I just created a video for my friends at fusor.net, talking about the fusion concept I am working on.
It relates to a specific type of IEC, and might be of interest to some of you.
While I applaud your enthusiasm, I’m not sure you really get how fusion works. You mention half way through (08:00) that you don’t consider the ion collisions as the cause of fusion! It is all about this!
ions need to be travelling fast enough that, if they happen to be on a collision path, that their kinetic energy will be able to overcome the electrostatic repulsion of their positive charges until they are close enough for the strong force to take over (or at least close enough for them to have a quantum probability of tunnelling through the remaining potential barrier)
If you are to get an appreciable number of fusion reactions then you need the plasma dense enough. If it is this dense (at the bottom of the potential well) then since most of the collisions will be elastic scattering at small angles, the plasma will begin to ‘thermalise’.
Of all the scattering collisions, some ions will end up with more energy & some with less. The ones with more can escape the potential well, some will then hit the grid or walls of the chamber, leaving just the ‘cool’ ones trapped
You can keep it hot by supplying ‘even hotter’ ions at the 100keV or so your power supply can deliver but the energy needed to do this will always be huge in comparison to the fusion rate.
Do you have any estimated figures of density & velocity distribution vs radial position?
jamesr wrote: While I applaud your enthusiasm, I’m not sure you really get how fusion works. You mention half way through (08:00) that you don’t consider the ion collisions as the cause of fusion! It is all about this!
James, thanks for your reply and valuable comments. I probably did not make myself clear in the video (it was a single take, one shot unedited recording), I do not disagree that particle collision is necessary to initiate fusion, what I do believe is that ionizing particles below ground potential, lowers the relative Coulomb barrier.
You point to a weakness in my design that I am also aware of, the issue of keeping the plasma lit, and hot enough to fuse. The Lawson criterion still applies, and enough particles need to be confined for long enough at a high enough temperatures to fuse, with the only difference in my thinking being that the numerical value of the criterion is lower when the plasma has been ionized at a negative potential.
If my first experiment fails, it will be because the plasma is too cold, and I will need to look at other ways to keep the plasma lit. Clearly if enough fusion reactions can be initiated, the plasma will be self heating, the question is at what point.
If anyone wants to help me with the numerical modelling I would be more than appreciative.
Remember water runs down hill, all we can do is tilt the hill a bit …
I’m a bit confused by the video and it seems like you are glossing over a few important points in the physics. You mention alpha particles being created in the fusor; are you using D-T fuel or a p-11B fuel? D-D fuel cannot produce 4He in a low pressure environment due to momentum conservations considerations. If you are referring to D-T or p-11B, why does the alpha particle only have 100 keV? Shouldn’t any particles created in the fusion event have their fraction of the change in binding energy (14.1 MeV in D-T) plus the potential difference of 100 keV. In that case, the 100 keV seems pretty small compared to the fusion yield. Could you clarify please?
asymmetric_implosion wrote: I’m a bit confused by the video and it seems like you are glossing over a few important points in the physics. You mention alpha particles being created in the fusor; are you using D-T fuel or a p-11B fuel? D-D fuel cannot produce 4He in a low pressure environment due to momentum conservations considerations. If you are referring to D-T or p-11B, why does the alpha particle only have 100 keV? Shouldn’t any particles created in the fusion event have their fraction of the change in binding energy (14.1 MeV in D-T) plus the potential difference of 100 keV. In that case, the 100 keV seems pretty small compared to the fusion yield. Could you clarify please?
Thank you for pointing this out, I was talking without a script, and could have used better terminology.
My experiment will be done with Deuterium fuel only so we are discussing D+D fusion, and the reactions as we know are D+D -> T + p and D+D -> 3He + n, in my talk I was referring to the positively charged fusion products, ie. the proton and the 3He nucleus.
If my predictions hold true, the fusing molecules will automatically orientate themselves so that the positive particle/nucleus (p or 3He) can go to ground via the flight tube, ie. up the potential energy slope. You can imagine the potential energy well is in the order of 100 Kev, while the fusion products have energies in the order of Mev, so they will have no trouble escaping.
A proton or an He3 nucleus with a charge of +1 climbing through a potential of 100 Kv liberates 100 Kev of power, so at 10^14 fusion reactions per second , you get !00,000 ev * 10^14 – 10^19 ev = 1.6 watt seconds. The fusion products are capable of climbing through a higher potential than that. The output is therefore limited only by our ability to insulate the cathode from ground.
I would go so far as to say that fusion is the only escape route for these trapped deuterons, and providing the temperature of the plasma can be maintained high enough, this escape route is the only way out, and nature will find it.
Steven Sesselmann wrote: what I do believe is that ionizing particles below ground potential, lowers the relative Coulomb barrier.
In a word… No.
All the potential does is create a E-field to accelerate the ions & electrons. At the point of closest approach any two ions in question will be right on top of each other so the the potential difference between them will be near enough zero. The E-field creating the ‘Coulomb barrier’ from their own positive charge is many orders of magnitude larger.
Lets do a few quick sums.
The deuterium cross-section at 50keV is roughly 0.02barns = 2E-30 m^2 This probability of fusion reaction corresponds to the area swept out by one ion as it approaches the other
take the square root of this to get the rough distance (to an order of magnitude) = 10^-15m – We need them to get this close to fuse (NB this is smaller than what is sometimes thought as the geometric size of a nucleus which corresponds to its scattering cross-section)
The electric field created by one ion at the position of the other at this distance is E=1/(4pi*eps0) *q/r^2 ~10^20 V/m
So in any case a few hundred thousand V/m Electric field created externally is insignificant to the field the ions must work against as they approach each other.
what I do believe is that ionizing particles below ground potential, lowers the relative Coulomb barrier.
