FAQ: Fusion ash/byproducts -Energy- How much?

It may be difficult to separate "theory" from "application," but let''s see if this helps facilitate the discussion.
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Richard Hull
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FAQ: Fusion ash/byproducts -Energy- How much?

Post by Richard Hull »

Note*** This FAQ is a long posting and full of good data. It is recommend that this be printed out. WARNING... I swear that if I ever have another person ask another question about what is contained here I will hunt them down and lash them to the fusor shell of Jon Rosentiel's fusor, turn it on full blast and walk away. RH

What happens during D-D fusion?..........

When doing D-D fusion in a simple fusor there are by-products or fusion "ash". They are the result of two equally probable reactions.

D+D = He3 + n + 3.27mev
D+D= T + p + 4.03mev

There is a never seen, "also ran" reaction possible which we will ignore as a non-issue.

D+D= He4 + gamma - Only one of this reaction for every 10,000,000 fusions!!!!

So, we will concentrate on the real expected ash from D-D fusion. The ash is in the form of the following by-products:

1. He3, Helium3 (rare, STABLE isotope of helium)
2. Tritium ( rare , UNSTABLE, radioactive isotope of Hydrogen)
3. Proton (protium) main energy bearer of fusion (~3mev)
4. Neutron (secondary energy bearer of the fusion (~2.45mev)

What is the "Ash" good for?..............

The Ash, plus excess energy is the actual leftover products from fusion. some of it is useful and other parts of it are not.

We will discuss the ash products in more detail........

1. Helium 3 - This is a rare and expensive isotope of helium that is made solely by fusion processes. It is stable and not radioactive. It is capable of fusion with itself and deuterium all of which are in the fusor already as it is produced. The possibility exists of burning it as well or extracting it for sale. We will analyze this possibility shortly as it also applies to precisely the same degree to the next fusion ash product............

2. Tritium - This is a rare isotope of Hydrogen. It, too, is a product of fusion only. It is radioactive with about a 13 year half life. It decays via very weak beta emission to form He3. Tritium and deuterium (already in the fusor) can fuse to form one of the easiest of all fusion reactions and produces one of the highest fusion energy yields (~14mev/fusion). Its ash is common, non-radioactive, helium 4. As with the He3 ash, we will investigate the tritium production rates in a simple fusor in a moment.

3. Proton (protium) - This ash product is in the form of a recoiling proton and carries a significant amount of energy, ( the greatest in the D-D fusion cycle ~3mev), from its related fusion event. It usually slams into the inner grid, but always winds up as common hydrogen gas by taking up a free electron. It, in theory could act as a poison of sorts as the fusor re-ionizes it and it speeds towards the center mixed in with deuterium ions. (useless, fusionless collisions) more on this later, as well.

4. neutron - This is a radioactive ash product in that it decays with a half life of about 10 minutes. It is also the only product that will definitely escape the fusor vessel while in operation. All neutrons will escape! These neutrons are one of the two main sources of fusion energy as they are classically decelerated in a water jacket, thereby raising its temperature. This is ultimately the end of the neutron story in D-D fusion.

Now to the math .......................................

We need to find out to just what degree hydrogen gas will poison the fusor and to what degree we produce secondary fusible and radioactive by-products or ash.

We commonly gauge fusor success rates by reporting the the isotropic neutron numbers measured from the fusor. This is very useful for when we double that number we get the number of actual fusions taking place in the fusor volume. Likewise, the D-D fuel is consumed at the rate of 4 deuterium atoms per neutron recorded or 2 deuterium atoms for every fusion.

Example:

Given - We have a 6" diameter D-D fusor backfilled and sealed off with pure deuterium gas to a pressure of 10 microns. It is seen to produce 1 million neutrons/sec or 2 million fusion events per second. Let us run it for a very long time for an amateur fusor....10 hours. Let us also say that it is consuming power from the wall outlet at the rate of 30kv @ 8ma or 240 watts.

Questions -

1- How much tritium are we making and how will this superb fusion fuel "boost fusion" as it reacts easily with the deuterium in the vessel (D-T reaction) as it continues to run?

2- How much He3 will we make that might also act as a poor, but real, booster as it burns via the D-He3 reaction? Or, if we choose, how much He3 can we make for sale via extracting it after the 10 hour fusor run?

3- How badly will the hydrogen gas ash build-up in the fusor and start to poison the reaction by being constantly re-ionized, accelerated to the poissor only to collide uselessly?

We now have given data and a series of questions that can be mathematically treated in an attempt to answer these questions.

What do we know based on the above?

1. Pressure in the fusor is 10e-2 torr.....or rounding off, about 10e-5 atmosphere, STP (Atmospheric pressure at standard temperature and pressure.)......................

As there are about 10e19 gas MOLECULES/cc at STP that would be 2X10e19 deuterium atoms/cc at STP (D2 is diatomic)..........

