examples.levelSet.electroChem.adsorption

This example tests 1D adsorption onto an interface and subsequent depletion from the bulk. The governing equations are given by,

$$\begin{array}{rl}{c}_t& =D{c}_{xx}\\ D{c}_x& =\mathrm{\Gamma }kc(1-\theta )\text{at }x=0\\ \text{intertext}andc& =c^{\mathrm{\infty }}\text{at }x=L\end{array}$$

and on the interface

$$D{c}_x=-kc(1-\theta )\text{at }x=0$$

There is a dimensionless number $M$ that governs whether the system is in an interface limited ($M\gg 1$) or diffusion limited ($M\ll 1$) regime. There are analytical solutions for both regimes. The dimensionless number is given by:

$$M=\frac{D}{L^{2}kcinf}.$$

The test solution provided here is for the case of interface limited kinetics. The analytical solutions are given by,

$$-D\ln (1-\theta )+kL{\mathrm{\Gamma }}_0\theta =\frac{kD{c}^{\mathrm{\infty }}t}{{\mathrm{\Gamma }}_0}$$

and

$$c(x)=\frac{c^{\mathrm{\infty }}[k{\mathrm{\Gamma }}_0(1-\theta )x/D]}{1+k{\mathrm{\Gamma }}_0(1-\theta )L/D}$$

Make sure the dimensionless parameter is large enough

>>> (diffusion / cinf / L / L / rateConstant) > 100
True

Start time stepping:

>>> currentTime = 0.
>>> from builtins import range
>>> for i in range(totalTimeSteps):
...     surfEqn.solve(surfactantVar, dt = dt)
...     bulkEqn.solve(bulkVar, dt = dt)
...     currentTime += dt

Compare the analytical and numerical results:

>>> theta = surfactantVar.interfaceVar[1]

>>> numerix.allclose(currentTimeFunc(theta), currentTime, rtol = 1e-4)()
1
>>> numerix.allclose(concentrationFunc(theta), bulkVar[1:], rtol = 1e-4)()
1

Last updated on Feb 14, 2025. Created using Sphinx 7.1.2.