FAQ - Losses - theory and operational issues of a fusor

It may be difficult to separate "theory" from "application," but let''s see if this helps facilitate the discussion.
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DrewMonroe
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FAQ - Losses - theory and operational issues of a fusor

Post by DrewMonroe »

Admin: This thread ultimately fulfilled a crying need that has been handled in bits and pieces in other FAQs. I have decided to take the liberty of turning this entire thread into a FAQ due to the volley of well thought out neophyte questions and the related answers, subsequent to this opening post...............RH

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Let me preface by saying that while I am relativley new and have only begun to order parts for my fusork, I have read many of the FAQ's, "Amatuer Nucleur Fusion" and "The World's Simplest Fusion Reactor". I have also thumbed through parts of Dolan's book. I just don't want you to think I haven't done any of my own research before asking these questions.

I am trying to understand what is physically happening during losses. For example, what happens when an Ion hits the grid? Does it pick up an electron bounce off and become neutral? Do ions hit the sides of the chambers or does the positive charge from the chamber being grounded keep that from happening? I think I read on a post when one does hit the chamber, additional electrons shoot into the chamber? Is that correct?

Richard Hull frequently refers to "Neutrals" being in the system. Are they just Deuterium atoms that for some reason the positive attraction of the chamber has not pulled away the electrons? Would there be any advantage to ionizing the Deuterium before being put into the chamber?

I also understand that some electrons escape from from the grid. I assume they rush toward the chamber? What happens when they get there? I have never seen a grid be insulated. I assume that that would lessen the electric field?

Could a chamber be made of an insulator? Ceramic for example? Would it still work if the Deuterium were ionized before being put into the chamber? How vital to the recirculation of the ions is the positively charged chamber?

Finally, what happens when alpha particles hit the chamber and grid? How much of the energy from the alpha particles hitting the chamber is transfered to heat?

I know this is a lot of questions, but I think it would go a long way to helping me understand the theory behind everything. I appreciate anyone who takes the time to answer any.

Thanks!
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Richard Hull
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Re: FAQ - Losses - theory and operational issues of a fusor

Post by Richard Hull »

Drew I appreciate your attempt to read as much as possible before posting. You have many questions for which the answers are murky and virtually all possible answers would be qualified and attached to many gotcha's with no gospel attached to any.

As I have posted before in many missives, Everything possible within the physics of the environment is going on at the same time in a fusor. If you can imagine it and the physics supports it, it is happening, but each process is going on to a greater or lesser degree that might only be guessed at.

Thus attempted answers given below will be stabs at the questions.

Q: "I am trying to understand what is physically happening during losses. For example, what happens when an Ion hits the grid? Does it pick up an electron bounce off and become neutral? Do ions hit the sides of the chambers or does the positive charge from the chamber being grounded keep that from happening? I think I read on a post when one does hit the chamber, additional electrons shoot into the chamber? Is that correct?"

A: Whew! Here goes....Losses are the premier job of the fusor. That's right wasting power to achieve a tiny fraction of a stated goal, fusion. Electron losses in colliding with and heating the shell is certainly the number one loss factor among many though not all possible electron loss factors. Ionic impacts and electron emissonal heating of the grid is a major loss, as well. Ions hitting the grid at high speed can remain ions in recoil, neutralize and recoil or, more rarely, be buried as deuterium atoms within the grid, itself. Useful, positive ions rarely impact the shell. However, fast neutralized deuterium atoms commonly impact the shell and either are buried or bounce back out as neutrals or ions. Fast neutrals can re-ionize themselves within the gas environment and also knock out electrons from the shell to ionize neutral atoms in the chamber gas. Buried deuterium atoms in the shell can be knocked back into the gas environment by fast neutrals and high speed electrons as either neutrals or deuterons. The whole business is a great big mess.

I will answer your remaining questions within your text below...........