This would make our fusor contain 2X10e19 X 10e-5 = 2X10e14 deuterium atoms per cc at the stated operating pressure.........

The fusor volume is 4/3 X Pi X 7.62^3 = 1600 cc..............

This makes our fusor charged with a grand total of 1.6x10e3 X 2x10e14 = 3.2 x 10e17 deuterium atoms.

2. As we are burning at 2 miilion fusions per second, we are losing dueterium atoms at the rate of 4 million atoms per second.

3. We are making one million atoms each of both helium 3 and Tritium every second that the fusor operates.

4. At the end of 10 hours continuous operation, we have a grand total of 10 X 60 X 60 X 1x10e6 = 3.6 x 10 e10 atoms EACH of both tritium and helium3. That is 36 billion atoms each.

5. At the end of ten hours we have used up (3.6 x 10e4) x (4 x 10e6) = 1.44 x 10e11 (144 billion) atoms of our deuterium fuel supply.

6. This leaves us with a mere 3.199999 x10e17 deuterium atoms left!!!!!!!...... Down from 3.2 x 10e17, our starting amount. For those too slow to catch the upshot of this.....we have used, effectively, NONE of the original deuterium fuel! Our fuel tank is still 99.9999% FULL!

7. We have expended about 2.4 kilowatt hours of electricity and produced about 2x10e6 x 7 x 10e6 x 3.6x10e4 X 10e-19 = 50 milliwatt hours of fusion energy, assuming all the energy of fusion is trapped in a water jacket. This makes the COP for fusion alone a rather dismal 0.00002. Still, if we can capture all the waste energy and all the fusion energy, the device is over unity to the tune of 1.00002 as a water heater.

For those not getting the picture already, we will go further and beat this dog to death.

The ratio of freshly produced tritium and He3 fuel to deuterium reactants in our vessel is now about 10 million to one. This is also the probability of fusion of those components if we were to start the fusor up afresh, right now!!! For every 10 seconds of renewed operation, we would produce only a single D-T fusion!!!
Actually, it would probably be greater than this by as much as 5 times or 5 D-T fusions every ten seconds (1 fusion every 2 seconds) due to the increased cross section of the D-T reaction.

The rate of D-He3 reaction would be many time less due to its reduced cross second at our low voltage fusor operation......Perhaps 1 D-He3 fusion every 100 seconds.

We can see that even after ten hours of operation at 2 million fusions every second in a tiny little 6" fusor. That we have made no real fusion fuels to aide us or boost fusion output.

For the same reasons as given for He3 and T, the hydrogen poisoning is a non-issue as only 1 in 10 million collisions might be expected to involve the worthless proton-D collision.

Continuing to beat the poor dog..........

We would have to operate the fusor for .5X10e7 seconds before we hit the ideal 50-50 D-T mix and were now a D-T fusor..... That is, after only 1 year and 5 months of 24 -7 operation, we would move from a 2 million fusion per second device to about a 200 million fusion per second device having morphed from D-D fusion to D-T fusion over this time frame.

Of course we have consumed over 2 million, 940 thousand watts of electricity and spent about $300 out of pocket (for electricity) in the effort. Woulda' been a lot better to spend about 100 dollars for about 10 curies of T in an ampoule and start off with the D-T fusor regime. (Of course, there is that AEC site license needed for the tritium buy).

Finally, as the dog draws its last breath................

The amount of He3 produced over the original 10 hour 2.4 kilowatt run, if totally recoverd and separated, and sold at current market rates, would bring in about 0.28 cents. With electricity costing about 10cents/kwh, we have lost 23.7 cents out of pocket in our bold, He3, make it-sell it, get rich quick scheme.

It was slow and painful, but that dog is now dead and hopefully resting peacefully.

So if you have any big ideas about getting rich quick or see a fusor producing even 1 billion neutrons/second, PUT IT BACK IN YOUR PANTS! NO ONE IS IMPRESSED... beyond giving you a big ole pat on the head and maybe three or four atta' boys.

Now for the reality check......

In the above calculations some small assumptions have been made that are not realistic, i.e., perfectly sealed fusor, 10 hour or 1.4 year run times, perfect, cost free extraction of He3 from the fusor run, etc. But all these have been pretty much in favor of the dreamer and ever hopeful fusioneer. In every case, the realities of operation would, at every turn, make the returns, profits and final values far below those calculated here (much worse than presented). So the above can be considered a super rosey picture. You might say an ideal or best possible scenario type of operation and results.

Pre-apologetic mea culpa:

I have tried my best to hold to capturing and handling all the powers of ten here. However every now and then one of those rascals gets away. The problem is that muffed powers of ten in a mathematically flowing discussion where there is a whole gang o' them, can multiply errors out of all reason real quickly. Especially, if the mistake is made early on in the effort.

