## #7 FAQ- mean free path

Richard Hull
Moderator
Posts: 12730
Joined: Fri Jun 15, 2001 1:44 pm
Real name: Richard Hull

### #7 FAQ- mean free path

Edit add in on Oct 11 2015 R.H This is a very important running discussion with lots of ideas and information. It should ultimately be read and mused over through all 4 pages.

Mean Free Path (MFP) is a very important concept in understanding the operation of the fusor. It is the very thing that allows the fusor to fuse. Without an acceptable mean free path within the reduced pressure of the fusor's deuterium gas fill, fusion would not be possible by the electrostatic acceleration-collider, operational, methodology used in the device.

In a recent discussion on mean free path of particles in a fusor a lot of wind has blown over the decks and I went on a quest.

The ideal gas law turns out to have lots o' gottchas and, as such, I have renamed it, in the fusor sense, as the "less than ideal gas law".

It turns out that due to a lot of bizzare goings-on in the fusor, the temperature of the gas is unknowable to a suitable accuracy and even the species content variables and relative abundance is also unknown!!! Makes for real easy calcs.

In spite of all of this negativism, there is a very handy calculator on line that is just fabulous and can answer many questions. It will let you do a hundred, "What if?" experiments. All you have to do is poke in some variables and, bingo! You have your mean free path using the variables based on species diameter, temperature and pressure all at hand using a more complicated equation than normally found for air molecules at STP. Go to...........

http://hyperphysics.phy-astr.gsu.edu/hb ... re.html#c3

And your are off to the races.

The hydrogen or deuterium molecule (neutral) is ~ 1 Angstrom in diameter.

The deuteron is about 1.7 femptometers in diameter (.00002 A)

Man! That first electron shell is big!

Mean free path is a relatively simple thing to figure even with the velocity dependent species thrown in. Unfortunately real life throws a lot of curves. Not all species are at the same temperature. There may be a banded or striated temperature layer or layers based on ion acoustic issues, resonances, etc. In some cases there may be many species.

For our purposes, the deuteron and proton or triton are all 100,000 times smaller than the smallest neutral. Unfortunately, they are sent out to do their work NOT amongst their peers, but sent to herd with mostly giant neutrals which are the real bulk particle content of the fusor. Like firing a BB through a wad of aircraft carriers.

Amazingly, the BB has nearly the same mass as the aircraft carrier, but a lot more kinetic energy. So a hit by a BB on a carrier can certainly impart great energy to it. Most of the time this will far exceed the 13 ev needed to ionize the D neutral. Now it is a deuteron. Depending on where it is in the fusor volume, and what energy it has at birth, it might join in the fray for fusion. Certainly ion-neutral collisional quantity and frequency are related to the energy of the neutral. Most neutrals are near 0 ev kinematically.

The mean free path of the deuterons through this mass would be on the order of the neutral's mean free path or slightly better. The more energetic the sea of neutrals as a whole, the less likely you are to collide with a neutral. (mean free path goes up). So it is a real mix of events species and energies.

The rarest of all hits would be a deuteron hitting a neutral near the shell. On these rare instances, it is almost certainly hit from behind by a deuteron heading towards the shell at ever dropping energy. (recirculating) Should such a hit take place with enough energy to ionize the neutral, it is most certainly bound to hit the sphere walls, but such collisions should be very rare. The general flux of deuterons is on a series of closed shell orbits through the center of the fusor.

It is important to keep in mind that regardless of where a deuteron is created in the fusor and by whatever method, it is trapped in a creational orbital shell plus or minus any vectoral kinetic energy it picked up in the ionization process. While a hit by a deuteron on a high speed neutral is possible, it is rare based on mean free path. A certain number of high speed neutrals are created "on the fly" as near matching velocity electrons on the same flight path co-join with deuterons. Such a possibility is most likely for decelerating, missed fusion deuterons exiting the inner grid to recirculate that are streaming towards the shell along with electrons naturally accelerated in that direction. This is probably related to the formation of the star rays visibility, as this is a geometrically enhanced, electrostatically formed high probability zone for like speed and direction electron-deuteron recombination that would give off light at the characteristic spectral frequency in the most dense neutral formation streams. These neutrals, so created, will certainly crash into the shell wall.

