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Neutron diffusion problem.

  1. Sep 29, 2007 #1
    It's part of my homework to find an analytical solution to a slab reactor with a reflector. The setup of the core is two slabs of equal thickness (-a to 0 and another 0 to +a) but different values for the diffusion coefficent, macroscopic absorption cross sections, and [tex]\nu \Sigma_f[/tex]. There is a reflector on each side and each has the same thickness (-b to -a and +a to +b) and material properties.

    As boundary conditions, I have that the flux is zero at the extrapolated distances from the reflector ([tex]\phi(X_{ex})=0[/tex], where [tex]X_{ex}[/tex] is the extrapolated distance). To link each part of the system together I have found that

    [tex]\phi_{r1}(-a) = \phi_{c1}(-a)[/tex]
    [tex]\phi_{r2}(a) = \phi_{c2}(a)[/tex]
    [tex]\phi_{c1}(0) = \phi_{c2}(0)[/tex]

    Using these conditions and solving the general time-independent diffusion equation for each section,I got

    [tex]\phi_{r1}(x) = A_1 sinh(\frac{x+X_{ex}}{L_r})[/tex]
    [tex]\phi_{r2}(x) = A_4 sinh(\frac{X_{ex}-x}{L_r})[/tex]

    for each reflector, where [tex]L_r[/tex] is the neutron diffusion length. In the core, I got

    [tex]\phi_{c1}(x) = A_2 sin(B_{c1}x)+C_2cos(B_{c1}x)[/tex]
    [tex]\phi_{c2}(x) = A_3 sin(B_{c2}x)+C_3cos(B_{c2}x)[/tex]

    where [tex]B = \frac{\nu\Sigma_f - \Sigma_a}{D}[/tex] (these three variables are defined in the problem).

    I then take the equations for flux of each half of the core and since they are equal at x=0, I find that [tex]C_2=C_3[/tex].

    Here's where I am somewhat stuck. I said the reflectors were symmetric about 0 so then


    and using the earlier mentioned continuity equations I get


    which is one equation with three unknowns (since [tex]C_2=C_3[/tex]).

    I'd like a little input into if I have all the conditions that link the sectors together and if my equations for the core are correct. I can't think of a reason not to have the [tex]Asin(B_cx)[/tex] term, but I can see that the [tex]Ccos(B_cx)[/tex] is necessary to prevent zero flux at x=0.
    Last edited: Sep 29, 2007
  2. jcsd
  3. Sep 30, 2007 #2


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    At each boundary, there is a flux and gradient of the flux for each energy group. The extrapolated BC at the outer edge of each reflector is OK. In dealing with asymmetric (material-wise) regions, one has to look at the current/gradient (J = -D[itex]\nabla\phi[/itex]) in the + dir and - dir, and these can be different.

    So in addition to flux, define the currents at the interfaces.
    Last edited: Sep 30, 2007
  4. Sep 30, 2007 #3
    Thanks, Astronuc. I did as you said and found the currents in each sector. For reference, I got





    Now, I know that at the reflector boundary, [tex]j_+(-X_{ex})=0[/tex] and [tex]J=j_++j_-[/tex] so I can calculate the current going the other direction at the edge:

    [tex]j_-(-X_{ex}) = \frac{D_rA_1}{L_r}[/tex]

    Now I've confused myself, I have current where flux is supposed to be zero. [tex]A_1=0[/tex] can't be true because then I'd have the solution that there is no flux across the slab. Is there something wrong with what I did/assumed?
  5. Sep 30, 2007 #4


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    That's fine. You CAN have current [ in the diffusion approximation ] when there is no flux.

    The current is proportional to the derivative of the flux. Just as in the center; you have a finite
    value for the flux, but the derivative of the flux [ and hence the current ] is zero by symmetry.

    When the flux hits zero at the extrapolated distance - it's not tending to zero asymptotically with
    a zero slope. No - it is going to zero with a finite slope; and hence current.

    For example, cos(x) goes to zero at 90 degrees. What is the derivative at 90 degrees?

    Well the derivative of cos(x) is -sin(x) and x is 90 degrees; so the derivative is -1.

    So as the value of x approaches 90 degrees; the value of the function goes to zero and the
    value of the derivative goes to -1.

    What's wrong with that?

    Dr. Gregory Greenman
  6. Sep 30, 2007 #5
    I thought current and flux were both the number of particles through a given area in a given time, the only difference being that current takes only particles in one direction?
    Last edited: Sep 30, 2007
  7. Oct 1, 2007 #6


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    The "directional flux" or "vector flux" that one solves for in transport theory is then number of
    particles through a given area....

    In the context of diffusion theory - you solve for the "scalar flux"

    The scalar flux is NOT the number of particles going in a given direction!!!