So, if any of you catch are error please report it so that this crucial posting can be corrected and I can be properly blamed. A few of you will follow up on this just for the latter joy alone, but in all events help me keep this thing accurate and true for a lot of newbies will have there bubble burst over these "hard sayins". If we are going to send them off to cry in their beer, let's do it well.

Richard Hull

P.S. Thanks to Jon Rosenstiel for his correction comments in creating this post. RH
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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Re: FAQ: Fusion ash/byproducts -Energy- How much?

Post by 3l »

Hi Richard:

Don't worry that is what the forum is good for.
I should know having dropped and mangled math in the past.
I keep getting emails on the fusor production stuff also.
I sometimes wish I could be curmudgendy but alas no.
Even pulsed where you get close to 10^9 fusions...the efficiency still sucks.
My answer is the same as yours Richard....until the power plug works it's all bull. The only possible way to improve fusion at high density and low speeds is to be a star unfortunately. That's why the esoteric exotic path is pursueded. It is my aim to goto thermonuclear fission where the odds are 1 in 200 in hydrogen and lithium. But to do that you must make 1mev protons in greater numbers than any time in history. That's when Ca entered the picture. Studying CA has illuminated pulse fusion difficulties in coupling electrical discharge to the ion pulse uniformly. Might be able to squeeze maybe 3 decimal points out of the efficiency problem but after that the method has severe limitations. I have not remotely given up. But beam beam collisional at higher vacuums seems to be the better way in achieving d-d fusion.
Back to Pontiac guys!

Happy Fusoring!
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Re: FAQ: Fusion ash/byproducts -Energy- How much?

Post by ex-engineer »

Richard -

It seems like this FAQ would be a nice place (though presumably adding another nail in the coffin) to include a discussion of what the odds are of an accelerated deuteron (pick a convenient voltage) fusing with a deuterium atom instead of just having an elastic collision - or series of them. In other words, even if we imagine that we manage to build a perfectly efficient ion gun, magnetically shield the grid from collisions, and accelerate every deuteron with full voltage, for every million accelerated deuterons, how many will fuse and how many will just end up all hot and bothered. I've been wandering around the ENDF files (after having gone through anything having to do with cross-section on both this and the old boards and looking through Jim Lux's couple of pages on the subject), and unless I'm making a mistake in interpreting it all (quite likely), the basic situation looks dismal -- before we start pointing out that perfect ion guns and magnetically shielded grids don't exist.

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Re: FAQ: Fusion ash/byproducts -Energy- How much?

Post by Richard Hull »

Matt has hit the nail on the head. It is all rather probablistic based on crossections. These are whole nuclei that we are trying to make do stuff that we want them to do. It is like trying to train a cat to fetch the paper, it can be done but how many have you seen do it?

The nuclei respond as if near the edge of heisenberg reality, and they are. When the world gets too fuzzy due to this effect, we are still capable of predicting OVERALL, GENERAL, macroscopic results, but we can never train the entire group of individual cats to do what we want, in lockstep.

Below this level, I think discussion of the subnuclear fleas is kind of a joke where we fool ourselves into thinking of their being of some ultimately useful purpose to man beyond enlightened philosophic discussions.

Again, the arrow points the wrong way for us. Nature only allows disassembly of matter , for the most part, by the hand of man.

e=mc^2 is ostensibly a one way street. (not truly reversable as in m = e/c^2 as a bulk physical process for matter particles) Only nature at the stellar level stands a chance of reversing the arrow and when she does reverse it, she still creates NO NEW MATTER PARTICLES, only mass defect. Why? Because charge can neither be created nor destroyed. This is why I opt for charge being a key player in the universe along with gravity.

Fusion is a mass defect operation and readily doable by man, but not in a fashion he can really control or win significant amounts of energy from in an economic fashion.

We are much better at disassembly operations (fission) and this operation is a net winner and producing a bunch o' grid power daily. The only reason we are good at it is that we are moving with nature's natural arrow for e=mc^2.

Remember, all fission is just, us, releasing trapped solar energy in the form of mass defect which was ultimately the result of gravity which created the mass defect around a bunch of already extant charged matter particles, originally.

We tend to use the terms mass and energy like they are truly interchangable at the charged particle (real matter) level. They are not!

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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Re: FAQ: Fusion ash/byproducts -Energy- How much?

Post by jst »

Whoa Horsey... The fact that fusors are very innefficient, is something we all realized quite a while ago. Richards analysis merely proves this in a nice concise package. It also points to what has been known for the better part of a century... we have nothing that could possibly contain the energy, heat or pressures associated with fusion!

The key problem with fusion as we currently practice it, is containment.

And this is not a problem we can solve with high energy ion guns triggering nucleii collisions. As energy levels go beyond a certain point, nucleii simply pass right thru each other as their wave functions take over. Fusion, like fission must be effected with particles operating within a certain energy band to permit capture and in a certain "contained" locale. Below that level they bounce, above that level they just go right thru each other.

