Diffusion mixture factor first cell

[Pak]

I am running the following diffusion reaction problem:

Solution 0
units mol/kgw
pH   7

Solution 1-5
units    mol/kgw
C    6.470e-06
pH    12.468 
Cl    7.27e-4
Ca    2.028e-02

Equilibrium_Phases 1-5
Calcite             0       130.7

TRANSPORT
-cells                 5
-lengths               0.00275    # dx
-shifts                500  #
-time_step             1e5   # dt
-flow_direction        diffusion_only
-boundary_conditions   constant, closed
-dispersivities        0
-correct_disp          true
-diffusion_coefficient    1.97752E-12

According to Phreeqc Manual: "USER'S GUIDE TO PHREEQC (VERSION 2). (Equations on which the program is based)" by David L. Parkhurst and C.A.J. Appelo. Dispersive transport (in the presented case, it is diffusive transport since there is not advection) in a central difference scheme is essentially mixing of cells. A mixing factor mixf is defined as:

mixf=D*dt/(n*dx^2) --->  Assuming n=1 (so long as the mixing factor is smaller than 1/3) the mixing factor of the presented problem will be --> mixf=1.97752E-12*1e5/(0.00275 ^2) which is roughly 2.615e-02;
If I am not wrong the sum of percentages of a mixture in PHREEQC after a diffusive transport must be equal to 1. For instance, in cell 2 the mixture for the new batch reaction after one time step will be:

Mixture 2
    2.615e-02 Solution 1   Solution after simulation 1.                           
     2.615e-02 Solution 3   Solution after simulation 1.                           
     9.477e-01 Solution 2   Solution after simulation 1.             
             
but in the first cell the mixture proportions are different:

Mixture 1.   
     5.230e-02 Solution 0   Solution after simulation 1.                           
     2.615e-02 Solution 2   Solution after simulation 1.                           
     9.216e-01 Solution 1   Solution after simulation 1.   

I am not really sure if that is due to a different space discretization (dx) or a different Diffusion coefficient or other reason. Does somebody know?

Thank you

[Pak]

I have found that multiplying the dx by sin(pi/4), namely the dx in the boundary is dx*sin(pi/4) will give the mixing factor of solution 0.

[dlparkhurst]

I think it is the constant boundary condition. The concentrations at z=0 are fixed, so gradients are defined by the slope from z=0 to the location of the first node, z=0.00275/2. Other gradients are defined by the distance between nodes, which is 0.00275 m.