# Nuclear fusion using plasma Part 2

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1. Sep 16, 2009

### timbot

I seem to be having a problem adding a reply to the end of the previous thread so I am starting a new one!

Thanks for everyone replying.

A couple of points. The first is that nobody seems to have tried two end-on plasmas directed at each other. A dense pair of plasmas fired at each other will make a hit 90% of the time. I have experimented with two welding flames directed at each other. (Dangerous and I don't recommend it). Very little of the flame gets through. Most 'splashes' sideways, that is most particles collide with one another.

The second point is that the individual average velocities required for these plasma flames to achieve fusion appears to be relatively low. (See my previous thread). The velocities can be steadily raised until a useful level of fusion is achieved. And by 'useful', I mean something above the minute and ephemeral - something continuous that can be used to generate large-scale electric power.

On the question of magnetic acceleration, doesn't a 'stripped' moving plasma have a magnetic field? If the surrounding magnetic field is intense enough, won't it hold the plasma? In fact, isn't this how a Tokamac works? Couldn't a straight-line linear magnetic field confine the plasma for the brief time required? If the plasma were held in accelerating 'pockets', wouldn't the plasma accelerate along with the pockets, until they are at a sufficiently high velocity to generate fusion?

I have designed such a magnetic plasma accelerator. I think continuous plasma fusion is technically feasible and commercially practical.

As for extracting the energy required, I suggest that a stream of gas, say helium, or water vapour, be directed at the point of impact at right angles to the plasma stream to generate super-heated gas for electricity generation. The deuterium can be extracted from the waste gas after the electric power has been generated.

2. Sep 17, 2009

### Staff: Mentor

Colliding beams have been considered, and no - it is not as simple as opposing flames.

Actually the process is quite complicated.

Um - no. Fusion takes place in what is essentially a vacuum with particle densities around 1014 particles/ cm3 is about 10-8 of the normal density of solids ~1022.

One has to compare the total process cross-section with that of cross-section of process that results in fusion. Most of the time, the nuclei are scattered, and a monoenergetic beam will scatter toward a Maxwellian distribution - with a low mean energy than the beam energy.

Based on the previous quote, I am skeptical.

Neutral beam injectors and their physics is well established.

3. Sep 17, 2009

### theCandyman

Please back up your claims with more than saying you did an experiment. How did you gather data for your experiment, how was it performed, ect?

You won't get a large amount of fusion by allowing two plasmas to pass each other in opposing directions. With a small cross section for two ions to fuse, you need to be able to provide MANY MANY chances for an ion to interact, a once through pass with another plasma isn't enough.

Plasma heating is mainly achieved by passing a current through it and using microwaves. One more is neutral particle injection, which is basically what you are suggesting - energizing particles with magnetic fields. I don't think it is used for much heating since it requires more energy, but it is a last little bit that can heat the plasma further when the other methods cannot.

4. Oct 8, 2009

5. Oct 9, 2009

### Staff: Mentor

Um, one needs to do a calculation of the pressure in the center of the sun, and then compare that to the pressure acheiveable by the strongest steady-state magnetic field we can produce AND the strongest structural material we can produce.

Don't even think about proton-proton fusion. At the least one should focus on d+d, or better yet p+11B.

theCandyman raises a critical issue - much of the energy of a plasma beam will be scattered before a fusion reaction is accomplished. One needs to compare the scattering cross-section with that of the fusion cross-section for the conditions one is proposing.

6. Oct 9, 2009

### QuantumPion

Furthermore the power density of the sun isn't particularly large. I thought it was something on the order of 100 W/m^3. The reason the sun makes so much energy is because it has a huge volume. To make commercial fusion viable, we need much higher power density.

7. Oct 9, 2009

### mheslep

http://fusedweb.llnl.gov/CPEP/Chart_Pages/5.Plasmas/SunLayers.html" [Broken] which produce energy from fusion at the rate of ~1 x 10^43 W for a few seconds. That's typically from a volume roughly the size of the earth, 1 x 10^21 m^3, or 1 x 10^22 W/m^3 which will run a few toaster ovens. Unfortunately this type of reactor also quickly unbinds itself.

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8. Oct 9, 2009

### Staff: Mentor

Yes. At ~15 million K (or 1.3 keV) the sun's core is rather 'cold' compared to what we need in fusion plasmas.
Ref: http://fusedweb.llnl.gov/CPEP/Chart_Pages/5.Plasmas/Sunlayers.html [Broken]

I was focusing on the pressure, P = nkT, where n is the particle density, k is Boltzmann's constant, and T is the plasma temperature, and then look at limiting pressure of the strongest materials. One simply does not make a laboratory plasma with the density of the sun. And one can simply forget p-p fusion in a terrestrial power plant.

