Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Nuclear fusion using a plasma jet

  1. Sep 13, 2009 #1
    As per a previous thread, if you fire two plasma jets of deuterium at each other at high enough relative velocity, the deuterons would collide and fuse into helium.

    Does anyone know what is the relative velocity in kilometers per hour necessary for deuterons hitting each other head on to fuse into helium?

    How is this calculated?
     
  2. jcsd
  3. Sep 14, 2009 #2
    As calculated in the previous thread, you're looking at velocities of ~500 km/s, whether for hydrogen or deuterium (square root of two or 40% lower for deuterium, which does not help all that much) and we don't know how to accelerate macroscopic objects to this kind of speed ...


    Note that there's no minimum fusion "threshold", but probability of fusion between two nuclei goes up quickly as you add more energy.

    There's an easy way to accelerate individual ions, you basically strip the deuterium atom of electrons and stick it into strong electric field, tens of kilovolts ... if you create a vacuum chamber with two concentric electrodes, apply 10 kV between electrodes, and put a small quantity of deuterium gas into the chamber, you _will_ see fusion events, but, to the best of our knowledge, this process can't be optimized to result in net production of energy.
     
  4. Sep 14, 2009 #3
    Hi Hamster143,

    The reason i keep hammering this point is that I have thought of a method to accelerate plasma jets to a very high speeds, using an accelerating magnetic field.

    The figure you mentioned is in the realm of feasibilty.

    Half of 500 km/s for each plasma jet is 250 km/s (directed against each other). Times 60% for deuterium is 150 km/s. That is 540,000 km per hour.

    Now if you want only 1% of the plasma jet to fuse, a statistician friend told me you can reduce the averge velocity to 25% of 540,000 km/hour, or 135,000 km/hour, assuming a normal distribution of particle velocities.

    A plasma accelerator which can accelerate plasma to an average speed of 135,000 km/hour is well within the realms of feasibility.

    I think continuous commercial plasma fusion is quite feasible if you direct two high speed plasma jets of deuterium againast each other.
     
  5. Sep 14, 2009 #4
    Thirty years ago, when I was a young researcher, I was told that one of our colleagues (Leopold Skripnik) had already proposed (theoretically) a “solid-state/beam thermonuclear reactor”. Indeed, he took a solid target and a fast beam of deuterons or so, I do not remember now. He calculated the efficiency of such a system. In a solid body the charged projectile loses its energy but can provoke an energy release due to fusion. He found that starting from some energy the losses are smaller than the gain but nobody believed it. In fact, it is a question of efficiency. Theoretically it is possible in a solid target. I do not know how about two colliding jets, where there is a region of their velocities to make the fusion energy release superior to projectile energy losses.
     
    Last edited: Sep 14, 2009
  6. Sep 14, 2009 #5
    I have never seen or read of magnetic acceleration of any charged particles except electrons. Are you talking about a "betatron" accelerator for deuterons? See
    http://en.wikipedia.org/wiki/Betatron
    I have seen and used 14.1 MeV neutrons from (I think) 300 KeV deuterons hitting a tritiated tantalum (or tungsten) target yielding D-T fusion, but it was extremely inefficient, like maybe 1 neutron per million deuterons or worse.
     
  7. Sep 14, 2009 #6
    300 KeV is too small energy for the projectile to get a reasonable efficiency. It should be increased by a factor of 100 or even more.
     
  8. Sep 14, 2009 #7
    So you're saying 30 MeV. Is this per beam in the center of mass, or deuterons on a fixed (stationary) target? What is the fusion cross section there? Considering that the energy release including neutrons in D-T fusion is only about 17 MeV, it is hard to visualize break-even power output with 30 MeV beams.
     
  9. Sep 14, 2009 #8

    Astronuc

    User Avatar

    Staff: Mentor

    Not for fusion. Look at the kinetic energies of the reactants for the Q values of the reaction and those are in the low MeV range, so it is impractical to accelerate deuterons much beyond 1 MeV in order to induce fusion. The optimal temperature for d+d fusion is around 500 keV - 1 MeV, and with diminishing returns one would probably want a temperature on the order of 300-400 keV.

    d + d -> T (1 MeV) + p (3 MeV), or 3He (0.82 MeV) + n (2.45 MeV)

    Colliding beams have been considered. With losses due to scattering, I believe the concept was found to be impractical.