Q: "Richard Hull frequently refers to "Neutrals" being in the system. Are they just Deuterium atoms that for some reason the positive attraction of the chamber has not pulled away the electrons? Would there be any advantage to ionizing the Deuterium before being put into the chamber?"*******
A: Neutrals are deuterium atoms that are slow, fast, relatively stationary or( "thermalized") due to any number of the afore mentioned processes. They are everywhere in the fusor and always compose 99.9999999999% of the bulk of the matter in the fusor chamber

Q: "I also understand that some electrons escape from from the grid. I assume they rush toward the chamber? What happens when they get there?"********
A: It is common for electrons to be shorn from the grid due to field and thermal conditions. (Already covered above) Assuming most electrons from the grid are accerated all the way to the shell, they can simply heat the shell due to impact, but many create x-rays. The x-rays under about 30kev in energy all hit the shell and exchange their energy creating electrons at the shell to produce heat or make deuterons in the gas.

Q: " I have never seen a grid be insulated. I assume that that would lessen the electric field?"*******
A: You could not and should not insulate the grid; it is a key electrical element in the system.

Q: "Could a chamber be made of an insulator? Ceramic for example?"*******
A: Back at you......WHY? It is a key electrical element within the system and a great heat sink, absorbing almost all of the energy put into the system as loss.

Q: "Would it still work if the Deuterium were ionized before being put into the chamber?"
A: We have an entire forum devoted to this (ion guns). Yes, it could be helpful, but no one has upped the fusion rate with one yet. Ion guns create a whole new expense and operational regime that can be costly in both time and money and is never used on the first pass at fusion here.

Q: "How vital to the recirculation of the ions is the positively charged chamber?"*******
A: It is absolutely critical and demanded to support the tiny amount of actual recirculation that does occur.

Q: "Finally, what happens when alpha particles hit the chamber and grid? How much of the energy from the alpha particles hitting the chamber is transfered to heat?" ******
A: 100% of the protons and helium3 atoms, (what you might be calling alphas), formed in the system will ultimately wind up as loss and heat. Classic alphas are high energy nuclear emissions of helium-4 atoms our fusors make helium-3 atoms.

Proviso and gotcha' associated with 100% of the above answers..................

You MUST be operating the fusor as is commonly operated by the bulk of the fusioneers to date. Any alteration of the operational regime can shift any or all answers given. (Please dear God, don't ask how unless you are prepared to give a rigidly precise and completely described regime with all new parameters and conditions!!!!!)

Obviously, there are many other ways to operate the fusor that would add to the cost and alter the operation while still doing fusion not covered in the the above answers.

Our way of fusing here is merely the simplest and cheapest method of construction within a gas fueled, plasmic environment to attain fusion thus far put forward for the amateur with limited resources to result in a useful fusion research tool.


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
The more complex the idea put forward by the poor amateur, the more likely it will never see embodiment
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Carl Willis
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Re: FAQ - Losses - theory and operational issues of a fusor

Post by Carl Willis »

Hi Drew,

The fusor is an example of a glow discharge apparatus. In some ways it is distinguished from other glow-discharge devices (for instance, it uses a highly-obstructed hollow-cathode discharge in a fusion fuel gas), but fundamentally the same phenomena that are at work in any ordinary discharge tube produce and maintain the discharge in the fusor. Glow discharge mechanisms haven't been discussed on our forums in detail, but Google and Wikipedia undoubtedly point to a lot of good information.

In a discharge, ions striking the cathode eject secondary electrons (and may be re-emitted as neutral atoms / molecules, re-emitted in a different charge state, or implanted). They also impart kinetic energy to the cathode that manifests as heat there. As the cathode heats up, electrons may be liberated by the thermionic effect. Thermionic electrons and those from secondary emission are accelerated toward the anode. On the way there, they may strike and excite or ionize gas molecules, which if ionized, will be pulled toward the cathode...and thus the cycle is repeated and the discharge sustained. Electrons arriving at the anode frequently have high energy, resulting in heating, liberating secondary electrons, and generating x-rays there.

The electric field in a fusor probably does no significant ionization itself. The charged species are essentially all formed as discussed above.

"Neutrals" refers to virgin deuterium molecules that admitted as fill gas, as well as neutral single atoms created in excitation events or fast-moving atoms and molecules resulting from neutralization of ions. Because neutral species are unaffected by electric fields, they cannot be directly accelerated to the high velocities that are important for fusion. However, they may participate in fusion reactions if they are moving fast enough relative to other fusible particles with which they collide.

