7 ms·

Fusion happens when hot enough hydrogen ions (=hydrogen atoms with the electron stripped so only the nucleus is left) are running into each other often enough.

by 317070 4y ago

Fusion happens when hot enough hydrogen ions (=hydrogen atoms with the electron stripped so only the nucleus is left) are running into each other often enough. So you need a high enough density of those that stay hot enough for long enough (the triple product, vertical axis) at a high enough temperature (horizontal axis) for enough atoms to fuse. At the boundary, enough energy will start being released from fusion reactions to keep the reaction going. at Q=1 you have theoretical break-even. Somewhere between Q=5 and Q=10 you could extract more heat than you put in. The heat produced by the reactions should be enough to sustain the reaction. At Q=inf it just keeps burning like a fancy camp fire. Just throw in fuel, remove ashes and enjoy the heat. Commercial fusion can only start being viable probably somewhere between Q=10 and Q=20. Do note that the projected future in OP's gif should contain way more attempts from various companies to reach that limit. Just mentioning sparc is a bit selective.

tifik 4y ago
Would it be somewhat correct to say that the Sun (or generally stars of similar type) is approaching that Q=inf at this stage of its life?
- 317070 4y ago
  Stars are at Q=inf, they don't have external heating. They are also on a weird place on this plot, considerably more to the top left side of the high Q's.
  - tifik 4y ago
    Had to have a second look at the chart after reading your explanation to understand the variables. Thank you!
- deleted 4y ago
  [deleted]
elevaet 4y ago
That's a great explanation, thank you. What does the black curve that intersects some of the Q curves represent?
- 317070 4y ago
  It's the theoretical breakeven event (Q=1) when you use high power lasers rather than the magnetic confinement of a tokamak. The NIF is doing research on this inertial confinement fusion, where they shoot a lot of power with lasers on a tiny pellet with hydrogen. So you get extreme densities and temperatures, but not a lot of confinement time, i.e. the ions cool down _really_ fast. That shifts the curve a little bit. In my opinion, it is a bit misleading to add this to the plot, because for this laser-based fusion, the theoretical breakeven lies really far from the practical, commercial breakeven. But as shown there on the gif, the NIF did manage to cross the Q=1 for inertial confinement fusion last year!
  - ncmncm 4y ago
    NIF is nothing other than dressed-up weapons research. But it is not more deceptive than other methods.
- sam 4y ago
  Author here. The black curve represents the hot-spot ignition condition for a laser inertial confinement fusion (ICF) experiment (like the NIF). This means that during the short period of inertial confinement, the self-heating exceeds all losses in the hot spot leading to an increase in temperature due to self heating. It only applies to the black 'x' points. The Q_sci^MCF contours correspond to scientific energy gain (ratio of fusion power to heating power crossing the vacuum vessel boundary) for a magnetic confinement experiment. For ICF we can't draw simillar Q_sci^ICF contours because the total fusion energy released depends on the degree to which the ignited hot-spot propagates a burn in the surrounding cold fuel. And this depends on other variables like the symmetry of the implosion which are not captured in this plot. If you're curious to read more about this check out Section III.F of the linked paper (pp.10-11).
stavros 4y ago
So does eg Q=100 mean that it'll still go out on its own at some point? Ie is Q a measure of duration/total energy output for a given energy expenditure?
- 317070 4y ago
  Yes, it means you will get out 100 times more energy than you put in, but at that point it does still go out if you don't keep supplying that external heat. At Q=infinity, you get out as much energy as you want compared to what you put in initially, but it keeps burning (as long as you add fuel and remove ashes, that is). At that point you pay an initial start-up cost and can fuse as much as you want.
  - stavros 4y ago
    I see, thanks! So this doesn't take into account the hydrogen, just the external energy/heat you put in? I guess it must, otherwise Q=inf would be free energy.
    - 317070 4y ago
      Correct. The energy balance comes from the fuel mass that is converted to energy with E=mc^2.
      - stavros 4y ago
        Right, thanks for the explanation!
    - willis936 4y ago
      You still have to run the cryogenic system for the superconductors and the pittance of power to keep the confinement coil power supplies. Q=inf means ignition: an entirely self-heated plasma. The collisions of the helium ash on the un-burnt fuel are enough to bring the fuel above the flash point. In this way it is like a campfire.
  - ncmncm 4y ago
ncmncm 4y ago
Commercial fusion will not be competitively viable at any Q. Even unlimited free heat is not valuable if it costs too much to make power from it. A volcano, earthquake, or hurricane releases many TWh of energy, but not usefully so, for reasons. Once you start to get enough neutron kinetic energy out, those neutrons have to be captured and their kinetic energy degraded to heat in thousands of tons of molten lithium pumped in big pipes snaking around inside your magnets, which you must then use to boil water and drive a turbine. The cost of operating such a plant would be at least 10x the equivalent fission plant. But fission is already not competitive, and gets less so every day. Part of operation would need to be purifying grams of tritium, daily, out of those thousands of tons of radioactive molten lithium, for the fuel needed to continue operating. That might not be possible. No one is even trying it yet. The power density of your D-T plasma would be much less than fissioning uranium (Th, Pu, etc.), which means you need a great deal of it, and a huge plant. If superconducting magnets squeeze it smaller, you have the problem that all this neutron flux has to come through a wall of limited area, destroying it in short order. Magnets so strong would need thousands of tons of steel to hold them in place, which would weaken rapidly under heavy hot neutron bombardment. So, the D-T fusion chased is a series of fascinatingly hard technical problems, but there is no plausible prospect of ever getting any commercially valuable power out. And, they are not even safer than a regular nuke; the thousand tons of molten radioactive lithium is hugely inflammable, even explosive, and the smoke turns into radioactive drain opener on contact with anything damp, e.g. lungs.
- jhrmnn 4y ago
  Are you suggesting that all those people doing fusion research don’t know about this, or that they make false promises to the public to fund their research?
  - ncmncm 4y ago
    At least many know and do not care. Many deceive themselves, as often seen here. Others don't think about it, considering it Somebody Else's Problem. Promises of commercial power are naked lies they will never be called to account for.
  - 4y ago

by 317070 4y ago

tifik 4y ago

Would it be somewhat correct to say that the Sun (or generally stars of similar type) is approaching that Q=inf at this stage of its life?

