D B10 fusion?

Discuss how polywell fusion works; share theoretical questions and answers.

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KitemanSA
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D B10 fusion?

Postby KitemanSA » Sat Nov 05, 2016 8:32 am

pB11 fusion is "aneutronic". Wouldn't that suggest that DB10 would also be? Also, since the D has twice to mass of the p in pB11, wouldn't that also suggest that it can happen at a lower energy?

I cannot find any info on this. Anybody?

hanelyp
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Re: D B10 fusion?

Postby hanelyp » Sat Nov 05, 2016 2:06 pm

With a D-B10 plasma I'd expect D-D fusions to dominate, the boron contributing more radiative losses than energy production to the net.
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D Tibbets
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Re: D B10 fusion?

Postby D Tibbets » Sun Nov 06, 2016 8:57 am

I thought I had a download of a nuclear reaction database that showed very many possible element/ isotope reactions, the cross section, energy yield and products, but alas, I can not find it or anything from a web search...

In any case, some speculation on the D-10B reaction. An extra neutron on the deuterium might improve the cross section, but remember that the Boron 10 has one less neutron than the Boron 11 and the plus 5 charge of the boron10 ion may be considerably more proton repellent than the Boron 11 isotope, or rather require the deuterium to get closer than the proton of 1H does to the Boron11 isotope. Then there are the various resonances to consider- which I know nothing about, except that the P-11B reaction has two, one narrow one at ~150 KeV and a broader one at ~550 KeV. The Deuterium (2H)- 11 B reaction probably has different resonances, if any and this may push the required KeV way up. In any case I have not seen this reaction listed as obtainable or useful for terrestrial fusion reactors. The next larger mass reaction I occasionally see is P-15N (part of the CNO cycle), which is also aneutronic.

If the D-10B reaction occurs it may make a carbon 12, but it may be a different isomer than that produced in the P-11B reaction. As such, it may not decay into nice alphas, but spit out a nasty gamma ray and settle into a stable Carbon12 isomer ... I wish I could find that database!

PS: The deuterium has ~ twice the mass of the proton, but at the same energy it is traveling at 1/4th the speed, and it is the speed more so than the inertia or momentum of the approaching particle that determines if a close enough approach occurs. The momentum scales linearly with mass, but the speed scales as the 1/square of the mass (with the same Z or charge).

Dan Tibbets
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ladajo
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Re: D B10 fusion?

Postby ladajo » Sun Nov 06, 2016 4:08 pm

A chart of the nuclides can help, I like this one:

https://www-nds.iaea.org/relnsd/vcharth ... tHTML.html

(As Posted on duplicate thread)
The development of atomic power, though it could confer unimaginable blessings on mankind, is something that is dreaded by the owners of coal mines and oil wells. (Hazlitt)
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ladajo
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Re: D B10 fusion?

Postby ladajo » Sun Nov 06, 2016 4:09 pm

I also think the inherent issue is that with D flying around, you would be creating all sorts of D-D in addition to whatever D-B10 you would get.
Where you did get a D-B10, it would be spitting out Be8 + an alpha, or three high energy alphas.
See Page 257, Table III
http://rspa.royalsocietypublishing.org/ ... 6.full.pdf

this other document can also help understanding, as it too explores the B10+D reaction, albiet in relation to B11+p.
http://rspa.royalsocietypublishing.org/ ... 9.full.pdf

Ivy Matt gets credit for digging these up.

(as posted on duplicate thread)
The development of atomic power, though it could confer unimaginable blessings on mankind, is something that is dreaded by the owners of coal mines and oil wells. (Hazlitt)

What I want to do is to look up C. . . . I call him the Forgotten Man. (Sumner)

KitemanSA
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Re: D B10 fusion?

Postby KitemanSA » Mon Nov 07, 2016 6:50 pm

D Tibbets wrote:PS: The deuterium has ~ twice the mass of the proton, but at the same energy it is traveling at 1/4th the speed, and it is the speed more so than the inertia or momentum of the approaching particle that determines if a close enough approach occurs. The momentum scales linearly with mass, but the speed scales as the 1/square of the mass (with the same Z or charge).
Ummm, no. It has 0.707 times the speed for the same energy.

2 x 0.707^2 = 1

Since it is the momentum that is significant (IIRC), and it should have 41% more momentum...

KitemanSA
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Re: D B10 fusion?

