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How many barns at least for a commercial nuclear reactor?

  1. Apr 19, 2016 #1
    The conventional commercial fission reactors use uranium-235 as fuel. Its cross section of (n, fission) reaction at thermal neutron is about 585 barns.
    My question is:
    Is there a known threshold of the cross section that makes a nuclear reaction not sustainable if the minimal cross section is not reached?
    For example, perhaps if the U-235(n, fission) = 580 barns, I guess it still works, but if only 100 barns, still work?
    It is only a must-have condition of sustainable chained reaction that the new generated neutrons should be more than used one, but it is not the full-fill condition. Big enough cross section is the another must-have condition, just wondering how big is narrowly big enough.
    For cancer therapy, the cross section of Gd-157 neutron capture is 253366 barns. Obviously it is too much, even it decreases to 585 barns, i.e. same with U-235, it should still work.
     
    Last edited: Apr 19, 2016
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  3. Apr 19, 2016 #2

    mfb

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    Those two things are not independent. The average fission yield from the neutrons of a fission reaction depends on the cross section.

    Fission reactors could work with significatly reduced U235 fission cross sections with a higher enrichment and heavy water. I don't have numbers (studies about parallel universes are rare...), but that should easily give a factor of more than 2.
     
  4. Apr 19, 2016 #3
    Do you mean the possible answer is 2 barns (e.g for 100% enriched fissile fuel)? Thanks.
     
  5. Apr 19, 2016 #4

    mfb

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    A factor of 2 (reduction of course) applied to 585 gives about 290 barn, but that value is just a guess.
     
  6. Apr 19, 2016 #5
    As per your guess, considering the the U-235 concentration of commercial fuel rod 5%, so the minimal cross section for any profitable reactor is circa 290*5%= 14.5 barns.
    Using above estimation, can we certainly predict all fusions reactor will never success?
    Because even the prominent D+T is only 5 barns, not to mention the long time boasted aneutronic p-B11 reactor only 0.8 barns at incredible temperature 650KeV.
    So, all fusion researchers should stop their work, in lest of wasting money?
     
  7. Apr 19, 2016 #6

    mfb

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    That calculation doesn't make sense at all.
    No, fusion reactors are completely different things, with different processes in different conditions.

    I don't get your obsession over cross section values. They are completely meaningless without looking at the cross sections of competing processes.
     
  8. Apr 19, 2016 #7

    Astronuc

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    One is grossly simplifying a complex physical problem. Light water reactors work with a spectrum of neutron energies, from the MeV range for fission neutrons to the thermal energies of fractions of an eV. When one takes into account all possible reactions, one neutron must remain from the each fission to cause one more fission, which is the definition of criticality.
     
    Last edited: May 8, 2016
  9. Apr 27, 2016 #8
    Try to clarify your question and its purpose, may be that will help find a suitable answer.

    I can not understand what you want to say from this part. What did you mean by mentioning the high cross section of Gd-157? I suppose Gd-157 is used as a localized gamma source to destroy the cancerous cells. The mechanism is as follows: the Gd-157 is attached to the infectious cells, it is bombarded with neutron flux, Gd-157 absorbs high number of neutrons which makes a subsequent high number of atoms produce gamma through (n,gamma) reaction, and eventually the high energetic gamma radiation destroys the cancerous cells. Note that in this specific application we need an isotope with (n,gamma) cross section as high as it can get!

    Note the need for (n,gamma) reaction here, while the need for fission reaction in the fission nuclear reactor.

    On the other hand, I suppose you need to differentiate between two key terms you may find in Nuclear literature: microscopic cross section (indication of the probability that a single neutron interaction with a single nucleus) and macroscopic cross section (indication of the probability that a single neutron will interact with the material - a bulk of nuclii).

    Simplifying stuff, the one threshold that a nuclear reactor designer would care about is that: CRITICALITY ! Criticality depends upon various factors including the neutron interactions cross sections of the fuel and the moderator -including fission-, the geometrical shape of the system and the mass composition of the system.
     
  10. Apr 27, 2016 #9
    Neutron is a special economy for commercial purpose.

    You emphasizing criticality is right, but if the cross section of U235(n,f) were only 0.001b, will you still interest in criticality? Sure, you won't, because of not economical.

    I guess the neutron economy threshold index is same for all applications, that is why I am interested in the unkown universal threshold value.

    In fact, for cancer therapy, the cross section of Gd-157 does not have to be so high, as large as no waste of neutron, then it is OK. For example, the other cancer therapy choice is Boron-10(n, alpha + gamma), its cross section only 3840b. That means the economic cross section must be less than 3840b, even the same level with fission reactor.
     
  11. Apr 27, 2016 #10

    mfb

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    It is not. It's like asking "what is the minimal area to play a game?" Well, for football you need more space than for chess.
     
  12. Apr 27, 2016 #11
    So, our question here is: what is the universal threshold value of microscopic cross section?

    Well, the answer is: the question is missing more data!

    Let me show you what I mean.

    I will assume that the question is about applications in fission nuclear reactors only.

    Now, assume I told you that the answer is 500 barn for (n,fission). Okay? So nice and direct.
    Few seconds later I would think.. " 500 barn for (n, fission) ! What about the (n,gamma) ? This nice isotope with the 500 barn may has high (n,gamma) reaction ! This means that it may absorb more neutrons than it emits! Then we can not use this isotope for nuclear reactors! And wait! What about the shape of the reactor? This will lead to leaking of neutrons out of the system! .. and so on".

    My point here is that you are asking about a factor that is not indicative.

    Nuclear Engineering is science combined with art. It is not that direct and stuff are not that straight forward. A lot of variables, and a lot of phenomenon to be taken in consideration for the design of each application/experiment. So, unfortunately, there is nothing that simple and that universal.
     
  13. Apr 28, 2016 #12
    When the cross section of (n, fission) is assessed, the pure (n, gamma) without followed fission is ignored, so the 2 data do not interfere.
    For thermal neutron incident, there must be X(n, fission) = X(n, gamma) + SF(Spontaneous Fission of new isotope which atom mass is increased with 1).
    Neutron leaking will degrade performance, so special design is needed.
     
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