# Whats the holdup with Fusion Power?

#### theCandyman

But that's not the worst case scenario, I doubt it even close. It's taken thousands of man hours worth of work to try and get ignition, so it's more than obvious it won't happen spontaneously.

#### GleefulNihilism

I'm an astrophysicist. When someone talks to me about fusion the poet in me always pictures a tiny star in a lab somewhere, . . .

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#### Coin

http://www.theoildrum.com/node/2164 [Broken] It is kind of long but it is worth the effort. Basically there are several very serious engineering hurdles to making a fusion reactor that can be used to actually produce power. Even once you can sustain a plasma, some of the parts involved in actually getting energy out of that plasma present multi-decade engineering challenges all by themselves! There is also the problem that operating a fusion reactor consumes some unusual substances like tritium, so you have to engineer your reactor to for example create more tritium as it goes... there's a timetable they expect to resolve all these issues on, but it is not trivial. Worth a look...

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#### Astronuc

Staff Emeritus
http://www.theoildrum.com/node/2164 [Broken] It is kind of long but it is worth the effort. Basically there are several very serious engineering hurdles to making a fusion reactor that can be used to actually produce power. Even once you can sustain a plasma, some of the parts involved in actually getting energy out of that plasma present multi-decade engineering challenges all by themselves! There is also the problem that operating a fusion reactor consumes some unusual substances like tritium, so you have to engineer your reactor to for example create more tritium as it goes... there's a timetable they expect to resolve all these issues on, but it is not trivial. Worth a look...
Thanks, Coin, that's a good article. Nice little summary of power input into ITER.

Power will be feed into the ITER plasma in three main ways: by transformer action causing up to 15 million amps to flow in the plasma; by neutral high energy beams of deuterium and tritium fired into the plasma; and by radio frequency energy fed in from antenna patches in the walls to excite resonances in the plasma, Transformer action is very efficient but necessarily pulsed. The other two forms of heating are less efficient but can be continuous. ITER is expected to generate 500MW of fusion energy output, with less than a tenth of that input power (Q>10) and hold that power for 400 seconds. Also it should generate 500MW output for an hour at an input of one fifth the input energy (Q>5). Although it is not stated as an aim, there is the hope that it might achieve what is called ignition where enough of the fusion energy remains in the plasma to keep the reaction going without the need of external input energy (Q = infinity). This will require higher plasma densities than needed with external energy input.
It would be desirable to have a continually operating plant. The power generation cycle is critical for a viable system, at least in todays environment.

Interestingly, it seems primarily based on DT reaction. The blankets and tritium generation/processing will add to the difficulties.

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#### gdp

Confinement time ala Larsen would apply to confinement approaches, inertial or magnetic. Confinement time does not seem to apply to any of the several beam - beam approaches (e.g. http://en.wikipedia.org/wiki/Inertial_electrostatic_confinement" [Broken]). That is, there's no intention to do ignition; they are purely 'driven' schemes. - Not that IEC has shown any possibility of power production
Todd H. Rider investigated such systems from a very generic (e.g., Kinetic Theory and 2nd Law) viewpoint in his Ph. D. Thesis in Nuclear Engineering, http://dspace.mit.edu/handle/1721.1/11412" [Broken]

His basic (and quite depressing) conclusions are as follows:

1.) "Nonthermal" or "nonequilibrium" plasmas (e.g. IEC, or "colliding beam" reactors such as "Migma") relax to "thermal" plasmas at a much faster rate than they undergo fusion reactions --- and the denser the plasma or beam is, the faster it relaxes. (This fact has been obscured in most IEC experiments because so far, the particles have been lost to the grids even faster than the thermalization timescale.)

2.) The 2nd Law of Thermodynamics implies that any attempt to maintain the beam or plasma in a "nonequilibrium" state will cost power --- and the further from equilibrium the system is maintained, the more power it will cost. Rider shows that under very general considerations, the additional power-gain from using a nonthermal plasma is always less than the power required to maintain the nonequilibrium state.

