12 ms·

A tokamak is a kind of fusion reactor, and basically the most prevalent. "100M degrees" (Kelvin) corresponds to 10 KeV (kilo electron volts), which is an impor

A tokamak is a kind of fusion reactor, and basically the most prevalent. "100M degrees" (Kelvin) corresponds to 10 KeV (kilo electron volts), which is an important figure to exceed for D-T fusion. D-T fusion which is the kind of fusion the ITER Tokamak (a forthcoming fusion reactor and international megaproject) intends to demonstrate. An older fusion experiment, JET (Joint European Torus) reached these levels, so this does not break new ground, but it is important if this Chinese Tokamak is going to provide data useful for ITER. I will note that it's rather unusual to refer to plasma temperature in Kelvin rather than in KeV. I edited this comment with a few more details to try to make it easier for laypeople to understand.

daxfohl 8y ago
What important happens at that temperature?
- NamTaf 8y ago
  As I understand, this represents sufficient energy to overcome the Coulomb Barrier [1] which naturally repels atoms apart. To cause fusion, you need to push particles together either hard enough or fast enough that they push through this repulsion and fuse. The repulsion is a product of the electrostatic repulsion of the positive charges of the nuclei (pushing the positive ends of two magnets together, essentially). The temperature of a gas is essentially a measure of the constituent particles' kinetic energy. Higher kinetic energy = higher temperature. 10keV represents enough kinetic energy for the D-T atoms to collide fast enough that they overcome the Couloumb repulsion and fuse together. [1]: https://en.wikipedia.org/wiki/Coulomb_barrier https://en.wikipedia.org/wiki/Coulomb_barrier
  - saagarjha 8y ago
    Except Wikipedia seems to put the barrier significantly higher, at 100 keV: https://en.m.wikipedia.org/wiki/Thermonuclear_fusion#Temperature_requirements https://en.m.wikipedia.org/wiki/Thermonuclear_fusion#Tempera...
    - piannucci 8y ago
      A system of particles in thermal equilibrium will contain a small fraction whose kinetic energy exceeds the average by a factor of 10. Also, the energy available in the center-of-mass frame of two colliding particles is higher if they happen to be moving in opposite directions. That could give you another factor of up to 2. In any event, I don't think you want your fuel fusing all at once!
      - mrpdaemon 8y ago
        In addition to this there is also a quantum effect called quantum tunneling [1] which allows for a very tiny probability for particles with insufficient energy to fuse upon collision anyway. [1] https://en.wikipedia.org/wiki/Quantum_tunnelling#Nuclear_fusion_in_stars https://en.wikipedia.org/wiki/Quantum_tunnelling#Nuclear_fus...
      - NamTaf 8y ago
        Expanding on this, this 10KeV temperature is the average of all particles, and there's a distribution around this average. Some will be higher, and thus more capable of colliding with high energy. In addition to that, whether two nuclei fuse is also dependent on how squarely they collide. A glancing blow intuitively allows both nuclei to push each other away a lot easier than if they experience a head-on collision. A graph of the fusion rate such as [1] shows that even at much lower average temperatures, fusion will occasionally happen when two higher-than-average nuclei collide head-on. As the temperature gets higher, this rate increases as more particles have enough energy and less require those head-on collisions. The peak rate for D-T according to this graph is about 70kEv. I can't speak to why the grandparent's link referenced 100kEv, as it's been a decade since I last studied this and I'm very rusty. [1]: https://en.wikipedia.org/wiki/File:Fusion_rxnrate.svg https://en.wikipedia.org/wiki/File:Fusion_rxnrate.svg
        fapjacks 8y ago
