# Steady state diffusion through three layers system

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

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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)

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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)

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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)

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(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

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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)

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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)

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(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)

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