3.) Furthermore, from an economic standpoint, any generating system that has to "recycle" a large fraction of its generated power just to keep itself running is a money-loser. (This is already a problem for "conventional" fusion reactors unless "ignition" can be achieved --- in which case the power will be "recycled" internally to the reactor, rather than externally (i.e., by running it back from the power conversion systems to the beam-drive or plasma heating systems, with inevitable losses and inefficiencies along the way).

4.) Finally, the increased bremmstrahlung losses faced by "advanced" fuels make "breakeven" highly unlikely for nearly all of them. The only reactions that are likely to "break even" appear to be D+T, D+D, and D+He3; for all other known reactions, http://en.wikipedia.org/wiki/Nuclear_fusion#Bremsstrahlung_losses_in_quasineutral.2C_isotropic_plasmas"

Furthermore, even D+He3 looks rather marginal, and pure D+D looks even worse. Despite D+D having a lower bremmstrahlung loss rate than D+He3 (bremmstrahlung loss rates scale as the mean of the squares of the nuclear charges), pure D+D has an even lower "gain" than D+He3, so that the ratio of bremmstrahlung loss rate to fusion power for D+D is much worse than D+He3, which in turn is much, much worse than for D+T. Under even the most optimist possible assumptions --- that power can be "recycled" at 100% efficiency (impossible!), and that the only loss mechanism is bremmstrahlung --- a "non-ignited" D+He3 reactor would have to "recycle" nearly 20% of its total generated power to make up for bremmstrahlung losses, and a pure D+D reactor over 1/3 of its total power, whereas a D+T reactor would only need to "recycle" a mere 0.75% of its total power. Factor in reasonable estimates of power-conversion, transfer, and beam-drive or heating efficiencies, and one finds that, even ignoring all other losses except bremmstrahlung and "power handling," absent "ignition," only D+T is likely to "break even." The loss rates for D+D and D+He3 are simply far too high to be practical --- let alone economically viable.

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#### Count Iblis

http://news.bbc.co.uk/2/hi/science/nature/3336701.stm" [Broken]

ITER was delayed because the participants couldn't agree on where to build it. When most were in favor of building it in France, the Iraq war started and the US didn't like that idea anymore:

The US has been against the French option because of France's opposition to the US-led invasion of Iraq.
What a way to make decisions on science and technology :yuck:

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#### ZapperZ

Staff Emeritus
2018 Award
http://news.bbc.co.uk/2/hi/science/nature/3336701.stm" [Broken]

ITER was delayed because the participants couldn't agree on where to build it. When most were in favor of building it in France, the Iraq war started and the US didn't like that idea anymore:

What a way to make decisions on science and technology :yuck:
But this isn't a decision on "science and technology". They are not designating some scientific/technological result to be valid or not. Politics has always played a role in choosing a site from a selected list of candidates. No one isn't aware of such a thing.

Zz.

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#### Astronuc

Staff Emeritus
http://news.bbc.co.uk/2/hi/science/nature/3336701.stm" [Broken]

ITER was delayed because the participants couldn't agree on where to build it. When most were in favor of building it in France, the Iraq war started and the US didn't like that idea anymore:

What a way to make decisions on science and technology :yuck:
Come up with 10 billion euros, and one can put a fusion reactor anywhere one likes.

ITER construction costs are estimated at 4.57B€ (at 2000 prices), to be spread over about ten years. Estimated total operating costs over the expected operational lifetime of about twenty years are of a similar order.
http://europa.eu/rapid/pressReleasesAction.do?reference=MEMO/05/226

Science 13 June 2008:
http://www.sciencemag.org/cgi/content/summary/320/5882/1405
This month, funders of the €10 billion ITER fusion project, which seeks to demonstrate that a burning plasma can be controlled to produce useful energy, face the daunting task of keeping the project's budget under control, as scientists present a wish list of design changes.

