A means to 'recycle' lost scattered energy.

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Chris Bradley
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A means to 'recycle' lost scattered energy.

Post by Chris Bradley » Thu Nov 20, 2008 7:21 pm


I'm just kicking this thread over for fear that this discussion will disappear under some 'Polywell' title, wtih which the following bears no association whatsoever. This is quite different to what the Polywell is attempting.



JamesC wrote:
> So I think I get it. In the thermonuclear version energy lost to scattering is not lost but just transfered to the heat and essentially recovered later on to some particle in the high end of the maxwellian distribution so the scattering losses are recycled.
>
> However in a beam version those losses cannot be recovered and are genuine losses and since the ratio of scattering probability to fusion probability is so biased toward scattering the average scattering loss exceeds the fusion gain even in an ideal scenario.
>
> Hence impossible even if we construct a PERFECT beam power device. If I had to describe such a device it would seem to look a bit like a particle accelerator with two monoenergic ion beams heading directly towards each other with some sort of containment field that would magically correct any scattering to straigten up a particles direction after a scattering event and then also reinject the energy the particle lost to scattering to get it back to fusion energy. Then upon fusion the newly fused particle would instantly disappear and the fusion energy released recovered without loss. This magic collision chamber would be long enough to ensure all particles entering it would fuse before exiting the other end.[Edit: Of course this device lives in a perfect vacuum]
>
> Ok.. so in this dreamland. Could this device produce net fusion power?
>
> The device has to put energy into each particle and energy to compensate each scattering event. How would I go about calculating if this ideal device produces net power? I think you had a go at something like this in your scattering post.




Chris Bradley wrote:
> Exactly! Top o'the class!!
>
> ...in one go you've summed up the issue and most of the solution I have suggested may be a route forward. The missing bit is just that instead of the beams being linear, they are circular and thus have no ends!
>
>
> JamesC wrote:
> > Could this device produce net fusion power?
>
> Regrettably the answer is 'unlikely', but something different is a step along a path nonetheless. Whether it is the right direction along the path.... ?!?





JamesC wrote:
> Great! time to pull out the calculator then.. I just have zero feel right now for the energies/losses we are talking about but your other post on scattering seems to be a good start!
>
> For a real world device I think the circular motion instead of the linear "thought experiment" version popped into my head also. Maybe it would take care of one other problem also which is that any neutrals/fused atoms would fling off the circular track "Instantly disappearing". The only problem is that a typical circular particle accelerator seems to need to be the size of france and cost a few billion but hey'.. maybe an electrostatic version instead of magnets could make it a bunch smaller. Again I just have no idea on the magnitudes of the forces here. Actually as I think about it having electronic control over the beam might help target a circular electrostatically controlled beam.. hmmm.
>
> I remember reading somewhere something about laser wave particle acceleration creating impressive linear velocities on a desktop but in any case none of it will be any use if even a lossless thought experiement cannot produce a net positive yield so I guess I better start with that!
>
> Thanks again, for helping to "bootstrap" me into fusion

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Re: A means to 'recycle' lost scattered energy.

Post by Chris Bradley » Thu Nov 20, 2008 7:25 pm

JamesC wrote:
> I just have zero feel right now for the energies/losses we are talking about but your other post on scattering seems to be a good start!

My preliminary calculations suggest that this whole process only shows potential net over-unity at over 200keV collision energy, and then it is only a very modest marginal gain - so in this configuration not likely to produce sufficiently 'over-unity' to positively drive a heat exchanger into net energy. Hence, even the tiny theoretical window of viability turns against the reality.

Still working on the numbers but my [potentially erroneous] calculations are currently saying that at that collision energy the *average* loss of energy given up to the other particle(s) is 17eV. (Some collisions are very large - unit fractions of the particle's energy - but there are only a couple of these likely to each chance of fusion, whereas there will be millions of little collisions.)

At this energy we are talking about 10E5 scatters with this average energy loss per fusion event, so you'd have to pump in some 2MeV of energy to keep that particle 'alive' [at its 200keV energy] long enough to get it to fuse.

You can see what is gained by this process of feeding back in the energy - a 200keV particle can only possibly loose 200keV in a fusor-type device, yet you need 2MeV to keep it going long enough to fuse even at these elevated particle energies. If a process can be set up that delivers this energy back into particles already accelerated within the device, so the particle might survive long enough to fuse.

