Another (simple) FAQ - DONE

[MSimon]

For "Best" I'd favor best Q, which given major losses vs. reaction rate would be at somewhat short of the energy for max cross section.

Ultimate Q for the resonance peak is about 22. For the cross section peak about 8. I can give you exact numbers if you want me to look it up.

[chrismb]

The paper to which I generally refer labels cross-section as sigma (σ). It further defines reactivity as a function that integrates cross-section and velocity across a temperature function and denotes it <σV>. I was putting f as density*density*cross-section*velocity. Is that wrong?

Not wrong, but possibly misleading.

The <σv> term you see is an 'averaged' fusion cross-section across the range of velocities. In a thermal plasma you can, supposedly, assume Maxwellian and thus (obviously!);

<σv> = (8/π)^1/2.m^−1/2.θ^−3/2 ([Integrate 0 to inf] {Eσ(E)exp(−E/θ)}.dE)


What you end up with is a reaction rate per unit volume of thermal plasma (just look at the units - m^6 on the bottom and m^3 on top, so it is a specific reaction rate by volume).

Given that in a beam process, like Polywell aims to be, most of the reactants aren't actually in 'the reaction volume' at any given time, so what volume do you attribute to a particle hurtling away from the reaction volume and due to come back for a new collision later?

it is much easier than this for beam-target; take the volume and density of your 'reaction volume', then for each beam particle work out rate=[v].[rho].[cs] then add them up (or multiply by the number of beam particles you have, if they are all mono-energetic).

[D Tibbets]

....
Given that in a beam process, like Polywell aims to be, most of the reactants aren't actually in 'the reaction volume' at any given time, so what volume do you attribute to a particle hurtling away from the reaction volume and due to come back for a new collision later?
...

Good point about reaction conditions.
A made up example.: A Polywell with a reactive core that is 10% the diameter of the Wiffleball, or 1/1000th the volume. Assuming most reactions occur within this core is interesting, but irrelevant. The times and densities involved are what counts. An ion would spend 10% of it's time in the core, if the velocity was averaged (like in a thremalized plasma). But because the speed varies - hot (at the selected fusion temperature) in the core, and cold and mostly fusion reaction insensitive outside the core. Because of confluence of all ions (hopefully) in the core the time spent outside the core is compensated for by the increased density in the core. ie: the core volume dwell time is small compared to total ion dwell time inside the Wiffleball, but the increased density compensates for this . Whether this is a near equivalence of if there is an advantage or disadvantage depends on some assumptions. The dwell time in the core is basically the core fractional radius squared multiplied by some constant to account for the variation in ion speed in the various regions. The density in the core is the core radius cubed. So the core has an advantage based on density alone. But, once the dwell time is considered it comes closer to a neutral effect.
Assumeing it is balanced, the ~60,000 fold energy density advantage that R. Nebel mentioned would need to be accounted for by the average density advantage (~ 1000 X) multiplied by the reactive advantage of monoenergetic ions (within the core). If this is ~ 100% / ~2% in a thermalized plasma , the final advantage seems reasonable.

This may account, at least in part, for R. Nebel's comment that the confluence (compactness of the core) is not critical (within some limit). As the core size goes up the density of reacting ions drops, but the volume (and thus the dwell time) of reacting ions increases an equivalent amount. I'm not sure, but I think this is inherent in a spherical system.

If the potential well decreases once the core reaches some small size (virtual anode formation), or the core approaches the diameter of the Wiffleball (thermalized in at least the angular or transverse directions*) the efficiency would suffer. Again an example of the knobs- or adjustments that are needed to acheive optimal performance.

* Based on my above argument, angular momentum is not necessarily all bad . I wonder if one of the more significant effects of increased angular momentum thremalization is that it strongly influences the rate of radial thermalization and/or the ability of restoring mechanisms like annealing to operate.

Dan Tibbets

[ladajo]

http://fti.neep.wisc.edu/static/TALKS/1 ... elroge.pdf

Some of this discussion thread is addressed by Joel in his paper.

In particular he addresses peak resonance around page 16.