If you were to coat the grid with a good insulator, eventually the surface of the insulator would attain a strong surface charge and the accelerating field and all current would disappear. But probably nothing is a sufficiently good insulator to have that outcome in a real fusor. The material would either suffer destructive electrical breakdown or would support significant conduction current, causing the fusor to operate much as it would with a bare metal grid. The same would hold true if the outer electrode were insulated. In a "demo fusor" in a glass bell jar, there are typically two concentric grid electrodes. Omitting the outer grid may work after a fashion because of potential difference between the cathode grid and the bell jar baseplate or because glass is actually a rather leaky insulator.

Alpha particles don't appear as a nuclear reaction product in any real amateur fusors. But if they did, they would generally behave in matter similarly to the charged nuclear products from ordinary DD fusion (protons, tritons, and He-3 or "helions"): stop abruptly in the first solid encountered and raise the temperature from the imparted kinetic energy.

Have you written an Introduction yet? I can't find it.

-Carl
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Richard Hull
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Re: FAQ - Losses - theory and operational issues of a fusor

Post by Richard Hull »

Carl has given an operational description which is great. Key to the process in a fusor is to distinguish between potential and current within the device. As a simple electrical device, (glow discharge device),we all know that the current multiplied by the applied voltage determines the energy expended at any given instant.

The voltage impressed across the device is strictly a source of acceleratory potential for current elements, (particles) within the fusor. All current elements created by the input power are electrons. Ultimately, other charged species can act as current elements. All current elements are subject to the acceleratory and decelratory action of the potential field established across the device. The field has little net effect on any truly neutral particle or atom.

The current drawn from the power source is strictly related to the number of charged current elements being created and neutralized each second or per unit time, their net created and neutralized charges being part of a simple closed electrical circuit. The losses and "work" done in the any circuit are ideally disappated in the "load". The load is typically and ideally the highest resistance element in any complete electrical circuit.

In our case, the fusor gas plamsa and discharge are the load. All the energy within the circuit is dumped into the gas environment initially as electrons and from there it is a great big complex mess of charge exchanges which offer high electrical resistance and, thus, the loss of all the input energy within the gas, chamber, fusor electrodes and its surroundings as heat. On average only 1 billionth part of the input energy is released as additional fusion energy which, itself, ultimately winds up as heat.

As I so often note the fusor for all real purposes is a total, net loss device.

Oh.....It also does a decent amount of fusion for amateur purposes and research.

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
The more complex the idea put forward by the poor amateur, the more likely it will never see embodiment
DrewMonroe
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Re: FAQ - Losses - theory and operational issues of a fusor

Post by DrewMonroe »

Thank you for such a detailed response.
 
The reason that I asked about using an insulator as a chamber is that I was thinking that any time deuterium hits the metal chamber, it would would lose kinetic energy in the form of heat. That atom then would have less energy which would make its crossection smaller, and therefore lower its chances of fusing. My thought was with a insulating material, it would lose less energy during the collision, therefore increasing its chances of fusing relative to a metal chamber. Of course this takes away acceleration caused by the positive field. I guess I was wondering if a cylindrical shaped fusor where the flat ends were positively charged and the the tube was made with insulating material would be more efficient.

I think my mistake has been thinking about the theory as too simple of a system. It seems while the device is simple, the result is a complex and randomized plasma soup, that has too much going on to think about as simply as I have.

Thanks again!
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Re: FAQ - Losses - theory and operational issues of a fusor

Post by DrewMonroe »

Thanks so much for the reply. I am going to do some more reading on glow discharge mechanisms and some reading on thermionic emissions.

Help me understand what you are saying about an insulator around the grid. I guess my thought is if two of the inefficiencies of a fusor are the escape of the electrons from the grid and the neutralizing of the deuterium ions from the collisions with the grid, wouldn't an insulated coating, however long it lasted, help with both of those innefficienies? Would an insulator weaken the electric field?

Thanks!
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Re: FAQ - Losses - theory and operational issues of a fusor

Post by DrewMonroe »

And my apologies for not introducing myself. I thought i had a couple years ago. I have post an introduction. Thanks again!
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Chris Bradley
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Re: FAQ - Losses - theory and operational issues of a fusor

Post by Chris Bradley »

Richard has wisely avoided putting emphasis on 'the grid' as playing any bigger part that anything else.

