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New Flux in a Sphere of moderator of Radius R

  1. Mar 7, 2006 #1
    I have a problem where I must show that neutron flux in the sphere, for a sphere of moderator of radius R. This is one of the problems (5.14.a) in chapter 5 of "introduction to Nuclear Engineering" by J. R. Lamarsh. I would like to include the equation in this request but I'm unable to paste it in.

    Anyone can tell me how do I go about solving this problem, I would appreciate it.
  2. jcsd
  3. Mar 7, 2006 #2


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    Is that the 3rd edition of Lamarsh. I have the 1st edition, but I just ordered the third. The problem numbers are different.

    If the problem is a point source then see this thread,

    If the problem is a distributed source,

    please show your work, i.e. the governing differential equation, i.e. the diffusion equation for spherical geometry, and the boundary conditions.

    Then we can assist you.

    BTW, this is a homework problem and actually belongs in the Engineering Homework section.
  4. Mar 7, 2006 #3
    Lucky mkhonza, you can use Latex to post equations. Read abou it here: https://www.physicsforums.com/showthread.php?t=8997.

    I will save you the time and post the problem.

    A sphere of moderator of radius R contains uniformly distributed sources emitting S neutrons/cc- sec.

    Show that the flux in the sphere is given by
    [tex]\phi = \frac{S}{\Sigma_a}[1 - \frac{R+d}{r}\frac{sinh(\frac{r}{L})}{sinh(\frac{R+d}{L})}][/tex]

    You can see my other thread, but since it deals with a single source, the problem is going to have a different differential equation and hence, a different solution for the flux.

    EDIT: S is divided by the macroscopic absorption cross section, not the microscopic absorption cross section. I cannot get a capital sigma in latex.

    EDIT 2: I fixed the sigma. Thanks, Grogs.
    Last edited: Mar 7, 2006
  5. Mar 7, 2006 #4
    In the 2nd edition, it's problem 5-16a. I can't speak for the 1st though.

    Try this:


    @Lucky mkhonza:

    This problem is going to look a lot like the infinite reactor with a point source. The differences are:

    1. You'll have a source term in the diffusion equation, so it will be inhomogeneous.

    2. Your boundary conditions will be different.

    The diffusion equation will take the form:

    \nabla^2\phi - \frac{1}{L^2}\phi = -\frac{S}{D}

    where S is a constant in this case.

    This is subject to the following Boundary Conditions:

    [tex]\lim_{\substack{r\rightarrow 0}} \phi(r) < \infty

    \phi(R+d) = 0

    You've got an inhomogeneous equation here. There are a few different ways to solve it, but I find the variation of constants easiest myself. The general solution will take the form of the homogeneous portion (the solution for a point source problem) plus the inhomogeneous part. The inhomogeneous part will take the form of your source term, or in this case, a constant. Solve for your inhomogeneous constant by substituting it back into the diffusion equation as [itex]\phi[/itex] then proceed to apply BC's and solve as you would any other problem.
    Last edited: Mar 7, 2006
  6. Mar 8, 2006 #5
    Thanks for all your replies...let met try

    Here is the Diffusion Equation (in spherical coordinates) for the problem

    [tex]\nabla^2\phi - \frac{1}{L^2}\phi = -\frac{S}{D}[/tex]

    general solution

    [tex]\phi = A \frac{e^{-\frac{r}{L}}}{r} + B \frac{e^{\frac{r}{L}}}{r} + C[/tex]

    A, B (homogeneous solution) and C (non-homogeneous solution) are constants.

    Substituting the above equation into Diffusion Equation gives C.

