I have questions.
1 - How do we avoid excess electrons from fusion building up in the core?
2 - How do we replace the electron energy lost from the plasma.
I did a little googling and read through the "Theory" forum titles here, where I found some information, but not exactly the answers. My question considers p-B11 fuels, where each fusion reaction releases 6 low energy electrons. These electrons are thermalized in short order and are expected to replace electron losses. See viewtopic.php?t=392. That is well and good, but is there a risk of their being to many electrons converting our quasi neutral plasma into a negatively charged plasma?
And for the second question, as I understand it in an operating reactor, the Well depth will need to be about 250 keV. Can that Well depth be maintained by the low energy electrons from the fusion reaction? If not then how do we maintain the Well depth without an excess of electrons in the quasi neutral plasma core?
Maintaining Quasi neutrallity - a question.
I'll take a stab.
1. Well, low-energy electrons at the top of the electrostatic will just roll down to the edges and bounce around. No problem there. Not sure where these electrons come from, though, and if they're liberated in the core they would be high-energy because of where they start.
Do alphas leave behind electrons? I thought the fuel ions didn't have any electrons to start with (i.e., they're fully ionized). I suppose the fuel ions would leave electrons behind as they ionize, though. It looks like they lose them all by 670eV. Not sure how their starting point compares to our hot electrons, but the well is a lot bigger so they're probably relatively cold.
2. We'll be pumping 5-10MW of hot electrons into it, and the plasma will be electron-rich anyway. I would guess you can toss cold electrons from the ions into the mix as a loss.
Not sure where the 250KeV number comes from. I thought we were looking at around 90KeV, but I can't recall exactly.
1. Well, low-energy electrons at the top of the electrostatic will just roll down to the edges and bounce around. No problem there. Not sure where these electrons come from, though, and if they're liberated in the core they would be high-energy because of where they start.
Do alphas leave behind electrons? I thought the fuel ions didn't have any electrons to start with (i.e., they're fully ionized). I suppose the fuel ions would leave electrons behind as they ionize, though. It looks like they lose them all by 670eV. Not sure how their starting point compares to our hot electrons, but the well is a lot bigger so they're probably relatively cold.
2. We'll be pumping 5-10MW of hot electrons into it, and the plasma will be electron-rich anyway. I would guess you can toss cold electrons from the ions into the mix as a loss.
Not sure where the 250KeV number comes from. I thought we were looking at around 90KeV, but I can't recall exactly.
The drive voltage depends on where on the curve you are operating.
The resonance peak (.1 barn and 50 KV nominal drive) or the absolute peak (1.2 barns - 200 KV nominal drive). In the first the gain is higher but the reactor is bigger vs. the second.
The resonance peak (.1 barn and 50 KV nominal drive) or the absolute peak (1.2 barns - 200 KV nominal drive). In the first the gain is higher but the reactor is bigger vs. the second.
Engineering is the art of making what you want from what you can get at a profit.
OK - Maybe I am missing something that the answer to this question will clear up. Given a hydrogen and a B11 in total isolation, strip one electron from the hydrogen and five electrons from the boron, slam them together, they fuse, creating three alphas, each lacking two electrons to be helium. The question is, are these 6 electrons the same 6 from an accounting stand point? That is, does the p-B11 fusion actually deposit 6 electrons in the plasma, or have we already stripped them off and deposited them by way of the electron gun?
Aero
Just to be sure I looked it up again. The p-B11 reaction produces three alphas. If the fuel is totally ionized the hydrogen and boron11 that is present have no electrons associated with them and no electrons produced in the fusion reaction (beta particles). As the alphas leave the magrid, they don't leave electrons behind directly, but they do reduce the positive ions in the system, so a relative excess of electrons would form. To compensate, fewer new monoenergetic electrons could be injected. Or, to compensate more new ionized hydrogen and boron could be injected. In otherwords, so long as you relpaced the ions lost to fusion (along with other ions that escape containment) the electricl balence would be unchanged.Aero wrote:OK - Maybe I am missing something that the answer to this question will clear up. Given a hydrogen and a B11 in total isolation, strip one electron from the hydrogen and five electrons from the boron, slam them together, they fuse, creating three alphas, each lacking two electrons to be helium. The question is, are these 6 electrons the same 6 from an accounting stand point? That is, does the p-B11 fusion actually deposit 6 electrons in the plasma, or have we already stripped them off and deposited them by way of the electron gun?
