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Fick's laws of diffusion - QuickField simulation example

An example is devoted to modeling the diffusion of water through a concrete wall. The diffusion equation (Fick's law) and the heat transfer equation have similarity. To simulate the diffusion process the heat transfer problem is calculated.

Engineering question

How to find Fick diffusion law simulation in solids?

Answer
Set up a plane-parallel QuickField Transient Heat Transfer problem for diffusion in solids and evaluate diffusion behaviour from computed field results.

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Typical applications
concrete diffusion media, porous construction materials, moisture transport structures

Fick's laws of diffusion

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Simulation problem

Problem Type
Plane-parallel problem of Transient heat transfer.

Geometry
Fick's laws of diffusion Estimate the water penetration into the concrete wall after 5 days exposure. Air Concrete Water 10 cm

Given
Initial concrete internal humidity: C0 = 10%;
Diffusion coefficient: D = 0.0001 cm²/s;

Task
Estimate the water penetration into the concrete wall after 5 days exposure.

Solution
The diffusion equation* (Fick's law) for the one-dimensional case is:
dC/dt = D * d²C / dx²,
where C(x, t) is the water concentration, D is the diffusion coefficient.

Heat equation** for the one-dimensional case is:
dT/dt = a * d²T / dx²,
where T(x, t) is the temperature, a is the thermal diffusivity.

The diffusion equation (Fick's law) and the heat transfer equation have similarity. The role of concentration in the diffusion equation is the same as a role of the temperature in the heat transfer equation. And the role of the diffusion coefficient D is performed by the thermal diffusivity, which is calculated as a = λ /Cρ,
where λ is thermal conductivity, C is specific heat capacity and ρ is density.
We can choose any values of λ, C and ρ to achieve the desired thermal diffusivity value. In the thermal problem, the following values were taken to specify the thermal diffusivity 0.0001 cm²/s: λ = 0.028 W/K-m, C = 1273 J/kg-K, ρ = 2200 kg/m³.
a = 0.028 W/K-m / (1273 J/kg-K * 2200 kg/m3) = 1e-8 m²/s = 1e-4 cm²/s

Result
After 5 days, the humidity of concrete on the air-side of the wall will be 56%, and the average volume moisture content will be 72%.
The distribution of humidity (temperature) inside the wall after 5 days exposure is shown in the figure.
humidity inside the wall

Reference
*Wikipedia: Diffusion
**Wikipedia: Heat equation

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