[jamesr]

Breakable wrote:

If they can build a fusion based neutron source they can easily build a conventional one.

In any case fusion power will enable cheap manufacturing of everything including fission reactors and nuclear bombs…

Not really… Fusion done the ITER or NIF way will be just as expensive as current fission reactors, if not more so. They just assume by the time they get fusion working everything else will have gone up in price.

[jamesr]

I did not see your #3 post above before posting this. In light of your second post, what do you think about the relative linearity we should expect in DPF devices since they do not depend on ‘ignition’ in the same way.

I don’t think either approach will be at all linear in their yield improvements.

Also since Focus Fusion never intends to use D-T using neutron count long term will be pointless (since we want as low a neutron count as possible with pB11). I think it would be better to quote yield as (Fusion energy released)/(Energy In), then later we can add figures for the recoverable portion of that energy, and finally the actually recovered portion.

In terms of fusion energy released I would expect less improvement between the first pB11 shots on FoFu-1 at full operating voltage, (so ignoring early low voltage test shots) and a final optimised pB11 shot in FoFu-1, than I would between NIF D-T shot currently and a D-T ignition at NIF.

NB. in ICF I take ignition as the point of starting a fusion burn wave, not necessarily more energy out than in. Whereas in MCF such as tokamaks, the ignition point implies a self sustaining reaction without the need for any more outside heating (although some may still be present to drive current for stability reasons)

[jamesr]

The NIF June update https://lasers.llnl.gov/newsroom/project_status/2011/june.php has more details on their progress.
From this you can work out figures. They quote, for example 4.5×10^14 neutrons from 1.33MJ of ultraviolet energy to the target, however the lasers are less than 1% efficient so they would have used around 330MJ of electricity.
So 4.5×10^14/330×10^6 = 1.4×10^6 neutrons/input_Joule

In all the recent DT shots they have not yet reached ignition, and only fuse a tiny fraction of the pellet. If they do, the yield will go up by several orders of magnitude.

There is a caveat though: Even if they reach ignition at the centre it is not yet clear whether the alphas produced heat the surrounding compressed plasma enough to create a fusion burn wave progagating out from the ignition point in the centre.
If the compressed plasma is either re-expands faster than the burn wave propagates, or the alphas fly out of the compressed region before depositing enough energy to heat the rest of the fuel to fusion temperatures then the whole thing will have been a bit of a waste.

If the burn wave succeeds in fusing more of the plasma than just the central ignition region then the neutron yield will go up by a few more orders of magnitude.

[jamesr]

Although interesting, the whole neutrons per Joule of input energy is pretty meaningless really for NIF unless they reach ignition at the centre of the pellet and have a burn wave that fuses enough of it before it blows itself apart.

[jamesr]

If you have a plentyful neutron source from D-T fusion then making Pu-239 from U-238 is fairly straightforward
U-238+n -> U-239 ->Np-239 + e -> Pu-239 + e
The beta decays in the reaction chains have half lifes of days, so you have to wait for the products to build up over time, but not too long otherwise they capture more neutrons and end up as something like Pu-240 which would spoil your plans and is then just as hard to separate out as the U-235/238 problem.
Also you need LOTS of neutrons to make a useful amount of Plutonium, which then can’t be used to breed the tritium needed to sustain the fuel supply for the D-T fusion.

Th-232 being and even-even nucleus is only fissionable by fast neutrons (like U-238 only worse), and is regarded as a fertile material to breed U-233 which is fissile.
Th-232+n ->Th-233 ->Pa-233 +e -> U-233 +e

But U-233 cannot easily be made into a bomb, although it has been done a few times. You need around 16kg of pure U-233 vs only 10kg of Pu-239 for critical mass. Pure Th-232 and U-238 cannot ever reach a critical mass on their own (they capture too many neutrons, rather than fissing, so cannot sustain a chain reaction).

Breeding fissle elements via fusion neutron capture, then chemically separating them from the fertile source is easier in some regards than separating U-235 and U-238, but we are a long way from anyone having that capeability. If they can build a fusion based neutron source they can easily build a conventional one.