Postby KitemanSA » Mon Nov 07, 2016 6:54 pm

ladajo wrote:A chart of the nuclides can help, I like this one:

https://www-nds.iaea.org/relnsd/vcharth ... tHTML.html
Sorry, how does that help determine the Xsection curve for D-B10 fusion?

ladajo
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Re: D B10 fusion?

Postby ladajo » Mon Nov 07, 2016 7:53 pm

It will not help with cross sections, other than providing isotope data. The other thing it can help with is exploring the potential decay chains.
In your case you are looking at D-D, D-10B, & D-8Be, as well as sub chains involving 0n1s and Alphas for the most part. What a mess.
Running a reaction with D and 10B creates all sorts of interesting possibilities.

The best way to build a cross section curve is via physical testing. See the linked papers. Build yourself a cloud chamber, and have at it.
The development of atomic power, though it could confer unimaginable blessings on mankind, is something that is dreaded by the owners of coal mines and oil wells. (Hazlitt)

What I want to do is to look up C. . . . I call him the Forgotten Man. (Sumner)

happyjack27
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Re: D B10 fusion?

Postby happyjack27 » Wed Nov 09, 2016 10:00 pm

looking up the atomic masses:

H1: 1.00782504
H2: 2.01355321

B10: 10.012937
B11: 11.009305

He: 4.002602

Consequently the kinetic energy in each reaction, assuming they both make 3 He, is:

Code: Select all

       reactants                              KE diff      products     
       h1           b11         h1+b11        diff         3he             he
       h2           b10         h2+b10        diff         3he             he
pb11   1.00782504   11.009305   12.01713004   0.00932404   12.007806   3   4.002602
db10   2.01355321   10.012937   12.02649021   0.01868421   12.007806   3   4.002602
                                       diff   0.00936017 
                                       
(all numbers are in amu - except the 3)



So, it looks like a d-B10 would produce a little over twice the energy per fusion than p-B11.

(0.01868421 amu * c^2 -- vs. -- 0.00932404 amu * c^2)

as regards energy input to reach fusion, i suspect this would depend entirely on the fusion cross sections at different velocities, and the energy required to achieve these velocities.

roughly i'd suspect the cross section is about the same. either case you're throwing a tennis ball (hydrogen isotope) at a tarp (boron isotope).

The energy to achieve that velocity, however - far more efficient to put that into the tennis ball than the tarp. and the h2 tennis ball is about twice as heavy as the h1 tennis ball. this means that the same velocity is about twice the kinetic energy (since ke = m*v^2)

All in all it seems about the same - a little under twice the energy for the same fusion rate. a little over twice the energy per fusion.

a small payoff, but the major difference seems to be that you use half as much reactants and produce half the helium for a given amount of energy.

same energy gain fraction, twice the fuel efficiency.

not sure fuel efficiency is all that important when you have to use a far rarer fuel in the first place...

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Re: D B10 fusion?

Postby kunkmiester » Thu Nov 10, 2016 12:45 am

I had a thought when I first heard of the Polywell, I'd had the thought that a civilization using it for power might use more advanced reactions as they exhaust fuels for the easier reactions. This would take larger reactors to get the well depth needed, but it would work. It sounds like there would be easier reactions than just climbing the mass ladder.
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KitemanSA
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Re: D B10 fusion?

Postby KitemanSA » Thu Nov 10, 2016 2:41 am

Still no data on the Xsection? It seems amazing that nothing is available.

D Tibbets
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Re: D B10 fusion?

Postby D Tibbets » Fri Nov 11, 2016 5:04 pm

KitemanSA wrote:
D Tibbets wrote:PS: The deuterium has ~ twice the mass of the proton, but at the same energy it is traveling at 1/4th the speed, and it is the speed more so than the inertia or momentum of the approaching particle that determines if a close enough approach occurs. The momentum scales linearly with mass, but the speed scales as the 1/square of the mass (with the same Z or charge).
Ummm, no. It has 0.707 times the speed for the same energy.

2 x 0.707^2 = 1

Since it is the momentum that is significant (IIRC), and it should have 41% more momentum...