Look what happened to the Superconducting Supercollider in Waxahachie, TX. They dug a $2 billion hole. During the design and the first construction stage, a heated debate ensued about the high cost of the project. In 1987, Congress was told the project could be completed for$4.4 billion, but by 1993 the cost projection exceeded \$12 billion.
http://en.wikipedia.org/wiki/Superconducting_Super_Collider#Cancellation

Detailed design and early construction work was proceeding on all major machine components. "The conventional construction for the first stage of the injection complex, consisting of the ion source and a linear accelerator stationed in a 250-meter tunnel, was complete." The first circular accelerator in the chain, the Low Energy Booster (LEB), consisting of a 600-meter circumference ring filled with resistive magnets, was designed and 90% of the tunnel complete. The next element in the sequence, the Medium Energy Booster (MEB), consisting of a ring of 4.0 kilometers in circumference, again using resistive magnet technology, was designed and excavation of the tunnel had started. The third and final accelerator before entering the large collider rings, the High Energy Booster (HEB), consisting of 10.8 kilometer circumference tunnel filled with superconducting magnets, was under design. Finally, for the 87.1 kilometer circumference collider ring, the excavation of seventeen shafts was complete, and the tunnel boring, begun in January 1993, had proceeded rapidly, with 77,065 feet (roughly 23 kilometers) completed by fall 1993.
http://www.hep.net/ssc/new/history/appendixa.html [Broken]

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#### mheslep

Gold Member
Todd H. Rider investigated such systems from a very generic (e.g., Kinetic Theory and 2nd Law) viewpoint in his Ph. D. Thesis in Nuclear Engineering, http://dspace.mit.edu/handle/1721.1/11412" [Broken]

His basic (and quite depressing) conclusions are as follows:

...
I find Rider helpful. He shows why some of the IEC and other ideas, as envisioned at the time must fail, and, like any good work, shows you where not to waste your time and the obstacles that must be overcome. He does not shut the door on everything fusion; towards the back of that thesis there are some work-around suggestions and a good quote from Mark Twain about the perils of 'knowing absolutely' that a problem can never be solved.

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#### beeresearch

I find Rider helpful. He shows why some of the IEC and other ideas, as envisioned at the time must fail,...
The Rider paper was written some time ago, and makes assumtions that may not apply to all fusion devices now.

Combinations of confinement metods may solve the problem, the Bussard polywell device uses magnetic and electrostatic, and my own S.T.A.R. reactor uses physical and electrostatic.

There are also a number of other inventions in the pipeline, which have not been published yet.

If we had blind faith in the Rider paper, we would have closed up shop long time ago.

Steven Sesselmann

#### mheslep

Gold Member
...If we had blind faith in the Rider paper, we would have closed up shop long time ago....
No, per the point I made above, even if one had 'blind faith' in Rider, there's no need to close up shop. Rider doesn't claim to close all doors, he even makes suggestions for alternatives.

#### mheslep

Gold Member
Monday the WSJ ran a front page interest piece on the the group of amateurs that construct basement/garage made fusion reactors based the Hirsch/Farnsworth inertial electrostatic confinement concept.
http://online.wsj.com/article/SB121901740078248225-email.html [Broken]

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#### beeresearch

Monday the WSJ ran a front page interest piece on the the group of amateurs that construct basement/garage made fusion reactors based the Hirsch/Farnsworth inertial electrostatic confinement concept.
http://online.wsj.com/article/SB121901740078248225-email.html [Broken]
Yes, it was a good article, and has generated quite a bit of interest in amateur fusion. I belong to this group and I know all these guys well, they are really dedicated and know everything there is to know about fusion.

Here is a link to a short video of my latest fusion reactor http://www.youtube.com/watch?v=XdEE0ry7Mxc"

Steven

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#### chayced

Thanks Steven, I've now found my new garage project. Just need to find a use for all those neutrons so I can convince my wife.

