius of about 3.5 cm with a 10T field. About 1/30th the radius of the coils in a proposed 100 MW machine (1 m radius). A 1 T machine would be 1/3rd the coil radius. And a .1T machine about 3.3 m gyroradius.
At 10 T it is a different machine. The transition is at about .35 T where the gyroradius = coil radius.
Some helpful urls:
http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=33
Tokamak oriented
Engineering is the art of making what you want from what you can get at a profit.
So until the machine gets above .35 T the fusion alphas will behave poorly? Smashing into the magrid and causing heat loss?
Above .35 T the fusion alphas will circulate and escape at predictable points?
What is the Be (Beryllium) fission reaction? http://en.wikipedia.org/wiki/Aneutronic_fusion didn't help me out there... Sorry, I just need to catch up with you here.
Above .35 T the fusion alphas will circulate and escape at predictable points?
What is the Be (Beryllium) fission reaction? http://en.wikipedia.org/wiki/Aneutronic_fusion didn't help me out there... Sorry, I just need to catch up with you here.
PB11= He4+Be8 = 3 He4 plus the usual energy.dch24 wrote:So until the machine gets above .35 T the fusion alphas will behave poorly? Smashing into the magrid and causing heat loss?
Above .35 T the fusion alphas will circulate and escape at predictable points?
What is the Be (Beryllium) fission reaction? http://en.wikipedia.org/wiki/Aneutronic_fusion didn't help me out there... Sorry, I just need to catch up with you here.
And yes. Below .35T you have high heat loads on the magnet casings. Above that (a factor of 10 is probably a good number where you can ignore alpha impingement) the heat load becomes insignificant (engineering wise). At 10T as long as your fusion is taking place farther than 10 cm from the coil casing the chances of an alpha hitting the casing becomes small.
This may be why Dr. B thought heating the coils would be a problem. He was thinking about 1T machines.
Engineering is the art of making what you want from what you can get at a profit.
I think you might need to make a correction. The gyroradius would need to be less than the machine radius minus the offest distance from the center where the Wiffleball border is (where the ions first start to be exposed to the coil induced magnetic field). If the wiffleball radius is 1/2 that of the coil radius, then the effective radius of the coils would have to be 1.25 or 1.5 times (?) the radius (without a Wiffleball) if at the same magnetic field strength. Also, since the magnetic field strength would increase as you approach the magnets there would be an uneven force on the ions- gyroradius would be elliptical instead of circular (I'm guessing). And, of course once the ion passes inside the Wiffleball ball border they would fly straight (in an isolated system) untill they reached the opposite Wiffleball border. So the path might look like a race track, with straight sides and rounded corners. Once the ions had enough speed to penitrate far enough past the Wiffleball border where it's gyroradius was less than the distance back to the Wiffleball border, it would be traped on a magnetic field line and/ or be heading through a cusp. At least, that is my current understanding...
Dan Tibbets
Dan Tibbets
To error is human... and I'm very human.
P-B11 fuses to form an excited carbon 12 isomer, which quickly breaks down (fissions) into an alpha and and excited beryllium nucleus. This beryllium isomer then quickly breaks down (fissions) into two alphas. These two alphas have a little more kinetic energy than the initial alpha.dch24 wrote:...What is the Be (Beryllium) fission reaction? http://en.wikipedia.org/wiki/Aneutronic_fusion didn't help me out there... Sorry, I just need to catch up with you here.
Dan Tibbets
To error is human... and I'm very human.
So, where are the fuel ions supposed to be injected, and at what energy? If outside the 'wiffleball' (not that I believe there is any chance of s sharp cutoff as seems to be implied) then those fuel ions won't make it (magnetron condition), whereas if they are injected at the edge of the wiffleball then a) the alphas can still strike these injectors directly and b) even if I did buy a distinct cutoff radius for the wiffleball, is it really going to hold at a very steady radius? Surely the wiffleball edge will undergo *some* potential variation of radius, thus washing over and back from the injectors?
If you were to try to inject neutrals, like in a tokamak, who's to say whereabouts they will actually get ionised? There will be some probabilistic distribution of where they get ionised, thus will thermalise. And before you say 'at the dense edge', then where do the 'first' fuel particles get ionised if the posited mechanism of the wiffleball hasn't already cranked itself up?
Soooo many unanswered questions, it is difficult to get a grip on anything so as to try to make suggestions for improvement.
If you were to try to inject neutrals, like in a tokamak, who's to say whereabouts they will actually get ionised? There will be some probabilistic distribution of where they get ionised, thus will thermalise. And before you say 'at the dense edge', then where do the 'first' fuel particles get ionised if the posited mechanism of the wiffleball hasn't already cranked itself up?
Soooo many unanswered questions, it is difficult to get a grip on anything so as to try to make suggestions for improvement.
I think the Wiffleball is first created with the injection of electrons, no ions or neutrals are injected untill after the wiffleball is established. I'm again guessing that neutrals would not ionize untill they reached the electron cloud within or very close to the Wiffleball. This would still have to be high up in the potential well for the machine to work. I'm not sure how injected ions would be contained within the magrid while having sufficient speed to reach the Wiffleball, you would think that they would go out the opposite side at the same energy or get caught up in the periferal magnetic field. It must/ might/ doesn't have something to do with the excess electrons in the system and/ or the interactions with the cusps, and/ or ............
Dan Tibbets
Dan Tibbets
To error is human... and I'm very human.
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It's going to be some kind of gee whiz kit. A direct energy conversion screen that catches the alphas streaming out of the cusps, thus protecting the injection guns, but that allows the injected ions/electron to go through in the opposite direction.
One of those hi-sci-fi techno naming moments ..... "space charge diode"?
energy flux diverter ... differentiated space-charge diverter ...
go to town .... then you've got to build the thing.
One of those hi-sci-fi techno naming moments ..... "space charge diode"?
energy flux diverter ... differentiated space-charge diverter ...
go to town .... then you've got to build the thing.