The reality is that beam-fusion devices like this (as opposed to 'thermonuclear') have a fatal flaw, and that is that fusion is a very rare form of collision. If you're interested in where the ion energies go, the two to get concerned about are Coulombic collisions and charge-exchange collisions.

The first is where the electrostatic fields cause the colliding particles to deflect each other away. This causes 'thermalisation' and the collective energies of the particles randomise and 'average out'. But seeing as most of the collisions are with cold, background atoms in fusor conditions, so the average rapidly tends towards 'cold'.

But the bigger route for ion losses still is the charge-exchange process - here, a fast ion will bump into a neutral background and the electron will end up getting swapped over. The result is one 'cold' ion and one 'fast' neutral (it was the fast ion, but it's now collected an electron!).

Both of these collisions have a reaction cross section of around 10^-15 to 10^-16 cm^2 at several to 10's of keV. One can therefore deduce the mean free path of an ion in a 10 micron fusor until it becomes neutralised and heads on into the shell: MFP = 1/(density).(cross-section) = ~1/(10^14).(10^-15~-16) = 10~100 cm.

The last paragraph explains why a fast ion has got about 1 metre of travel in it, on average, before it is neutralised and nails itself into the shell.

Meanwhile, the cross-section for fusion is ~10^-26 cm^2. That means the probability of an ion becoming neutralised is 10^10 times more likely that a fusion event, if an ion were to meet up with another atom. **There** is your main loss of fusible ions - it is the very nature of the fusion collision itself that means it's *the collision* that is the main loss mechanism!!

Many ions may collide with the grid or wherever else, but this itself does not show why fusion reaction rates cannot be 'improved' over a given eifficiency. What *does* show that fusion reactions can't be improved is how quickly an ion is lost whilst undergoing the very collisions we want to encourage that might lead to fusion, but more usually lead to scattering and neutralisation!

An ion would have to travel, on average, a staggering 10^6~10^7 kilometres before it fuses with a background atom in a 10 micron fusor, yet will be neutralised within 1 metre. The fusions we see are the statistical anomalies, the 'significant outliers' to those statistics, the ones that get lucky. A fusor is hammered with power to get quadrillions of ions flying around, and just enough of those win their lotteries that they can be detected!
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Re: FAQ - Losses - theory and operational issues of a fusor

Post by Dennis P Brown »

"An ion would have to travel, on average, a staggering 10^6~10^7 kilometres before it fuses with a background atom in a 10 micron fusor, yet will be neutralised within 1 metre."

Wow - I never realized! Great post!
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Richard Hull
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Re: FAQ - Losses - theory and operational issues of a fusor

Post by Richard Hull »

Chris' last posting here is sage and his last two sentences say, in a short and sweet manner, what I have said over many past posts. It is worth taking out of context and is quoted again here.....

"The fusions we see are the statistical anomalies, the 'significant outliers' to those statistics, the ones that get lucky. A fusor is hammered with power to get quadrillions of ions flying around, and just enough of those win their lotteries that they can be detected!" (CB)

If you think the fusor is doing something amazing, you are right, but for all the wrong reasons. It works because we are forcing the terribly improbable to occur enmass. This is certainly clever, but not necessarily of value to future energy needs. For all the newbs.......If I might paraphrase Dante..... All hope abandon, ye who enter here and seek fusion energy within.

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
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Re: FAQ - Losses - theory and operational issues of a fusor

Post by Dan Tibbets »

I havn't read all of the replies, so my apologies if I am redundant. At extremely simplified terms, with a cathode at high voltage an electron is freed from the metal and flies away from the wire at the accelerated energy from the voltage. As the wire heats up this electron emission increases dramatically due to thermionic effects. The electrons have <1/60 the momentum of the ions (or neutrals) in the wire so they escape much more rapidly. Ions, if they escape are quickly pulled back to the surface of the wire, where they are grounded/ neutralized. So for practical reasons electron emissions from the electrode dominate. With an anode the electrons that may have been emitted are mostly quickly pulled back provided they were at the surface of the electrode. A little distance such as with a passing electron changes things some. Because of this an anode will not create free charge carriers- electrons or positive ions as a general rule. Things can be manipulated to change this to a degree and now you have an ion gun.