#### gdp

Monday the WSJ ran a front page interest piece on the the group of amateurs that construct basement/garage made fusion reactors based the Hirsch/Farnsworth inertial electrostatic confinement concept.
http://online.wsj.com/article/SB121901740078248225-email.html [Broken]
No one has ever claimed that the "Farnsworth Fusor" was hard to make.

The problem with the FF is that it consumes several BILLIONS of times more power than it produces --- and there are very strong arguments based on fundamental physical laws, and in particular the 2nd Law of Thermodynamics, that the FF will NEVER be able to "break even." It is a cheap, portable neutron source --- but it will NEVER be a net producer of energy.

You might as well try to heat your house by burning sand.

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#### mheslep

Gold Member
No one has ever claimed that the "Farnsworth Fusor" was hard to make.

The problem with the FF is that it consumes several BILLIONS of times more power than it produces --- and there are very strong arguments based on fundamental physical laws, and in particular the 2nd Law of Thermodynamics, that the FF will NEVER be able to "break even." It is a cheap, portable neutron source --- but it will NEVER be a net producer of energy.

You might as well try to heat your house by burning sand.
I don't believe its a 2nd law limitation. Grid impacts are the main loss IIRC.

#### gdp

I don't believe its a 2nd law limitation. Grid impacts are the main loss IIRC.
No, grid losses are merely the current dominant loss mechanism. But even if grid-losses could be impossibly reduced to zero, the 2nd Law still kills you two ways:

1.) The coulomb collision rate is many times larger than the fusion reaction rate. Therefore, the plasma will thermalize before it reaches breakeven density. By the 2nd Law, to maintain a nonequilibrium particle distribution costs power --- and from the 2nd Law, Rider has shown that under very general conditions, attempting to maintain the plasma out of equilibrium costs more power than one gains from the nonequilibrium distribution.

2.) Even if all other loss mechanisms were reduced to zero, the plasma will still emit Bremsstrahlung Radiation (X-rays from collisions between the ions and the electrons). Rider has show that for every fuel combination except D+T and possible D+D, bremsstrahlung losses will greatly exceed the fusion power. Therefore, unless one can find a magic way to convert X-rays into energy and recycle that energy with near-100% efficiency (which is forbidden by the 2nd Law!), the reactor cannot produce more power than it consumes.

#### mheslep

Gold Member
No, grid losses are merely the current dominant loss mechanism. But even if grid-losses could be impossibly reduced to zero, the 2nd Law still kills you two ways:
Ok, but that would be _one_ way, not two. X-rays from the collisions are not dependent on the system entropy.

gdp said:
1.) The coulomb collision rate is many times larger than the fusion reaction rate. Therefore, the plasma will thermalize before it reaches breakeven density. By the 2nd Law, to maintain a nonequilibrium particle distribution costs power --- and from the 2nd Law, Rider has shown that under very general conditions, attempting to maintain the plasma out of equilibrium costs more power than one gains from the nonequilibrium distribution.

2.) Even if all other loss mechanisms were reduced to zero, the plasma will still emit Bremsstrahlung Radiation (X-rays from collisions between the ions and the electrons). Rider has show that for every fuel combination except D+T and possible D+D, bremsstrahlung losses will greatly exceed the fusion power. Therefore, unless one can find a magic way to convert X-rays into energy and recycle that energy with near-100% efficiency (which is forbidden by the 2nd Law!), the reactor cannot produce more power than it consumes.
Nebel, formerly of Los Alamos, recently on an IEC machine w/ no grid:
Nebel said:
...1. The theory says that you can beat Bremstrahlung, but it's a challenge. The key is to keep the Boron concentration low compared the proton concentration so Z isn’t too bad. You pay for it in power density, but there is an optimum which works. You also gain because the electron energies are low in the high density regions.
...
4. The machine does not use a bi-modal velocity distribution. We have looked at two-stream in detail, and it is not an issue for this machine. The most definitive treatise on the ions is : L. Chacon, G. H. Miley, D. C. Barnes, D. A. Knoll, Phys. Plasmas 7, 4547 (2000) which concluded partially relaxed ion distributions work just fine. Furthermore, the Polywell doesn’t even require ion convergence to work (unlike most other electrostatic devices). It helps, but it isn’t a requirement.
The title of that paper: Energy gain calculations in Penning fusion systems using a bounce-averaged Fokker-Planck model. That is, after Rider, Chacon et al went through the very difficult Fokker-Planck path and found one doesn't necessarily need a mono energetic distribution to get net gain. Nebel is referring to B-P fusion here.
Here's the recent, complete conversation. Most of it is about the feasibility of magnetic confinement of the electrons which form the electrostatic well in their particular design, but there's some discussion of Bremm. and equlibrium issues as well.
http://www.newmars.com/forums/viewtopic.php?f=29&t=5395&st=0&sk=t&sd=a&start=20 [Broken]