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9. Oct 10, 2009

### Orion1

Lawson criterion and triple product...

The Lawson criterion is an important general measure of a system that defines the conditions needed for a fusion reactor to reach ignition, that is, that the heating of the plasma by the products of the fusion reactions is sufficient to maintain the temperature of the plasma against all losses without external power input.

Lawson criterion:
$$n_{\rm e} \tau_{\rm E} \ge L \equiv \frac{12}{E_{\rm ch}}\,\frac{k_{\rm B}T}{\langle\sigma v\rangle}$$

For the D-T reaction, the physical value is at least:
$$n_{\rm e} \tau_{\rm E} \ge 1.5 \cdot 10^{20} \; {\rm s}/\mbox{m}^3$$

In the case of the D-T reaction, the charged fusion products energy is $$E_{ch} = 3.5 \; \mbox{MeV}$$.

The minimum of the product occurs near $$T = 25 \; \mbox{keV}$$.

A still more useful figure of merit is the "triple product" of density, temperature, and confinement time, $$n_{\rm e} T \tau_{\rm E}$$. For most confinement concepts, whether inertial, mirror, or toroidal confinement, the density and temperature can be varied over a fairly wide range, but the maximum_pressure attainable is a constant.

Triple product:
$$n_{\rm e} T \tau_{\rm E} \ge L T \equiv \frac{12k_{\rm B}}{E_{\rm ch}}\,\frac{T^2}{\langle\sigma v\rangle}$$

For the D-T reaction, the physical value is about:
$$n_{\rm e} T \tau_E \ge 3.75 \cdot 10^{21} \; \mbox{keV s}/\mbox{m}^3$$

This number has not yet been achieved in any reactor, although the latest generations of machines have come close. For instance, the TFTR has achieved the densities and energy lifetimes needed to achieve the Lawson criterion at the temperatures it can create, but it cannot create those temperatures at the same time. ITER aims to attain both objectives.

ITER is expected to confine plasma discharges for $$\tau_E = 500 \; \mbox{s}$$ seconds, at a density of $$n_{\rm e} = 10^{20} \; \mbox{m}^{-3}$$, and a temperature of $$T = 20 \; \mbox{keV}$$, which is about 17 times the solar core temperature at $$T_c = 1.171 \; \mbox{keV}$$. It is anticipated that ITER will produce about ten times more fusion power than input power. As of June 17, 2009, the total price of constructing the experiment is expected to be in excess of €10 billion. The first plasma operation is projected to be in 2017.

ITER triple product criterion:
$$LT \leq n_{\rm e} T \tau_E \leq 10^{24} \; \mbox{keV s}/\mbox{m}^3$$

Reference:
http://en.wikipedia.org/wiki/Lawson_criterion" [Broken]
http://en.wikipedia.org/wiki/ITER" [Broken]
http://en.wikipedia.org/wiki/Sun#Core"
Computer simulations of the ITER fusion reactor - Federico D. Halpern

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10. Oct 11, 2009

### theCandyman

For inertial confinement, I think the pressures can be quite high. The NIF can achieve more than 40 Mbars - but the targets are destroyed in the process.

Another note, there have also been some propositions to make ITER smaller than the initial design.

11. Oct 11, 2009

### mheslep

After the quench and explosion http://news.bbc.co.uk/2/hi/science/nature/7626944.stm" [Broken]with the LHC last month, I wonder what are the consequences of a quench in the gigantic super conducting magnets planned for ITER. Is there a design feature of some kind that prevents all of that LI^2 energy from dissipating catastrophically?

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12. Oct 11, 2009

### Staff: Mentor

I believe there is a quench detection system and mitigation strategy, but I can't readily find the details. There is some mention of magnetic quench herein -
http://www.fec2008.ch/preprints/it_1-3.pdf [Broken]

It may involve a low amplitude AC current superimposed on a much larger DC current. The change in AC would signal a change in the magnet's resistivity.

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13. Feb 5, 2010

### maxmc1027

Tell you guys what. I have a vacuum pump, and a clear tube already made to these specifications, I have a few videos up on youtube and I was researching this option before i saw it. The type of magnets i would be using would be a toroid shape with the poles inside of the metal(no poles exposed, meaning the electric field will actually have a "pinching" effect on the ions, or electrons, depending on polarity and direction.) Two things I AM lacking, are a power supply (probably two neon signs, 12kv @ 60ma rectified). The theory is to accelerate two beams of ionized deuterium, confined by the special electric field just before collision and see what happens. I'm not expecting practical fusion energy, but the design would be very similar to a 1.5 meter diameter Farnsworth fusor(in some sort of tube version). Would this help out at all if I made a video?

14. Feb 9, 2010

### mheslep

There's amateur fusor forum over here:
http://www.fusor.net/
Folks there have been building them for years.