    The best way of thinking of the scalar flux would be the total pathlength of all the neutrons in an
    elemental volume divided by the size of that volume, per unit time.

    When you calculate neutron transport with a Monte Carlo code, for instance, that is EXACTLY
    how you tally the flux. You keep track of how far your simulation particle goes - direction independent -
    that is if the neutron goes to the left 2 cm and then to the right 2 cm after a collision - that's
    a total of 4 cm [ the fact that the directions are opposite does NOT cancel. ]

    Recall what the expression for a reaction rate is - it is the product of the cross-section and the
    scalar flux. The cross-section tells you the probability per unit pathlength of a reaction. The
    scalar flux gives you the total pathlength. So their product is a reaction rate.

    Understanding the difference between a "scalar flux" and a "vector flux" is VERY important in
    transport theory. A scalar flux is NOT the rate of flow of neutrons through an area. It just happens
    to have a name similar to the vector flux which is the number flowing through an area.

    But the scalar flux [ which is what you solve for in diffusion theory ] is a VERY different animal
    from the flow of neutrons through an area. It is NOT that.

    Dr. Gregory Greenman
  8. Oct 1, 2007 #7
    Thanks for clearing up my misconceptions on flux. Now about the current, I have a few questions.

    The current is proportional to the derivative of this scalar flux, which is in turn proportional to the pathlength. Then what exactly does it mean to have a negative change in pathlength? That the number of neutrons is reduced at a more significant rate than the total pathlengths of each neutron in your volume of interest are adding together? I am guessing that scalar flux can be constant if the pathlength is long enough and the medium has a low absorption cross section?

    Thanks for your patience with me.
  9. Oct 1, 2007 #8


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    I'm not sure what you are asking. In what context do you have a negative change in pathlength?

    Let me take a stab at it. Suppose you run a problem, and get a flux i.e. pathlength for a given region.

    Now you replace some material with something else - and the flux / pathlength decreases.

    That could either be because the material absorbed some neutrons - so you had few neutrons.

    It could also mean that you had just as many neutrons, but the new material was a non-absorbing
    scatterer and the energy of the neutrons is lower, so their speed is lower, hence lower pathlength
    traveled per unit time.

    Can you give me an example of the situation that still is of concern or question?

    Dr. Gregory Greenman
  10. Oct 1, 2007 #9
    I was thinking that in my assignment the reflector would result in a lower flux compared to the core. So in a comparison of one volume in the core to an adjacent volume in the reflector, one of these regions would have a shorter pathlength. As neutrons pass from one of these volumes to the other, they join the volume with the lower pathlength - so a decrease in flux from our standpoint. Is this kind of consideration incorrect?
  11. Oct 1, 2007 #10


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    Are you sure you are not confusing "pathlength" and "mean free path"?

    What I mean by "pathlength" is the total distance traversed by all the neutrons in a given volume
    in a given time. If you take that total pathlength and divide by the volume and the time; then the
    result is the scalar flux. That's how you tally flux in a Monte Carlo code for example.

    The scalar flux multiplied by the volume and a time interval gives you the total distance traveled
    by all the neutrons in that volume and in that period of time. Since the cross-section is the
    probability per unit length of getting a reaction; the product of the cross-section and the total
    pathlength [ i.e flux X volume X time ] is the number of reactions for that cross-section.

    But pathlength - and hence scalar flux - doesn't have to do with how many neutrons are crossing
    a given area per unit time. That is given by the vector flux.

    Dr. Gregory Greenman
  12. Oct 1, 2007 #11


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    This might be something of a useful reference.

    Elementary reactor theory
    http://lpsc.in2p3.fr/gpr/PPNPport/node20.html [Broken]

    I'm not sure what is meant by path-length, unless as Morbius indicated that one means mean free path.

    There is the displacement from the source which can be any direction. Reflectors would likely have different scattering properties than the core simply because they don't necessarily contain fissile material.
    Last edited by a moderator: May 3, 2017
  13. Oct 1, 2007 #12
    Don't forget your extrapolated distance b_ext after a and b; zo = 1 / 3D; b_ext = a + zo. Also, though I wanted to, I do not believe that the core can be homogenized. Lamarsh says that unless the distance in fuel is significantly less than your mean free, your collision probability is too high and therefore not realistic for flux calcs. I have noticed however, that the flux profile becomes flattened in reflected reactors. Seeing as how the Diffusion coeffcients and Macroscopic XSs are so close, and the cores and reflectors so large, I think that the profile may have a near-zero derivative for a while in the middle.

    If anyone can post any help about writting a MATLAB script for this problem I would definitely appreciate it.
  14. Oct 2, 2007 #13


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    "Pathlength" is definitely NOT the mean free path.