[jamesr]

Aquariumnerd wrote: does the photoelectric absorption always result in the release of electromagnetic waves of a different wavelength

You have it backwards. The photoelectric effect is the absorption of a photon by an atom, kicking out an electron, not the release of a photon. As such the absorption cross section rises with the number of electrons in the material the x-ray photons are passing through, ie the atomic number of the element. (The electron will just bounce around depositing its energy as heat (lattice vibrations) until it recombines with another atom).

The absorption cross-section is roughly proportional to atomic_number^4/(photon_energy^3). Since Beryillium is only Z=4, it is the lowest atomic number material that is strong enough to make vacuum chamber windows from.
This datasheet shows even Beryllium absorbs most photons below around 1.5keV
http://www.profluxpolarizer.com/PDF/Windows/Duraberyllium.pdf

If the window was normal glass (silicon dioxide) then the aborption would be much higher, and lead glass would absorb almost all x-rays. Just to be complete, the absorption curves of materials also have spikes where if the energy of the photon is just over the threshold to kick out one of the inner electons from an atom the probability of absorption goes up, these are known as the K,L & M edges corressponding to the energy of the first three atomic shells of an atom.

[jamesr]

Aquariumnerd wrote: thanks man. Could the xrays them self possibly used for imaging :/ With a number of sensors containing a photo phosphore and photo-diodes and collimator in-between etc or something :/ physics is all coming back to me :] Would the absorption by the walls be too much to produce a good image like this. Wouldn’t IR be cheaper anyway and better quality. Could someone explain to me, is the emittion of xrays anything like pair production, so i know its caused by the photoelectric effect but is there any way of pinpointing the virtual point at which the xray came from without a sensor all around the device :/ :/ Seems mad but just asking because it sounds possible. I just like to ask these things if u get me.

From http://en.wikipedia.org/wiki/Bremsstrahlung:

Bremsstrahlung (German pronunciation: [ˈbʁɛmsˌʃtʁaːlʊŋ], from bremsen “to brake” and Strahlung “radiation”, i.e. “braking radiation” or “deceleration radiation”) is electromagnetic radiation produced by the deceleration of a charged particle when deflected by another charged particle, typically an electron by an atomic nucleus. The moving particle loses kinetic energy, which is converted into a photon because energy is conserved. The term is also used to refer to the process of producing the radiation. Bremsstrahlung has a continuous spectrum, which becomes more intense and shifts toward higher frequencies when the energy of the accelerated particles is increased.

In a plasma the electrons are constantly scattering through small angles, emitting relatively low energy 10-100eV photons, as the angle of scatter increases more of the energy is transferred, and the photon is in the 10’s keV range.

There is very little radiation in the IR->visible region, since this corresponds to the kind of energies of bond vibrations (IR) and electrons dropping down from excited states in bound atoms (visible). In a fully ionized plasma there is simply no mechanism for emission in this range. The plasma only glows with the characteristic redish purple at the coolest places, where some recombination to neutral hydrogen is taking place (ie. where the temperature is less than a ~10eV).

The X-rays can be used for imaging directly. Since they cannot be focus easily, the best way is with a simple pin-hole camera. That is a metal plate with a small hole in and a piece of photographic film a short distance behind. You can use B/W photographic paper directly leaving it in a sealed envelope so as to not expose it to light, then develop as per normal.

However, only the highest energy x-rays will have a chance of getting through the window, due to photoelectic absorbtion. Unless the window is made from something like Beryillium (not possible in an amateur setup). So, with the window acting as a high pass filter the image formed will only be of the hottest parts of the plasma. The combination of Xray and visible image therefore gives you lots of information about the temperature distribution.

Other X-ray interaction mechanisms only become important at even high energies. 100keV->2MeV for Compton scattering, and >1.02MeV for Pair production can (ie the rest mass of an electron and position)

[jamesr]

How about having a laser inside the chamber that can scan the edge for imperfections, then give high power pulses to blast away any peaks. So you’ll end up eroding it all faster, but the laser system could fire & repolish the surface in-situ, then all you need to do is evacuate the chamber to get rid of the waste.