        How does one control the chain reaction of the fusion process? I understand fission reactors using control rods to absorb some of the neutrons to prevent those neutrons from hitting other fissile particles, but this seems like a harder problem. Those particles fusing at a "low" temperature in the distribution of equilibrium cause additional fusion reactions at higher temperature thresholds in other particles because of the energy released, if my mental model is correct? And assuming the fuel is gaseous, the idea of a control/absorptive retarder seems like a much harder problem. Edit: Oh! Maybe they reduce the strength of the magnetic containment field? Which reduces pressure inside the reaction chamber and thus reducing temperature?
  - thaumasiotes 8y ago
    > you need to push particles together either hard enough or fast enough that they push through this repulsion What's the difference between "hard" and "fast"? What does pushing "hard" mean?
    - vbezhenar 8y ago
      I guess hard is something inside the star where matter is compressed by gravity and fast is when you're accelerating matter. In the end it's the same, only you can't use gravity for small reactors to replicate star technology yet, so you must accelerate particles. SciFi likes to talk about antigravitation. But supergravitation would be cool as well :)
    - semi-extrinsic 8y ago
      They may be trying to differentiate between high temperature as "fast", like in this case, and high pressure as "hard". Though both can lead to a particle kinetic energy above the Coloumb barrier, in the former case the time between collisions will be lower (AFAICT), which may be important. I'm not a plasma guy though.
    - garmaine 8y ago
      GP isn't using standard terminology, but I would presume "hard" was in reference to more mass, which results in higher kinetic energy at lower temperatures.
    - consp 8y ago
      In my general understanding of these things: Fast -> High kinetic energy since they won't acquire mass this is basically speed Hard -> More attempts to make it happen (higher particle interaction) But I might be wrong since plasma is weird stuff.
    - NamTaf 8y ago
      Sorry, I should've been much clearer, that was really badly worded as I was distracted also doing my day job :) As a sibling comment pointed out, I wasn't using any specific terminology as I was trying to talk at a layman level. There's not really a difference. I should've said 'hard enough and fast enough'. Conceptually, there's two main methods of fusion - inertial confinement, and magnetic confinement. In inertial confinement, lasers are shot at the plasma to squeeze it together so the pressure (and thus temperature) increases until fusion occurs. This is also what happens in stars, except they use gravity not lasers. Conceptually, this is what I alluded to when pushing it 'harder' In magnetic confinement, the pressure/temperature is increased by shooting electrical currents through the plasma among other techniques that i'm not as familiar with. The volume hasn't really decreased, but the kinetic energy and pressure of the plasma has increased so it's the same principle, but this is conceptually what I alluded to when i said pushing it 'faster'. In either sense, the idea is to somehow increase the plasma temperature which increases the kinetic energy of the particles enough that they overcome the electrostatic barrier. It was just really badly worded by me and I apologise for that!
- khuey 8y ago
  You overcome the natural repulsion between deuterium and tritium and fuse them together.
rococode 8y ago
You seem to know what you're talking about - do you know what the end goal of the work in this field is? Is it working towards nuclear fusion as an energy source?
- NamTaf 8y ago
