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Permeability is usually measured on flat specimen and it generally has these units: [amount of stuff]-[distance through which it diffused]/[surface area diffused through]-[time to diffuse]-[driving pressure].

Assuming you use a flat specimen for your material characterization, the surface area used is obvious and is the same on both sides of the specimen. With a tube, however, the surface area is different inside and out. Presumably the actual surface you should use is some virtual surface between the inner and outer surfaces because the stuff inside has room to spread out as it diffuses through, but I'm honestly not sure where it should be. I could also see the argument that you use the inner surface as it's the smallest and therefore rate limiting.

Stated another way, in the diagram I've drawn, what is the r between r_inner and r_outer such that diffusion of blue into the environment is equal in both configurations?

diffusion path graphic

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  • $\begingroup$ If I had to guess, it's probably the average radius of the annulus, but that's probably loaded with some assumptions about diffusion behaving very linearly and I'm not convinced my intuition is right here. $\endgroup$
    – William Grobman
    May 29, 2018 at 21:16

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For the case where the wall thickness is small compared to the diameter of the cylindrically-shaped container, you can use the inside area of the tube with little error.

For the case where the wall thickness is of order ~tube diameter, a reasonable estimate would use the average radius to calculate an "average" area.

To get a better estimate in the case of a thick wall and small tube diameter, you can either find a solved problem that furnishes an expression for the 'effective" area to use, or do a finite-element model of the problem.

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  • $\begingroup$ What type of average would you suggest? Could be as simple as a midpoint or weighted by the extra material in the outer layers by just integrating. $\endgroup$
    – William Grobman
    May 30, 2018 at 0:29
  • $\begingroup$ try simple midpoint first, but be prepared to go to finite element modeling for accuracy. $\endgroup$
    – niels nielsen
    May 30, 2018 at 1:37

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