Resultant particles hitting ion or electron Injectors

[Robthebob]

Is this a problem? If it is, how are the ion or electron injectors shielded against the resultant particles like alphas, neutrons, etc?

[krenshala]

My understanding of the idea was to have the injectors outside, imparting just enough energy to the ions/electrons to get them through the magrid and into the well. Once in the well, the physics of the polywell would do the rest.

[D Tibbets]

I'm not sure how the injection guns interact with the fast fusion ions comiing out the cusps either. The electrons have to be injected with alot of energy so that they can reach near the center despite the repulsion of the electron excess within the grids (this is what drives the potential well). The ions can be injected with just enough energy to fight thier way throuh the cusps, too much energy would presumably be bad(?). Of course the energy for speeding up the electrons can come from the positive charged magrid. It would take a very large positive potential (or magnetic field) on the ion guns to deflect the mega electronvolt fusion ions, and strong magnetic fields to protect the electron guns. This may be something conviently left to the engeneers to resolve once the concept is proved.


Dan Tibbets

[Robthebob]

Yeah, come to think about it, the electron injectors wont pose a problem, they just go in and either circulate or form the WB.

The thing I guess I was worried about is the position of the ion injectors, they have to be placed below the potential well of the electrons; or the ions will just fly out of the polywell. Or I'm not understanding this correctly?

Seems like the ions will have to put into the system without any velocity or it will reach farther out of the potential well on the other side and have a chance of hitting the ion injectors.

[Robthebob]

Okay, if the injection guns dont have to be placed close to the polywell, then there's no problem. I was just confused.

What effects do the ions have to fight against in order to get into position? I was under the impression that once the electrostatic potential is in place, with or without formation of the WB, then if you inject ions into the system with some sort of velocity entering into the potential well, then it will preserve that velocity when it reach the other side and escape the potential well on the other side, assuming no collision has happened, which I think isnt too bad to assume, because the electrons are too slow at this point.

I guess with the WB, since the electric field inside the WB would be zero... maybe? Then, there may be a duration when the ions are not undergoing acceleration, and therefore, wont reach that far on the other side.

[Aero]

I think the safest assumption is that the acceleration/deceleration force felt by the ions is symmetrical along any radial line WRT the center of the polywell.

[D Tibbets]

...I guess with the WB, since the electric field inside the WB would be zero... maybe? Then, there may be a duration when the ions are not undergoing acceleration, and therefore, wont reach that far on the other side.

There is apparently no magnetic field within the Wiffleball. But, there is an electrostatic field present. This is what is created by the drive energy from the electron injection, and makes the potential well that the ions fall back and forth in. There are arguments about the nature of this electrostatic field, including it's shape and durability.

[EDIT]. It has occured to me that you might be referring to no electrostatic field within the central zone past all the electrons (much smaller voulume than the Wiffleball). Assuming that there is such a zone,at least in approximation, then the ions would be on thier own The only force they would experiance in this area would be from other ions as all the electrons are outside of this central core area; and the subsequent formation of a virtual anode would be due to the inertia of the ions. This ignores other interactions that might be occuring like the ions tugging electrons along with them due to the ions much greater inertia in this area (hot ions and cold electrons.


Dan Tibbets

[Robthebob]

oh yeah, so I dont wanna make another thread, hope people will see this question.

As for the problem that polywell will thermalized (really, everything thermalizes), how expansive will just the reactor be?

Let's say that this machine thermalizes in like 10 seconds, just a number, and takes 20 seconds to prep to start over, we will need 3 machines for power to be generated all the time.

If just the reactor isnt too expansive, as long as the allowed time of fusion isnt some crazy small number like less than a second with prep time 10 times more than that, then I guess it wont be a problem.

Oh yeah, how frequent do the ions collide with the electrons in the WB?

I'm guessing that's the primary factor in determining how fast polywell thermalizes.

[D Tibbets]

As for the problem that polywell will thermalized (really, everything thermalizes), how expansive will just the reactor be?

Let's say that this machine thermalizes in like 10 seconds, just a number, and takes 20 seconds to prep to start over, we will need 3 machines for power to be generated all the time.

If just the reactor isnt too expansive, as long as the allowed time of fusion isnt some crazy small number like less than a second with prep time 10 times more than that, then I guess it wont be a problem.