    I was admonishing Candyman not to CONFUSE pathlength with mean free path.

    Another way to interpret the scalar flux is to say that it is the total pathlength of all neutrons in a
    given volume per unit time.

    This is exactly how what is called a "pathlength tally" for flux works in a Monte Carlo transport code.

    In a Monte Carlo code, you are explicitly tracking simulation particles - each representing some
    number of real particles - called the weight of the particle. Now as you explicitly track each simulation
    particle from collision to collision and from material boundary crossing to material boundary crossing;
    you keep a running sum of the distance the particle moved. [ This is the NOT a vector quantity -
    that is with positive contributions for movement to the right and negative to the left. NO - it is ALL
    positive contributions. ]

    In a dynamic problem, time is discretized into time steps. For a given time step, and within a given
    discretized volume - you keep track of the distance traveled by each of the simulations neutrons in
    that volume. At the end of the time step; you multiply the total path length traveled by a neutron
    within that volume, and multiply by the weight of that simulation particle. You sum that quantity for
    ALL neutrons that spent time in that volume during the time step. You then divide this grand total
    by the size of the volume, and the length of the time step. THAT will be the scalar flux in that volume.

    Again consider the expression for a reaction rate per volume per unit time. That is given by the
    product of the scalar flux and the macroscopic cross-section. The macroscopic cross-section is
    the probablity per length of a reaction. So what quantity when multiplied by the probability per unit
    length will give you the reaction rate?

    Can you see that the quantity that needs to be multiplied by the probability per length has GOT to
    be the total distance traveled by ALL the neutrons in that given volume per unit time. That is the
    pathlengh I'm talking about. The pathlength divided by the volume and the time in which that
    pathlength was accumulated is the scalar flux.

    Dr. Gregory Greenman
    Last edited: Oct 2, 2007
  15. Oct 2, 2007 #14
    No, I understand what you meant (and your last post really cleared things up). I supposed I worded it badly when I said "total the pathlengths", which is what I thought you were supposed to do when you said that "[t]he best way of thinking of the scalar flux would be the total pathlength of all the neutrons in an elemental volume divided by the size of that volume, per unit time" earlier.

    What I'm trying to say is, looking at a comparison of two volumes next to each other, one yielding a different pathlength than the next, can you say what is happening to the flux (the number of particles per unit time per unit area, which I think is what I'm looking for) knowing that there is a change in the value of the pathlength in the successive volume?

    If there is an increase, you can say that there's a source. If it's decreasing, in the problem I laid out in my first post, you're in the reflector. How is this used to find the flux?
  16. Oct 2, 2007 #15


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    If we have two region of space that are equal in volume and one has a greater pathlength - then that
    one will have the greater flux.

    The total pathlength L is equal to the product of the scalar flux times the volume of the region.

    L [ neutron-cm ] = scalar flux [ neutrons per square cm per sec ] * Volume [ cubic cm ] * time [ sec ]

    If you multiply this total path length L by a cross-section; you will get the number of reactions.

    Dr. Gregory Greenman
  17. Oct 3, 2007 #16


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    Yes, that is my understanding of path length. I just wasn't sure if that was the intended reference.

    Well, using the relationship in Morbius's last post,

    Flux = (L / V)/[itex]\Delta{t}[/itex]. One would sum the pathlengths of individual neutron in a volume V and divide by that volume and time increment, which is what one would do with a Monte Carlo program.

    Otherwise, one solves the diffusion equation (or transport problem) analytically or numerically depending on the problem. The transport problem would be solved numerically.

    It's not only the absorption cross-section, but the total cross-section as well. Neutrons can scatter in and out of energy bands as well as spatial. The neutron population will fall off at the edge of the reflector or boundary of a bare core simply because there is nothing outside to cause the neutrons to scatter back into the reflector or core.
  18. Oct 3, 2007 #17


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    Yes - this is also an important point. At the edge of the reflector or boundary where you have nothing
    to scatter the neutrons back in - then you are definitely NOT in the diffusion regime.

    The whole assumption with diffusion theory is that the angular dependence of the vector flux is linearly
    anisotropic - that is the vector flux can be written as Phi(mu) = a + b*mu - where "a" is proportional to the
    scalar flux, "b" is proportional to the current, and "mu" is the cosine of the angle of the neutron's
    direction with respect to some coordinate axis.

    If you have a boundary, or edge; and vacuum outside - then clearly if you do a polar plot of the
    directional or vector flux - it drops suddenly to zero for directions that go from vacuum to core.

    Hence, in the vicinity of the edge - diffusion theory doesn't hold. However, by use of the
    extrapolation distance - which is derived from transport theory - one can "fudge" diffusion theory
    to give a better answer than it would by itself.

    Dr. Gregory Greenman
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