> The only problem is that a typical circular particle accelerator seems to need to be the size of france

These are very very slow particles compared to those 'nucelon' smashers. At 200keV you need merely 10 cm or so of radius, and fractions of Teslas of magnetic field strength. The electric and magnetic fields required are all very 'do-able' - theoretically!!

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Re: A means to 'recycle' lost scattered energy.

Post by inflector » Thu Nov 20, 2008 8:11 pm

It is precisely this sort of focusing of beams and collection of scattered energy that I hope to accomplish with the positively charged tetrahedrons in the fusor I described here:

viewtopic.php?f=14&t=6833#p42516

The basic idea being that the positive charge on the tetrahedral grid will serve to focus the scattered ions back into a line. The three forces coming in at 60 degrees from each other will serve to center the ions. The three corners of the tetrahedron will serve like an electric notch and attenuate circular motion around the axis.

Once the ions lose their energy because of the combination of the positive grid repelling force and the negative inner grid's attractive force, the ions will be lined up in the same center line which goes from the outer vertex of the tetrahedron to the center of the fusor's inner grid. Lower energy ions will stop closer to the inner grid, higher energy ions will stop further out.

It is my belief that some of the collisions will take place between the higher energy ions bringing the lower energy ions up to greater speed. Then somewhere closer to the center of the fusor, all these ions will start colliding with accelerated ions coming in the opposite direction from the tetrahedron on the opposite side. The shape of the tetrahedrons will serve as a constant aligning force nudging scattered ions back into position after minor collisions.

- Curtis

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Re: A means to 'recycle' lost scattered energy.

Post by JamesC » Thu Nov 20, 2008 10:29 pm

Good stuff, so (theoretically) if a configuration can get these scattering particles back on track you will be investing 2MeV to yield a fusion from between 5 and 15MeV which looks like a (narrow) window. Of course that 17eV is the actual loss not what it might actually cost to get the particle back on track.

Ok some quick questions.

It seems like 200keV is not that hard to achieve so why not make it 300keV. Does the power needed go up exponentially?

At 200keV you said 10E5 scatters are needed. How is this calculated?

I want to try to figure out what the limits are of two highly focused beams, how do I do this. Lets say you throw everything possible at creating a highly focused beam, how can I figure out the probability curves for fusion. Clearly this is the first key parameter for this hypothetical "dream machine". I am going to try to build high level models of two versions of the dream machine, energy recapture and particle reguiding.

This is really interesting stuff. I wonder if I will ever get any other work done ever again!

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Chris Bradley
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Re: A means to 'recycle' lost scattered energy.

Post by Chris Bradley » Thu Nov 20, 2008 10:35 pm

Curtis Faith wrote:
> It is my belief that some of the collisions will take place between the higher energy ions bringing the lower energy ions up to greater speed.

This will happen, but remember that this slow particle has already given up some of its energy in its other [thermalising] interactions, so the new 'full' energy particle will then, in turn, have a little bit of its energy sucked out, etc., etc... this IS the process of thermalisation. To keep the energy in the particles requires an active input of energy, more than just the movement of charged particles in an electric potential gradient.

If you pour marbles into a bucket, the ones you are pouring in never impart their energy back to the slower ones such that they can bounce out to their original height again. They may bounce up and down for a while, with slowly diminishing energy until they lie 'energy dead' at the bottom. This is a precise facsimile of ions in the fusor. Both gravity and the electrostatic field cannot do overall 'work' on the 'particles', all such particles have a given kinetic+potential energy and, statistically speaking, this sum tends to go only one way! This is why Carl points out how difficult it is to pull particles back on a coherent track. Something 'active' needs to be done and to date those processes employed are inefficient.

But the actual direction of that track can, possibly, be influenced by the means you have described. I think it may be an improvement to the fusor for further reasons. I meant to get back to your original post on this at some stage..

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Chris Bradley
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Re: A means to 'recycle' lost scattered energy.

Post by Chris Bradley » Thu Nov 20, 2008 10:56 pm

JamesC wrote:
> Good stuff, so (theoretically) if a configuration can get these scattering particles back on track you will be investing 2MeV to yield a fusion from between 5 and 15MeV which looks like a (narrow) window.

or 2.5MeV in the case of recovering neutrons from DD. Very, very narrow.

In fact, I think too narrow, hence I say it may be 'a step' - but I would not go to say 'viable' until I see it working BETTER than the theory*!!!


>
> Ok some quick questions.
>
> It seems like 200keV is not that hard to achieve so why not make it 300keV. Does the power needed go up exponentially?