Oh yeah, how frequent do the ions collide with the electrons in the WB?

I'm guessing that's the primary factor in determining how fast polywell thermalizes.

Hopefully, the Polywell will never thermalize. Rember that this is a dynamic situation. Consider each individual electron. Using numbers related to WB 6 an electron entered the potential well at ~ 1 billion centimeters per second, say that results in an average speed of 500 million cm/s. The electron claimed lifetime within the magrid with recirculation was ~100,000 passes. Assuming the Wiffleball was ~ 25 cm wide, the electron would have one transit every ~ 1/20 millionth of a second, which works out to be a total lifetime within the machine of ~ 5 milliseconds. So long as the thermalization time for this electron is some significant amout of time over over this number the single electrons and thus the total population of the electrons never thermalize past some tolerable amount. Essentially, all of the electrons are being replaced every 5 ms. I don't know what the lifetime for the non fusing ions is, I suspect it is in the same neighborhood as the electron lifetime in terms of passes. Presumably the same thermalization limits apply. In a larger machine the transit times/ lifetimes would be longer (size increase divided by the proportionate increase in drive energy).
If the machine ends up needing to be pulsed for whatever reason, I'm guessing that the 'run time' will be in the 10's of ms and the recovery time may be a similar amount of time (things are happening fast). Some Big capacitors would help to even out the output.

I don't know how the ion-electron collisions compare to like particle collisions. I think that momentum transfer is most efficient when the particles have similar masses and charges so the ion- electron collisions may not contribute much(?).


Dan Tibbets

[D Tibbets]

RobtheBob, with your questions, if you havn't already done so, you should read (several times) the paper below. Keep in mind that these are the thoughts of Dr. Bussard and there are unconfirmed or contested facts, which fuels alot of the discussions here.

http://www.askmar.com/ConferenceNotes/2 ... 0Paper.pdf



Dan Tibbets

[chrismb]

I don't know what the lifetime for the non fusing ions is, I suspect it is in the same neighborhood as the electron lifetime in terms of passes.

Ah, therein lies the trouble.

In viewtopic.php?p=13258#13258 I used data verbatim from a paper. Which was tough, because when I read that there would be lower-than-1E-12 vaccum on the outside and a core at 1/10th atmospheric just a few feet away, I was ROTFLMAO.

Anyways, let's be serious here with some likely chamber desities. Let's say an ion will pass through medium which, on average, has a density of other ions of around 1E20/m^3. That'd be pretty good going by any fusion experiments to date.

Let's say we've got a 1m vessel with an ion source radius of 0.5m and a drive voltage potential for that radius of 20keV.

I'm gonna put DD into this theoretical reactor as that's what will need to be tested to prove concept, I'm presuming it won't get to p11B unless neutrons can be shown out of DD. I'm also going to be highly generous and say that every interaction will be a 'head on' so that the experiment gets the benefit of the full 80keV cross-section for DD = 20millibarns.

So the mean distance to a fusion event would be 1/[(1E20/m^3).(2E-30m^2)]=5billion metres. In a 2m reciprocation, there and back, within the ion radius that'd be 2.5billion reciprocations. A deuteron accelerating from stationary to 20keV (in a 20kV/0.5m field) accelerates at a=(4E4[V/m]).(1.6E-19[C])/3.2E-27kg=2E12m/s^2. Using 0.5m = 0.5at^2, so t (for that quater cycle) = 7E-7s, or 2.8E-6s/reciprocation.

2.5billion such trips later (to get an 'average' of one fusion event) means that the deuteron has to have a residence time, without loss of energy, of {2.5billion [reciprocations] x 2.8E-6[s/reciprocation]} = 7000 seconds, about 2 hours.

[D Tibbets]

[Edit- deleted]

[D Tibbets]

I don't know what the lifetime for the non fusing ions is, I suspect it is in the same neighborhood as the electron lifetime in terms of passes.

Ah, therein lies the trouble.

In viewtopic.php?p=13258#13258 I used data verbatim from a paper. Which was tough, because when I read that there would be lower-than-1E-12 vaccum on the outside and a core at 1/10th atmospheric just a few feet away, I was ROTFLMAO.

Anyways, let's be serious here with some likely chamber desities. Let's say an ion will pass through medium which, on average, has a density of other ions of around 1E20/m^3. That'd be pretty good going by any fusion experiments to date.