Dave Cooper has suggested that the fusor, when operated, will have an ever increasing energy level for the neutrals. I think this would be the case if they recycled or were trapped in the volume in some way where they could not be destroyed or absorbed. Unfortunately the neutrals at some energy stasis level should all just be wall bound and die. This is the final fate for virtually all neutrals of mean free path exceeding the diameter of the chamber. (~150mm) which would be a mean temperature of 650 degrees kelvin or an energy of 0.05 ev. Collisions with neutrals of this type by fast deuterons would just ionize the neutrals so involved as effectively as electron ionization by impact.

Neutral-neutral collisions by .05ev neutrals would do nothing. Yet, we would expect the chamber to be pretty much devoid of statistically significant neutrals of energies greater than .05ev, just due to mean free path issues allowing wall collisions from any point in the chamber at 10 microns!

It all comes back to the fact that the fusor fuses in "velocity space" and not in a contained neutral plasma at a given unifrom temperature. This makes volumetric assumptions a bit dicey.

A lot of this discussion changes if one uses ion guns. These sort of force the issue in that a stream of pure, fully directed deuterons are sent towards the inner grid region as a current. Thus, fusion is seen to increase dramatically. For all missed deuterons from ion guns, the above scenario is still valid as their connection to the gun is lost after the first past. They just become part of the soup.

Thanks to Dave Cooper and Frank Sanns for inspiring me on this quest of trying to understand the more than difficult fusor ion-neutral environment. We are all each other's best mentors and critics.

Richard Hull
Progress may have been a good thing once, but it just went on too long. - Yogi Berra
Fusion is the energy of the future....and it always will be
Retired now...Doing only what I want and not what I should...every day is a saturday.

Richard Hull
Moderator
Posts: 12730
Joined: Fri Jun 15, 2001 1:44 pm
Real name: Richard Hull

### Re: FAQ- mean free path

A follow up to this older post............

A 6" fusor has a volume of ~4/3 (Pi) (7.5) cm. or ~1762 cc. At 10 microns, the gas molecule density would be about 10e14 molecules per cc.

A typical fusor will then have about 1.7 X1017 molecules in it at the normal 10 micron fusing pressure. If the fusor is operated at a current of 10ma then there are would be a maximum of 6X10e16 ions created every second. This means we are effectively ionizing only 35% of the gas in the chamber. This leaves 65% of the material in the chamber at or near room temperature. It will slowly rise in temperature, but will never reach any significant figure in the ionic sense or in any sense that the mean free path among the bulk of the material in the chamber would be significantly impacted.

Neutrals can certainly have energies greater than their own ionization potentials, but not in and amongst there own similarly sized bretheren that are, as a bulk, at much colder temps.

I sure hope this back of envelope math is tolerably correct.

The upshot is that the mean free path in a fusor is well above the simple room temperature figure for the gas pressure, but no where near as high as figuring on the hot ion temperatures attained in the ideal ion created at the wall and accelerated through the full potential of the device. This is mostly due to the fact that in our simple fusors ions are more likely to be created in the vacinity of the central grid (sadly) due to the high field gradient induced by the short radius of curvature of the inner grid wires. Near wall ionization is created totally by the return electron current streaming towards the shell.

It is important to note that the electron MPF is very great in a fusor. It is also a virtual given that 1/2 of the input energy is spent at the outer shell due to direct input energy, first ionization, electron impact. Note* the outer shell certainly dissapates far in excess of 1/2 the input energy if not nearly 100% based on fast neutral impact, recoil ionization electrons, compton electrons, photo electrons, etc. A tremendous amount of the lost kinetic energy in the deuteron cycle system is converted to wall electron collisional losses.

Richard Hull
Progress may have been a good thing once, but it just went on too long. - Yogi Berra
Fusion is the energy of the future....and it always will be
Retired now...Doing only what I want and not what I should...every day is a saturday.

DaveC
Posts: 2346
Joined: Sat Jun 30, 2001 5:13 am
Real name:

### Re: FAQ- mean free path

Richard - Your math is good and the points being pursued are very useful.

This is one element in the calculation you demonstrated, that probably a bit off. Its a subtle point. The number of ions per second that the 10 mA figure represents is certainly correct, as is the approximate number of gas molecules for the 10 micron pressure. It just that one can't ratio the ions/sec to the total gas molecules as a fraction of the total number of gas molecules being ionized. Since the gas ions are in motion, the number per second actually bears no relationship to the total number of molecules. It could easily be many times the total number per second, without violating any physical laws. It DOES tend to spin the brain, but there are many things within the fusor that do that, as we are all learning.