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#### beeresearch

Your extracts are correct, the simple problem facing most IEC systems is that the electrostatic field can not penetrate a thermal plasma, and that is of course because it becomes conductive. So once that happens you can no longer confine the plasma and wahlah fusion stops.

This might not be a problem in my S.T.A.R. device, for a couple of reasons. the novel design features a hollow cathode centered inside and insulated from a spherical anode.

Apart from being physically confined by the cathode, the cathode has an amazing ability to syphon off electrons. This happens when the plasma becomes thermal.

When a cold surface is in contact witrh a thermal plasma, the cold surface is bombarded by charged particles, and because electrons in a thermal plasma move much faster than the ions, that surface is hit many more times by negative chaarges than by positive charges.

It is natural for a conducting surface to attemt to reach thermal equilibrium with the plasma, but in a hollow sphere, this is not possible, because the negative charges are constantly absorbed by the inside surface and moved to to the outside surface (as in a Van DeGraaf generator).

So once again, as long as I can retain an ion beam in the accellerator tubes, I don't care if the plasma inside the cathode becomes thermal.

http://www.beejewel.com.au/research/images/Reactor.gif" [Broken]

Steven

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#### gdp

Ok, but that would be _one_ way, not two. X-rays from the collisions are not dependent on the system entropy.
No, two ways:

1.) The IEC concept, and in particular the "gridless" IEC concept, depends crucially on maintaining a disequilibrium state between the electrons and ions, in order to create the electron space-charge potential well that the ions fall into. Due to coulomb scattering of the electrons off the ions, the electron and ion distributions will necessarily relax toward mutual thermodynamic equilibrium, which relaxation creates entropy. Maintaining a disequilibrium state between the electrons and ions necessarily costs power, to make up for the energy lost to the entropy being produced as the distributions continuously try to relax back toward equilibrium. Rider has proved using a 2nd Law argument that the power consumed to maintain the electron/ion disequilibrium exceeds the gain in fusion power obtained from operating the IEC reactor in an electron/ion disequilibrium state; this portion of Rider's argument does not depend on nor involve any specific loss mechanism --- it depends only on kinetic theory, and the 2nd Law of Thermodynamics.

2.) Since according to the 2nd Law, maintaining a disequilibrium state costs power, a portion of the reactor's output must therefore necessarily be "recycled" to maintain the disequilibrium state, over and above the power required to make up losses. By the 2nd Law, this power-recycling process cannot be 100% efficient. Furthermore, since an IEC reactor does not operate in an "ignited" mode, power must necessarily also be recycled to make up for losses in the reactor itself (of which the single largest loss is bremsstrahlung). Rider shows that for any fuel combinations except D+T and D+D, both the amount of recycled power and the losses incurred during power recycling will be prohibitively high (i.e., a large fraction of the reactor's total power output) --- and therefore, an IEC reactor cannot reach economic breakeven, even if somehow it does manage to achieve scientific and engineering breakevens.

