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Ficks laws of diffusion

# Fick's laws of diffusion

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.

**Problem Type**

Plane-parallel problem of transient heat transfer.

**Geometry**

**Given**

Initial concrete internal humidity: *C0* = 10%;

Diffusion coefficient: *D* = 0.0001 cm^{2}/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 * d*^{2}C / dx^{2},

where *C(x, t)* is the water concentration, *D* is the diffusion coefficient.

Heat equation** for the one-dimensional case is:

*dT/dt = a * d*^{2}T / dx^{2},

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 /*C*ρ,

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 takenselected to specify the thermal diffusivity 0.0001 cm^{2}/s: λ = 0.028 W/K-m, *C* = 1273 J/kg-K, ρ = 2200 kg/m^{3}.

*a* = 0.028 W/K-m / (1273 J/kg-K * 2200 kg/m3) = 1e-8 m^{2}/s = 1e-4 cm^{2}/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.

**Reference**

*Wikipedia: Diffusion

**Wikipedia: Heat equation

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