This is why I'm calling for the use of WFA based electron beams at near cosmological energy levels. The objective is to present a lepton wave wall at the surface of the microstar with the specific purpose of containment. Imagine a bunch of firehoses spraying high pressure water at a beach ball to keep it located in the air. We're talking a force field with 100MEV energy levels that appears like a wall of ~10^14+p.s.i. capability. Thats a very hot, very high pressure container, which exceeds anything found in any stellar mass I'm aware of ( there aren't many cosmic masses that exceed 10KEV ).

If I'm right, and based on Richard's rather nice summary of energies involved, this model will contain the entire nucleii mass as a plasma microstar, including all by-products including neutrons! No small achievement in its own right for us to pull off.

And it is the ability to contain the entire fusion plasma including byproducts and all, that is the big deal here. That capability significantly raises the dwell time of any reaction, from the sub-picoseconds of a collision to micro or even milliseconds. Do that, and Richard's nicely calculated near none-existent efficiency number is amplified by a factor of 10^3 to 10^6. At which point, things look much better.

In my "sewer pipe" proposal, I take an "all good things must come to an end" philosophy. When the plasma microstar is good and cooked the WFAs closest to the drift tube deflect their beams away from the drift tube, causing a rupture in the microstar which launches the plasma down the drift tube. The dirft tube would contain conventional MHD energy extraction technology, fusion fuel injection, and a waste product extraction system that I haven't got a clue on... yet... Thats why I call it a sewer pipe solution ;) .

And if we ever make this work, won't it be a prize embarrassment to all the magnetic confinement crowd :) to have some cheap contraption made of steel "sewer pipe" to actually work?
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Re: FAQ: Fusion ash/byproducts -Energy- How much?

Post by DaveC »

Good review Richard, Thanks. The numbers seem to be correct within a factor fo 10 and that's just about "exact" for theory in Physics.

I would just like to add the point, that we do not actually "contain" anything in the fusors. All the fusing is done on the fly.

Deuterons that miss swing around and either collide with something or loop around and have another pass through the collision region, although not necessarily along the same trajectory.

Now the degree to which the grid can "focus" ions or electrons, so that they pass through without colliding with the cage... to that same degree the total current will be reduced.

If the ions were to circulate without collisions, the input power would be simply the number of ions times the energy each ion acquires, divided by the time it took for that to happen. That's it. The fields casuing circulating are conservative and thus require no additional energy. Power would go to zero.

The relatively large currents in the real world fusor are connected with ions terminating their flights on metal electrodes. With mA level currents, this is telling us that about 10 ^12 electrons/ions are being collected through collision. If a couple million fusions take place, we are getting ppm level collision percentages, but with an energy increase of 3MEV versus 20KeV of original energy.Thus some 150 X energy increase, but only in ppm levels. Thus energy conversion is in the 150 ppm range of the original energy inputs, or about 0.015%.

So.... to get anything respectable, like 20 to 30% conversion efficiency, far fewer unwanted collisions must occur. And therein lies the tale....

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Re: FAQ: Fusion ash/byproducts -Energy- How much?

Post by Adam Szendrey »

Hi,

Just wondering...A fusor with a "fusing" efficiency of 100 % (all the energy that goes in is used to create fusion reactions) would be VERY dangerous.
I mean a fusor with an extremely low efficiency can produce lethal x-ray radiation/neutron flux. A 100 % efficient fusor (again this does not mean that energy out= energy in, it means that all the input energy is used for fusion) it would produce several orders of magnitude more radiation/heat, making it a very unpleasant companion in a couple hundred meter radius.

But...i'm a bit confused here...high energy byproducts of fusion will collide (or not) on their way...so in an even low efficiency fusor, why doesn't a sort of "thermonuclear chain-reaction" takes place, as the multi MeV particles collide with other particles (fuseable particles), and create more high energy particles, etc.? They have a long way to collide before they hit the inner grid...they simply miss the other particles? The plasma density is too low?
Maybe there is a critical palsma density. Maybe the runaway events are such scenarios where the sufficient plasma density is reached?
During such a runaway the the grid structure desintegrates, and thus the required electrical field system collapses, disappears, and all fusion reactions stop.

Adam
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Re: FAQ: Fusion ash/byproducts -Energy- How much?

Post by Richard Hull »

Dave is correct! I find more and more fusioneers here that think we are confining or containing something! It isn't so!

We are creating a "focused collision zone". Nothing more.

The fusor, as we are operating it, is a "spherically focused, particle accelerator - collider. The inertial confinment is a bit of a misnomer. It is true that there is an increased density of matter in the center of the device, but this is due to the mass collider action such that there is a greater "mass presence" in the center of the device.

Now, Adam has asked about dangers. If all the energy input to the device was turned into fusion events then we would create a huge and prodigious output that would kill, in just seconds, anyone near it.

