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    Steady state diffusion through three layers system

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    Consider a system that consists of a flat layer of material of thicness L that creates radioactive particles surrounded on each side by a very thic layer of material that absorbs the partices that are produced by the radioactive layer.... see attached

    © BrainMass Inc. brainmass.com October 5, 2022, 1:40 pm ad1c9bdddf
    https://brainmass.com/math/functional-analysis/steady-state-diffusion-through-three-layers-system-577351

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    A condensed version of the solution is provided below. Please see attached for full workings, with equations.

    (a) The equations are:
    (1.1)
    Since the system is invariant under translation in the xy plane, the concentration profile is a function of time (t) and altitude (z) only.
    In a steady state, the concentration profile, by definition, does not change with time:
    (1.2)
    The steady state concentration profile is a function of z alone So the equations for the steady state become standard ordinary differential equations:
    (1.3)

    For equation (1.3.1) we get:
    (1.4)


    Since are constants, integrating twice with respect to z yields the general solution:

    (1.5)
    Where the I subscript indicates this is the concentration for the inner layer and are arbitrary constants. Remember that

    Equation (1.3.2) is a second order equation with constant coefficients:
    (1.6)
    Since we define real k such that and the equation becomes:
    (1.7)
    We "guess" a solution in the form
    (1.8)

    Plugging it into (1.7) we get:

    (1.9)
    So the general solution is a linear combination of the two possible solutions:
    (1.10)
    Where are arbitrary constants. The O subscript denotes the solution for the outer layer.
    To summarize, the most general solutions for the steady state diffusion equations are:

    (1.11)

    And remember that we defined the positive constants:
    (1.12)


    (b)
    Now, we note that the solution for the inner layer is a quadratic function of z (that is, a parabola).
    If we require that the solution is even in z, that is
    (1.13)
    We get:

    (1.14)
    So the even solution is:
    (1.15)

    The general solution for the outer layer is:
    (1.16)
    For the solution to be physical, it must be finite as .
    Since the term explodes as and the term explodes as

    So if we divide the space into two subspaces, namely the upper layer where and the lower layer where we get two physical solutions:

    (1.17)

    (c)
    We have three unknowns, namely the constants and we have four boundary conditions.
    The continuity of the concentration at yields:

    (1.18)
    The continuity of the concentration at yields:

    (1.19)

    The continuity of the flux at

    (1.20)
    The continuity of the flux at

    (1.21)
    So the equations for the constants are:

    (1.22)

    We have to add an implicit condition where at z=0 the concentration is finite, that is:
    (1.23)

    (d)
    Examining the equations (1.22) it is obvious that
    (1.24)
    And that (remember that where is the finite concentration at the origin):

    (1.25)
    So the constant values are:
    (1.26)
    Where is the constant steady state concentration at the center and
    (1.27)
    So the solution for this system is:
    (1.28)

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com October 5, 2022, 1:40 pm ad1c9bdddf>
    https://brainmass.com/math/functional-analysis/steady-state-diffusion-through-three-layers-system-577351

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