Also, I would like to suggest that the neutral gas temperatures ARE quite high - but are not in equilibrium - once the fusor plasma forms. The wall temps are the thermal sink to which the energetic ions and neutrals dump their energy. The walls being massive relative to the dilute gas, (perhaps some 6 to 8 orders of magnitude more molecules in the fusor walls than in the gas) are thus barely heated, while the excited gas molecules are at ionization energies of 13 eV on up to keV..... until they collide with the cold wall. Even when too hot to touch, the fusor wall is barely 0.050 ev. Thus the recoiling gas molecule, will retain, most of its energy.

Why doesn't the fusor simply burn up?? The gas density is too low. The negative polariity grid wires are another thing, however. The grid is the target of every ion. That some miss this target is evidenced by the central concentration or "poissor" star. But most ions end up meeting the grid. It gets seriously hot, as a result.

Here's a fascinating thought to ponder: If the central grid structure were to be so designed that few if any ions struck it, then this type of fusor could have an almost zero real current flow, even at high voltages! The ions would either collide with a neutral and simply do a molecular ionization routine, or... collide with another ion and possibly fuse, or eventually collide with the fusor wall.

The fusion process conserves charge, so the only real input current would be wall current and whatever equivalent real energy input is needed to drive the light output. Instead of KW input, the power could be a few tens of watts.

I know it is possible to design an electrode structure that does not intercept very many ions/electrons. Actually we all know this is possible, since every CRT uses such an electron gun. The focus voltages are provided by the lightest of HV divider circuitry, since ... the focus electrodes draw almost no current! Only the CRT face collects electrons, after they have given up some of their energy as light, in colliding with the phosphor.

I have also been working on the electron( or ion) spatial density, considering a beam of say circular cross section. When the velocity associated with the accelerating voltages is taken into account, the ions are remarkably widely spaced, for the currents commonly seen in the Fusor. For Deuterons, at 20keV, there are only 4.5 per cubic micron, per microamp. So a 20mA current if entirely in a single ion beam, the ion density is a mere 90,000 per cubic micron. Thats 14.4 femto-coulombs per micron cubed. I was actually using a 1 um diameter beam, so these are extremely dilute charge particle beams.

What looms out of this to me, is this question: How does ANY fusion happen??

Clearly, these discussions are making us all think a bit more, and as Richard Hull has noted, we are thus learning by the effort.

Dave Cooper

Frank Sanns
Posts: 1750
Joined: Fri Jun 14, 2002 6:26 pm
Real name: Frank Sanns
Location: Pittsburgh, PA USA

### Re: FAQ- mean free path

Thank you gentlemen for summing up role of MFP and Ion currents in a couple of threads. In these threads is the answer to the wasted mechanisms and energies that keep fusion efficeincy so very poor. It is not that fusion is not happening, it is that we unnecessarily waste the overal energy that we put into the system. We do not fully utilize the energy once we get the particles up to energy and we are dumping scads of wasted energy into unproductive ion currents.

Dave, I will be anxious to see the fusion results that you can achieve with your fine beam ion guns. I also have contiuned to think about why fusion is occuring and I would be really interested to see the neutron numbers form an experiment where the ion guns are rotated through a wide set of angles relative to each other. I know diametrically opposed should be the best since that is the maximum energy but I can think of other mechanisms that could explain fusion at other angles. Just one more degree of freedom to add to the plethora already there! Thanks guys.

Frank Sanns

DaveC
Posts: 2346
Joined: Sat Jun 30, 2001 5:13 am
Real name:

### Re: FAQ- mean free path

Thanks for a good summary.. Frank.... I hope it won't be too long before I can get something built. With the emphasis being on effective collisions I expect to run at high current density, but very low currents.

I am trying to get into my head what it will take for a dense beam, that does not dissipate itself in wall or grid collisions.

We shall see..

Dave Cooper

Richard Hull
Moderator
Posts: 12730
Joined: Fri Jun 15, 2001 1:44 pm
Real name: Richard Hull

### Re: FAQ- mean free path

Thanks Dave for the corrections to my first post. I will ammend my thoughts accordingly and bow to your long history in this area. I am an engineer and was working from a simple current carrier basis. I forgot that there is some recirculation and the ion lifetimes might be such that the ion numbers might exceed the production currents on a second to second basis.