mheslep said:
Nebel, formerly of Los Alamos, recently on an IEC machine w/ no grid:
nebel said:
..1. The theory says that you can beat Bremsstrahlung, but it's a challenge. The key is to keep the Boron concentration low compared the proton concentration so Z isn’t too bad. You pay for it in power density, but there is an optimum which works. You also gain because the electron energies are low in the high density regions.
I am extremely skeptical of this claim by Nebel. Bremsstrahlung scales as the square of the ion charge, so bremsstrahlung off Boron is 25 times worse than bremsstrahlung off D or T, and six times worse than bremsstrahlung off He3. Since the fusion power scales as the product of the proton and boron ion densities, trying to beat bremsstrahlung by running a "lean mix" (lowering the boron ion concentration relative to the proton concentration) necessarily also decreases the output power, so it is a self-defeating strategy. To achieve the same power, a "lean mix" reactor will require a proportionally higher core volume. Since bremsstrahlung power scales with ion number, at a rough estimate, I would expect that for the same bremsstrahlung loss rate, a p+B reactor would require on the order of 25 times lower boron concentration than proton concentration, requiring a core volume roughly 25 times larger for the same output fusion power, i.e., at least roughly three times larger reactor radius. Raising the reactor radius makes everything more difficult, since contrary to Bussard's claims, it is not possible to make the magnetic field also scale with radius: For superconducting coils, the maximum B is set by the critical field strength of that superconductor, not by the dimensions of the coil, while for normal coils, electrical resistive losses in the coils will become prohibitive as they get larger, plus the coils will become more and more difficult to cool.

nebel said:
4. The machine does not use a bi-modal velocity distribution. We have looked at two-stream in detail, and it is not an issue for this machine. The most definitive treatise on the ions is : L. Chacon, G. H. Miley, D. C. Barnes, D. A. Knoll, Phys. Plasmas 7, 4547 (2000) which concluded partially relaxed ion distributions work just fine. Furthermore, the Polywell doesn’t even require ion convergence to work (unlike most other electrostatic devices). It helps, but it isn’t a requirement.
Red Herring. The 2nd Law limit on IEC comes from the necessary disequilibrium between the electron and ion distributions --- not from the secondary disequilibrium between ion species. Two-stream instability is a collective effect that increases the thermalization rate of the plasma --- but even if two-stream and other instabilities were somehow completely eliminated, the unavoidable coulomb collisions between the electrons and ions will still cause their energy distributions to relax toward equilibrium with each other, generating entropy during the process. To maintain the electron/ion disequilibrium will cost power. Rider shows that maintaining this disequilibrium will cost more power than will be gained from operating at an electron/ion disequilibrium.

#### gdp

So once again, as long as I can retain an ion beam in the accellerator tubes, I don't care if the plasma inside the cathode becomes thermal.

http://www.beejewel.com.au/research/images/Reactor.gif" [Broken]

Steven
Since the probability that two ions will scatter when they collide with each other is many orders of magnitude larger than the probability that they will undergo a fusion reaction, you cannot in fact maintain the beam inside the accelerator tubes: After no more than a few collisions, the scattered ions will leave the acceptance apertures of the accelerator tubes and be lost.

Google on "beam-beam scattering" for details; it's the second largest loss mechanism in a colliding-beam machine after beam/residual-gas scattering, and the primary loss mechanism in a high-luminosity collider --- which is one of the several reasons why the designers of high-energy physics machines are moving toward single-pass linear colliders with beam energy recovery, rather than circular colliders. (BTW, I used to design particle accelerators for a living...)

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#### beeresearch

Since the probability that two ions will scatter when they collide with each other is many orders of magnitude larger than the probability that they will undergo a fusion reaction, you cannot in fact maintain the beam inside the accelerator tubes: After no more than a few collisions, the scattered ions will leave the acceptance apertures of the accelerator tubes and be lost.
Yes, I am aware of the scattering problem and I am hopeful that some of the scattered ions will reflect off the inside walls of the cathode and get another go at fusing. This may or may not work.

