Have ANY projects got a future?

[Steven Sesselmann]

Chris,

We should get together a list of ideas that "Should" be tried.

And then we can cross off those items that have been tried and that did not work.

This would save some of the repetitive work that is being done.

Steven

[Steven Sesselmann]

Reply to: viewtopic.php?f=16&t=7428#p53943

Dave,

This was well written, and does indeed see through some of the smoke.

I would like to make one more point...

With the Fusor, we are creating a potential energy well (PEW), using electrostatic fields, and to maintain this well, we need to input a continuous amount of energy, and this is why we clock up such high losses.

In theory, at least, there is no reason why it should require any energy to hold a potential energy well open. Energy is required to open the PEW in the first place, but not to hold it open.

The losses that are being measured are leaks!

A second factor that would be an essential factor to successful energy gain is, that in the bottom of a PEW, fusing atoms suddenly move towards eachother, small distances, but huge velocity. This coalescing has the effect of further lowering the potential in the center. Create enough fusions in the centre of a potential energy well, and your batting average will be vastly improved.

The dueterium atoms will fuse, like water down the bath drain!

Steven

http://www.beeresearch.com.au

[001userid]

One idea that I have been looking at for several months, involves superconducting ions. It is another out on a limb idea, but it makes alot of sense.

If superconducting electron paths can be made at room temperature, why not superconducting ion paths?

The path of a pitch is defined in solid state. No vacuum is required to maintain a mean free path, as this is also defined in the solid state.

I see two options in the configuration, either crossing paths or various paths converging on the same point. There is very little information on the subject of ion conductivity. Where to start?

Joe Sal

[Dan Tibbets]

I had abandoned this thread earlier, but after rereading the post below and thinking about it I cannot resist giving some counter arguments to the ' polywell is BUSTED ' arguments.

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From Frank S.-
"Please explain to me that if all the deuterium or any other fuel that you choose is FULLY ionized and traveling at 100 kv and there are no electrons for any loss, how can even this be anywhere near break even no matter what the size?

I also just did a quick calculation on the maximum theoretical energy output from a 3 meter diameter polywell device. There are 14,000 liters which is 3.8 E 26 atoms at 10 mtorr. Assuming ~3 MeV per fusion or 1.5 MeV per atom, that gives 390 MW. The theoretical calculation you reported was 100 MW. This is frightfully close to the theoretical physical law breaking reality of the universe. That is tough to accept. This also assumes a 100% full and complete turn over of all of the spent fuel and 100% new fuel every second without fail. On top of that, how can this scale to the power of 7??? It is already nearly at the theoretical maxium for the energy contained in an atom.

Hell, in an atomic or hydrogen bomb the efficiency of mass conversion is only fractions of a percent of the starting materials and you think polywell will surpass this too by orders of magnitude as you have reported?

Call me pig headed but I am not seeing it. Way too much pie in the sky and outrageous and unsubstantiable numbers are being thrown around. There is no way in this universe that any of this can even be close to the truth. Look at the facts guys, I say the polywell myth is BUSTED! "

Frank Sanns
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Concerning the energy yield per fusion:
A more accurate and complete assessment of the possible energy yield per D-D fusion is described in this quote from page 7 of this pdf

http://www.askmar.com/Fusion_files/EMC2 ... plants.pdf

"The result of these observations was that the average
energy of one net DD reaction could be made to vary
from the low value of 3.65 MeV arising from the basic
DD fusions alone, to 10.24 MeV for burn of the 3He
directly produced "DD 1/2 cat, to 14.63 MeV for additional
burn of the 3He resulting from decay of the tritium produced.
Capture of the neutrons from the neutron branch of the
associated DD reactions could also add energy to the plant
system, through their use to cause fission in a 10B loaded
blanket,,,"

You only considered the energy contribution from the neutron produced. You also have to include the kinetic energy from the proton, helium 3 and/or tritium that are produced. And, as described above, recycling the appropriate fusion products can yield even higher yields from the 2 parent deuterium atoms.


Concerning the density of the ions, given as 10-5 torr (10 microns), is uncertain I think. Certainly the vacuum outside the magnets is low, I'm uncertain what the average density inside the coils is, but it is presumably more. The ions are injected into the interior via an 'ion gun'. There have been estimates that the density at the focused core approaches 1 atmosphere as the ions converge and pass through the center region. Also, the volume you gave is true for a sphere, but the actual volume would be slightly to moderately more depending upon how many magnets end up being used. WB6 and 7 were truncated cubes. In any case the ions are confined to a smaller volume within the magnets (Magrids), so I'm uncertain what the actual effective volume would be. 14,000 liters is probably a good ballpark figure. At one atmosphere (STP) there are 6.02 E 23 gas molecules in 22.4 liters, so in 14,000 liters there would be 3.7 E 26 molecules ( or in this case ions) . Assuming 10 micron pressure (~1/100,000th of an atmosphere) there would be 3.7 E 21 ions present (round up to 4 E 21).