I can't imagine a scenario where ions would hit the walls, short of negative ions. Maybe this could be explained. I only see neutrals and electrons hitting the walls. Ions in orbits that are close to the wall (created there) could experience an extra push in the direction of the wall from a neutral, I suppose.

What do you figure might be a realistic average gas temperature in the inter-grid zone of the fusor? I realize in non-equilibrium conditions it is a crap shoot. 500ev? 1000ev? What might the neutrals temp rise to? It is obvious that even if the neutrals and gas temps are so high that MFP goes to meters, then about 100% of the neutrals are wall slammers as are the electrons.

I, too, wish you luck in devising a grid that doesn't chew up energy. I have long dreamed of a wall ionization system using high field ionization due to thousands of micro needle implantation at the shell walls where you might literally expect 100% wall coverage for ion production at this ideal point.

RIchard Hull
Progress may have been a good thing once, but it just went on too long. - Yogi Berra
Fusion is the energy of the future....and it always will be
Retired now...Doing only what I want and not what I should...every day is a saturday.

DaveC
Posts: 2346
Joined: Sat Jun 30, 2001 5:13 am
Real name:

### Re: FAQ- mean free path

Richard - some excellent questions, for which I have one some (somewhat) educated guesses.

Starting with the easiest one, the non-intercepting grid or lens system. This one is tractable with one major caveat: That the ions arrive within a certain limited central angle of the lens axis.
This may become a serious limitation in the free wheeling environment of the spherical fusor, but probably can be pracitcally managed in a linear collider type geometry that we have all chewed on in the past.

As to the average gas temperature or energy... that is quite a bit more nebulous to me .. at the moment. A first guess is that given a reasonably long mean free path, say cm, then ions once created will collide only once or twice on their way to the center.
If we assume those collisions to be capable of producing another singly ionized ion from a neutral, then the first ion is only slowed slightly and arrives with an energy representative of where in the radial dimension of the fusor, it was first created.

Lacking any specific information on which radial shells might be preferred ionization regions, we could just assume that neutrals could be ionized almost anywhere, and then head off toward the negative polarity at the center. This means that for a given HV potential, the ions will have a range of energies - from barely ionized to full multi-KeV energies... So.. a first guess would be the average energy would at the radial potential where the fusor volume is 50% of the total volume.

Since the Fusor is a spherical shell, then the mean radius is just the cube root of 1/2 time the fusor radius -which works out to be (0.5)^0.333 or about 0.7937 *Radius. So, in round numbers the average ion energy should be something like 79% of the maximum KeV being applied.... maybe.

But the average neutral temperature probably is time dependent. One could imagine that given sufficient operating time, the neutrals would approach equilibrium with the ions. The time to do this woould be longer at low pressures and shorter at the higher end.

Just estimate how long, we would need a plausible hypothesis for the energy sharing. If we follow an ion around, we will get an idea what to expect. Since a deuteron and a neutral deuterium are nearly the same mass, in a head- on collision ( we ignore A fusion event here, for simplicity). the energies of the deuteron and neutral are exchanged. The neutral gets greatly accelerated, and the deuteron is strongly braked.. (bremstrahhlung here???)

Obviously most collisions will not be direct, so the energy exchange will less exciting But assuming they all were direct collisions, then we have this process cooling the ions, but probably NOT reducing their number...while in between collisions they regain some or all of their lost energy.

Since there is no restriction that a neutral cannot be more energetic (after a collision), then one can expect that given sufficient time, the equilibrium energy would be some fraction of the mean ion energy. The energy lost to the walls will be larger at higher pressure and temperature so the gas inside has to drag the fusor shell along with it in temperature.

If the Fusor's thermal impedance to the ambient were calculated, then the temperature rise can be predicted. I think we all know what this indicates... the thing gets very very hot...

I will tackle the remaining questions you raise... when my head clears...from these.

Good exercise.. I think we are getting some good insights..

Dave Cooper

Richard Hull
Moderator
Posts: 12730
Joined: Fri Jun 15, 2001 1:44 pm
Real name: Richard Hull

### Re: FAQ- mean free path

Thanks very much Dave for the pearls of wisdom.