If you are injecting neutral gas, it would be a different story. As the gas is ionized within the magrid, fuse, and then leave, the electrons would be left behind. This would presumably have bad consequences on your ability to inject new monoenergetic electrons to maintain the potential well. In the current machines the portion of produced fusion ions to fuel ions is so tiny that it can be ignored (~10e-10 or smaller ?). In a high output steady state commercial reactor you probably could not use a gas puffer. If the reactor was pulsed, or with some other method to purge the excess electrons, it might work. I suspect this would increase input energy needs and decrease the time dependant fusion yield. But, in large Polywells there could sugh a high Q that such compromises may be tolorable and still maintain profitable power production.
Dan Tibbets
To error is human... and I'm very human.
MSimon:
eV is energy, volt is electric potential ... mixing up units for brevity breeds confusion.
Do you mean electron-volt (eV) here when you abbreviate to volt (V)?The resonance peak (.1 barn and 50 KV nominal drive) or the absolute peak (1.2 barns - 200 KV nominal drive). In the first the gain is higher but the reactor is bigger vs. the second.
eV is energy, volt is electric potential ... mixing up units for brevity breeds confusion.
I meant exactly what I said. Volts.icarus wrote:MSimon:Do you mean electron-volt (eV) here when you abbreviate to volt (V)?The resonance peak (.1 barn and 50 KV nominal drive) or the absolute peak (1.2 barns - 200 KV nominal drive). In the first the gain is higher but the reactor is bigger vs. the second.
eV is energy, volt is electric potential ... mixing up units for brevity breeds confusion.
Engineering is the art of making what you want from what you can get at a profit.
I understand ion guns can be complex beasts, but my limited understanding is that basically neutral gas or solids are first ionized (heated) into a neutral plasma, then the ions are acellerated/ guided by magnetic fields and or electrodes out of the gun, while the electrons hit something, and are carried to ground through a wire (or to an electron gun if you are talking about an ion drive in a spaceship).TallDave wrote:In regards to D Tibbets' point: does an ion gun inject pre-ionized, er, ions?
Presumably, electron guns are simpler because it is much easier to extract electrons from a surface (thermionic effect) with heat compared to the much more massive left behind ions, so everyone is happy so long as you have some current flow to replace the emitted electrons.
Dan Tibbets
To error is human... and I'm very human.
MSimon:
I wish others were .... so someone else is talking about eV's when they shouldn't be?
The drive voltage is in volts, the energy of the particles, as in 0.5*m*v^2 is eVs ... why don't people just use joules for particle energy to avoid the source of confusion? why the trend to jargonise it beyond comprehension?
Precisely. You seem to be quite exacting in other areas so I expected your terminology to be correct.I meant exactly what I said. Volts.
I wish others were .... so someone else is talking about eV's when they shouldn't be?
The drive voltage is in volts, the energy of the particles, as in 0.5*m*v^2 is eVs ... why don't people just use joules for particle energy to avoid the source of confusion? why the trend to jargonise it beyond comprehension?
TallDave:
http://en.wikipedia.org/wiki/Electron_volt
"By definition, it is equal to the amount of kinetic energy gained by a single unbound electron when it accelerates through an electrostatic potential difference of one volt."
Myself, I think the electron-Volt is a vulgar unit that stems from the inability to reduce electromagnetic phenomena to more fundamental geometric relationships whereby they would involve only units of length and time ... with appropriate constants .... it does more to advance confusion than understanding .... imho.
Indeed you should. If a 250kV well accelerated an ion it would be an ion volt .... a 250kV electric potential well accelerates an electron to 250keV ... electron volts.Does a 250KV well accelerate an ion to something like 250KeV? I should find the correlation.
http://en.wikipedia.org/wiki/Electron_volt
"By definition, it is equal to the amount of kinetic energy gained by a single unbound electron when it accelerates through an electrostatic potential difference of one volt."
Myself, I think the electron-Volt is a vulgar unit that stems from the inability to reduce electromagnetic phenomena to more fundamental geometric relationships whereby they would involve only units of length and time ... with appropriate constants .... it does more to advance confusion than understanding .... imho.
eV = V*Z for a fully ionized atom. Z = the number of charges on the atom.TallDave wrote:Does a 250KV well accelerate an ion to something like 250KeV? I should find the correlation.
I'm not sure how the e got in there, but I remember it being relevant in some context. I should probably go look it up, but exam on Monday...
Engineering is the art of making what you want from what you can get at a profit.