Clay Kinetic Model

[taa18]

Hello

I have been trying to model kaolinite stabilization with lime. For kaolinite dissolution rate I am using literature data which suggests taking exponent of H+ activity raiesd to the power -0.472 but phreeqc is not taking a negative number. I assigned this number to a and phreeqc is assgining some sort of value to a, I am trying to understand what specifically that number could be but I just have no clue. Please help.

Kaolinite
  -start
  10 pk_H = 11.1249 #data from marty 2015
  20 pk_w = 13.9586
  30 pk_OH = 12
  40 a = 0.472
  40 rate = 10^-pk_H * ACT("H+")^0.777 + 10^-pk_w + 10^-pk_OH *ACT("OH-")^a     
#                          Integrate...
 
  50 A0 = parm(1)
  60 V = parm(2)
  70 moles = (A0 / V) * (m/m0)^0.67 *rate* (1 -  SR("Kaolinite"))

 
  80 save moles * time
 -end

[dlparkhurst]

You have two line number 40s. The second is overwriting the first (which defines a). Change the second line number to 45.

[taa18]

Thank you so much for your reply! I did notice my mistake, what I don't understand is that phreeqc is somehow getting the value of a, where is it getting the value? Is it alkalinity?

[dlparkhurst]

All variables are initialized to zero for numbers or empty string for strings.

[taa18]

One more question regarding the kinetic equation parameters, I want to know what the software is putting in for m and m0 for my kinetic parameters but I have 3 kinetic parameters. How can I use the user punch command to tell the software to feed the data to a spreadsheet for a specific kinetic parameter? When I put m/m0 it gives the values for the last kinetic parameter.

[dlparkhurst]

If you are trying to debug your RATES expression, you can add PRINT statements in the RATES script, like:

Code: [Select]

90 PRINT TOTAL_TIME, m/m0

Note that the RATES script is executed many times during a kinetics calculation, so you will get multiple lines in the output file for one kinetics calculation. Also, the times of the print will not be monotonic and the last execution of the script may not be at  the end of the integration period.

M and M0 are local to the specific RATES definition (Kaolinite) and are not available in USER_PUNCH. The Basic function KIN("Kaolinite") can be used in USER_PUNCH and is equal to the value of M at the end of the time step. There is not a function to retrieve M0, but you can save it with a PUT function and retrieve it with a GET. If you add this statement to the RATES definition it will be available in USER_PUNCH:

Code: [Select]

100 PUT(M0, 1)
The time (TOTAL_TIME) that the value of M0 is stored is irrelevant because M0 is constant throughout the calculation. You can then retrieve the value in USER_PUNCH to print M/M0 as follows:

Code: [Select]

xxx PUNCH KIN("Kaolinite) / GET(1)


[taa18]

Okay, thank you so much!

[dlparkhurst]

The only thing I noticed is that you reached equilibrium with Montmorillonite, which stops the reaction. If Montmorillonite is more soluble, I think the reaction continues to a lower pH.

[taa18]

I am having trouble to interpret the sat'd phases figure that PHREEQC spit out from the montmorillonite dissolution precipitation model. The figure shows an uptick in amount of water consumed. Why is there an uptick? Shouldn't it be decreasing? Does that indicate an error or am I misunderstanding something?

[dlparkhurst]

I don't see a problem with the mass of water changing. I'm not going to try to write the balanced reaction, but you are dissolving Montmorillonite, and precipitating quartz, gibbsite, albite, and talc. Looks like the mass of water tends to increase as a result.

[taa18]

Okay, thank you so much for the explanation. I just wanted to double check with you.

[taa18]

Follow up question on bentonite model, is it possible to add kinetic parameters to secondary minerals in PHREEQC? If yes, is there any manual I can refer to?

[dlparkhurst]

You can add KINETICS and rates in the same way you do for other kinetic reactions. However, if there is none of the kinetic reaction and the SI < 0, rate would be zero. You probably need a special rate for 0 <= M < threshold and SI > 0 because surface area is 0 or very small. You probably have another rate for M >= threshold and SI > 0 or SI < 0 (dissolution). It is good if the rate for m > threshold is continuous and differentialble at m = threshold.