When an electron boils off of the surface of a cathode, it flies away with KE equal to the voltage of the electrode. This KE is transferred in part to other electrons when another electron is encountered- such as with a neutral gas atom/ molecule. This will lead to ionization of that atom. The KE of the emmitted electron is great enough that it may lead to the ionization of hundreds of atoms. This is what leads to the profusion of negative and positively charged free charge carriers that are born away from the surface of the electrode. The positive ions are attracted to the negative grid so long as the ion is at a greater radius than the grid. Inside the grid Gauss Law changes things. If the ion hits the grid, two primary things happen. The ion transfers much of it's KE to the wire as heat, and the ion picks up a electron from the continually replaced electron supply in the wire to become neutral with some small retained KE. This might represent an average event, but a range of possibilities can occur, - lower energy transfer, total energy transfer with the ion embedding in the wire, sputtering of other neutrals , ions and electrons off of the wire, each with their own share of the KE, etc... The important point is that ions created away from the electrode due to cascading secondary ionizations from the high energy electrons, will then be accelerated by the negative voltage on the cathode grid towards the center (depending on the symmetry of the grid, the exact center may not be the result). As the ion approaches closer to a particular wire it will curve more towards that wire, and may hit it. This leads to the transparency term. If on a single pass the ion has a 1/10 to 20 chance of hitting a wire the transparency is referred to as ~ 90-95%.

I believe there are two loss mechanisms involved. If the KE of the ion is lost when it hits the cathode wire (or it is up scattered so that it can hit the shell). This is one loss mechanism. The other loss mechanism is the free electrons traveling outward hitting the wall. Basically for each ion created (in deuteriums case with a Z of one) there is one free electron created, plus the electrons emitted from the cathode. Because of the transparency issue, each ion lasts 10 to 20 times longer than the an electron (this in terms of passes or distance traveled, with the speed of the ions and electrons considered the time duration of the electrons and ions is further modified. So the total electron current - primary and secondary electron current, may be ~ 1000 times greater than the ion current. Thus it is the electron current that accounts for most of the losses.

You might insulate the cathode grids to increase the ion lifetimes in terms of ion passes. Note that insulation refers to magnetic insulation, not electrostatic insulation like with a rubber insulating coating. This is good in terms of ion losses, but as pointed out above, electron losses are already dominate.

The Elmore Tuck Watson Fusor concept reverses things, the electrons are now making multiple passes limited by the transparency of the anode grid, along with thermalization issues. In this situation the primary electrons are created with an independent E-gun located outside the anode grid. The electrons accelerate towards the anode grid, drift once inside the grid until they pass out the other side, then are again accelerated back towards the center. While inside the anode grid, the electrons build up a virtual cathode and this can become dominate over the electron emitting grid/ e-gun electrons as a source of the space charge, because each electron now has a lifetime of 10-20 or more passes instead of just 1. It can be viewed as the supplied electron current being 1/10 less for the same space charge effect (virtual cathode) or 10 times the virtual cathode strength at the same electron current. The ions are born or somehow injected at low energy inside the positive anode, so they are trapped by the virtual cathode just as they are trapped by the real cathode in they typical fusor. But, now the ions may have a lifetime of unlimited passes- except limited by thermalization issues. This is good for the ion lifetimes, the ion based current is decreased considerably, but the electron losses through hitting the anode grid is still a major problem. Again magnetically insulating this anode grid may mitigate the electron current losses.

The amount of gains necessary to push a Fusor's performance to the point where a Q greater than one might be obtained is tremendous (often a ~ one billion fold improvement in efficiency is given) , so the magnetic shielding of the anode grid has to be very good. Also, thermalization issues must be controlled to an uncertain extent. This is the approach claimed for the Polywell. Even then the goal is not met. Such things as recirculation while maintaining the magnetic shielding, density boosting mechanisms, and impedance of thermaliation need to be addressed. I should note that recirculation of electrons is the whole ETW purpose, the tricky part comes from needing to allow for this while also having good magnetic shielding of the grid, and especially the density boosting properties. Basically, the goal is that you can pump electrons in faster than they leak out (or hit the grid) to a higher limiting density and having this high density lead to increased fusion rates to useful levels- that exceed the electron losses and provides for meaningful excess power for utilization.

Dan Tibbets
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