MCT does not reproduce LRM with linear binding via exchange

[jbreue16]

Expected Behavior

An MCT with two channels, where one channel acts as the solid phase with transport disabled and the other acts as the liquid phase, with transport, should be able to reproduce the results of an LRM with linear binding, when exchange rates are choseing according to adsorption and desorption.

Actual Behavior

The simulations yield different results

So there is either a Bug or an error in the setup. Once this is fixed, this would make a good addition to our tests.

[jbreue16]

@AntoniaBerger

Can you please create a branch with the corresponding test script ?

[AntoniaBerger]

Unlike the LRM, the MCT model takes into account the volume structure of the exchange for each collum. In the two-channel case above, the equations are given by $\frac{\partial c^\ell}{\partial t}=- u\frac{\partial c^\ell}{\partial z}+D_\text{ax} \frac{\partial^2 c^\ell}{\partial z^2} + k_d c^s -k_a c^\ell A_1 / A_0$ $\frac{\partial c^s}{\partial t}= - k_d c^s+ \color{red}{k_a c^\ell A_1 / A_0}$.

The general LRM equations differ in the second equation i.e: $\frac{\partial c^\ell}{\partial t} + \frac{1}{\betat} \frac{\partial c^s}{\partial t}=- u \frac{\partial c^\ell}{\partial z} + D\text{ax} \frac{\partial^2 c^\ell}{\partial z^2}$ $\frac{\partial c^s}{\partial t}=k_a c^\ell - k_d c^s$.

If we normalize the adsorption factor of the MCT (i.e. $k_a^{MCT} =k_a^{LRM} \cdot A_0 / A_1)$ the LRM and MCT align. An other case is if $A_0 = A_1$ (see below). Where $t_R$ represents the analytical retention time. In the gerneral case this is not the case (see one examole below).

[jbreue16]

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