My latest reactor will be operational in about two weeks, and I hope to be able to run a series of experiments between 20 KV and 100 KV to get a measurement of the fusion rate.

Most of my early experiments were dogged by various problems and the reaction rate never exceeded 500,000 fusions/sec, a Q of the order 1e-10, but I hope to improve on this in the next round.

I will do some reading on the subject you suggested.

Steven

#### gdp

Yes, I am aware of the scattering problem and I am hopeful that some of the scattered ions will reflect off the inside walls of the cathode and get another go at fusing. This may or may not work.
It will not. Moreover, the Universe does not respond to "hope."

My latest reactor will be operational in about two weeks, and I hope to be able to run a series of experiments between 20 KV and 100 KV to get a measurement of the fusion rate.
It will be many, many orders of magnitude less than the loss rate.

Most of my early experiments were dogged by various problems and the reaction rate never exceeded 500,000 fusions/sec, a Q of the order 1e-10, but I hope to improve on this in the next round.
First, I suspect you are grossly overestimating your Q.

Second, even if your Q is correct, a low Q is to be expected because the coulomb scattering cross-section is many, many orders of magnitude larger than the fusion reaction cross-section, and no more than one or two collisions is sufficient to remove the ion from the beam.

The peak of the D+T fusion cross-section is about http://en.wikipedia.org/wiki/Tritium#Controlled_nuclear_fusion" for the D+T or D+D reaction is technically divergent, but can be regularized by inserting a maximum impact parameter on the order of your tube radius, and a maximum scattering angle on the order of the angle subtended by the tube aperture as viewed from the collision region. You will discover that the coulomb cross-section is much, much larger than 5 barns.

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#### beeresearch

I suspect you are grossly overestimating your Q.
I have set up a java page for anyone with a BTI bubble detector to calculate their Q here.

http://www.beejewel.com.au/research/fusion_calculator.htm" [Broken]

At the bottom of this page is a link that takes you to a list of various fusion results.

Steven

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#### mheslep

Gold Member
No, two ways:
Yes I see now, two.

...I am extremely skeptical of this claim by Nebel. Bremsstrahlung scales as the square of the ion charge, so bremsstrahlung off Boron is 25 times worse than bremsstrahlung off D or T, and six times worse than bremsstrahlung off He3.
.
Well in a perfectly neutral system. Bremmsstrahlung is proportional to electron density, electron temperature, and the ratio of electrons to ion Z. These virtual cathode systems are by definition not perfectly neutral, as the electron/ion ratio > 1 sets up the electrostatic well.

Since the fusion power scales as the product of the proton and boron ion densities, trying to beat bremsstrahlung by running a "lean mix" (lowering the boron ion concentration relative to the proton concentration) necessarily also decreases the output power, so it is a self-defeating strategy.
Only to a point, as Nebel suggested with the 'optimum' qualifier, as the power gain function is not linear in all its parameters.
Red Herring. The 2nd Law limit on IEC comes from the necessary disequilibrium between the electron and ion distributions --- not from the secondary disequilibrium between ion species. Two-stream instability is a collective effect that increases the thermalization rate of the plasma --- but even if two-stream and other instabilities were somehow completely eliminated, the unavoidable coulomb collisions between the electrons and ions will still cause their energy distributions to relax toward equilibrium with each other, generating entropy during the process. To maintain the electron/ion disequilibrium will cost power. Rider shows that maintaining this disequilibrium will cost more power than will be gained from operating at an electron/ion disequilibrium.
As I understand it, though Rider/Nevins correctly point out the 2nd law issues in play, there are two areas where they fall short: 1) the electron confinement times for a virtual cathode device are shorter than the thermalization/collision time with ions so that the electron temperature never has the opportunity to rise enough to cause unsustainable Bremmstrahlung, 2)their mathematical treatment of collisionality is inadequate. That is, the FP model performed by Chacon et al 2000 improves power gain (Q) by 5 to 10x over that predicted by Nevins. Take this last part up with Chacon et al.

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