Homepage Forums Education Fusion reactor comparison chart

Viewing 15 posts - 1 through 15 (of 32 total)
  • Author
    Posts
  • #1273
    Avataropensource
    Participant

    Is there a comparison chart out there that shows all of the working reactors real numbers in terms of how much more energy they need to be producing to break even?

    #11023

    opensource wrote: Is there a comparison chart out there that shows all of the working reactors real numbers in terms of how much more energy they need to be producing to break even?

    Well, I attempted to start such a list some time ago, but few were willing to report their results.

    Find it here…

    http://www.fusor.net/board/view.php?bn=fusor_construction&key=1212188288

    It is still active, and if anyone wants to report, I will update the list.

    Steven

    #11033
    Avataropensource
    Participant

    …Does that mean that the best reactor is “1.97e-08” (watts?) below breaking even?!

    #11034

    The list is not comprehensive, it just lists IEC reactors, but yes, for IEC that’s about the range.

    Tokamak reactors such as JET in the UK, have demonstrated a Q of around 0.75, and ITER is expected to show a Q more than 1.

    Steven

    #11035
    AvatarLerner
    Participant

    I think its best to use the ratio of total fusion pwoer divided by total (wall-plug) input energy. If we are talking about pure D feul, where msot comparisons can be made, the best tokamka results and the best DPF results are about the same at 10^-5 or so. Even with DT JET is nowhere near net energy.

    #11036
    Avataropensource
    Participant

    The reasons as to why the distance from creating net energy is so great aren’t very obvious to me.

    Can this be put more simply? To be clear, having a Q of 1 means a reactor can sustain itself, but not create a net amount of energy. A reactor of Q = .75 is producing 75 percent of its own energy for sustenance, but requires another 25% from another power source in order to yield fusion reactors of any sort…?

    How does pB11 factor into these results?

    #11040
    AvatarIvy Matt
    Participant

    It’s my understanding that the fusion energy gain factor (Q) is what is also known sometimes as “scientific break-even”, and compares an input of electrical energy to an output of fusion energy in the form of kinetic energy of particles or electromagnetic radiation. So, in order to power itself, a fusion reactor running at a Q of 1 would need to convert the fusion energy back to electricity with 100% efficiency, which is impossible. Although Q = 1 would be a big scientific breakthrough, a fusion reactor that generates electricity would have to have a Q of greater than 1 to power itself, let alone to produce surplus electricity. A Q of 20 is about what is needed for a practical tokamak reactor.

    Regarding JET’s Q of 0.75, if you check the fusor.net thread Steven Sesselmann linked to, you’ll find Richard Hull’s criticism of JET’s energy accounting in his June 4, 2008 comment. Apparently they counted the energy required to heat the plasma, but not that used to produce the magnetic fields, run the vacuum pumps, etc. The point is that an economical reactor design would need to take these things into account. Not to discourage anyone too much, but I suppose you could compare it to the difference between breaking the sound barrier and producing an economical supersonic airliner.

    The p+B11 reaction doesn’t really change the accounting of the output energy, which is still in the form of kinetic energy of particles and electromagnetic radiation. What it does change is the way that output energy could be converted back into electrical energy. Charged particles (p+B11) and neutral particles (D+D, D+T) are converted to electricity using different methods, and the method (direct conversion) for charged particles is expected to be considerably more efficient than the method (steam turbines) for neutral particles. That said, any reactor that uses the D+T or D+D reactions also produces some charged particles as well as electromagnetic radiation, and direct conversion was originally posited as a method to obtain additional electrical energy from such reactions. However, it’s my understanding that in conventional tokamak designs the charged particles are intended to recirculate to keep heating the fusion plasma, which would preclude them being used for direct conversion. As for electromagnetic radiation…I’d love to see a tokamak + “onion” reactor design. That would certainly be interesting.

    #11041
    AvatarRezwan
    Member

    Ivy Matt wrote: I suppose you could compare it to the difference between breaking the sound barrier and producing an economical supersonic airliner.

    Great analogy!

    #11045
    Avataropensource
    Participant

    Since theoretically this process can scale to ‘better’ than fission reactors (energy density), what is not getting unleashed in these fusion reactors so that they don’t even breakeven? For example, is it because the atoms aren’t fusing at a high enough rate, not enough of them are fusing, the energy output of each fused atom isn’t great enough…? What category do you feel the most energy is lying dormant?

    #11048

    Theoretically all fusion reactors produce net energy…

    Since most of the reactor designs are intended for converting fusion energy into heat, one could theoretically say that all the input energy is also converted to heat, so if you input one 1MW to get one watt out, you could theoretically recover 1,000,001 watt, ergo a Q of 1.000001, however the conversion of thermal energy to useful electric power is at best 60%, so we go back to the drawing board.