In a word… No.
The electric field created by one ion at the position of the other at this distance is E=1/(4pi*eps0) *q/r^2 ~10^20 V/m
James, before you throw the baby out with the bathwater, stop for a minute and consider the observers potential.
The equation above is valid for particles generated at ground potential, ie, the experimenter and the particles origin is the same.
Not so when the ions are generated below ground potential, in the case of my FICS fusion reactor, the plasma holds a negative potential with respect to the observer, and I think one needs to multiply the equation above by 10^-5, when observing the event from ground potential.
I don’t mean to put a complete dampener on your plans, I just wanted to highlight the folly of your concept.
It really doesn’t matter where they were ionized. Once the ions have been oscillating back & forth in the well millions of times, scattering off all the others & mixing their energies & velocities, their origin is meaningless. Plasmas need to be understood by their statistical distributions and how they vary in space.
The potential is a macroscopic scalar quantity found by solving Poisson’s equation (ie integrating the charge density over the entire volume, with appropriate boundary conditions, eg the walls of the chamber are set to 0). It is a function of position, not of any particular particle.
The gradient of the potential at each point is then the local electric field.
In a plasma, by definition, macroscopic properties like this should only be used on scales larger than the Debye length: http://en.wikipedia.org/wiki/Debye_length
Below the Debye length the closest few dozen ions & electrons have to be accounted for individually rather than just considering the bulk, statistical properties like temperature. But this is still on the scale of nanometres to micrometers.
Fusion happens on a much smaller scale still. As I said before, as two ions become close enough to fuse, they will be essentially be at the same point in space. So they will be at the same potential. The only significant influence on one ion is the other ion.
jamesr wrote: The only significant influence on one ion is the other ion.
I agree with most of your post, just this last part that I would like to single out…
In my hypothesis the above quoted statement is false, there is a third factor which can not be ignored, namely the observer. A particles preferential potential is a function of its mass, and as such, the preferred potential of a deuteron is above ground potential. We know this, because if you drop a deuteron it does not fall down….it floats up!
Therefore by dragging it below ground potential we are taking it further away from its preferential potential, the deuteron will in my opinion do anything it can, to get back to where it belongs, and as sure as water runs down hill, nature will find a way, and we know that the only way out is fusion.
This is the equivalent of putting the deuteron (king) in check, and the only way out for the deuteron (king) is fusing (towering).
An experiment that might interest you and clear up some of this discussion. In 2007 a paper was published from Sandia used deuterium fuel gas in the Z-machine. The fuel gas was stored at the cathode potential (-4 MV, much closer to the peak in D-D fusion cross section than 100 keV.). This is common practice in Z-pinch machines. No increase in radiation yield is observed as your hypothesis suggests. Similar experiments are conducted on plasma focus devices without any significant increase in radiation yield. You can argue that the pinch devices are fast pulse machines and the fusor is a DC device, but the time to bring the fuel gas to the local potential is a fraction of a nanosecond while the pulse duration is >100 ns.
I wish you luck in your experiment but based upon the existing data you will not see an increase in fusion yield over storing the fuel at anode potential or some other potential.
Have you come across this. http://www.oup.co.uk/pdf/0-19-856264-0.pdf it is a useful excerpt that goes over the basics.
From this you will see on page 5 that the potential barrier is of the order of 1MeV the barrier is measured as a function of distance, r.
What you claim, if I am understanding you correctly, is that by lowering the potential at large r, that you shift the whole graph down and so lower the height of the barrier.
What I’m saying is that if you lower one, you lower both. It doesn’t matter what reference level you pick for what the potential is at large r, it is the same for both, the height they need to climb is the same.
If your theory was true, you could setup an experiment with the potential lower than 1MV with respect to your ‘observer’ and the barrier would be completely shifted out of the way, and make fusion trivial. This is clearly nonsense.
Not a simple 1:1 relationship, ionizing the atom at 10^-5 volts potential might lower the coulomb barrier by a factor as follows…
The mass energy potential of +1 hydrogen ion at ground potential, less the negative potential divided by the total mass energy at ground potential.
So I think the coulomb barrier might be lower by a factor of ((10^9) – (10^5)) / (10^9) = 0.9999
Not much in the scheme of things, but maybe enough to see an improvement in the fusion rate.
Some images from the construction of my F.I.C.S. fusion invention.
Before asking questions read the discussion of fusor.net here http://www.fusor.net/board/view.php?bn=fusor_theory&key=1326524523
All right, I’ve read the thread on fusor.net, watched the video, and read/skimmed the patent application. I look forward to the results of your experiment. I don’t consider myself qualified to offer a critique on your invention, but I do have a question: According to your patent application, the preferred embodiment of your invention uses deuterium fuel. Is there any particular reason deuterium is preferred, apart from the relative ease of fusing deuterium as opposed to heavier elements?
Ivy Matt wrote: Is there any particular reason deuterium is preferred, apart from the relative ease of fusing deuterium as opposed to heavier elements?
Thanks for your question, Apart from the availability and the known ease of fusing Deuterium, no, there is no particular reason, the other fuels have also been mentioned in the patent. Clearly a fuel having a positively charged fusion product is preferred, as those reactions contribute in the charge separation process.
I think my idea of using the kinetic energy of the fusion ash to separate and amplify a charge through an electrical potential is novel, and if the process works, it will offer another way to convert fusion energy into electricity directly.
The p + B11 reaction is troublesome, due to it’s small cross section, and so far as I am unaware, no amateur scientist has achieved p+B11 fusion in any measurable way or form, although I do have a friend who has been trying for the last 10 years. Who knows, one day he might succeed.