  Sustainable high-temperature plasma is what we need to create fusion as an energy source. Current fusion processes rely on pulsing a very short instant of high temperature. This creates a brief instance of fusion, but takes heaps more energy than that quick moment of fusion releases. Thus, it's a net negative of energy. One way to do this that the US uses for nuclear weapons research is to zap a small bit of gas with a tonne of high energy lasers all at the same instant. A high temperature plasma represents a continuous supply of fusing atoms. The current research at ITER, this place, etc. are attempts to create a persistent environment for fusion. If they can do that, then it creates an environment where research can focus on 1) reduce the energy required to hold it at that temperature (which includes limiting how much plasma leaks out, since leaking plasma drops the temperature), and 2) work out ways to extract the energy created by fusion. As I understand (and I could be wrong, it's been years since I last read about it), ITER plans to generate a net-negative energy situation (i.e. it'll never produce energy, just consume it) but hopes to create a sustainable plasma field at temperatures that cause fusion.
  - zaarn 8y ago
    ITER will never produce electric energy but they plan to achieve Q=5, ie, output 5 times more energy than put in. The more important work on ITER is work around enabling actual power stations using Q=5 and developing the tech to maintain operating fusion reactors (remote robots, etc.)
    - therein 8y ago
      At that point can't you just store the output current in supercapacitors for later use anyway? You can even feed it into the machine itself. Supercapacitors easily have better than 20% efficiency for each cycle.
maaaats 8y ago
> "100M degrees" (Kelvin) Off topic, but this distinction made me laugh. Like the difference between Kelvin and Celsius would throw everything off.
- marcosdumay 8y ago
  Well, there is almost certainly somebody that read the number and thought about Fahrenheit.
  - jdironman 8y ago
    I'm that somebody.
  - hkt 8y ago
    Do scientists use farenheit? Genuine question
    - muhbags 8y ago
      Never. It's either no units (theoretical physics), eV (particle physics) or Kelvin/Celsius for experimental physics.
      - ravar 8y ago
        Technically there is Rankine if you want to do thermo in imperial units.
        EForEndeavour 8y ago
        Technically, Rankine and Fahrenheit are not imperial units.
        godelski 8y ago
        I've only known engineers to use Rankine and they hate it. I've never known a scientist to use it.
    - deleted 8y ago
phkahler 8y ago
Didn't old-school CRT televisions "heat" electrons to more than 10KeV?
- randlet 8y ago
  I don't know about CRT's but 10KeV is not particularly energetic. A CT unit may operate in the ~100-200kV range (so max energy of 200keV) while radiation therapy units operate in 6MV-18MV range (so electron accelerated up to 18x10^6 eV). I don't know much about fusion, but my guess is the 10keV is impressive because it is a self-sustaining fusion reaction rather than being impressive because of the absolute energy of the reaction? edit: someone down the page mentioned that containment is the issue. In a CRT you are just accelerating an electron across a few thousand volt potential and slamming it into the screen.
  - phkahler 8y ago
    >> edit: someone down the page mentioned that containment is the issue. Right, and that was my understanding. It's easy to accelerate a particle or a stream of them to high energy. It's another thing entirely to contain a gas/plasma at those energies. The distinction becomes more obvious when they flip back and forth between impressive sounding temperatures and simple KeV measures.
- jsjohnst 8y ago
  Considering that multiple particle colliders can accelerate particles over 1 TeV (LHC topping 13 TeV), yes, 10KeV isn’t a very high number taken out of context.
- sfjlksjfdldas 8y ago
  Sure, but it costs too much energy to break even (although there are some cute devices that are being researched)
- DennisP 8y ago
  Fusion is easily achievable in your garage. The hard part is getting out more power than you put into it. https://en.wikipedia.org/wiki/Fusor https://en.wikipedia.org/wiki/Fusor
deleted 8y ago
[deleted]