    Clearly a fusion scheme that takes advantage of direct charge to current conversion will have a huge advantage. Focus fusion promises to do this, as well as my soon to come FICS fusion.

    Steven

    #11050
    Avataropensource
    Participant

    The next questions (in addition to the ones in my last post) for me are:
    1) What percentage of the energy output are the x-rays? That is, in terms of the output of heat and various energies the fusion reaction gives off, what percentage of that energy is represented by x-rays.

    2) Secondly, is the 80% figure for the onion capturing the x-rays realistic?

    Is the answer that x-rays aren’t a majority of the power output of the fusion reaction (maybe because heat is too) – and additionally – that the onion can’t capture nearly 80% of the x-ray’s power? If that’s the case, then FF needs a different design.

    I know that FF has been tested, so what are the numbers for the various forms of energy – and their proportions to one another – that the fusion reaction gave off?

    P.S. This raises the question for me:
    “What are all of the various types of energy a fusion reaction can produce?” heat, high energy rays (x-rays), etc… So can the primary byproduct of a fusion reaction be on a much lower band of the EM spectrum? Would this help conversion to electricity? (I found something called EMP fusion, but it doesn’t seem to be what I was looking for.)

    #11062
    Avatarvansig
    Member

    if i recall correctly, the model, with various different parameters, predicts from about a third to half of input energy being emitted in x-ray.

    the 80% figure for the onion capturing the x-rays might be unrealistic. if we capture its energy with photo-voltaic cells, we can get 10 to 20%. if we capture its energy with a heat engine, we can get about 55%. (but i have not fully explored the possibilities).

    the present set of parameters for attempting break-even is going to try mostly to capture energy from the exit beam. since it consists of a pulse of charged particles, it can be transformed at high efficiency (today’s high voltage transformers routinely achieve 98% or better).

    #11063
    Avataropensource
    Participant

    vansig wrote: if i recall correctly, the model, with various different parameters, predicts from about a third to half of input energy being emitted in x-ray.

    If about 1/2 of the energy from the fusion reaction is x-rays, then of what form is the rest of the energy?

    vansig wrote:
    the 80% figure for the onion capturing the x-rays might be unrealistic. if we capture its energy with photo-voltaic cells, we can get 10 to 20%. if we capture its energy with a heat engine, we can get about 55%. (but i have not fully explored the possibilities).

    Why is the hypothesized (80%) capture for x-rays so much higher than the actual efficiency for photo-voltaic cells? I thought both were capturing energy using the same mechanism…

    Furthermore, don’t the outputted x-rays vary widely in frequency?

    vansig wrote:
    the present set of parameters for attempting break-even is going to try mostly to capture energy from the exit beam. since it consists of a pulse of charged particles, it can be transformed at high efficiency (today’s high voltage transformers routinely achieve 98% or better).

    Is the “exit beam” the one-third to one-half figure you stated earlier as the percentage of energy output in the form of x-rays? What percentage of the fusion reaction creates charge particles?

    #11064
    Avatarzapkitty
    Member

    vansig wrote: if i recall correctly, the model, with various different parameters, predicts from about a third to half of input energy being emitted in x-ray.

    the 80% figure for the onion capturing the x-rays might be unrealistic. if we capture its energy with photo-voltaic cells, we can get 10 to 20%. if we capture its energy with a heat engine, we can get about 55%. (but i have not fully explored the possibilities).

    Nope. The efficiency of photoelectric conversion of a photon is directly proportional to the energy of the photon. That’s why infrared-range photoelectric converters are not nearly as efficient as visible-range converters. X-ray conversion will be even more efficient than visible light conversion. The trick is catching enough of them first.

    There’s no great mystery about X-ray photoelectrics… it’s just that before FF there was no reason to investigate it as a power source.

    And the heat from an FF unit, once removed from the core, would be what is considered “low grade” heat in industrial terms… not efficient for turbine use.

    vansig wrote: the present set of parameters for attempting break-even is…

    … not being attempted for FoFu-1 (what a name 🙂 ) The only coils in the drift tube are for measurement, not power conversion.

    What [em]will[/em] be attempted is a demonstration of scientific feasibility. If they can show that an FF-style DPF produces enough extra energy to make a generator feasible then it’s game, set and match… and without having to build an actual generator on top of everything else they have to do.

    #11065
    Avataropensource
    Participant

    There seem to be some basics to straighten out here. Can you reply to my questions as well zapkitty?

Viewing 15 posts - 1 through 15 (of 32 total)
  • You must be logged in to reply to this topic.