[TallDave]

Simon,

Ah yes, thanks. I thought the temperature reference was funny so I cited that when I couldn't find the resonance number. Good times.

50KV for p-B11 looks right to me. Reminds of the long-ago discussions of different accelerations for different ions.

Kiteman: You may want to not give "energies" for fuel mixes, but well depths instead. For a well of 50KV, you have different energies for p and B11 because they have different charges; i.e. the particle species may be something like monoenergetic but Z+1 protons and Z+5 boron ions won't ever have the same energy in the same well. My humble advice is: ignore partial ionization (eV vs KeV says this is like expecting to find ice in a volume of boiling water), and give the energies of the respective species at the optimal (guesstimated) well depths, which I'm thinking can't end up being all that far from the 65KV given above after fuel optimization and etc. I do recommend including Dan's excellent sentence below:

Considering that due to efforts to minimize bremsstrahlung, there is an excess of protons in the mixture, so figuring out what effective accelerating electric field would be needed to gain significant advantage from the resonate peak is complicated.

Also, here's a fun chart of ionization energies, which the forum doesn't like:

http:// en.wikipedia.org/wiki/Ionization_energies_of_the_elements_(data_page)

My understanding is the horizontal sum is the total energy for complete ionization (the CRC #s are in eV), so for B11...

8.29803 + 25.15484 + 37.93064 + 259.37521 + 340.22580 = 670.98

[KitemanSA]

You guys are confusing the heck out of me!

Simple question:

How do I edit my answer?

What specific wording / equation changes should I make? Please "quote and edit".

Thank you.

[D Tibbets]

Simon,

Ah yes, thanks. I thought the temperature reference was funny so I cited that when I couldn't find the resonance number. Good times.

50KV for p-B11 looks right to me. Reminds of the long-ago discussions of different accelerations for different ions.

Kiteman: You may want to not give "energies" for fuel mixes, but well depths instead. For a well of 50KV, you have different energies for p and B11 because they have different charges; i.e. the particle species may be something like monoenergetic but Z+1 protons and Z+5 boron ions won't ever have the same energy in the same well. My humble advice is: ignore partial ionization (eV vs KeV says this is like expecting to find ice in a volume of boiling water), and give the energies of the respective species at the optimal (guesstimated) well depths, which I'm thinking can't end up being all that far from the 65KV given above after fuel optimization and etc. I do recommend including Dan's excellent sentence below:

Considering that due to efforts to minimize bremsstrahlung, there is an excess of protons in the mixture, so figuring out what effective accelerating electric field would be needed to gain significant advantage from the resonate peak is complicated.

Also, here's a fun chart of ionization energies, which the forum doesn't like:

http:// en.wikipedia.org/wiki/Ionization_energies_of_the_elements_(data_page)

My understanding is the horizontal sum is the total energy for complete ionization (the CRC #s are in eV), so for B11...

8.29803 + 25.15484 + 37.93064 + 259.37521 + 340.22580 = 670.98

It doesn't change the final energy that much- I'm surprised it takes hundreds of eV to pull off the last electron from the boron nucleus, but I suspect the 340 eV is the energy to pull off all 5 electrons, not just the last one. ie: by the time you reach 339 eV, the first 4 electrons are already gone (actually, the first 4 are gone by the time you pass 260 eV).

Dan Tibbets

[hanelyp]

How do I edit my answer?

Probably should break the question up into parts. 'Best' reaction rate and Q for p-B11 and D-D. Start with simple numeric estimates, then expand into how those numbers were derived and what considerations other than energy / well depth apply.

[KitemanSA]

How do I edit my answer?

Probably should break the question up into parts. 'Best' reaction rate and Q for p-B11 and D-D. Start with simple numeric estimates, then expand into how those numbers were derived and what considerations other than energy / well depth apply.

You forgot the part where:

What specific wording / equation changes should I make? Please "quote and edit".

[ladajo]

Here is a fairly simple question. Any takers?

What ion energy is best for deuterium fusion, tritium deuterium fusion, or proton boron 11 fusion?