The real problem is that in using the kilowatt we have to create ions. So let us say that we create a full kilowatt of ions with our kilowatt of electricity input (impossible). Still, based on their energy, they can never, ever all do fusion as fusion is a cross sectional probability thing.

So...... we will, just for grins, make a second impossible assumption and say that we not only create a perfect kilowatt of ions, but that all collide and fuse.

A kilowatt is 1000 joule seconds. Therefore, this relates to an ion current of 10e 22 particles of 1 ev energy each. These won't fuse! You see the problem!!? We will have to supply a reasonable assumption of energy per particle to even warrant fusion! To peak out on the D-D curve at about 1 mev, that would leave 10e16, 1 mev duetrerons in collision. Again, assuming the riduculous perfect fusion of these particles, this would mean 10e16/2 fusions per second or 5X10e15 fusions.

Questions --

What would be the shell absorbed energy?
What would be the neutrons flux from the device?

The last question is easy to answer rather quickly. As there are 5X10e15 fusions/sec, there will be 2.5x10e15 neutrons per second leaving the fusor. A hapless fusioneer at a 1 meter distance from the fusor would suffer a flux of aproximately (2.5x10e15) / 4 X Pi X (100^2) = 2x10e10 n/cmsq/second.

This would kill in seconds!! Remember, THIS IS A FLUX! 20 billion neutrons hitting every square centimeter of your body every second. Jon Rosentiel's best run was a flux of about 20-40 neutrons/sqcm/sec.at one meter. (Jon still holds the amateur record)

Now onto the question about the energy absorbed in the shell. We have, to again, total all the energies of all the trapped reactants. We will assume that all their energies are delivered to the wall of the fusor. That would be He3 ~ .82mev + P ~3.02mev + T ~1mev. this would convert to 2.5x10e15 X 4mev + 2.5x10e15 X .8mev or 1 x10e22 + 0.2 x10e22 = 1.2 x10e22 ev absorbed by the shell. This converts back to 1200 watts into the shell. assuming a water jacket and slowing of the bulk of the emitted neutrons. you might figure on another 650 watts in hot water. So, for the 1kw in we got almost double out.

The bottom line is we were burning about .001cc of D2 @stp per second. Creating a deadly neutron hazard for any nearby operator and putting out about double our input energy with an ion current of only about 1ma.

I think I might have done eveything correctly here.


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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Re: FAQ: Fusion ash/byproducts -Energy- How much?

Post by 3l »

Hi Richard :

You have done good.
I don't know how many emails I've answered to this very problem of so called containment. In fact ALL machines in the thermonuclear fusion realm are colliders pure and simple.
That includes the H bomb.

Happy Fusoring!
Larry Leins
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Re: FAQ: Fusion ash/byproducts - endoergic d+d rxns

Post by john_h »

Here is a little calculator I found at the Los Alamos site. It is called qtool.

http://t2.lanl.gov/data/qtool.html

When I entered d+d, it reported 6 "open channels," the 3 we have been talking about and three more: One is the null reaction d+d => d+d, and two are deuterium disintegrations:
d+d=> n+p+d Q=-2.2259
d+d=> 2n +2p Q=-4.4491

The calculator reports in addition to Q-Values a "Threshold" value that appears to a multiple of the absolute value of the Q value.

I have a couple of questions.

1) What is the meaning of threshold?
2) Where do you look to learn the likelihood or cross-section values for these reactions? Is there a standard reference? A link on the net?

Many thanks, John Harris
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Re: FAQ: Fusion ash/byproducts -Energy- How much?

Post by jst »

Just to point out a few issues here;

1) A fusor, probably will never produce more than about 50KW of energy. The mircostar is extremely small, and I don't think you can get that much fuel in or energy out. I would love to see one of these produce 4MW, but that seems extremely unlikely given the size of the microstar.

2) At 50KW, a fusor would discharge something like 81megaCuries/sec! Some form of containment isn't an option... it is mandatory. That thing will deliver a lethal dose in much less than 1 second. At the experimental stage, we're looking at 6 feet of water or 20 feet of sand as a shielding medium in all directions ( gotta big swimming pool? )!

3) Current fusors are "by collision" devices; correct. In fact as Richard(?) said this is nothing but a low energy acclerator in the form of a sphere. The sphere increases the probability of collision, which would otherwise be very small, and difficult ( Uncertainty Principle here ).

4) Current fusors do not offer any "containment" capability; Only a partially true statement. Current fusors inject ions into a microstar. They contain and locate the microstar within a limited region of space ( just like a cylinder in a gas engine does ), and the associated plasma and fuel . However, it DOES NOT contain fusion byproducts, including any and all alpha, beta, gamma and neutron outputs ( in excess of the driver voltage of the fusor ) ... and this is precisely the source of fusor "inefficiencies". As there is no positive feedback into the process, because all input energy and all output energy is effectively lost, current fusors are at best laboratory curiosities suitable for lambasting the multibillion dollar efforts of "mainstream" fusion researchers.