daxfohl 8y ago

What important happens at that temperature?

NamTaf 8y ago
As I understand, this represents sufficient energy to overcome the Coulomb Barrier [1] which naturally repels atoms apart. To cause fusion, you need to push particles together either hard enough or fast enough that they push through this repulsion and fuse. The repulsion is a product of the electrostatic repulsion of the positive charges of the nuclei (pushing the positive ends of two magnets together, essentially). The temperature of a gas is essentially a measure of the constituent particles' kinetic energy. Higher kinetic energy = higher temperature. 10keV represents enough kinetic energy for the D-T atoms to collide fast enough that they overcome the Couloumb repulsion and fuse together. [1]: https://en.wikipedia.org/wiki/Coulomb_barrier https://en.wikipedia.org/wiki/Coulomb_barrier
- saagarjha 8y ago
  Except Wikipedia seems to put the barrier significantly higher, at 100 keV: https://en.m.wikipedia.org/wiki/Thermonuclear_fusion#Temperature_requirements https://en.m.wikipedia.org/wiki/Thermonuclear_fusion#Tempera...
  - piannucci 8y ago
    A system of particles in thermal equilibrium will contain a small fraction whose kinetic energy exceeds the average by a factor of 10. Also, the energy available in the center-of-mass frame of two colliding particles is higher if they happen to be moving in opposite directions. That could give you another factor of up to 2. In any event, I don't think you want your fuel fusing all at once!
    - mrpdaemon 8y ago
      In addition to this there is also a quantum effect called quantum tunneling [1] which allows for a very tiny probability for particles with insufficient energy to fuse upon collision anyway. [1] https://en.wikipedia.org/wiki/Quantum_tunnelling#Nuclear_fusion_in_stars https://en.wikipedia.org/wiki/Quantum_tunnelling#Nuclear_fus...
    - NamTaf 8y ago
      Expanding on this, this 10KeV temperature is the average of all particles, and there's a distribution around this average. Some will be higher, and thus more capable of colliding with high energy. In addition to that, whether two nuclei fuse is also dependent on how squarely they collide. A glancing blow intuitively allows both nuclei to push each other away a lot easier than if they experience a head-on collision. A graph of the fusion rate such as [1] shows that even at much lower average temperatures, fusion will occasionally happen when two higher-than-average nuclei collide head-on. As the temperature gets higher, this rate increases as more particles have enough energy and less require those head-on collisions. The peak rate for D-T according to this graph is about 70kEv. I can't speak to why the grandparent's link referenced 100kEv, as it's been a decade since I last studied this and I'm very rusty. [1]: https://en.wikipedia.org/wiki/File:Fusion_rxnrate.svg https://en.wikipedia.org/wiki/File:Fusion_rxnrate.svg
      - fapjacks 8y ago
        How does one control the chain reaction of the fusion process? I understand fission reactors using control rods to absorb some of the neutrons to prevent those neutrons from hitting other fissile particles, but this seems like a harder problem. Those particles fusing at a "low" temperature in the distribution of equilibrium cause additional fusion reactions at higher temperature thresholds in other particles because of the energy released, if my mental model is correct? And assuming the fuel is gaseous, the idea of a control/absorptive retarder seems like a much harder problem. Edit: Oh! Maybe they reduce the strength of the magnetic containment field? Which reduces pressure inside the reaction chamber and thus reducing temperature?
- thaumasiotes 8y ago
  > you need to push particles together either hard enough or fast enough that they push through this repulsion What's the difference between "hard" and "fast"? What does pushing "hard" mean?
  - vbezhenar 8y ago
    I guess hard is something inside the star where matter is compressed by gravity and fast is when you're accelerating matter. In the end it's the same, only you can't use gravity for small reactors to replicate star technology yet, so you must accelerate particles. SciFi likes to talk about antigravitation. But supergravitation would be cool as well :)
  - semi-extrinsic 8y ago
    They may be trying to differentiate between high temperature as "fast", like in this case, and high pressure as "hard". Though both can lead to a particle kinetic energy above the Coloumb barrier, in the former case the time between collisions will be lower (AFAICT), which may be important. I'm not a plasma guy though.
  - garmaine 8y ago
    GP isn't using standard terminology, but I would presume "hard" was in reference to more mass, which results in higher kinetic energy at lower temperatures.
  - consp 8y ago
    In my general understanding of these things: Fast -> High kinetic energy since they won't acquire mass this is basically speed Hard -> More attempts to make it happen (higher particle interaction) But I might be wrong since plasma is weird stuff.
  - NamTaf 8y ago
    Sorry, I should've been much clearer, that was really badly worded as I was distracted also doing my day job :) As a sibling comment pointed out, I wasn't using any specific terminology as I was trying to talk at a layman level. There's not really a difference. I should've said 'hard enough and fast enough'. Conceptually, there's two main methods of fusion - inertial confinement, and magnetic confinement. In inertial confinement, lasers are shot at the plasma to squeeze it together so the pressure (and thus temperature) increases until fusion occurs. This is also what happens in stars, except they use gravity not lasers. Conceptually, this is what I alluded to when pushing it 'harder' In magnetic confinement, the pressure/temperature is increased by shooting electrical currents through the plasma among other techniques that i'm not as familiar with. The volume hasn't really decreased, but the kinetic energy and pressure of the plasma has increased so it's the same principle, but this is conceptually what I alluded to when i said pushing it 'faster'. In either sense, the idea is to somehow increase the plasma temperature which increases the kinetic energy of the particles enough that they overcome the electrostatic barrier. It was just really badly worded by me and I apologise for that!
khuey 8y ago
You overcome the natural repulsion between deuterium and tritium and fuse them together.

rococode 8y ago

You seem to know what you're talking about - do you know what the end goal of the work in this field is? Is it working towards nuclear fusion as an energy source?

