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Main >> Applications >> Sample problems

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. Water penetration into a concrete wall after five days of exposure, representing diffusion depth and concentration distribution according to Fick's laws.

Engineering question

How to calculate the water penetration into the concrete wall after 5 days exposure?

Answer
Use the equivalence between Fick's diffusion equation and the transient heat equation, map concentration to temperature and diffusion coefficient to thermal diffusivity, and calculate moisture penetration after 5 days.

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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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