MCT to reproduce LRM with linear binding via exchange

Chromatography Modeling
1

We had problems reproducing a LRM with linear binding using the MCT. I’ve thus written down the equations to see if theyre really equivalent.

MCT model

Assuming there are no reactions, the MCT model equations are given for all channels l\in\{1,\dots,N_k\} and components i\in\{1,\dots,N_c\} as

\frac{\partial c_{i,l}^\ell}{\partial t} = - u \frac{\partial c_{i,l}^\ell}{\partial z} + D_\text{ax} \frac{\partial^2 c_{i,l}^\ell}{\partial z^2} + \sum_{k=1}^{N_k} e^i_{kl} c_{i,k}^\ell A_k / A_l - e^i_{lk} c_{i,l}^\ell,

with N_k being the number of channels, e^i_{kl} being the exchange rates for component i from channel k to channel l, c_{i,k}^\ell being the liquid phase component i in channel k and v, D_\text{ax} can be independently set for each channel.

Replicating the linear LRM with a two channel MCT

The equations for a LRM with linear binding are given as

\frac{\partial c^\ell}{\partial t} + \frac{1}{\beta_t} \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

We try to replicate the above by an MCT with two channels, where one channel accounts for the liquid phase and the other accounts for the solid.
First, we have to setup the correct volumes, cross section areas of the two channels.
In the LRM we have \beta_t = \frac{\epsilon_t}{(1-\epsilon_t)} and the volume of liquid and solid phase is given by V \epsilon_t and V (1-\epsilon_t) respectively, with V being the volume of the column.
We get the two cross section areas of the channels by A \epsilon_t and A (1-\epsilon_t) respectively, with A being the cross section area of the LRM column.
So we get the relation A_2 / A_1 = (1-\epsilon_t) A / \epsilon_t A = \frac{1}{\beta_t}

For the second channel, we set v=D_\text{ax}=0 and set exchange paramters to the binding parameters, i.e. e_{12} = k_a, e_{21} = k_d.
This way, we get

\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 A_2 / A_1 - k_a c^\ell \\ \frac{\partial c^s}{\partial t} = k_a c^\ell A_1 / A_2 - k_d c^s

We add \frac{1}{\beta_t} times the second channel equation to the first channel equation and substitute \beta_t for the cross section divisions and get

\frac{\partial c^\ell}{\partial t} + \frac{1}{\beta_t} \frac{\partial c^s}{\partial t} = - u \frac{\partial c^\ell}{\partial z} + D_\text{ax} \frac{\partial^2 c^\ell}{\partial z^2} \underbrace{+ k_d c^s \frac{1}{\beta_t} - k_a c^\ell + k_a c^\ell - \frac{1}{\beta_t} k_d c^s}_{=0} \\ \frac{\partial c^s}{\partial t} = k_a c^\ell \beta_t - k_d c^s,

which is slightly different from the LRM equations.

There is no difference between the two, when \beta_t=1, i.e. when \epsilon_t = 0.5.

The only way to simulate a LRM with linear binding using the MCT is thus by accounting for \beta_t in the adsorption parameter, i.e. set

k_a^\text{MCT} = k_a^{LRM} \frac{1}{\beta_t}.

This makes sense from a modelling perspective since the binding fluxes are computed from the binding rates w.r.t a reference volume. That reference volume is the solid volume in our chromatography models. For the exchange however, the reference volume is always the volume of the efflux channel. That is, the reference volume for the adsorption flux in the MCT is the liquid phase channel. To model an LRM using the MCT, we thus have to modify the exchange rate k_a^\text{MCT} so that the MCT exchange flux and the LRM adsorption flux match.

@AntoniaBerger can you please double check the math and, if you agree, verify the result via simulation?

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2

This looks all fine to me and the simulations also agree with you.

Thank you!

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3

The deviation looks beautiful and is exactly what I would have expected.

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