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Homework Help: Diffusion in different regions

  1. Feb 25, 2015 #1
    1. The problem statement, all variables and given/known data
    A slab of radioactive material of thickness L lies in the x-y plane surrounded by a material that can be thought of as extending to +/- infinity from -L/2 and +L/2. The system is in steady. Find the general solution for the diffusion equations in each region(ok). Find the form of the general solution in Region 1 that is an even function and in 2 that is physically meaningful(ok). Find the general solution that satisifes the boundary conditions(?)

    2. Relevant equations
    Region 1
    [tex]0=D_1 \frac{\partial^2 c}{\partial x^2}+H[/tex]
    for [itex]\left| z \right|<L/2[/itex]
    Region 2
    [tex]0=D_2 \frac{\partial^2 c}{\partial x^2}-Rc[/tex]
    for [itex]\left| z \right|>L/2[/itex]

    3. The attempt at a solution
    The general steady state solutions for each region are
    [tex]c=Ce^{\lambda z}+De^{-\lambda z}[/tex]
    [itex]\lambda=\pm \sqrt{\frac{R}{D_2}} [/itex]

    In region 1 to be even A=0, for region 2 the concentration must tend to zero as z approaches +/- infinity, so


    De^{-\lambda z} & z>\tfrac{L}{2}\\
    Ce^{\lambda z} & z<-\tfrac{L}{2}

    The flux is the same on both sides and concentration continuous. I am having difficultly picturing these, what is happening at z=0 and z=+/-L/2.
    Any pointers would be greatly appreciated.
  2. jcsd
  3. Feb 27, 2015 #2


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    Staff Emeritus
    Science Advisor

    At the outer boundaries, are they nonreentrant, i.e., once a particle leaves, it's gone, so there is no return current.

    What is the boundary condition at the interface? Think about the current.
  4. Feb 28, 2015 #3
    At z = L/2, ##D_1\frac{dc}{dz} = -\frac{HL}{2}##. This is the other boundary condition that follows from integrating the differential equation once.

    Also, at z > L/2, express c as : ##c(z) = Ec^{-λ(z-\frac{L}{2})}##

    The mass fluxes in the two regions must match at z = L/2.

    The continuity of concentration at the boundary gives:

    Physically, what's happening is that, by symmetry, there is no mass flow across the boundary z = 0, and all the mass generated between z = 0 and z = L/2 has to pass through the boundary at L/2. This is HL/2. This mass flow is then reacted away in the region beyond z = L/2.

    Last edited: Feb 28, 2015
  5. Mar 4, 2015 #4
    Just seen your replies.
    I ended up down the lines you have mentioned. Needed to think about the flux at the boundary and the equations being continuous at the boundary.
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