The comment that the D-D reactions only converts <1% of the mass into energy is given. I'm uncertain about your energy yield for the complete (100%) conversion of matter though. Complete conversion of two deuterium ions would yield ~ 4,000 MeV. I based this on the rest mass of the electron, which is 511,000 eV, and since protons and neutrons are ~ 2,000 times heavier...
Converting electron volts to Joules taken from this site-

http://www.unitconversion.org/energy/el ... rsion.html

gives ~ 2 E -19 Joules = 1 eV. So 4,000 MeV would give ~ 6 E -10 Joules, which over one second would give the same number of Watts. Adjusting your estimate for the numbers of atoms in the 3 meter Polywell to 4 E 21 (see above) atoms/ions in the machine at any given time, divided by two (pair of deuterium atoms per reaction) gives a total of ~ 2 E 21 reactions possible.
2 E 21 reactions X 6 E-10 Joules per reaction gives 1.2 E 12 Joules, and over one second that would be 1.2 E 12 Watts, or 1,200 Gigawatts. One of us is off in his calculations. Using the fusion yield instead (~ 4 MeV per reaction) gives a result of ~640 Megawatts.
As a check, plugging numbers into this calculator to get 2 E 21 fusions per second yields ~ 1.8 Gigawatts.-

http://www.beejewel.com.au/research/fus ... ulator.htm

Mm... this is ~3 times higher than my calculation. I'll assume its's more accurate, so ~ 1/20th of the ions present would have to fuse per second to give ~ 100 Megawatt output.

Is this rate reasonable? The lifetime of an electron in the WB6 has been reported as ~ 0.1 microsecond, in which time it makes ~ 100,000 transits. Assuming the ions have a similar lifetime, and makes a thousand or more transits in that time, and using the often claimed need for 10,000 transits for a reasonable chance for a fusion reaction to occur, then things would balance out well. Guessing that the density is higher and that the number of transits made by an ion is greater would relax the conditions further. Keep in mind that the electron and ion losses are replaced so the numbers involved would be steady state.



Dan Tibbets

[Frank Sanns]

The atoms do not anhailate, the just fuse and the energy that I quoted is correct. All of the other energy that you talk about was waste input energy or at least non fusion productive energy so it should not be counted twice.

Secondary nuclear ash will burn but its build up rate is really slow and it will not be a significant part of the equation.

I will not be so bold as to say busted but I do not see any significant changes to think it would be anything other than pie in the sky.

Frank Sanns

[Dan Tibbets]

No, annihilation energies are not relevant, but you included them in your post so I included both, with hopefully accurate yields, for comparison. Your claim that annihilation only produced a few 100's of MeV for the number of aviable atoms is way off based on my included calculations. So argument from that standpoint is negated (assuming my calculation is more accurate).
Actual energy yield for the two dominate DD reactions yields an average of 3.65 MeV per fusion, so by rounding down to 3 your calculated results for the aviable fuel fusion calculation was low by a factor of 1.2. And, it was significantly below both my calculation, and the source I referenced.
The nuclear ash recycling is just that. The high speed fusion products quickly leave the reaction space with minimal chance of reacting with a deuterium on the way, and fly to the walls, where they give up there energy as heat, etc. The He 3 and tritium would be recovered in the vacuum pump exhaust gasses , processed and reintroduced as new fuel.
This might or not be economically viable if the system works at all, but it does allow you to recover more energy if the base DD reaction is only providing marginal gains (or losses) and the radioactive tritium would need to be isolated and used or stored in any case.


Dan Tibbets

[Chris Bradley]

I think there are things that could be said about either calculation, your respective two taking on 'opposing' ends. In reality, no, I don't think it is reasonable to presume you're going to consume 1/20th of the reactor fuel per second. This is a couple of orders of magnitude better than tokamaks yet we're talking about a device that hasn't yet *proved* a single fast-fast neutron according to the 'advertised' details.

Also, you realise you are saying that you are going to be able to fully populate the volume of the reaction vessel by material injected by ion guns alone, in 20 seconds! That full mass of material passes through the ion guns to displace what is already in there? 20 seconds....hmmm...4E21 ions/20s = min. 32 Amp ion guns. Has that ever been achieved before?

Sure, you can go thru the calcs and if you believe that Polywell really can generate a continuous multi billion kelvin central ball of reaction volume at 1 atmosphere without any neutrals or other contaminants therein, then you *could generate* the numbers being talked about here. But that's a heck of an *if* without some real, substantial, repeatable, traceable, verifiable diagnostics and results, plus real evidence of the confinement time of the ions (that is, total energy confined/power loss, not this figure for electrons).

The other part which needs to really get some evidence behind it is how well you're ever going to make use of the T and 3He. JT-60 has run DD up to a point that, theoretically, it could've exceeded net if it had been running DT (it's not set up to do so). Yet even when it was running DD so well, it still didn't 'self-generate' enough T to actually make *much* of a difference to the overall neutron output.

So, Dan is probably right if you believe in atmospheric-pressure billion-K reaction volumes devoid of any contamination and driven by multi-amp ion guns, but Frank provides the reality check.

Frank S. wrote:
> I will not be so bold as to say busted

Stop backtracking, Frank! You've already stated a view, stick to your guns! It's for someone to prove you're wrong, not the other way around. Let's see the colour of those Polywell neutrons, then those of us who couldn't foresee Polywell's greatness will all happily backtrack and shall confess "mea maxima culpa".