I got Q~10 and 1.6MJ per shot with slight modifications of LPPX.

[MTd2]

V0=45
L=18
A=5
B=2.8

Slightly modified:

C=150 ( Instead of 120 )

Now, these are the unusual parts

P=4500 ( about 6.5x times the sea level pressure)
z=0.4

The yield in neutrons for the old parametrization ( which is the same, apparently, for T-D) is 1*10E15N or about 10KJ. That means an yield of 1.6MJ per shot with decaborane.

[Lerner]

Practical problem–that makes the electrode shorter than the insulator(2.8 cm). Also, too high density means too much energy goes into ionizing the gas–which Lee’s model does not take into account, I believe.

[MTd2]

But you used 200 Torr to achieve your desirable yield! Why not 4500 Torr?

The reason it is that I have to force Lee`s program to run at 200 Torr, it simply refuses to compute. It says pinch cannot form above 20 Torr. So why do I have to stop at 200 Torr?

[Lerner]

The variable z is the length of the anode in cm. The shorter you set it, the greater the pressure, since the slower the sheath can run during the pulse rise time. But there are physical limits. The electrode has to be longer than the insulator and , as I said, if you put too little energy in per unit mass, too much is consumed iin just ionization. That could affect us as well with the shortest electodes–7 cm.
Lee’s limit of 20 torr is arbitrary. Just elimiante the line inthe code that says”stop” and it runs fine, as I assume you found out.

[MTd2]

But isn`t the limit of 200 Torr arbitrary too? Why not setting a bit higher pressure higher yields? Won`t you try it? Have you estimated a function from when ionization disturbs?

BTW, the z=0.4 (cm) for z is arbitrary. Setting up a lower value up to zero (cannot write 0, but 0.00001, or something, you get the idea), doesn`t change the yield. So, isn`t there an insulator tha can do the same job with just 0.3 cm?

[Lerner]

I did not set any such limit. But you are ignoring the ionization energy that I mentioned. Roughly speaking ,at 14 cm electrode length , the energy of each ion in the sheath is around 12 times more than the ionization energy of D, 14eV. At 7 cm, it is only 3 times as much, since the sheath is moving only half as fast, meaning that maybe 1/4 of the energy goes into ionization–maybe too much. At 3 cm the pulse would not have enough energy to inoize the whole sheath–no sheath would form.

[MTd2]

I asked you about the limit of 200 Torr to the ionization, but you explained me about the relation to the length of the cathode. Why stopping at 200Torr, but not 400Torr, 600 Torr, or any other higher value? But thanks for explain me that relation but I also would like some literature on that.

[Lerner]

I did answer your question, but evidently not clealry enough. There is no 200 torr limit. You must decrease the lenght of the anode, here labled z, when you increase the pressure, because the sheath rund more slowly, so it have moved a shorter distance by the time the pulse peaks. Roughly density goes up as the square of the length goes down. So, since you can’t make the anodes much shorter than 7 cm on this machine, this sets a rough upper limit on the pressure. But you can play around beyond 200 torr. Ultrahihg pressures are out for the reasons stated here and in the last message. With too high presure, you devote all your energy to ionization–there is none left over for the motion of the sheath.

[MTd2]

Oh. I got it! I want to keep playing with the yields. So, how much beyond 45kV can you increase without having to change the machine?

[Lerner]

None. That’s the maximum. To go higher, we need to put things in oil baths, with major changes in the physical structure, and a very significant down time.

[MTd2]

With z=14, the yield increased until pressure was about 190Torr and saturated at around 3,5E13 neutrons and remained constant until 250 Torr when it began to decrease until the device stopped working at 350 Torr.

I was playing with z=7. The yield increased until pressure was about 500Torr and saturated at around 10E14 neutrons and remained constant until 800 Torr when it began to decrease until the device stopped working at 1050 Torr.

I reduced the massfr to simulate a smaller drag on the of the mass and massfr, by a quarter, to 0.1 from 0.14 for z=7 and the only important thing that changed was the saturation interval, that went from 500-800 Torr to 800-1000Torr.

I further reduced currf and currfr to 0.5 from 0.7, to simulate a less intense sheath, there was no neutron saturation, just a peak at Torr 350, with 3.310E13 Neutrons.

So, both z=7 and z=14 give similar neutron yields, by optimizing conditions. So, is there space for some optimization between 7<z<14 or is it already optimized with 3.4*10E13 neutrons?

Even with this yield, you have about 50KJ in x-rays when using decaborane. What is the need for the axial field? Does it really triplicate the yield? That means 150KJ in x-rays.

[Author]

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