Whats the holdup with Fusion Power?

In summary, controlled thermonuclear fusion is a complex and challenging problem that involves achieving high temperatures and pressures, confining a large number of atoms in a plasma, and dealing with energy loss and instabilities. It requires both temperature and pressure for a long enough time, and a significant number of atoms to produce useful amounts of energy. Various approaches, such as magnetic and beam-beam confinement, are being explored, but there are still many obstacles to overcome before practical fusion energy production can be achieved.
  • #36
beeresearch said:
...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.
 
Engineering news on Phys.org
  • #37
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]
 
Last edited by a moderator:
  • #38
mheslep said:
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
 
Last edited by a moderator:
  • #39
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.
 
  • #40
mheslep said:
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.
 
Last edited by a moderator:
  • #41
gdp said:
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.
 
  • #42
mheslep said:
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.
 
  • #43
gdp said:
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]
 
Last edited by a moderator:
  • #44
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
 
Last edited by a moderator:
  • #45
mheslep said:
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.
 
  • #46
beeresearch said:
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...)
 
Last edited by a moderator:
  • #47
gdp said:
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
 
  • #48
beeresearch said:
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.
 
Last edited by a moderator:
  • #49
gdp said:
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
 
Last edited by a moderator:
  • #50
gdp said:
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.
 
  • #51
mheslep said:
gdp said:
...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.

Sorry, no. IEC systems, while they do have a very slight charge imbalance, nevertheless do still satisfy the "quasineutrality" condition to an excellent degree of approximation, as shown by Rider in his thesis.

mheslep said:
gdp said:
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.

I am still very skeptical, and I'd want to see the data. IIRC, Rider explicitly shows that in a quasineutral plasma, the bremsstrahlung loss rate and fusion power depend on the densities of the ion species in the exact same way, so that the ratio of bremsstrahlung losses to fusion gain is a constant, independent of any monkey-business with the ion mixture. Anything that decreases bremsstrahlung losses should therefore decrease the fusion power by the exact same fraction.

Has Nebel published any of these claims in a refereed journal, or is it the only source for Nebel's claims the blog exchange between Nebel and Carlson?

mheslep said:
gdp said:
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,

Rider deals with this. In effect, one is "refrigerating" the electrons by removing them from the system before they can equilibriate. Since the electrons are "cold" compared to the ions, the ions therefore continuously lose energy to the cold electrons through coulomb collisions, producing entropy, and requiring that additional power be recycled to maintain the ion distribution. Rider finds that the power expended to maintain the disequilibrium will exceed the additional gain from operating at disequilibrium.

There is no escape from the 2nd Law.

mheslep said:
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.

I have downloaded the paper, but have not yet had time to read it. However, I note already that according to their abstract, they are performing an "optimistic" (their term!) calculation that explicitly neglects electron-ion collisional interactions --- and neglecting electron-ion interactions is simply not physically realistic in these disequilibrium systems.

Even worse, they appear to be treating the electron distribution as a fixed, prescribed "background" that generates a potential-well of two assumed forms: square-well and parabolic --- neither of which are particularly physical.

It is like attempting to estimate the performance of an automobile by neglecting road and air friction, and concluding that top speeds of 900 mph should be possible. Well, of course you can get unphysically good results, if you throw away the most important physical limiting factors!

A truly "self-consistent" calculation of fusion gain in an IEC device will need to explicitly treat the continuous transfer of energy from the ion to the electron population, rather than ignoring collisions and treating the electron population as merely a fixed prescribed background that is unaffected by the ion population as Chacon and Miley appear to be doing in their paper.
 
  • #52
Chacon et al do indeed account for losses due to electrons radiating and escaping the system. The quote above refers to simplifications made for the FP calculations only. The electron losses are calculated in the familiar closed form way. I'll get back to this in detail later time permitting...
 
Last edited:
  • #53
gdp said:
...Has Nebel published any of these claims in a refereed journal, or is it the only source for Nebel's claims the blog exchange between Nebel and Carlson?...
Google scholar for "R. Nebel iec" will quickly show his publications in the area, but I am not sure what you are looking for. The Chacon et al paper is the most relevant to that blog discussion.
 
  • #54
Hi,

I think the time delay has a lot to do with the pressure, heat issues, but mostly because the only really large scale attempt, in Europe anyway, is happening in France. Yes, I am English.

Utwig
 
  • #55
Rider finds that the power expended to maintain the disequilibrium will exceed the additional gain from operating at disequilibrium.

Sure, but isn't that statement dependent on his other assumptions about gain, which the Chacon paper argues are not accurate?
 
  • #56
However, I note already that according to their abstract, they are performing an "optimistic" (their term!) calculation that explicitly neglects electron-ion collisional interactions --- and neglecting electron-ion interactions is simply not physically realistic in these disequilibrium systems.