For DD I would guess it goes up approx. linearly through to 400keV, then actually ramps off, as per the reactivity curve I have plotted in another post. Remember this is collision energy, so 400keV would be equivalent 800kV drive voltage in a fusor. (i.e. half the drive voltage from what you see on that graph I plotted)

>
> At 200keV you said 10E5 scatters are needed. How is this calculated?

For Coulomb scattering losses, I've turned Rutherford scattering equation into an energy transfer equation between interacting particles then integrated in two axes the impact parameter up to the Bohr radius for hydrogen. I get a long winded equation with logs in it, then multiply out to get an equivalent cross-section to compare with the fusion cross-section. For recombination and ionisation losses I am using the cross-sections from 'Massey', which may be an old out of date text now, but I don't think the universe has changed much since his day. (That's not to say I've interpreted the data right.) There are multiple caveats in these approaches and I'm still mulling over it all.

>
> I want to try to figure out what the limits are of two highly focused beams, how do I do this. Lets say you throw everything possible at creating a highly focused beam, how can I figure out the probability curves for fusion. Clearly this is the first key parameter for this hypothetical "dream machine". I am going to try to build high level models of two versions of the dream machine, energy recapture and particle reguiding.

Theoretically, that's quite simple and you can read through;

http://fds.oup.com/www.oup.co.uk/pdf/0-19-856264-0.pdf

and you should be able to figure out the rest. You seem to be on the ball with this stuff.

best regards,

Chris MB.


*(Well, it is permissible to live in hope with such outcomes - it has certainly happened with tokamaks. No one yet really knows why tokamaks adopt the 'H-mode', which you can see in images where there is little mass transport at the edge and a tight 'surface'. This was a surprise to tokamak researchers but means tokamaks have been able to perform 'better than expected'.)

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Re: A means to 'recycle' lost scattered energy.

Post by JamesC » Sat Nov 22, 2008 4:43 pm

>the motion of the fast particles must be continuous and cannot afford to reciprocate back and forth

Yes, at some point they would have zero speed as they oscillate which would seem an ideal time to form a neutral especially when it's living in an electron cloud! ( remember I have only very knowledge of these things which is why I am asking so many questions! )

So really to minimise the neutral count which seems by all accoutns to be a bad thing, you need to have a minimum number of electrons lurking around ( no electron cloud! ), as close to a vacuum as possible ( no gas! ), and continously moving fast ions.

I am slowly working up my "idealised dream" machine model and have some more questions.

This scattering loss which was calculated as ~17eV per scatter event on average in a previous post. This is derived by integrating the probability of a scattering angle vs the loss at a scattering angle .. something like the attached picture ? So after the collision the sum of the energies of both particles is less by the scattering loss which is the loss that needs to be actively injected again "somehow". Is that loss symmetrical? or will one particle loose more than the other. The change in direction should be symmetrical if both particles have the same energy/velocity?

What is the nature of the energy released, is this an X-Ray, a photon or something else. Is there any chance to recover any of this energy. Will this lost energy be a directed energy or lost in a random direction?

Thanks!
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Re: A means to 'recycle' lost scattered energy.

Post by Chris Bradley » Sat Nov 22, 2008 7:20 pm

Your diagram is as I envisage the calculation, and the total loss at all angles is the integral under the curve. Yes. And, indeed, the fusion cross-section is comparable only with the extreme angles of incidence of the scattered particles.

In terms of 'energy loss', what I am aiming to calculate isn't 'loss' in the sense of a photon or whatever. I'm simply talking about how much energy is transferred to a stationary nucleus off of which the beam nucleus impinges and scatters. So the target picks up this energy. But you are right as this process is not 100% and there would be EM emissions. However, it'll be fairly minor (overall), excepting for the very large angle scatters associated with large energy losses which are rare.

I suppose it may be better to talk about 'energy transfers'. But I'm essentially playing safe and being pessimistic here, and assuming that this thermalisation loss (for this is what it is we are talking about) is lost and gone and will end up as low-quality heat somewhere in the system. It is this energy we have to replace by pumping energy in.

Now, in terms of getting rid of un-ionised materials, we are at a point where we can go two ways. We can a) either accept, and therefore aim and deal with, a situation where these fast ions are passing through a volume of, essentially, cold neutral gas and we have, then, both ionisation losses and coulomb scattering losses to accommodate, or b) where we try to keep the back ground as ionised and as low a density as possible.