NamTaf 8y ago
Sustainable high-temperature plasma is what we need to create fusion as an energy source. Current fusion processes rely on pulsing a very short instant of high temperature. This creates a brief instance of fusion, but takes heaps more energy than that quick moment of fusion releases. Thus, it's a net negative of energy. One way to do this that the US uses for nuclear weapons research is to zap a small bit of gas with a tonne of high energy lasers all at the same instant. A high temperature plasma represents a continuous supply of fusing atoms. The current research at ITER, this place, etc. are attempts to create a persistent environment for fusion. If they can do that, then it creates an environment where research can focus on 1) reduce the energy required to hold it at that temperature (which includes limiting how much plasma leaks out, since leaking plasma drops the temperature), and 2) work out ways to extract the energy created by fusion. As I understand (and I could be wrong, it's been years since I last read about it), ITER plans to generate a net-negative energy situation (i.e. it'll never produce energy, just consume it) but hopes to create a sustainable plasma field at temperatures that cause fusion.
- zaarn 8y ago
  ITER will never produce electric energy but they plan to achieve Q=5, ie, output 5 times more energy than put in. The more important work on ITER is work around enabling actual power stations using Q=5 and developing the tech to maintain operating fusion reactors (remote robots, etc.)
  - therein 8y ago
    At that point can't you just store the output current in supercapacitors for later use anyway? You can even feed it into the machine itself. Supercapacitors easily have better than 20% efficiency for each cycle.

maaaats 8y ago

> "100M degrees" (Kelvin) Off topic, but this distinction made me laugh. Like the difference between Kelvin and Celsius would throw everything off.

marcosdumay 8y ago
Well, there is almost certainly somebody that read the number and thought about Fahrenheit.
- jdironman 8y ago
  I'm that somebody.
- hkt 8y ago
  Do scientists use farenheit? Genuine question
  - muhbags 8y ago
    Never. It's either no units (theoretical physics), eV (particle physics) or Kelvin/Celsius for experimental physics.
    - ravar 8y ago
      Technically there is Rankine if you want to do thermo in imperial units.
      - EForEndeavour 8y ago
        Technically, Rankine and Fahrenheit are not imperial units.
      - godelski 8y ago
        I've only known engineers to use Rankine and they hate it. I've never known a scientist to use it.
  - deleted 8y ago

phkahler 8y ago

Didn't old-school CRT televisions "heat" electrons to more than 10KeV?

randlet 8y ago
I don't know about CRT's but 10KeV is not particularly energetic. A CT unit may operate in the ~100-200kV range (so max energy of 200keV) while radiation therapy units operate in 6MV-18MV range (so electron accelerated up to 18x10^6 eV). I don't know much about fusion, but my guess is the 10keV is impressive because it is a self-sustaining fusion reaction rather than being impressive because of the absolute energy of the reaction? edit: someone down the page mentioned that containment is the issue. In a CRT you are just accelerating an electron across a few thousand volt potential and slamming it into the screen.
- phkahler 8y ago
  >> edit: someone down the page mentioned that containment is the issue. Right, and that was my understanding. It's easy to accelerate a particle or a stream of them to high energy. It's another thing entirely to contain a gas/plasma at those energies. The distinction becomes more obvious when they flip back and forth between impressive sounding temperatures and simple KeV measures.
jsjohnst 8y ago
Considering that multiple particle colliders can accelerate particles over 1 TeV (LHC topping 13 TeV), yes, 10KeV isn’t a very high number taken out of context.
sfjlksjfdldas 8y ago
Sure, but it costs too much energy to break even (although there are some cute devices that are being researched)
DennisP 8y ago
Fusion is easily achievable in your garage. The hard part is getting out more power than you put into it. https://en.wikipedia.org/wiki/Fusor https://en.wikipedia.org/wiki/Fusor

deleted 8y ago

[deleted]