Dividing a column into more cells preserving kinetic behaviour - A0/V

[peterwadeuk]

Greetings good persons,

I am trying to model a column of soil and I need to modify the guess for surface area of rock added to the soil.

I have created a stripped-down example reproducing my problem - I am not able to reproduce the time-to-exhaustion of weathering of the added rock invariant of the number of cells I define.

If I model the entire column as one cell, I introduce "plagioclase" at 0.2470 moles per litre (the example has one litre pore water and a porosity of 0.25).

The surface area of the rock containing plagioclase is 693.9 and the proportion of the surface area that is plagioclase is 0.4661.

I have attached the input files describing my irregularity.

If I run the problem with one cell, plagioclase is exhausted after 28 steps at 63,446,900 seconds.
If I run the problem with 5 cells, with the amount of plagioclase in the cell at 20% of one cell, and area and volume (computing a rate multiplier of A0/V) also reduced to 20%, exhaustion of plagioclase occurs at the same step number (28), but at 12,689,400 seconds, meaning that it reacted faster than in the one-cell problem.
If I run the problem with 8 cells, with all values changed to 1/8th of the single-cell problem, I find that exhaustion of plagioclase occurs at step 28, at 7,930,860 seconds.

I am clearly erring in my description of the kinetic problem.

For a start, I am defining the kinetic block with the amount in moles of mineral that there is in the cell, and then I am subjecting it to a kinetic expression that scales according to the input area and volume, which, being divided by the same number to represent smaller cells, is invariant from one problem to another, making the plagioclase in the smaller cells react quicker.

I would be very grateful to have my conceptual error with respect to the construction of the kinetic block pointed out.

All the best,
Peter

[dlparkhurst]

I think it is that you should not change KINETICS definition as you increase cells. Even though you decrease the length of the cells, each cell still has ~1 L water, so A/V should not change. The rates will be the same regardless the number of cells, but the time of contact will decrease in each cell as the number of cells increase.