From Page 16 of Joel Rogers paper:

"The b=1 condition fixes the product of ion energy(Ei) x ion density(n). Fusion rate is proportional to n2 which falls with rising energy as E-2. Velocity U rises as E0.5.
To make n2<sU> stationary (and therefore Q maximum) requires s to have slope E0.5/E-2 = E1.5.
Maximum fusion rate occurs at the ion energy where the slope of the DD yield curve is E1.5, indicated by the point where the 1.5-slope line kisses the DD curve."

Base on the same page charts which I cannot insert.

[TallDave]

Dan,

I thought so too the first time I looked at this issue a ways back, but I recall my original source said the total energy was cumulative across that row, so I had to go back and fix my calculation.

This is less definitive but seems to say that as well:

More generally, the nth ionization energy is the energy required to strip off the nth electron after the first n − 1 electrons have been removed.

http://en.wikipedia.org/wiki/Second_ionization_energy

Emphasis added.

It is a bit hard to tell for any particular chart whether the values given are cumulative or specific to that elelctron, but I assume the custom is to give the specific value, not the cumulative.

UPDATE: Ah, ok this is pretty definitive here:

The second, third, etc., molar ionization energy applies to the further removal of an electron from a singly-, doubly-, etc., charged ion.

So there you go.

http://en.wikipedia.org/wiki/Molar_ioni ... e_elements

Also, you can see some of the other elements have values that can't be cumulative (because they're less than the sum of the previous energies).

Anyways, as you say it doesn't matter much because even in a fully Maxwellian distribution of ion energies at 65KV the proportion of ions at <1 KV is probably small enough to ignore (who wants to count the sigmas?), let alone a relatively narrow distribution.

[chrismb]

You guys really are having your cake and eating it!!!

As an ion heads towards the 'outer shell' of the supposed wiffleball, wherein it 'anneals', what energy does it then have??? Enough to maintain a 5+ ion state???!!

After it drops back down to 1+ ionisation state by pulling in electrons to neutralise it (it is not only going slow, remember, it stops before falling backwards!) (if not fully neutralised where it drifts off), so then the acceleration on it is a 1+. it picks up a bit of speed and then you find it HAS TO HAVE A COLLISION to loose another electron. Sure, the energy is only XeV, but it is nearly as likely to collect an electron than drop one. And how often is it going to collide with something else so as to loose these extra electrons?

One moment I'm being told that these ions only collide with stuff at the dense centre, the next the wishful-thinkers are telling me it has enough collisions to shed these extra electrons! Which is it?

I recommend (a) you familarise yourselves with Saha's equation, (b) you familiarise yourself with the cross-sections for ionisation and recombination collisions, (c) you apply all of this to some particle dynamics' calculations to work out when, where and how probably it is likely to acheive a 5+ state. Until then, I highly recommend you stick with the presumption that 5+ boron ions will be a significant minority.

The answer to the original question is, as near as darn it 'the energy of peak cross-section for that particular reaction'. And that is that.

[D Tibbets]

You guys really are having your cake and eating it!!!

As an ion heads towards the 'outer shell' of the supposed wiffleball, wherein it 'anneals', what energy does it then have??? Enough to maintain a 5+ ion state???!!

After it drops back down to 1+ ionisation state by pulling in electrons to neutralise it (it is not only going slow, remember, it stops before falling backwards!) (if not fully neutralised where it drifts off), so then the acceleration on it is a 1+. it picks up a bit of speed and then you find it HAS TO HAVE A COLLISION to loose another electron. Sure, the energy is only XeV, but it is nearly as likely to collect an electron than drop one. And how often is it going to collide with something else so as to loose these extra electrons?

One moment I'm being told that these ions only collide with stuff at the dense centre, the next the wishful-thinkers are telling me it has enough collisions to shed these extra electrons! Which is it?

I recommend (a) you familarise yourselves with Saha's equation, (b) you familiarise yourself with the cross-sections for ionisation and recombination collisions, (c) you apply all of this to some particle dynamics' calculations to work out when, where and how probably it is likely to acheive a 5+ state. Until then, I highly recommend you stick with the presumption that 5+ boron ions will be a significant minority.