    Another consideration is the neutrality of plasma. Usually injected deuterons are neutralized, otherwise the local positive charge can induce plasma instability as well as dispersion of the deutron beam. So either one collides neutral deutron beams or one has to add similar current so electron beams to the reation volume.
     
  10. Sep 14, 2009 #9
    I really do not know all the numbers - I never did it myself. But what is necessary to reach is to obtain the thermo-nuclear reaction for sure and recuperate as much of lost energy as possible. I do not know details of his calculations. Anyway, all energy release is then used in a heat machine with a known efficiency so there may be some positive output.
     
  11. Sep 14, 2009 #10
  12. Sep 14, 2009 #11
    Yes but one has to diminish the losses too. The loss cross section may decrease faster with E than the reaction cross section. Besides, I do not know what target and what projectile Leopold used in his calculations.
     
  13. Sep 14, 2009 #12
  14. Oct 8, 2009 #13
  15. Nov 16, 2009 #14
    Would the d-d cross section graph be the same for magnetic confinement as for inertial confinement? What about cold fusion? In fact I'm struggling a bit with the concept of cross-section. It doesn't seem to mean just the physical measurement of the area of a 'cross-section'. Correct?

    It seems from some things I've been reading that the cross-section is dependent on the process, or derived from the experimental reaction rate rather than the physical dimensions of the deuteron. (E.g. http://hyperphysics.phy-astr.gsu.edu/Hbase/nuclear/nucrea.html#c3; http://www.iupac.org/goldbook/R05169.pdf; [Broken] http://fds.oup.com/www.oup.co.uk/pdf/0-19-856264-0.pdf [Broken] at p.3)

     
    Last edited by a moderator: May 4, 2017
  16. Nov 16, 2009 #15
    We speak of scattering cross sections or of reaction cross sections. It is not an atomic size squared. It is a process-dependent thing and is determined as a ratio of the reaction output within a certain solid angle to the incident particle (projectile) flux.
     
    Last edited: Nov 16, 2009
  17. Nov 16, 2009 #16
    Last edited: Nov 16, 2009
  18. Nov 19, 2009 #17
    So the cross-section of d-d would be different for a tokamak than for the NIF or cold fusion? I'd like to try to compare these. Does anyone know of good sources online or a primer of sorts? Or a source that explains why cross-section is not atomic radius squared? Thank you greatly.

     
  19. Nov 19, 2009 #18
    See Fig 1.4 on pdf page 6/9 in for cross section plots of D-D and D-T reactions.
    http://www.tdr.cesca.es/TESIS_UPC/AVAILABLE/TDX-0114104-103202//05CAPITOL1.pdf [Broken]
    10-28 m2 is 10-24 cm2 = 1 barn is more than geometric for a proton (~ 60 millibarns).

    (It is very difficult to find appropriate physics papers that are not pay per view)

    Bob S
     
    Last edited by a moderator: May 4, 2017
  20. Nov 19, 2009 #19

    Astronuc

    User Avatar

    Staff: Mentor

    One would have to compute the temperatures of the d-d plasma under NIF or cold fusion conditions.

    One thought on cold fusion was that the palladium atoms, or rather the electron field around the palladium nucleus, somehow allowed the deutrons to approach each other so that the fusion reaction could occur at lower temperature.
     
  21. Nov 19, 2009 #20

    mheslep

    User Avatar
    Gold Member

    http://books.google.com/books?id=7k...age&q=deuterium coulomb cross section&f=true"

    The fusion scheme described by the OP is typically called beam-beam or accelerator based fusion. These approaches can produce fusion, but they are hopelessly inefficient. The problem is that the coulomb (or Rutherford) scattering cross section is 1000x, maybe 10000x greater than the fusion cross section for a given beam/plasma energy, as indicated in Figure 11.3 of the reference. This means for every collision that succeeds in producing fusion, another 1000 will 'bounce' away. These bounced ions have begun to 'thermalize', or trend towards the average energy (temperature) of all the particles in the system, at which point we no longer have a beam with which to smash into something.

    Say the beam energy per ion is 50 keV, well up towards the sweet spot for D+T fusion cross sections. Then every successful D+T fusion releases 17 meV, but on average it required ~1000 x 50keV, or 50meV to be wasted.
     
    Last edited by a moderator: Apr 24, 2017
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Nuclear fusion using a plasma jet
  1. Nuclear fusion (Replies: 7)

  2. Nuclear fusion (Replies: 1)

Loading...