Indeed, the analysis on this board which itemises why current fusors can never sustain fusion applies just as well to those multi-billion dollar projects with magnetic confinement, for exactly the same reasons... and we note that magnetic fields are nowhere near as strong as the blast coming from 45KEV fusors.


In short summary, my position is, that fusion can never be made to work unless there is a positive feedback mechanism in the fusion plasma region ( i.e. the microstar ). Since fusion energies are in the realm of minimum 4MEV up to 22MEV, any containment protocol must necessarily exceed 22MEV to keep both the microstar and its fusion products in the same locale.

The only known "containment" mechanism for neutrons is collission with another ( of any type ) more energetic particle. Therefore, a fusor must produce an ion shower of greater than 22MEV to stop (say ) 81megaCu/sec of neutron radiation. ( Perspective -- Remember, 1AMP is 10^18 electrons, and a MilliAmp is 10^15 electrons ). We have our work cut out for us.

Currently, the only technology anyone has that can produce MEV level ion showers is the CA - WFA based technologies that Larry has such a strangle hold on. Larry's recent proposal to inject a strong electron beam into a fusor microstar to produce MEV range protons is spectacular in its simplicity. If we can cascade such a device into a high ( 100 MEV ) energy beam ( preferably of electrons ) then something approaching full contaiment might be achieved.

While I don't expect 100% efficiency ( and therfore 100% containment ) the ability to extract 1/3 of a microstars energy ( i.e. 16KW ) would not be an insignificant result. But it would never produce megawatt or gigawatt power levels.
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Re: FAQ: Fusion ash/byproducts -Energy- How much?

Post by Adam Szendrey »

Joe,

The size of the microstar is not necceseraly a limiting factor. Remember that at higher energies a more powerfull containment field will result in the same size, BUT ofcourse energy (plasma) density increases.

As for cascading CA. I don't know exactly what do you mean by that but remember that a 4 MeV (these numbers are used as simple examples) proton can accelerate an electron up to 2 keV, and not to several MeV (electrons are about 2000 times lighter).
So if you produce a 4 MeV proton beam and ram it into an electron plasma it will result in a 2 keV electron beam (though i'm not sure that would work), or more like a neutral beam of 4 MeV protium atoms (i would guess that electrons recombine with protons as they match speed).
So i don't know how you want to cascade it.
But if the ion plasma is not a proton plasma , instead it's something heavier, say nitrogen plasma, than the resulting particles will have much higher energy.
Nitrogen has 7 protons and 7 neutrons. Thus it is 14*2000 times heavier (very crude approximation indeed) than a single electron.
Thus the result is a 56 MeV (!!) ion beam...Quite a meanie!
Sounds a bit irrealistic...a 56 MeV ion beam (or again rather a neutral beam, ofcourse it will only be neutral if ions have a long enough path to match speed) from a 2 keV electron beam?
A nitrogen nucleus is 14 times heavier than a single proton, but it is 7 times "more positive" too.
So i would guess that a nitrogen nucleus will require a path twice as long as a single proton does, to match speed with the electron that is hauling it.
Oh and ofcourse a nitrogen atom will need a lot more energy to be totally ionized!

Adam
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Re: FAQ: Fusion ash/byproducts -Energy- How much?

Post by jst »

Good points. Never the less, the likelyhood that we'll be forming large microstars, capable of feeding that much fuel thru them does seem a bit remote at best from where we are now.

As for cascading, I agree with your analysis. Larry has conveniently paved the way, for some pretty basic tests. They would be;

1) Does it actually work as advertized? Does a 2kev electron beam actually punch out a 4MEV proton beam from a spherical microstar of probably 2KV? Would that be a 8MEV beam if dueterium ions are used? Test?

2) If the microstar is more like a filament, does that still stay at 4MEV, or does it escalate the longer the plasma path is?

3) Just to make sure, exactly what does happen if a proton is fired at an electron plasma microstar? Experimental evidence?

From my point of view, I see WFA as achieved by a particle being fired into a string plasma filled with standing waves. The CA effect in a fusor I see as one lobe of the standing wave. A 2000X amplification looks pretty promising in that light. But without experimentation this is all bluesky...

P.S.-- BTW, since CA is a conservation of momentum thing, just keep out of the beam and radiation should be close to background.

P.P.S. -- Speculaton... do you suppose that an asymetric current inside a fusor might cause 4MEV something to come out of the fusor in a specific direction? Could this be the source of anomalous radiation events? Bet you most people don't have the fusor instrumented at all points of the compass...
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Re: FAQ: Fusion ash/byproducts -Energy- How much?