317070 4y ago
Stars are at Q=inf, they don't have external heating. They are also on a weird place on this plot, considerably more to the top left side of the high Q's.
- tifik 4y ago
  Had to have a second look at the chart after reading your explanation to understand the variables. Thank you!
deleted 4y ago
[deleted]

elevaet 4y ago

That's a great explanation, thank you. What does the black curve that intersects some of the Q curves represent?

317070 4y ago
It's the theoretical breakeven event (Q=1) when you use high power lasers rather than the magnetic confinement of a tokamak. The NIF is doing research on this inertial confinement fusion, where they shoot a lot of power with lasers on a tiny pellet with hydrogen. So you get extreme densities and temperatures, but not a lot of confinement time, i.e. the ions cool down _really_ fast. That shifts the curve a little bit. In my opinion, it is a bit misleading to add this to the plot, because for this laser-based fusion, the theoretical breakeven lies really far from the practical, commercial breakeven. But as shown there on the gif, the NIF did manage to cross the Q=1 for inertial confinement fusion last year!
- ncmncm 4y ago
  NIF is nothing other than dressed-up weapons research. But it is not more deceptive than other methods.
sam 4y ago
Author here. The black curve represents the hot-spot ignition condition for a laser inertial confinement fusion (ICF) experiment (like the NIF). This means that during the short period of inertial confinement, the self-heating exceeds all losses in the hot spot leading to an increase in temperature due to self heating. It only applies to the black 'x' points. The Q_sci^MCF contours correspond to scientific energy gain (ratio of fusion power to heating power crossing the vacuum vessel boundary) for a magnetic confinement experiment. For ICF we can't draw simillar Q_sci^ICF contours because the total fusion energy released depends on the degree to which the ignited hot-spot propagates a burn in the surrounding cold fuel. And this depends on other variables like the symmetry of the implosion which are not captured in this plot. If you're curious to read more about this check out Section III.F of the linked paper (pp.10-11).

stavros 4y ago

So does eg Q=100 mean that it'll still go out on its own at some point? Ie is Q a measure of duration/total energy output for a given energy expenditure?

317070 4y ago
Yes, it means you will get out 100 times more energy than you put in, but at that point it does still go out if you don't keep supplying that external heat. At Q=infinity, you get out as much energy as you want compared to what you put in initially, but it keeps burning (as long as you add fuel and remove ashes, that is). At that point you pay an initial start-up cost and can fuse as much as you want.
- stavros 4y ago
  I see, thanks! So this doesn't take into account the hydrogen, just the external energy/heat you put in? I guess it must, otherwise Q=inf would be free energy.
  - 317070 4y ago
    Correct. The energy balance comes from the fuel mass that is converted to energy with E=mc^2.
    - stavros 4y ago
      Right, thanks for the explanation!
  - willis936 4y ago
    You still have to run the cryogenic system for the superconductors and the pittance of power to keep the confinement coil power supplies. Q=inf means ignition: an entirely self-heated plasma. The collisions of the helium ash on the un-burnt fuel are enough to bring the fuel above the flash point. In this way it is like a campfire.
- ncmncm 4y ago

ncmncm 4y ago

Commercial fusion will not be competitively viable at any Q. Even unlimited free heat is not valuable if it costs too much to make power from it. A volcano, earthquake, or hurricane releases many TWh of energy, but not usefully so, for reasons. Once you start to get enough neutron kinetic energy out, those neutrons have to be captured and their kinetic energy degraded to heat in thousands of tons of molten lithium pumped in big pipes snaking around inside your magnets, which you must then use to boil water and drive a turbine. The cost of operating such a plant would be at least 10x the equivalent fission plant. But fission is already not competitive, and gets less so every day. Part of operation would need to be purifying grams of tritium, daily, out of those thousands of tons of radioactive molten lithium, for the fuel needed to continue operating. That might not be possible. No one is even trying it yet. The power density of your D-T plasma would be much less than fissioning uranium (Th, Pu, etc.), which means you need a great deal of it, and a huge plant. If superconducting magnets squeeze it smaller, you have the problem that all this neutron flux has to come through a wall of limited area, destroying it in short order. Magnets so strong would need thousands of tons of steel to hold them in place, which would weaken rapidly under heavy hot neutron bombardment. So, the D-T fusion chased is a series of fascinatingly hard technical problems, but there is no plausible prospect of ever getting any commercially valuable power out. And, they are not even safer than a regular nuke; the thousand tons of molten radioactive lithium is hugely inflammable, even explosive, and the smoke turns into radioactive drain opener on contact with anything damp, e.g. lungs.

jhrmnn 4y ago
Are you suggesting that all those people doing fusion research don’t know about this, or that they make false promises to the public to fund their research?
- ncmncm 4y ago
  At least many know and do not care. Many deceive themselves, as often seen here. Others don't think about it, considering it Somebody Else's Problem. Promises of commercial power are naked lies they will never be called to account for.
- 4y ago