The answer to the original question is, as near as darn it 'the energy of peak cross-section for that particular reaction'. And that is that.

I'm not sure the speed of the ion is the whole story. I'm guessing the space charge is very important. Also, keep in mind that the edge region where the ions are slow, is the region where the electrons have the highest energy. The relative electron- ion speeds are still high.

Dan Tibbets

[TallDave]

Enough to maintain a 5+ ion state???!!

A whole 671eV? In a 65KV well? Wishful thinking indeed.

One moment I'm being told that these ions only collide with stuff at the dense centre,

You must have missed the whole part about the edge having a much higher collison cross-section.

(it is not only going slow, remember, it stops before falling backwards!)

They don't stop, they just have zero radial velocity. They're going to have some amount of transverse velocity, probably small in relation to 65KV but large in relation to 671eV.

Until then, I highly recommend you stick with the presumption that 5+ boron ions will be a significant minority.

Huh? You really think a large majority of ions are going to be below 671eV? In the core of a 65KV well? Yeah, I'm going to stick with Bussard's usecs and full ionization, tyvm.

As the neutral gas filled the machine interior, fast injected
electrons created ionization in this gas. The ion and electron
densities produced by this fast ionization were too low to
drive the system to the electron beta=one condition.
However, the low energy electrons resulting from this
ionization rapidly cascaded with additional neutral atoms,
being driven by electron/electron collisions with the
incoming injected fast electrons, and made still more low
energy electrons. The cascade time e-folds at a rate of
1/(no)(sigmaizn)(veo), where (no) is neutral density,
(sigmaizn) is ionization cross-section for low energy
electrons at speed (veo). Typically, for no = 1E13 /cm3 (i.e.
ptorr = 3E-4 torr), veo = 1E9 cm/sec (Ee = 100 eV), and
sigmaizn = 1E-16 cm2, the cascade e- folds with a time
constant of about 1E-6 sec (one usec). Thus all of the
neutral gas is ionized and the system is filled with low
energy electrons in only a few usec. Wiffle Ball trapping
works very effectively here. If all the electrons were still at ca. 100 eV, the surface beta would be about beta = 0.01, at B
= 1000

However, the low energy electrons are heated by fast
collisions with incoming fast injected electrons. The
Coulomb energy exchange time for this process is also about
1 usec. Thus the device will reach beta = one conditions
when the mean electron energy is about 2.5 keV, in ca. 20
usec. Beyond this point excess electron density will be
driven out beyond the beta = one limit; the field will have
expanded as far as it can within MHD stability limits.

This process uses “cold“ electrons to start, with “hot“
electrons as drives, to yield a beta = one population of “hot“
electrons. Of course, while the terms —cold“ and hot“ imply
Maxwellian temperature distributions, these systems do not
exhibit this on the time scales of interest. This is called the
“two-color“ electron startup mode, and will work for any
machine which is e- driven and supplied with neutral gas
input at the proper rate. This is the preferred method of
startup for reactor-scale systems.

The overall result is that a deep potential well is provided in
a few tens of usec, and the ions formed by ionization are
trapped within this well, heated by the fast e- injection to
well depth energy, and thus yielding fusion.

Bussard studied the ionization issue extensively (it was one reason he built PZLx-1). If you have a paper that suggests Bussard's number is way off, I'd be curious, but it seems safe to ignore this in the FAQ. It seems unlikely neutrals can survive more than usecs.

[chrismb]

Bussard studied the ionization issue extensively (it was one reason he built PZLx-1). If you have a paper that suggests Bussard's number is way off, I'd be curious, but it seems safe to ignore this in the FAQ. It seems unlikely neutrals can survive more than usecs.

You really just don't seem to understand, and I am clearly very bad at explaining some very basic nuclear collision theory:

Just because a 1+ ion has accelerated into a 65kV well, it doesn't mean that all the electrons simply jump off the damned thing! ions have to be in collision with something for them to loose any more electrons.

Why would electrons jump off an ion just because it is travelling fast????