D B10 fusion?

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

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KitemanSA
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Re: D B10 fusion?

Postby KitemanSA » Sat Nov 12, 2016 9:11 pm

hanelyp wrote: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.
I was under the impression that the switch was to B11 heavy to reduce the number of protons which lose all the energy. In that case, the same would work to reduce the side chains.

All of which is interesting, but STILL doesn't provide a cross-section graph!

D Tibbets
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Re: D B10 fusion?

Postby D Tibbets » Sun Nov 13, 2016 2:47 am

The production of neutrons by D-D side reactions also produces tritium in further subsequent side reactions- well may be. In the Polywell at least, the tritium and proton produced from D-D fusion are above the potential well energy (presumably, unless you are going to extreme KE injection energies). They leave the system. As pointed out multiple times in various threads, the Polywell is not an ignition machine - the fusion products do not contribute significantly to further fusion or plasma heating. They have to be collected externally, processed and reintroduced as new injected fuel ions.

The same would presumably hold for neutrons. Despite higher neutron absorption by eg: 10B, there is very little Boron in the reaction volume. The plasma may have a density of perhaps one thousands of an atmosphere (10^22 particles/ M^3). That this relatively low density of particles can lead to significant fusion is helped by the density squared scaling, but also critically by a confinement time represented by the number of passes across the machine(along with the speed). At least 10 thousand passes within a few meter diameter machine is needed at 10^22/ M^3 densities. A neutron only has one pass, or rather probably only 1/2 pass as most fusion produced neutrons are released in the central core of a working Polywell reactor. This is a confinement time/ distance difference of at least 20,000. The neutron captures by 10Boron or any other isotope in the reaction space is extremely tiny in relation to the parent fusion rate. This would hold for any fusion product reaction , even ignition plasma where the fusion produced ions may still be magnetically and/ or electrostatically confined at times >> greater than absolutely unconfined neutrons.

The 1/2 cat D-D reaction has been described by Bussard and others(?). Here there is a 10B blanket surrounding the reactor which captures the escaping neutrons, with heat and and tritium production contributing to the final desired energy balance. I don't know the thickness of this solid blanket, but an exercise with guessed numbers may illustrate the possible contribution of neutron absorption inside the plasma. Assume the 10B solid blanket is several inches thick. A gas of the same intercepting mass/ number of atoms would have to be compressed about 700 fold (700 atmospheres) to have the same density. As the plasma is probably about 0.001 atm (10^22particles/ M^3 at STP) results in a final difference of ~700,000. For the plasma to absorb as many neutrons as the solid exterior blanket, the plasma reaction volume would need a radius of perhaps 1,000,000 inches or about 30 Kilometers.
I have a nuclear fallout calculator wheel buried in a box somewhere, but I recall that perhaps several feet of dirt will protect against fallout radiation- mostly gamma rays and neutrons, but hundreds of yards or more of air is required. And the typical plasmas are at a thousandth of an atmosphere, or in a Tokamack example perhaps a millionth....

Keep in mind that we are talking density, not pressure- so don't get too confused by my jumping between the units. One atmosphere of gas at STP has density of about 10^25 particles per cubic meter. Heating to a higher temperature gas, like a plasma, may result in a much higher pressure in a confined space (PV=nRT), but the comparative densities are the same. This may be misleading if relationships are not accounted for- such as MIT's recent claim that their Tokamak reached over one atmospheric pressures. At their temperatures, this was still the mundane ~ 10^20 particles per cubic meter (still better than 10^19 particles per cubic meter of some Tokamak results)

Dan Tibbets
To error is human... and I'm very human.

D Tibbets
Posts: 2769
Joined: Thu Jun 26, 2008 6:52 am

Re: D B10 fusion?

Postby D Tibbets » Tue Nov 15, 2016 3:55 pm

KitemanSA wrote:
hanelyp wrote: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.
I was under the impression that the switch was to B11 heavy to reduce the number of protons which lose all the energy. In that case, the same would work to reduce the side chains.

All of which is interesting, but STILL doesn't provide a cross-section graph!


My impression is that the preferred P-11B mixture has a preponderance of protons in order to minimize Bremsstruhlung losses. Perhaps 10 protonsfor each 11B. This is to reduce the Z 5 Boron from producing 25 times as much Bremsstruhlung radiation as would a pure proton plasma at the same density. Bremsstruhlung scales as the square of Z of the target ion. By maintaining the same final density of ions, but having (eg:)90% of the total as protons and only 10% Boron would maintain about the same fusion rate but reduce Bremsstruhlung losses to an average of ~ 3 units per particle/ ion. An even mixture would be about 13 units. Pure protons would yield 1 unit, pure Boron would yield 25 units.- The sum of the two in the mixture times the Bremsstruhlung yield of each particle. An even mixture of 5 protons at 1 Bremsstruhlung units each plus 5 Borons at 25 units each = 130 units/ 10 particles. Proton heavy mixture of 9 protons at 1 unit each plus 1 Boron at 25 units =34 units/ 10 particles. Pure protons (or any other hydrogen isotope)would be 10 units per 10 particles.

In the D-3He reaction the reasoning is different. Here the heavier 3He is in excess. This is to suppress the D-D side reactions that produce neutrons. The additional Bremsstruhlung losses from the 3He is modest because it only has a Z of two.

Dan Tibbets
To error is human... and I'm very human.


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