FWIW, that was raised at T-P as well. Nebel's answer was:

At the risk of putting my foot in my mouth, the usual answer is that ion-electron collisions are much smaller than ion-ion collisions. It's much easier for particles of the same mass to transfer momentum to one another. This is the same effect as shooting pool with a cue ball that weighs the same as the other balls vs. shooting pool with a heavy cue ball. It's really hard to stop that heavy cue ball (it's just the combination of conservation of momentum and energy).

The best discussion I've seen of this is in chapter 4 of Glasstone and Lovberg (Controlled Thermonuclear Reactions, Robert E. Kriger Publishing Co. 1975.) but I suspect that it is out of print. Generally, calculating collisions of ions with electrons is more complicated than electron-electron collision, ion-ion collisions, or electron-ion collisions. The general rule of thumb is that electron distributions and ion distributions will equilibrate at a much faster rate than they will transfer energy to each other.

Anyways, to answer the OP: the reason fusion power is taking so long is that no one has a design that is economically viable. Even with some optimistic assumptions, ITER/DEMO's plant power density is way too low to compete with light-water fisison reactors -- and fuel for those won't run out for at least 1,000 years. IEC/FRC are a bit more promising in this regard, but they haven't seen much funding yet and have their own problems to resolve.
 
Last edited:
<h2>1. What is fusion power and why is it important?</h2><p>Fusion power is a type of energy that is produced by fusing together two or more atomic nuclei to form a heavier nucleus. It is important because it has the potential to provide a virtually limitless source of clean and renewable energy, without producing greenhouse gases or long-lived radioactive waste.</p><h2>2. Why is fusion power taking so long to develop?</h2><p>Fusion power is a complex and challenging technology that requires extremely high temperatures and pressures to initiate the fusion reaction. Scientists are still working on finding ways to sustain the reaction and harness the energy produced. There are also significant engineering and economic challenges that need to be overcome before fusion power can become a viable energy source.</p><h2>3. What are the current obstacles in achieving fusion power?</h2><p>One of the main obstacles in achieving fusion power is the need for a reliable and efficient way to confine and heat the plasma (a state of matter in which atoms are stripped of their electrons) to the necessary temperatures. Another challenge is finding materials that can withstand the extreme conditions inside a fusion reactor, such as high temperatures and intense radiation.</p><h2>4. When can we expect fusion power to become a reality?</h2><p>It is difficult to predict an exact timeline for when fusion power will become a reality, as there are still many technical and scientific challenges that need to be overcome. However, there are several large-scale international projects, such as ITER and the National Ignition Facility, that are making significant progress towards achieving fusion power. Some experts estimate that commercial fusion power could be available within the next few decades.</p><h2>5. What are the potential benefits of fusion power?</h2><p>The potential benefits of fusion power are numerous. It could provide a nearly limitless source of clean energy, reduce our dependence on fossil fuels, and help mitigate the effects of climate change. It could also have a significant impact on global energy security and could potentially create new industries and job opportunities. Additionally, fusion power does not produce long-lived radioactive waste, making it a much safer option compared to nuclear fission power.</p>

1. What is fusion power and why is it important?

Fusion power is a type of energy that is produced by fusing together two or more atomic nuclei to form a heavier nucleus. It is important because it has the potential to provide a virtually limitless source of clean and renewable energy, without producing greenhouse gases or long-lived radioactive waste.

2. Why is fusion power taking so long to develop?

Fusion power is a complex and challenging technology that requires extremely high temperatures and pressures to initiate the fusion reaction. Scientists are still working on finding ways to sustain the reaction and harness the energy produced. There are also significant engineering and economic challenges that need to be overcome before fusion power can become a viable energy source.

3. What are the current obstacles in achieving fusion power?

One of the main obstacles in achieving fusion power is the need for a reliable and efficient way to confine and heat the plasma (a state of matter in which atoms are stripped of their electrons) to the necessary temperatures. Another challenge is finding materials that can withstand the extreme conditions inside a fusion reactor, such as high temperatures and intense radiation.

4. When can we expect fusion power to become a reality?

It is difficult to predict an exact timeline for when fusion power will become a reality, as there are still many technical and scientific challenges that need to be overcome. However, there are several large-scale international projects, such as ITER and the National Ignition Facility, that are making significant progress towards achieving fusion power. Some experts estimate that commercial fusion power could be available within the next few decades.

5. What are the potential benefits of fusion power?

The potential benefits of fusion power are numerous. It could provide a nearly limitless source of clean energy, reduce our dependence on fossil fuels, and help mitigate the effects of climate change. It could also have a significant impact on global energy security and could potentially create new industries and job opportunities. Additionally, fusion power does not produce long-lived radioactive waste, making it a much safer option compared to nuclear fission power.

Similar threads

Replies
18
Views
873
  • Nuclear Engineering
Replies
2
Views
1K
  • Nuclear Engineering
Replies
19
Views
2K
Replies
129
Views
7K
Replies
7
Views
2K
  • Nuclear Engineering
Replies
3
Views
1K
  • Nuclear Engineering
3
Replies
70
Views
8K
Replies
7
Views
2K
Replies
6
Views
1K
  • Nuclear Engineering
Replies
14
Views
3K
Back
Top