    [tex]C = \frac{SL^2}{D}[/tex]

    therefore: [tex]\phi (R+d) = A \frac{e^{-\frac{R+d}{L}}}{R+d} + B \frac{e^{\frac{R+d}{L}}}{R+d} + \frac{SL^2}{D} = 0[/tex]

    Applying boundary conditions

    [tex] B= -Ae^{-\frac{2(R+d)}{L}} - \frac{SL^2(R+d)}{D}e^{-\frac{(R+d)}{L}}[/tex]

    Substituting B

    [tex]\phi = \frac{A}{r}e^{-\frac{r}{L}} - \frac{A}{r}e^{\frac{r}{L} - \frac{2(R+d)}{L}} + \frac{SL^2(R+d)}{Dr}e^{\frac{r}{L} - \frac{(R+D)}{L}} + \frac{SL^2}{D}[/tex]

    From here I used Fick's Law to solve for A but I get a mess of terms which at the end I don't get the correct answer.

    I will appreciate if anyone comments on my work above.
    Last edited: Mar 8, 2006
  7. Mar 8, 2006 #6
    I believe your particular solution will also have a r term in it. I calculated the particular solution as follows.

    [tex]\phi_p = Er + F[/tex]
    Guess the solution.

    [tex]\phi_p = \frac{Er}{L^2} - \frac{F}{L^2} = -\frac{Sr}{D}[/tex]
    Solving for the constants.

    [tex]E = \frac{L^2S}{D}[/tex]
    E is solved, there is no constant in the answer so F must equal zero.

    [tex]\phi(r) = A\frac{e^{-\frac{r}{L}}}{r} + B\frac{e^{\frac{r}{L}}}{r} + Er[/tex]
    The flux for a multisource sphere.
    Last edited: Mar 8, 2006
  8. Mar 8, 2006 #7
    I followed your suggestion but I'm getting a different E than yours. I believe you used the spherical coordinates

    [tex]E = \frac{L^2Sr}{D(2L^2 - r^2)}[/tex]

    I will carry on to see if I could get the correct answer.

    Thanks to all......
  9. Mar 8, 2006 #8
    Try this form for a general solution:

    [tex]\phi = A \frac{\sinh{-\frac{r}{L}}}{r} + B \frac{\cosh{\frac{r}{L}}}{r} + C[/tex]

    I was taught that if the inhomogeneous part is a polynomial, your solution should be of the same order (0 in this case,) so I agree with the 'C' you posted above. You can simplify that to:

    [tex]C = \frac{SL^2}{D} = \frac{S}{\Sigma_a}[/tex]


    [tex]L^2 = \frac{D}{\Sigma_a}[/tex]

    With sinh and cosh, when you apply the [itex]\lim_{\substack{r\rightarrow 0}} \phi(r) < \infty[/itex] BC, you'll get B=0, so you're left with:

    [tex]\phi = A \frac{\sinh{-\frac{r}{L}}}{r} + \frac{S}{\Sigma_a}[/tex]

    Solve for A using the extrapolated boundary condition, and you should be able to get the answer you're looking for after a little bit of algebra.
  10. Mar 8, 2006 #9
    If the nonhomogenous part of the equation does not have an r term in it, it will disappear out of the equation after you apply Fick's law though. Is the current at the center of the sphere zero?
  11. Mar 8, 2006 #10
    It should be. You've got a uniformly distributed source, so if you pick an infinitely small sphere, i.e., a point, you have no source there. At the center of the sphere, you've got the same amount of material -> the same source in every direction, so there should be no net current.

    If you substitute a constant into the diffusion equation:

    [tex]\nabla^2 C- \frac{1}{L^2} C = -\frac{S}{D}[/tex]

    the first term is zero and you end up with:

    [tex]-\frac{C}{L^2} = -\frac{S}{D}[/tex]

    which makes sense. If you have an r in your source (inhomogeneous) term, the flux will actually increase as you move away from the center of the sphere, which makes no sense for a uniformly distributed source.

