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Point Kinetics

  1. May 22, 2007 #1
    Consider a reactor being held sub-critical, Keff < 1, by control rods. The reactor has heavy water reflector and has been operatored for some time prior to the current state (we got Photo Neutrons). Keff is then bought to
    1.

    It seems to me, for Keff = 1 the neutron population should be increasing even if we have allowed enough time for Delayed and Photo N's to reach equalibrium. We have a source of neutrons from spontaneous fission in the fuel mix, 238 235, that would cause this to be the case. If my understanding of the point kinetics equations is right, then, with a constant source term there would be an increasing neutron population and the increase would be linear(???). The rate of increase would be imperceptable to an operator.

    For constant neutron population, constant power, at any operating power level, Keff < 1. Granted, very very close, but still less than one.

    Is my thinking correct?
    Can any provide an analytical solution (1 average delayed group with only thermal neutrons) with constant source term?

    Kind Regards
    PW
     
  2. jcsd
  3. May 23, 2007 #2

    Astronuc

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    Staff: Mentor

    No. The point of keff = 1 is that the neutron population includes delayed and source neutrons. The population due to prompt fission neutrons is simply n - delayed - source.

    For keff = 1, the number of neutrons produced directly by fission, delayed neutrons, and source must balance the neutrons being absorbed (to cause more fissions, or parasitically captured) and leaking from the system.

    Careful on the use of spontaneous fission which implies that the fissile nuclei undergo fission without an external (incident) neutron. Certaily in a fresh core with natural or enriched U, the bulk of fissions are from U-235, with some fast fission in U-238. As the reactor operates, some of the U-238 converts (is transformed to) Pu-239 (and higher istopes) which also experiences fission.

    Sure, but one should be able to derive it relatively easy. This would be a standard homework problem in reactor physics.
     
    Last edited: May 23, 2007
  4. May 25, 2007 #3
    I am new to this subject and texts that I have read suggest that with an independent source, independent of neutron flux, that there are a multitude of equalibrium (dn/dt = 0 ) neutron population levels for Keff < 1.

    For each distinct value of keff < 1 there is distinct value of neutron population, n is proportional to -Source/(keff-1). Indeed, I have an old British text that states something along the lines; "then, in the critical reactor, n and rho are related by the condition

    n = - (Source*tau)/(k*rho). " tau being the mean lifetime.

    Again, many texts, state that U238 "spontaneously fissions", so there is always an independent source in a core containing U238. Then, if Keff = 1 gives dn/dt = 0, what many in the industry call "critical", there is another value of keff that gives dn/dt = 0, for the same neutron population level (power).

    For example, if at start-up, Keff < 1, we double power a number of times the reactor can be at a very large power, call it P1, with Keff < 1, dn/dt = 0, with out ever going "critical " during the process. If we are at a larger power, P2 and Keff = 1 and dn/dt = 0 we can reduce P2 by inserting control rods causing Keff < 1, then at P1 remove control rods to re-establish dn/dt = 0. Which Keff value are we at, 1, or something less than 1?

    What I am having trouble seeing is that, for a distinct power level, there are two values of Keff (one being 1) that give dn/dt = 0.

    Yours in confusion
    rlr
     
  5. May 26, 2007 #4

    Astronuc

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    Staff: Mentor

    OK, I think I see the source of confusion (no pun intended).

    If there is a source, which is usually of constant strength, there will be a constant n(t), i.e. n(t) is proportional to S, whether k < 1 or k = 1. Only if k > 1 will n(t) increase. This also implies that the fissile source of neutrons is very low, i.e. [itex]\Sigma_f\phi[/itex] << S.

    If [itex]\Sigma_f\phi[/itex] > S, then if the reactor keff becomes < 1, then dn(t)/dt < 0, and the reactor will decrease in power until either keff = 1, which could mean the source of negative reactivity is removed, or it continues until the n(t) is again directly proportional to S, i.e. the neutrons from fission << S.

    A reactor can only increase in power, i.e. dn(t)/dt > 0 iffi keff or if the source strength is increased (e.g. by adding more sources).
     
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