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

Conversion vs breeding. Thermal and fast reactors.

  1. Feb 27, 2013 #1
    Hi there,
    I have came across a question in past exam papers that basically asks...

    What do the terms Conversion and breeding mean? I put conversion is the change from a fissionable element to a fissile element such as U238 to Pu239. When more of these fissile products are created than used (conversion gain > 1) breeding is achieved.

    Then the question goes on to ask "Explain why conversion is usual in thermal reactors but breeding is normally restricted to fast reactors".

    My best guess for breeding being more likely in fast reactors is that fast neutrons have a KE greater than 100Mev. The activation energy for U238 1.749Mev, therefore any energy between 100Mev and 1.749Mev will not be of the right energy to cause fission, therefore the neutron combines with the nucleus and the 2 beta decays can occur to breed fissile material.
    The proportion of energies lower than 1.749Mev will be lower than the proportion higher. Therefore conversion rate > 1 thus breeding occurs.

    As for the thermal reactor having conversions but no breeding, I have no idea and would appreciate some help please.

    Thank you.
     
  2. jcsd
  3. Feb 27, 2013 #2
    You still get some conversion in thermal reactors, just not enough to replace the fissile material lost.
     
  4. Feb 27, 2013 #3

    Astronuc

    User Avatar

    Staff: Mentor

    Firstly, fast neutrons do not have energies greater than 100 MeV! Fast neutrons are born as fission neutrons, the fission neutron spectrum is in the low MeV range.

    The most probable fast neutron energy is ~0.7 MeV and the average energy of fission neutrons is ~ 2 MeV. The neutron spectrum in a fast reactor is mostly between 10 keV to 10 MeV, and primarily between 20 keV and 500 keV. Pu-239 has a relatively low fission cross-section which is comparable in magnitude to the capture cross-section of neutrons in Pu-239 and U-238. In addition, there is usually a portion of the core, which has lower fissile content (compared to the driver fuel), which serves as a reflector and seed region. Fast neutrons in that region are more likely captured by U-238.

    In an LWR, Pu-239 has a relatively high fission cross-section in the thermal range - higher than U-235. The U-235 and U-238 are chemically intermingled in the enriched uranium (in the form of UO2). The U-238 is converted in the fuel, primarily in the rim region of the fuel pellets, and it is there that is subject to the same thermal flux as the U-235. As fuel operates, the buildup of Pu-239, Pu-240, Pu-241, shield the inner part of the fuel pellets from the thermal flux (self-shielding), and more fissions take place in the Pu-235 than in the U-235. Basically, thermal reactors burn Pu-239 almost as quickly as it is produced, but not quite. At an exposure of about 50 GWd/tU, in fuel with an initial enrichment of 5% U-235, nearly two-thirds of the fissions occur in Pu-239/Pu241, and the U-235 content has decreased substantially. The production of Pu-239/-241 is not sufficient to offset the build-up of fission products and depletion of U-235.
     
  5. Feb 28, 2013 #4
    Appologies! I meant to say 100 Kev. Thank you for your answer. Much appreciated!
     
  6. May 7, 2013 #5

    DEvens

    User Avatar
    Education Advisor
    Gold Member

    The buzz phrase there is neutron economy.

    To keep a reactor critical you can't lose too many neutrons. Every neutron that gets used in producing a converted nucleus is lost to the reaction. So, to produce more converted nuclei than are used up in the reactor, you actually have to use up more than one neutron per fission.

    A typical light water moderated thermal reactor isn't sufficiently economical with neutrons to be able to spare that many neutrons and still stay critical. Light water is a good moderator, but it also absorbs neutrons to form deuterons.

    One solution is to reduce the moderation and adjust the reactor so that it stays critical with faster neutrons. You stay away from energy regimes where your moderator absorbs strongly, and the neutrons have fewer moderation-scattering interactions so fewer chances to get absorbed.

    Another solution, not frequently used, is to use a different moderator. Heavy water isn't quite as efficient a moderator as light water, but it absorbs neutrons less. Not zero, note. The absorption reaction for heavy water produces Tritium, and that's a challenge. Eventually you find you need to remove the Tritium from the moderator in order to stay in your operation limits. Then you've got Tritium sitting around.

    But heavy water moderated reactors can have higher neutron economy. A group (At Chalk River? Not sure.) built a Thorium based, heavy water moderated, thermal breeder. There have also been some designs (not built as far as I know) of mixed Uranium-Thorium, heavy water moderated, thermal breeders.
    Dan
     
  7. May 14, 2013 #6
    Hey what is so bad about having tritium around? It's used for fusion no?
     
  8. May 14, 2013 #7
    To be honest this entire thread is confusiong.. how are conversions and breeding different?? They both turn fertile material to fissile material no?
     
  9. May 14, 2013 #8
    I think the main difference is whether your breeding ratio is greater than 1.

    In a breeder reactor, you can breed at least as much, if not more fuel than is consumed.

    In non breeder reactor types, you typically cannot breed more fuel than is consumed.

    A typical LWR will have bred plutonium at a fixed rate during its life, and by the time you replace the fuel, the oldest fuel bundles will typically have at least as much plutonium fuel as they do uranium.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Conversion vs breeding. Thermal and fast reactors.
Loading...