I was stunned that there might be over 79% of the main energy in neutrals once equilibrium is reached. If this is the case, there should be virtually no collisions of neutrals or ions compared to the population!! Given a radius of 15cm, and assuming only 60% of the acceleratory potential of 30kv is the gas temp at equilibration, then that would put a large cadre of ions and neutrals at the 18kev temp range which would be about 200 million kelvins. This forecasts an MFP amongst themselves and other more energetic ions and neutrals at a minimum of 32 kilometers for 1.2 angstrom neutrals and over 150 kilometers for sub angstrom deuterons!!!!!!

The above would mean a virtual infinitude of recirculations for ions and nearly a zero chance of collisions within the device for equilibrium neutrals or virtually 99.9999999% of all thermally equilibrated neutrals slam the walls.

I am not being argumentative here, just running the math.

Restated, the vast majority of thermally equilibrated neutrals are wall heaters with little possibility of collision with their peers and even less with ions. Likewise, in a 200 million degree gas environment 100,000 recirculations for a giving ion of similar energies is in the 50% probability range! Actually, the ions have to come to rest and cool after an inner grid miss and then re-accelerate so they are more vulnerable during the cooler than 200million degree phase of their recirculatory eliptical orbits.

If I remember aright, MFP refers to the statistical probability that at a given gas density and a given temperature 63% of the particles will go freely unobstucted by collision for the stated distance.

Wow!

Again, I do realize that there are a number of slower gas particles and ions in the device. I think we can assume these are mostly near the shell and the inner grid. As we have discussed many times. There are numerous temps of ions and neutrals throughout the device. I just never thought that even 10% of all species at any one time would even approach 50% of the input energy! You have said that at equilibrium, a significant number of ions and neutrals might have as much as 79% of the input energy reflected as their temperature. this knocked me for a loop and implies a nearly infinite recirculatory period and near 100% neutral wall collisions for the bulk of these equilbrated neutrals.

Richard Hull
Progress may have been a good thing once, but it just went on too long. - Yogi Berra
Fusion is the energy of the future....and it always will be
Retired now...Doing only what I want and not what I should...every day is a saturday.

DaveC
Posts: 2346
Joined: Sat Jun 30, 2001 5:13 am
Real name:

### Re: FAQ- mean free path

Richard - thanks for your much valued thoughts, and for drawing some interesting conclusions on this.

I think it is early on my part to be advancing anything with dogmatism.. Please consider these just ideas for discussion.

One thing your latest comments brought out to me.. was the significance of the enormous differences in size of the Deuteron with its double weight nucleus and no electron, and the Deuterium atom. My analysis of last night... really doesn't address that point. I had just assumed they DO collide.

It may be only a "probability of collision" issue, but if that probability is very low, as the huge MFP indicates, then the thermalization in the gas mix may be so very gradual as to be neglible. That would leave things in the basic non-equilibrium condtion, we have come to assume thus far.

There are many nooks and crannies here to be investigated.

Dave Cooper

Frank Sanns
Posts: 1750
Joined: Fri Jun 14, 2002 6:26 pm
Real name: Frank Sanns
Location: Pittsburgh, PA USA

### Re: FAQ- mean free path

The distance from the center is not as big a factor as it may seem. It would seem that if an ion were only a short distance from the inner grid that those ions would "waste" fusor energy. Actually there are very few of them just because of geometry. A sphere contains 89% of its volume in the outer half radius. Only 11% of the volume is in the inner half radius. As long as you are running far enough up on the cross section curve (ie 30KV or higher), the wasted energy from the close ions is almost a non issue.

I hate to keep bringing up RF driven or pulsed but that is an effective way to keep energy from getting lost in unproductive electron and ion currents. Think of a drift tube in a linac. Works great. The inner grid can be just like that.

Also, it does not take ion current per se for fusion. Think of two situations. The first and most common way is what we all seem to want to do....crank up the power to the fusor. This primarily makes glutonous ion and electron currents and we try for the highest neutron numbers and of course they come. However, this is by far the least efficient use of power.
The second way is electrostatics. That is the principle that I think is the most effective use of energy for a fusor. You put a potential on a grid and all charged particles move. If they are moved in a way that they can collide then this is the best use of energy. Only the energy for charge separation is used and no wasted ion and electron currents here. I know, I know there will be a current when the charges separate but that is only during the cycle change or to throw the switch on and is the minimum energy to get a charged particle up speed.

Frank Sanns