Examples

Here are some examples to help you work out how to use the models and then to apply them to your own problems

Fickian Diffusion

EXAMPLE 1: How long does it take for the metal under the coating on a car to realize it started raining?

D0 = 1e-8 cm2/s
L = 163 microns
While one could do this as an absorption study, another way is to find the breakthrough time for a free film. Based on this time, you had better have a good primer to keep the water away from the metal.
The answer is 21.3 minutes

EXAMPLE 2: How long would it take for the lacquer in the above example to saturate?

This requires blocking the second side and allowing enough time (suggestion set the time at the maximum of 1000 min and see what happens)

The answer is indeed about 1000 minutes (about 16 hours) as can be seen by the square root of time plot.

Concentration Dependent Diffusion

If you expose a completely dry, but hydrophilic polymer film, such as polyvinyl alcohol, to water vapor such that it absorbs to a viscous liquid how long might this take neglecting any surface resistance?

EXAMPLE 3: Absorption of water from dry film (neglecting surface resistance):

L (dry) = 298 microns
C = 0.50 volume fraction
D0 = 1e-8 cm2/s
D30% = 8e-6 cm2/s

The half-time for absorption is calculated as with the process being completed at about 30 seconds. Do you believe this? No. So what is wrong? The surface resistance has been neglected. Would you try to dry a wet film at 50% solids and calculate this as a diffusion problem? Please do not. See below in the Surface Resistance section for a better result.

Surface Resistance

EXAMPLE 4: The continuation of the water absorption example in the Concentration Dependence section involves keeping the same data while adding a surface resistance with an h equal to about 1e-5 cm/s.

The result now shows the typical sigmoidal absorption curve with a time delay at the start. The half time for the absorption is given as 42.3 minutes. These results are in general agreement with those found in [Hansen, CM. “The significance of the surface condition in solutions to the diffusion equation: explaining “anomalous” sigmoidal, Case II, and Super Case II absorption behavior, European Polymer J., Vol. 46 (2010) 651-662] modeling the results of [Hasimi, et al. Transport of water in polyvinylacohol films: effect of thermal treatment and chemical crosslinking, European Polymer J. Vol. 44 (2008) 4098-4107.] where the half-time was found as 38.4 minutes using a somewhat more accurate procedure.

EXAMPLE 5: Solvent absorbing into polymer film - linear with time.

This example was intended to closely model the absorption of methanol into PMMA at 30°C with a 1 mm free film, where experimental data confirm a linear uptake with time. The limitations of time and the app have meant that the present example, while inspired by the above, is not a copy of it.
The following settings have been found useful:

L = 42 microns
C = 0.25 volume fraction
D0 = 2e-10 cm2/s
D30% = 9e-6 cm2/s, h = 1e-6 cm/s.

Check the boxes “Abs.”, “Blocked”, and “lin. T”. Use TimeOut equal to 1 min. or perhaps more depending on the speed of your computer.

The absorption curve is essentially linear with time (the settings could perhaps be adjusted more to give a more perfect straight line). The absorption curve shows a delayed start as a form of sigmoidal absorption when a square root of time plot is used. The concentration gradients appear mostly controlled by diffusion at 10 minutes, with mixed control 50 minutes, and complete surface control at 100 minutes.

Example 6: SUPER CASE II

As stated in the text, Super Case II is found when the rate of uptake is more rapid than linear with time. Such an example can be demonstrated with the following settings and a running time of 5 minutes.

L = 10 microns
C = 0.5 volume fraction
D0 = 3e-11 cm2/s
D30% = 9e-6 cm2/s
h = 3e-6 cm/s
time = 1e-12 minutes.

Calculating for one minute shows there is diffusion resistance at this time as well as surface resistance since the surface concentration is limited to 30% of the final value. After 3 minutes the concentration gradient profile is flat showing the surface resistance is now in control. It should be possible to play with these variables, especially C, to get a straight line on a linear time plot. Hint: With C = 0.4 and the above settings, there is a reasonable approach to a straight line, but improvements could be made. It is emphasized that this is only meant as a demonstration, and that the settings are not representative of any known practical situation.

Example 7: Petrol Permeation

One can estimate how much petrol can permeate a 0.5 mm thick container wall made of LDPE. Would this estimate justify using such a container?

In the concentration dependent section enter the following values:

L = 505 mm
C = 10% (approximately correct)
D0 is estimated at 1e-8 cm2/s

For the D30% in order to make the diffusion coefficient rise a factor of 10 over the concentration range of interest one must enter a value 3 orders of magnitude higher than the D0 used, this being 1e-5 cm2/s.

The curvature in the concentration gradient shows the concentration dependence of the diffusion coefficient. Whether or not this is a suitable container depends on circumstances, but with a breakthrough time under 2 hours, this would not be a promising candidate. It should also be noted that Petrol consists of many components and this would probably model the most volatile of these since these generally diffuse most rapidly. This example is not intended for other than demonstration of a type of calculation.

The official site of Hansen Solubility Parameters and HSPiP software.