[jamesr]

The bulk of the radiation would, I think, be bremsstahlung x-rays. So if you’re running at say +/-20kV on the power supply, then electrons accelerated upto 40keV can emmit a photon of upto that energy when they scatter off an ion. Of course this is a lot softer than the other sources if you actually get any nuclear reactions, and so mostly won’t penetrate the chamber walls. But they shouldn’t be ignored as they could give you a serious dose if you were standing in front of the camera windows.

You should build in interlocks – so you can only turn it on from outside the room/behind concrete wall.

[jamesr]

The lead-lithium combination would likely be used in tokamak and inertial confinement schemes. So from this aspect it is the same as all the other big-ugly MCF & ICF designs

The lead is not to protect from the neutrons, as a heavy element its pretty useless at that. It can protect against some gamma and hard x-rays that penetrate the first wall, and result from the various nuclear reactions with the lead itself and the lithium. Its primary use is to breed more neutrons.

To make enough tritium to sustain your fuel supply you need to breed more neutrons to react with the lithium since you will never capture all the high energy neutrons from the D+T reaction. Some you can breed if you have a Beryillium wall, but not normally enough. When a neutron hits a lead nucleus it has a chance of knocking out some more neutrons.

Molten lead makes a good coolant since it is liquid over a large temperature range and does not have to be at high pressure, unlike water used in PWR fission reactors.

[jamesr]

Have a look at these simulations:
http://www.astro.virginia.edu/VITA/ATHENA/implosion.html
http://www.astro.virginia.edu/VITA/ATHENA/blast.html
(click on the right hand images to play the animations)

They are not exactly the situation you describe (given the simulations are in a rectangular box), but they do show how interactions between initially smooth compression waves lead to instabilities and turbulent mixing.

What I think would happen in your scenario is that the slight imperfections in the waves travelling towards and away from the centre would interact and become unstable.
Further, this instability and subsequent mixing would reduce the peak denisty capable to well below fusion requirements.

[jamesr]

OK so maybe I didn’t explain myself very clearly… So here’s another way of looking at it:

If your oscillations in density & temperature grow in amplitude, at some point the radial gradient from a region of high to low density will grow beyond the point where the simple linear equations cease to fit, and non-linear interactions become important.

Generally there are only two classes of instability involving gradients: Rayleigh-Taylor; where the driving force is parallel to the gradient, and Kelvin-Helmholtz; where the driving force is perpendicular to the gradient (anything else is just a mix of the two). So although it is not exactly the same scenario as the classical Rayleigh-Taylor description, characteristics such as equation of the growth rate is the same.

If you want the peak density/temperature to be in the range for fusion then the gradient away from that peak must be very steep – too steep to be stable.

[jamesr]

Hi, and welcome to the forum!

Its good to think about novel ways of achieving the Lawson criteria. The trouble is the density, temperature & confinement time product is a very large number. In a spherical compressive model as you propose, the slight asymmetry would introduce Rayleigh-Talyor like instabilities way before the compressive wave could reach the centre. So peak density & temperature would be far lower than needed.

If it were that simple, the spherical imposive method of detonating plutonium bombs, would also work for fusion. To ignite a fusion burn wave at the centre of a spherically compressed volume you need the compression, heating and ignition to occur faster than any of the instabilities have time to develop. The instability grows exponentially in time, but is proportional to the size of the initial perturbation. Or in other words your large sphere being driven at the outside would have generate a spherical wavefront perfect to less than one atom width.

By increasing the size of the sphere to increase the ratio of driving oscillation to proposed peak oscillation in the centre, the wave takes longer to reach the centre and so reduces the tollerence on the sphereical accuracy. The exponential growth of the instablilty will always be faster then the cubic growth in geometric scaling.

On the other hand if by some miracle you could get it to work – having that much fusionable fuel in a sphere would make a very big bomb. The few mm size of the pellets, in a 10m radius vacuum chamber used in devices like NIF to a large extent due to the limited load the walls could take when it detonates. If as in your case the rest of the volume was full of ‘fuel’ then you would blow up the city!

[jamesr]

So does anyone have any more details? I would think that any laser based system would be far heavier, and less efficient than a DPF based thruster.

[jamesr]

Interesting. I guess it kind of relates to the Parker Spiral, formed by the interaction of the Sun’s magnetic field with its solar wind, and the Sun’s observed surface rotation.

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