In the case of a), we would then be describing a fusor. This is essentially the fusor's operation and is where most of the fusion comes from fast ions colliding with the nucleii of some background neutral medium. I am thinking that, again, elevating the collision energy limits, but never eliminates, the electron interactions but we are then talking >500keV for net energy gains by this treatment. It all gets yet further suspect and 'hairy'.

The other mode, b), which you are thinking for your 'dream machine' is where you keep two beams going that are brought together to collide and with no background materials. This is not possible and there have been recent threads discussing this very point. However, in the case we are talking about, you can, at least, do better than (a). So you may not get to have a 'pure' coulomb-collision-only device but you can bias it in your favour by running two colliding beams and pulling the vacuum as low as you can go.

Or the other option for the (b) scenario is where you hold up a continuous beam of ions running through a background medium of ionised fuel. Again, you get nothing for free, and here you still have the electrons running around and will happily generate fast neutrals to compromise your efficiency, but also you will then have to balance the degree of ionisation with the density (and thus the reaction rate) because this resultant plasma will screen electric fields. So if your device uses electric fields to maintain the beam, then it may stop operating under this (b) schema. I have put forward an argument, a few times on this forum, that a successful beam confinement system, such as we are talking about, will have to employ both electric and magnetic fields so I feel this would prove to be a limitation.

So, there are a few options there but I don't see any that I'd necessarily want in a 'dream machine'. Reality comes to bite - as usual!!

But keep thinking and questioning. Maybe there is a route through!!

best regards,

Chris MB.

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Re: A means to 'recycle' lost scattered energy.

Post by Carl Willis » Sun Nov 23, 2008 12:18 am

Hi James,

>This scattering loss which was calculated as ~17eV per scatter event on average

How come? This quantity is dependent on some unspecified particle energy, for one thing. But more important is that this kind of number seems to be fundamentally arbitrary to me, because it depends on an arbitrarily-chosen minimum scattering angle for the purposes of calculation. I put a handwritten note at the bottom of the page showing how you'd presumably calculate your ~17 eV number, to illustrate why the minimum scattering angle (and the particle energy) matter. Notation:

E is the incident particle energy. I presume it to be a single constant value.
E' is the scattered particle energy, ranging over E'max to E'min according to conservation of energy and momentum
<E'> is the average scattered particle energy.

So what you call the average scattering loss is then (E - <E'>), Line 1. This formulation you already described ("...derived by integrating the probability of a scattering angle vs the loss at a scattering angle") so I won't dwell on it other than to mention that my notation is sigma_s(E,E') for the differential cross-section with respect to E' (equivalent to the probability of scattering into a particular angle); and sigma_s(E) is the total scattering cross section at all viable angles. For charged-particle, Coulomb scattering like we are talking about, sigma_s(E,E') is called the "Rutherford cross section". It's come up a lot on here. One common formula for it is on Line 3.

The problem is that in the real world, the maximum scattered particle energy, E'max, is equal to E (or equivalently, the minimum scattering angle is zero degrees). The electrostatic forces responsible for Coulomb scattering act on EVERY incident particle, no matter what the impact parameter, because the forces have infinite range. This makes the total cross section infinite and the scattering loss undefined, unless you artificially put an upper bound on E'max even though lots of particles are scattered into any small delta(E') below E'max. If we didn't have a Coulomb potential here, if we were dealing with the scattering of neutrons by the strong force for instance, the problem would be more meaningful (in fact it often finds application in neutron-shielding calculations). But here, you're going to have to provide some serious convincin' that this 17 eV number actually means something!

-Carl
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Re: A means to 'recycle' lost scattered energy.

Post by JamesC » Sun Nov 23, 2008 10:04 am

>This scattering loss which was calculated as ~17eV per scatter event on average

>>How come? This quantity is dependent on some unspecified particle energy, for >>one thing.

I am not putting myself forward as a physicists and actually I have no idea why the loss is calculated as 17eV. I was just trying to get my head around the result suggested by Chris Bailey in this post.

viewtopic.php?f=15&t=7183#p51011

I thank you for your mathematical addition and I need to study that since the math doesn't come naturally to me. I am trying to boil it down to the simplest possible terms. Alot of the math seems to act over the statistical result of the simplest interactions. I want to understand that simplest interaction and build up from there.

Anyhow, actually in the last post by Chris I see my understanding of this is wrong and there is no actual scattering energy loss like I understood it.. which is a good thing but I am going to add to this in the next post.

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