[MTd2]

OOPS! Sorry! 😀

“For Ei =300keV, this implies B>14GG for p, B>3.5GG for α , and B>1.3GG for 11B.”

http://arxiv.org/ftp/physics/papers/0401/0401126.pdf (page 18)

14GG is what is necessary for protons! For borons is 1.3GG!

I got my hopes too high!

[MTd2]

I cannot read that paper but to achieve relativistic speeds would be necessary way too much energy, even to set up the EM potential of the electrodes, which would be enormous.

[MTd2]

http://data5.blog.de/media/104/3644104_8c143c10ee_m.jpeg

Not a good cake, right?

[MTd2]

There is a DPF device that has parameters much larger than LPPX:

(4.1mF, 300kV, 4MA, 187kJ, 400ns time to peak current, dI/dt ~ 1013 A/s)

http://www.scielo.br/scielo.php?script=sci_arttext&pid=S0103-97332004000800054

Speed 2, at Leopoldo Soto lab, one of the friends of LPPX, I think.

There could be some sort of collaboration. I guess they can achieve a few thousand of Joules with their Speed 2, in case LPPX doesn`t succeed. At least, the quantum suppression of bremsstrahlung could be demonstrated.

[MTd2]

Francisl wrote: If 45 KV is not high enough, can we use a 30 to 50 KV deuterium particle beam to bombard a 45 KV pinch? Would that provide the 75 to 95 KV energy to probe the voltage levels that would be effective?

Not really, because what causes the fusion is the dense current of the pinch.

[MTd2]

The papers are just about the interval up to 50KV, and this is clearly not enough since the LPPX is 45Kv. 90KV is clearly the ideal, but I don`t know the lower value. I`d have to play with the values and I`d have to play with the simulation spreadsheet first. Let me make an estimative though. Saturation occurs when all wasted energy is due the dynamical resistance. So, that means an asymptotic value of V=RI. For an optimized configuration of 100KV, see table 3, ( https://focusfusion.org/index.php/forums/viewreply/7483/ ) is about 1.5MA, we achieve a pinch saturation at 1.5MA @ 35KV. Given that we want saturation to occur at 3MA, we must double the voltage, to 70KV. Give a bit of safety margin, to avoid the flatten curve before saturation, we could say 75KV. So yes, your guess is right.

[MTd2]

Aeronaut wrote: Long distance transmission lines double the voltage to reduce the current by 50%. Using P=(I^2) R, doubling the transmission voltage results in 25% of the losses. Not sure how this relates to raising voltage to raise current…

There are 2 things here. The wasted energy in collisions and the free electrons. The wasted energy will go like P=(I^2)*R anyway, like in a wire. This is a fraction of the electrons that effectively behave like they were in a resistive medium.The increase of I of an input just makes things go even more wasted.

But note that “P” is not the total energy of the capacitor banks, just the part where it is wasted. So, by increasing the total voltage AND keeping BOTH total resistance (which would happen naturally anyway) and total current CONSTANT, is the same as diminishing the current that goes through the resistive process. So, not only R goes down, as also I (resistive) go down so that conservation of energy is respected. So, more current will be available to be absorbed by the pinch.

[MTd2]

The reason for this saturation is caused by the design of the DPF. When a EM magnetic pulse runs through the cathode, it charges and push a thin layer of gas ahead of it. The gas becomes hot, a plasma, and moves fast. This means that there is transfer of energy between the pulse and the plasma mainly in 2 ways: inducing a current in the plasma and by giving it kinetic energy. The latter is a dynamical source of resistance since the speed of the plasma is high, and mostly doesn`t vary much, independently of the variation of the parameters of the DPF device, at least in the window of 15-50KW: up to 100km/s, at the tip of the cathodes, comparable to the Sun`s plasma velocity. This dynamical resistance has about 7 mili Ohm. So, the bigger the current, the bigger the loss of energy. The remaining current for the plasma pinch is small, given that most of it was lost in heat.

The solution to this problem is the same that is used for long distance electrical lines, meaning, using higher voltage to minimize losses.

[MTd2]

According to table 1, table 3 and Figure 7 of this paper:

http://www.plasmafocus.net/IPFS/2010 Papers/Energies PP.pdf

Saturation occurs for 1.5MA. So, if you discharge more than 1.5MA, at low voltages, the pinch won’t achieve more than 1MA, and it doesn’t matter what configuration is used.

[MTd2]

Aeronaut wrote: Sing Lee’s equivalent circuit diagram and labels rely heavily on the word ‘fraction’. The backplate’s inductance and current fractions are significant contributors to the machine’s performance or lack thereof. Have any of his students reported any experiments to reduce backplate inductance?

It was tested in several different kinds of DPF devices. The limitation is inherent to all kinds of Sing Lee`s variations. What was not tested though it was the regime of 90KV. Las Vegas DPF will attempt to study this region.

[MTd2]

According to Sing Lee, if you are using the Sing Lee model, no matter what you do, you will face the same limitation. The use of the axial field is the way to find a loophole inside Lee`s theory. But it seems that the axial field just enhances slightly the saturation limit. What I see as a solution is just to double the voltage. Everything is kept the same, the energy stored in the banks is enough. The axial field should be 10x stronger due the faster speed of the arriving current, but that would be a great thing since it would be above the background magnetic noise.

[MTd2]

Lerner is aware of this issue!!!

http://denseplasmafocus.org/index.php/forums/viewthread/443/

I wonder how he will overcome this.

[MTd2]

BTW, the best LPPX can achieve right now is less than 10Joules, even with Goldilocks point found…

[MTd2]

For the purpose of LPPX, that is 300J for peak D-D reaction, following the figure 8 of

http://www.plasmafocus.net/IPFS/2010 Papers/2010 Pp2 IJER.doc

the LPPX device should have the following specs:

around 600KJ per shot, or 60KJ-100KJ IF the Goldilocks point is found, with the usual 3MA(0.3-0.5MA) in the pinch, but 9MA(0.9M-1.5MA) discharged by the banks. But the voltage should be increased to 90KV. That would be a great thing since if things scale well to Boron given that a net positive energy could be obtained even with down to 50% of efficiency.

At 50KJ, the best LPPX can achieve right now, would require 150MJ per shot to achieve the objectives…

[MTd2]

Francisl wrote: The question is at what point has the plasmoid been squeezed enough to produce economical fusion?

No, it hasn`t. Lee`s model predicts that the DPF device cannot give away enough energy since the current saturates.

[Author]

Posts