Post by Adam Szendrey »

A couple more thoughts:

The electron beam approaching the ion plasma might be deflected/scattered by the inner negative grid.
This is another point for making an axial tubular grid arrangement, so both the gun and the grid's axis is on the same line (the grids would be open at both ends ofcourse, like a simple tube).
I have an idea for cascading that might or might not work.
The following arrangement should be imagined:
An electron gun a positive plasma (protons for eg.) and after that a negative plasma of a heavier element than a single proton (deuterium for eg.).
Now as the electorn beam accelerates the protons (given that the path is not long enough for recombination to occur) reach the negative plasma and haul the heavier neg. deuterium ions with them.
Some concerns: Since protons are "big" and they hit a negative plasma containing similarly big ions (bigger), collisions take place reducing efficiency, or resulting in total failure.
The protons "steal" the electrons from the negative deuterium ions and nothing else happens. Though 4 MeV particles might not recombine with relatively stationary particles...

IF this some way works the result "should" be a 8 MeV beam of deuterium ions, 4 MeV protons, and 2 keV electrons...Now there are some possibilities here....the extra electron from the deuterium ions (neg.) "jump over" on the protons, or the protons recombine with electrons from the original beam...or both.

It is much easier to accelerate positive deuterons with the original 2 keV electron beam to get a 8 MeV deuteron beam, but that does not contain a "secondary" ion beam of protons.

Adam
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Re: FAQ: Fusion ash/byproducts - endoergic d+d rxns

Post by Richard Hull »

The two other reactions are variants of the Oppenheimer-Phillips reaction! This reaction is unlikely below about 4-6 mev (terrible cross section). These reactions are sometimes referred to as "stripping reactions".

These reactions are ENDOERGIC! These are big energy LOSERS! You are literally ripping matter appart and DEFEATING the "nuclear" bonds. You are not fusing here, just tearing assunder.

At one time, on the old songs list, it was put forth by some rather un-informed naysayers against amateur fusion in a fusor that the neutrons observed from the fusor were the result of "stripping" and not fusion. Of course this put me on a quest to contact good and knowledgeable physicists. I posted on the results found via discussion and in the excellent book I reported on in the Books and refs forum, "Fast Neutron Physics".

The upshot is that no form of stripping reaction is even possible on a statistical basis until about 4 mev and becomes a bit more common after 6 mev. In a stripping reaction you are hitting with such force that you are just shearing the deuterons into component parts and not really doing any fusion at all!

Finally, it is stupid to suggest stripping as opposed to fusion at amateur energy levels! All we are ever capable of IS FUSION! This is based on simple cross sections at our voltages.

The threshold levels mentioned are those energies where these two latter reactions have some chance of occuring on a statistically viable level.

FOR WHAT IT"S WORTH...................

It is also most interesting that on just about every neutron generator manufacturer's literature and web-page, they list the reactions used to generate their neutrons via the D-D or D-T reactions, BUT NEVER MENTION THAT THIS IS FUSION!!!

Ya' think they might not want to step on delicate toes? OR Avoid the "F" word as in 'deadly' nuclear reactions?

You would think that someone out shopping for an "N" generator would not have to be coddled to or sheltered as they would be well aware of the reaction being fusion. But maybe not.


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Re: FAQ: Fusion ash/byproducts - endoergic d+d rxns

Post by ex-engineer »

John -

Here's a link to the ENDF data for cross sections on the two fusion reactions of interest, along with the (far) more prevalent elastic collision cross section: http://t2.lanl.gov/cgi-bin/endf?0,0,/in ... teron/H/dd

I'm assuming that the data is collected in a "black box" way such that a 1 MeV deuteron that actually has an elastic collision or two before eventually fusing is still counted as a fusion event. Someone who knows more about cross section measurement than I do (which would be approximately everyone who knows anything about it at all) might be able to set me straight on that.

My read on the data is that for every 12 1 MeV deuterons shot into a cloud of deuterium, ~2 will fuse. So your best case fusion 'gain', assuming lossless ion generation and acceleration, would be (10 + ~7)/12. It's tough to run a steam-driven generator on that...

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Re: FAQ: Fusion ash/byproducts -Energy- How much?

Post by ab0032 »

I really enjoyed this whole post. Really fun to read, nice demonstration of what is going on and putting things in the proper perspective.

I dont know if I am missing something here, but there is one point I cannot follow regarding the D-T and how to get to 50%. Is there something wrong with the powers of 10?

Richard Hull wrote:
> This would make our fusor contain 2X10e19 X 10e-5 = 2X10e14 deuterium atoms per cc at the stated operating pressure.........
> The fusor volume is 4/3 X Pi X 7.62^3 = 1600 cc..............
> This makes our fusor charged with a grand total of 1.6x10e3 X 2x10e14 = 3.2 x 10e17 deuterium atoms.

