[MTd2]

Why isn’t it expected?

[MTd2]

This is a sequence of events that more or less describe what I am thinking.

http://img689.imageshack.us/i/plasmoid1.jpg/
http://img683.imageshack.us/i/plasmoid2.jpg/

In the Lee’s model, the pinch is formed by the equilibrium of the compressing radial plasma and the self reflected supersonic waves at the z axis. The mantle of the plasmak is formed by the pinch itself of the Lee Model. The filaments cut through the pinch, though and rich the center of the plasma. The instability is created in the middle compact plasma and it sucks the matter content of the pinch, forming the plasmoid.

The plasmoid lasts even after after the pinch is. According to Lee’s model, the pinch lasts about 20ns, whereas the emission of Ions lasts 40ns.
There is strong neutron emission in the begining, while D+D is being burned, and in the end, after He3 is depleted.

Note that this looks like a little bit the the main sequence star from its burth until supernova.

Pinch looks like the initial nebula. D+D correspond to a star in the main sequence. D+He correspond to the Giant branch, where as I explained above there is a tendency of He to cool and expand the system, and in this case, avoid the plasmoid contraction. And the final emission represents the supernova stage, where the He cannot hold anymore the EM(instead of gravity), which ends in the explosion of the plasmoid/star.

[MTd2]

I was thinking of what would be a good visual description of this situation. The picture I have in mind is the Plasmak, but without the contain layer:

http://www.prometheus2.net/ICC_2002_POSTER.pdf

In this case, there are 2 superimposed torus: an he3 torus and a deuterium torus. The he3 will be nested inside the deuterium, but given that the bremsstrahlung of He 3 is 4 times more intense than deuteriums, it will have a relatively shorter tail than deuterium and will tend have a large cross section around the center because it will be more concentrated.

[MTd2]

You can arrange 1 mol of He in a cube of 28cm of size. 0.1 trillionth of that means a cube about 2400 smaller or of 10micrometers. That’s for an ideal gas, that is, 1atm and 300K. At constant pressure it swallow to 1cm^3 at 100KeV. There is compression of course, but considering that ideal gas fails at 2atm in general, because it starts to show incompressibility factors, we see that He3 can be quite a factor in scattering Deuterium.

[MTd2]

Now, if this works, it is great news for when 1.2MA is reached. Tritium will be trapped and given that it has the same high mass of He3, but low charge, it will sit in the outer layer. The good news it is that D+T have the greatest cross section at 80Kev – 200KeV, which makes its location on the colder zone a great advantage to increase fusion rates. It is actually the highest at least among the lightest nucleus.

[MTd2]

A more stable situation is He3 closet to the axis and Deuterium closer to the border. But what happens it is that colder He3 will be in contact with hotter D. The D+ He3 below 100K is too low, so He3 will just cool without fusion, but if any of them fuses, it will just emit and alpha + proton at very high energies.

So, what will happen is that the He3 layer and D will cool each other and tend to expand adiabatically , given that the plasmoid is mostly opaque to xrays. This will tend to reduce the cross section for D+D. The good side of this it is that the shear between layers of different layers and the plasmoid will last longer and given the conteracting negative pressure to the compression.

[MTd2]

Even though the number of He3 is small than D, it is likely that the probability of fusion with Deuterium is hugely enhanced by 2 factors:

1. It is 1/3 slower than deuterium at thermal equilibrium because it is heavier.
2. It loses energy 4x faster than deuterium by Bremsstrahlung radiation.

Being slower means that it will collide much more frequently with D than Deuteriums collide between themselves. Any random 2 deuteriums will most of the time have a much smaller relative radial speed difference than between a D and He3. Losing energy by bremsstrahlung very fast means that it will lose speed faster and this just will increase its probability to collided with a D.

So, He3 has much much more collisions than any Deuterium. Plus, its cross section at 150-200KeV is almost 10x higher than D+D.

[MTd2]

Do you know a book with tables about cross sections for different elements, like proton or deuterium + element? It might be that D+B10 cross section is much higher than D+D at its peak value.

[MTd2]

Nothing really. I just stumbled on that on wikipedia. I am still trying to figure out his idea. But there are cool animations on his website! 😀

[MTd2]

Rezwan wrote: Thanks Ivy Matt! I’ve posted this on the website. By the way, I desperately need to flesh out the rest of the “Aneutronic Contenders” section, so other aneutronic endeavors are welcome – collect links, some analysis. Start a new thread for each.

Thanks again!

Do theoretical designs count as a contender?

[MTd2]

What is the source of this information:

“The primary reaction produces three alpha particles”. This is the same result of boron – 11, which was my question, but I couldn’t find the source.

Also, the pinch of boron, according to Eric’s theory, is 10 times denser than lead. So, Boron-10 could work as a shield for neutrons.

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If is arbitrary, why the insistence with 5MW? I think this is at least a very good estimative.

[MTd2]

30KJ of net energy at each firing x 180 ~ 5MW
Alright, you say 1000Hz, so he expects 5KJ converted for each firing. But, why this value, then? Why not 500Hz or 2000Hz or whatever?

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Yes, it’s older. But maybe, based on that, you could make a good summary of the new things available on the new paper.

[MTd2]

I found this paper. The content seems to match the abstract.

http://fusionenergy.lanl.gov/Documents/MTF/Why_MTF/Why-MTF-Comments.html

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