    EDIT: OK, I see the problem. In this equation:

    [tex]\phi_p = \frac{Er}{L^2} - \frac{F}{L^2} = -\frac{Sr}{D}[/tex]

    you shouldn't have the 'r' in your source term. That's a linearly dependent source term, not a uniform source term, which should just be a constant. If the right-hand side is [itex]-\frac{S}{D}[/itex] you end up with E = 0 instead of F.
    Last edited: Mar 8, 2006
  12. Mar 8, 2006 #11


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    Assuming 'uniformly' distributed source, uniform BC and symmetry, yes, current is zero at center.
  13. Mar 9, 2006 #12
    Aha...you guys rock.

    [tex]\phi = A\frac{\sinh({-\frac{r}{L})}}{r} + \frac{S}{\Sigma_a}[/tex]

    Solving for A using the extrapolated boundary condition, this is what I get.

    [tex]A = -\frac{S}{\Sigma_a} \frac{(R+d)}{ \sinh (-\frac{R+d}{L})}[/tex]

    Substituting A

    [tex]\phi = \frac{S}{\Sigma_a}[1 - \frac{R+d}{r}\frac{sinh(-\frac{r}{L})}{sinh(-\frac{R +d}{L})}][/tex]

    using the following relations of Sinh

    [tex]\sinh({-\frac{r}{L}}) = -\sinh({\frac{r}{L}})[/tex]


    [tex]\sinh({-\frac{R+d}{L}}) = -\sinh ({\frac{R+d}{L}})[/tex]

    I get the answer I have been looking for.

    [tex]\phi = \frac{S}{\Sigma_a}[1 - \frac{R+d}{r}\frac{sinh(\frac{r}{L})}{sinh(\frac{R +d}{L})}][/tex]

    Thank you so much guys...you have made my day.
    Last edited: Mar 9, 2006
  14. Mar 9, 2006 #13
    Arrgh! I just noticed that minus sign in the SINH term. It shouldn't be there. I cut and pasted the latex from the exponential form and forgot to take it out. :blushing: It doesn't change the answer in this case, but in a non-symmetrical problem, I think it would. That general form should read:

    [tex]\phi = A \frac{\sinh{\frac{r}{L}}}{r} + B \frac{\cosh{\frac{r}{L}}}{r} + C[/tex]
  15. Mar 9, 2006 #14
    Will it make any difference if someone used your first suggestion besides considering non-symmetrical problem? that's using..

    [tex]\phi = A \frac{\sinh{-\frac{r}{L}}}{r} + B \frac{\cosh{\frac{r}{L}}}{r} + C[/tex]

    Obviously using

    [tex]\phi = A \frac{\sinh{\frac{r}{L}}}{r} + B \frac{\cosh{\frac{r}{L}}}{r} + C[/tex]

    one get directly to the solution

    [tex]\phi = \frac{S}{\Sigma_a}[1 - \frac{R+d}{r}\frac{sinh(\frac{r}{L})}{sinh(\frac{R +d}{L})}][/tex]

    Thanks Grogs

    Another thing...How do I include someone's QUOTE on my reply?
    Last edited: Mar 9, 2006
  16. Mar 9, 2006 #15


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    Hit the quote button below the particular post you wish to quote, but it will not catch other quotes in that post.

    Alternatively, copy the text, paste it and put [Q=person] at beginning and [/Q] following text, where Q=quote.

    Also and , where S=sup for superscript and S=sub for subscript
  17. Mar 9, 2006 #16
    Well, like I said, it shouldn't make a difference on this problem. My only concern is that, if it really does make a difference on *some* problems - and I honestly can't say for certain if it does, your instructor might take off a point or two because the equation is technically wrong. I'm probably in a similar class to what you're in right now (I'm taking reactor theory) and I've got a pretty picky instructor, so I try to avoid all those little errors so I don't get lots of little deductions on my HW problems.

    I guess a better way of putting it is I know the
    [tex]\phi = A \frac{\sinh{\frac{r}{L}}}{r} + B \frac{\cosh{\frac{r}{L}}}{r} + C[/tex]
    form works, so that's what I use. :smile:
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