So we have 3.2x10^17 D, to get to 50-50, we have to options, if we want to stay at the same density, we have to run til we have 3.2x10^17 divided by 2 tritium and refill the deuterium lost, as we need 4 d for one t, or we burn 4/5 of the d , keep one fifth and produce 1/5 t from the 4/5. Then we end up with 2/5 of the pressure we started with, which is then less then 10 micron. Either way, we need a long time to get there, if we produce 1 million t per sec.

To produce 1.6x10^17 t at a rate of 1mio/sec or 10^6/sec, we need to run 1.6x10^11 sec. (To produce 1/5 of 3.2x10^17 takes almost as long.)

> Continuing to beat the poor dog..........
> We would have to operate the fusor for .5X10e7 seconds before we hit the ideal 50-50 D-T mix and were now a D-T fusor..... That is, after only 1 year and 5 months of 24 -7 operation, we would move from a 2 million fusion per second device to about a 200 million fusion per second device having morphed from D-D fusion to D-T fusion over this time frame.

So I get 10^11 sec and not 10^7. Since a year has about 3x10^7 sec, you will have to run about 3.000 years, til one approaches 50%

Did I miss some change in assumptions regarding fusion rate or something else?

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Re: FAQ: Fusion ash/byproducts -Energy- How much?

Post by Dustin »

Great summary Richard,

The only problem I see is:

3. Proton (protium) - This ash product is in the form of a recoiling proton and carries a significant amount of energy, ( the greatest in the D-D fusion cycle ~3mev), from its related fusion event. It usually slams into the inner grid, but always winds up as common hydrogen gas by taking up a free electron. It, in theory could act as a poison of sorts as the fusor re-ionizes it and it speeds towards the center mixed in with deuterium ions. (useless, fusionless collisions) more on this later, as well.

The 3Mev proton will not (in all probability) hit the grid, it will continue in whatever direction it initially had and probably embed in the shell (or pass through it).
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Re: FAQ: Fusion ash/byproducts -Energy- How much?

Post by Richard Hull »

All calcs were idealized and no intergration of the resultant fuel/ash change over time was taken into account. The thrust is, many years would be needed..... so forget all about using the ash as fuel in a fusor device. That was the bottom line.

Newbies are famous for looking at the D-D fusion ash, T and 3He, as new fuel. It ain't happening. We have our hands full getting the 100% load of D to fuse. The ash is nothing.

Finally, 99.999% of the protonic ash would never be attracted to the inner grid. A 50kv grid is not going to turn a 3mev proton. All would slam into the wall and 0.0000000% would pass through the chamber walls.

Due to shell bombardment, hydrogen embrittlement would take place over long periods of time and some of the buried hydrogen would start to pop back out into the gas environment once suface loading reached some saturation point as bombardment continued. Zero protium makes it to the outside would except through the vacuum exhaust ports.

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Fusion is the energy of the future....and it always will be
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Re: FAQ: Fusion ash/byproducts -Energy- How much?

Post by Dan Tibbets »

R.H. Just to throw in a complication. There is a proposed utilization of the tritium and He3 ash for additional fusion. This proposal by R. Bussard applies to a Polywell. As mentioned the fusion ash has way too much energy to be retained by the potential well. It would be neutralized as it hit the walls, and if not embedded, it becomes part of the neutral gas that has to be pumped out of the vessel. It could then be processed and reintroduced inside the magnetic/ electrostatic containment magrids as new fuel. In theory, since the D-T yields ~ will be ~ 6 times as much per fusion, the net fusion output of the machine may be as much from secondary D-T fusion as the original S-S reactions.

Of course this is based on the D-D Polywell being at least marginally successful with D-D fusion, and the fusion production would be in the MW instead of microwatts such as found in a gridded gas discharge fusor. As you pointed out the percentage of tritium available for further fusion (even if you could prevent it from embedding in the wall) would be tiny compared to the deuterium density in the fusor. Any tritium that is in the fusor space (after it has lost most of it's KE, and/ or charge exchanged, etc. may fuse faster than the D-D baseline, because while the tritium is extremely rare, each would be colliding frequently with the much more prevalent deuterium. This mixture modifies the fusion rate, though I'm uncertain of the formula to describe the interaction. It would be similar to mixing excess protium or deuterium with B11 or He3 to reduce the Bremsstrulung, while not slowing the fusion rate by a similar proportion. In the fusors case of course this would not make much (if any) difference in the total amount of D-T fusion, but the tritium present would not last as long as the raw tritium density would suggest. In otherwords the ~ 5-10 D-T fusions per minute may be misleading , if id is based only on the tritium density alone.

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Re: FAQ: Fusion ash/byproducts -Energy- How much?

Post by Richard Hull »

Anything is possible. Breeding fusion fuel is quite viable, Provided tens or hundreds of thousands of watts are on hand to power the fusion breeder to make a substantial quantity of ash that is both extractable and usable. Again, however, never in anything we, as amateurs, will ever produce.

There are many fusion dreams with none of them coming true, yet.

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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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