Correct, or rather incorrect for my statement.
The magnitude of the velocity difference is smaller . Momentum = mass * velocity would seem to favor the deuterium, but I am still not sure, even with correction for my error. Reviewing momentum in a basic Physics text. does mention Momentum= mass * velocity, but then expands on this to include a delta time component. Something going faster will penetrate deeper into a target. The most obvious example, at least for me, is bullet terminal ballistics. A faster lighter bullet is better at penitrating armour. There are multiple considerations, but this is the basic premise. Delta time in the deuterium instance means that the deuterium approaches the target slower. As such, it is exposed to the decelerating electromagnetic force longer. As such, it will decelerate more with all other things being equal. With increased momentum, this may compete against this trend. More momentum means it decelerates more slowly while slower speed means it has a longer time to decelerate as it approaches the target particle while the repulsive electromagnetic effect increases via the inverse square law. Where the balance would be , I don't know, but the solution is complex and probably needs integration to get an acurate answer. I don't know the solution. The only thing I have to go by is that this reaction is not listed as a possible fusion reactor fuel mix.

Apparently the D- P10 (or rather 10B- I'm trying to conform to the standard nomenclature more) can result in three alphas, Several gamma decays (presumably without subsequent fission into alphas?))were also described in the chart linked by Idaho. I don't know the frequency of these side reactions. As such there is gamma radiation , still no neutrons, but also less exothermic energy as the 10B to 12C difference in nuclear binding energy is less than the 10B to 3* 4He difference.

The energy yield is perhaps mildly to moderately less and the input temperature is presumably more, possibly much more, so the net yield is less and again presumably, more difficult to obtain.

AS pointed out in another post, the D-D side reactions producing neutrons is unavoidable. These side reactions could be reduced by having a predominace of Boron ions in the plasma- like the scheme to reduce neutron production with D- 3He fuel. But, this resulte in an increased proportion of high Z Boron relative to the fusion yield. The Btremsstruhlung costs are more sevear. With P-11B reaction you can have a prodominance of protons and supress the Bremsstruhlung and not have a penalty of increased neutron production.

I'll end my speculations here. Without an actual fusion cross section profile reasonable conclusions are impossible.


Dan Tibbets
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hanelyp
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Re: D B10 fusion?

Postby hanelyp » Fri Nov 11, 2016 8:04 pm

Another consideration: It's been suggested that shifting the balance in p-B11 plasma to proton heavy has a favorable effect on fusion/loss rates. Doing this with D-B10 would further shift the fusion balance towards D-D with the boron serving mostly to increase radiative losses.
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happyjack27
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Re: D B10 fusion?

Postby happyjack27 » Fri Nov 11, 2016 9:52 pm

D Tibbets wrote:AS pointed out in another post, the D-D side reactions producing neutrons is unavoidable. These side reactions could be reduced by having a predominace of Boron ions in the plasma- like the scheme to reduce neutron production with D- 3He fuel. But, this resulte in an increased proportion of high Z Boron relative to the fusion yield. The Btremsstruhlung costs are more sevear. With P-11B reaction you can have a prodominance of protons and supress the Bremsstruhlung and not have a penalty of increased neutron production.

Dan Tibbets


From searching I've come up with mostly "B10 is a great neutron absorber". So at least there's that - a lot of your neutrons are probably just going to convert your B10 to B11. But then there goes your fuel.

But what's the branching ratio of H2+H2->He3 + N vs H2+H2->H3 + H? If you're producing just as many protons as neutrons, then presumably you'll have a bunch of pB11 fusion going on as well. Not sure which would be in excess, though - you'll probably lose a lot of neutrons, and a lot of the protons will probably take place in other fusion reactions. And then there's also H2+H3...

Not clear on the fusion rates and branching ratios here, if any of that is even significant.
Last edited by happyjack27 on Fri Nov 11, 2016 10:10 pm, edited 2 times in total.

ladajo
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Re: D B10 fusion?

Postby ladajo » Fri Nov 11, 2016 10:06 pm

10B is a neutron vacuum, and used extensively as such in the nuclear fission trade.
I think the great drama in this approach are the associated and downstream reaction chains creating a mess.
The primary benefit of 11B is the very low production rate of 0n1s. Where with 10B, you are looking at D-D right up front, in addition to the D-10B and subsequent chains (where you'll get some fun stuff). I agree that the ratios are not clear, however the papers linked above by me give some insight to that, as does some manual labour walking the possible reaction chain paths on the chart of the nuclides. The big clue there is the comparison of half lives for the unstable isotopes in the possible chains.
The development of atomic power, though it could confer unimaginable blessings on mankind, is something that is dreaded by the owners of coal mines and oil wells. (Hazlitt)

What I want to do is to look up C. . . . I call him the Forgotten Man. (Sumner)


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