The X-Ray output should be 150KJ and not 30KJ.

[MTd2]

I did simulations over a considerable range of values and I always find that the optimal yield for neutrons is aroud 3.4*10E13 neutrons, which means 300J for D-D. The axial field should multiply that by 3, giving 900J and the boost of 160X of Boron gives the yield of 150KJ.

[delt0r]

Your numbers seem off. Even with DT fusion (ie 14MeV neutrons) 3.4e13 neutrons is only about 76J of neutrons. With DD that goes down to less than 50J giving a total fusion yield of less than 100J. This is assuming a full DD burn (DD->He4, H,n).

[MTd2]

I`ve been using 1*10E11 Neutrons meaning 1J in lots of other posts. No one has corrected me and even Lerner assumed that as right (he kept discussing with me as there was nothing wrong). Let`s see:

http://en.wikipedia.org/wiki/Fusion_power#D-D_fuel_cycle

A neutron from D-D has 3.93*10E-13 J. So, 3.4*10E13 neutrons yields 13.3J. Tripling that gives 40J. So, the total Boron fusion has 6400J.

But, there is a catch here. Nowadays, LPPX goes above 100KJ, probably the mean temperature nowadays is 150KeV. Weirdly, the results from the updated Sing Lee`s model does not follow D-D curve, but of D-T fit. So, maybe at such temperatures the Tritium branch is preferential and most of the tritium generated goes to generate high energy neutrons.

That means 76 x 160J= 12160J. Which is the scientific demonstration of the technology (10KJ) a planned. The 3X of the axial field makes it achieve 36480J per shot , which is the prototype demonstration expected within a few years. (33KJ).

[MTd2]

BTW, I didn`t find any paper about the cross section of the tritium branch when increasing temperature.

[delt0r]

The branching ratio is not temperature dependent for DD fusion, its always 50-50 (its a standard model thing). Also its generally assumed that all the T produced is then burned at the rate its produced (about 10-100 times higher cross section) and this gives a 14MeV n (so half the neutrons have about 14MeV the other half about 2.5 MeV). However people generally differed on how much of the He3 is then burned. Often the approximation that no He3 is burnt at all.

I still don’t follow your numbers. Why do you triple your 13.3J? And why do you assume that Boron H yield is so much higher? Even at optimal temperature the cross section is many times lower than DD, or even DHe3.

[MTd2]

I triple the numbers because according to Lerner, by giving angular momentum to the filaments that will form the plasmoid, will increase the mass and density of the plasmoid. Also, the reason why Boron – H will be used is the very reason of why LPPX exists, that is, Lerner expects that it will have compression rates thousand folds higher than other types of fusion. You can find paper about it on LPPX website and this website.

But I really cannot understand. or believe, why the branching ration is a constant if all other cross sections are constant. I cannot find any study about this so, I cannot believe you unless you show me a proof of that.

Anyway. It might be that Lerner is giving 24MeV to each neutron seen, since the objective is counting the total energy given away by the fusion process. The number of neutrons is a diagnostic tool anyway,to study the pinch.

[delt0r]

The *branching* ratio is 50/50 to about 10e-5 accuracy. This is predicted from the standard model and verified by a *lot* of experimental evidence. You can learn the standard model, and atomic models yourself if you don’t believe it, you don’t even need to go that advanced in the theory in fact. Remember for atomic models (strong force) 150KeV is not really that much hotter than 1KeV. Note we are not talking about the fusion cross section (reaction rate) that’s dominated by tunneling and the electric force, we are talking about the branching ratio “after” a fusion event that’s dominated by the strong force.

There is the DD-He4+2gamma, but this is extremely rare. You must have 2 gammas to conserve all the relevant quantum numbers. So you end up with a fine structure constant squared in the branching ratio and thus you have 50-50 for DD->T+p and DD->He3+n with almost zero contribution from DD->He4.

It may gain some temperature dependence when you get to the 1GeV per nucleon and higher ranges. But probably not. Since the *fusion* branching ratio is still via the same process (you form a He4* intermediate).

[MTd2]

delt0r wrote: Note we are not talking about the fusion cross section (reaction rate) that’s dominated by tunneling and the electric force, we are talking about the branching ratio “after” a fusion event that’s dominated by the strong force.

Yes, that’s right. I forgot about this, heh… So, I conclude that Lerner is counting the total energy yield, which is diagnosed by the number of neutrons. Given that tritium is completely consumed in the fusion, or otherwise Lerner would have noticed radioactive waste, we have that:

17.6MeV from D-T + neutron ( 14.1 MeV)

7.3 MeV from D-D + neutron ( 2.45MeV)

So, that is 24.9 MeV for every 2 neutrons, which gives 12.45 MeV per Neutron.

This is close to the D-T neutron yield, so for final energy collection, this is almost the same as collecting neutrons from D-T, meaning, within the range of the final objective, around 33KJ.

[vansig]

MTd2 wrote: I`ve been using 1*10E11 Neutrons meaning 1J in lots of other posts. No one has corrected me and even Lerner assumed that as right (he kept discussing with me as there was nothing wrong).

Neglecting to look into correctness is not evidence of correctness.

By the way, the expression, “1*10E11” is a confusing mixture of two forms of a quantity.
FORTRAN and its successors use the E to mean “×10^”, so syntactically, it would be clearer
if you would write either 1E11, or 1×10^11, instead.

the problem is, that parsers for these languages are liable to parse your expression, and yield a value that is a factor of ten away from what you expect.

[MTd2]

The keyboard here at home suffers from configuration issues, heh. I cannot make it to work properly. I just type what is consistent with what excel uses.

[Lerner]

No, friends, the DT logic is not right. The fraction of DT that gets burned depends on the density. When we can detect DT netutrons (not so far in this experiment) we can use the DT/DD fraction to measure the density.

[delt0r]

Lerner wrote: No, friends, the DT logic is not right. The fraction of DT that gets burned depends on the density. When we can detect DT neutrons (not so far in this experiment) we can use the DT/DD fraction to measure the density.

I sorry but i don’t follow. So you are saying that we assume that the T produced from DD fusions are typically not subsequently burnt, or that because you don’t detect DT neutrons, that we then make the assertion?

If the T produced from DD reactions are confined I would assume they would fuse quickly since the cross section is so much higher even at 150KeV. If this is not the case, why not?

[MTd2]

delt0r wrote: If the T produced from DD reactions are confined

Hmm. Makes sense that the Tritium formed is not confined. It comes from the fusion reaction with a temperature of 1MeV. This is higher than what is required to keep confined by the plasmoid magnetic field, since the average temperature is 150KeV.

http://en.wikipedia.org/wiki/Nuclear_fusion#Criteria_and_candidates_for_terrestrial_reactions

[MTd2]

But if there is not significant energy going out the system, how to achieve the desired result?

[Lerner]

Currently we are burning a very small fraction of the total D in the plasmoid because the density is relativley low. The tirium is trapped, and a larger fraction of that burns, but it is still a small fraction. At higher currents, we will burn more of both.

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