Let's say we've got a 1m vessel with an ion source radius of 0.5m and a drive voltage potential for that radius of 20keV.

I'm gonna put DD into this theoretical reactor as that's what will need to be tested to prove concept, I'm presuming it won't get to p11B unless neutrons can be shown out of DD. I'm also going to be highly generous and say that every interaction will be a 'head on' so that the experiment gets the benefit of the full 80keV cross-section for DD = 20millibarns.

So the mean distance to a fusion event would be 1/(1E20).(2E-30)=5billion metres. In a 2m reciprocation, there and back, within the ion radius that'd be 2.5billion reciprocations. A deuteron accelerating from stationary to 20keV (in a 20kV/0.5m field) accelerates at a=(4E4).(1.6E-19)/3.2E-27=2E12m/s^2. Using 0.5m = 0.5at^2, so t (for that quater cycle) = 7E-7, or 2.8E-6 for one round trip.

2.5billion such trips later (to get an 'average' of one fusion event) means that the deuteron has to have a residence time, without loss of energy, of 7000 seconds, about 2 hours.

I may be exposing my ignorance, but if so I'll learn from it.

As I interpret your formula you are saying that mean distance to a fusion event = 1 / density X Barns. The mean free path decreases as the density increases, but the collision rate would increase, so shouldn't the density be in the numerator? Or, to use the Lawson criterion...

http://hyperphysics.phy-astr.gsu.edu/HB ... on.html#c1

...your number reflects this fairly closely nt= ~ 10e15 s/cm3 for deuterium. But the n (density) then needs to be moved to the other side of the equation so the final number should be distance traveled for a fusion event (= your number) divided by density, or ~ 10 meters. This would be ~ 5-10 passes through your machine. When less generous interactions than you granted are present there is still alot of wiggle room befor you reach the ~ 10,000 passes needed for effective fusion that I have seen quoted. The corresponding required lifetimes (in seconds) of the ions at 20KeV well depths are quite small ( millisecond time scales?).

If I'm completly confused ( only a short distance need be traveled befor that reaction can occur ) there still seams to be a problem with your final calculation. 2.5 billion reciprocations / ~2.8 million reciprocations / seconds = 900 seconds. This is nearly an order of magnitude lower.


Also, keep in mind that in an ideal Polywell the ions only escape when they are at or beyond the top of the potential well, so while the ions would need to be replaced, thier energy would be mostly conserved. Bussard mentioned this in his Velencia paper (and elsewhere?). The confinement time of the electrons are important from an energy balance perspective, but the major concern for ion containment is the need to maintain adiquit internal densities for practical fusion rates while preventing arcing problems that could occur when the ions escaped faster than they could be removed with pumping.


Dan Tibbets

[chrismb]

As I interpret your formula you are saying that mean distance to a fusion event = 1 / density X Barns. The mean free path decreases as the density increases, but the collision rate would increase, so shouldn't the density be in the numerator?

I've not offered any equations with 'density' and 'reaction rate' in it. But if the density increases, the MFP to fusion decreases, the number of reciprocations across the space decreases (what you're talking about is embedded in that 'line') so the time to fusion decreases.

When less generous interactions than you granted are present there is still alot of wiggle room befor you reach the ~ 10,000 passes needed for effective fusion that I have seen quoted. The corresponding required lifetimes (in seconds) of the ions at 20KeV well depths are quite small ( millisecond time scales?).

Sorry to dissapoint. It is several billion passes required before 'an average' fusion event.

If I'm completly confused ( only a short distance need be traveled befor that reaction can occur ) there still seams to be a problem with your final calculation. 2.5 billion reciprocations / ~2.8 million reciprocations / seconds = 900 seconds. This is nearly an order of magnitude lower.

I've calculated a) how many reciprocations are needed before an 'average' fusion event, and b) the time-of-flight of one of those reciprocations. Total time to fusion is therefore (a) x (b). Not sure what you are thinking of with your denominator. I will go back and edit in the units, it seems to have caused confusion without them.

[Betruger]

Sorry to dissapoint. It is several billion passes required before 'an average' fusion event.

Isn't this direct contradiction of Dr Nebel's explicit statement of the 10k passes figure?