two kinetics for mineral dissolution

[Selene Eliana Patani]

Hi everybody. I'm Selene, a new environmental engineering PhD student. I'm new with Phreeqc and I have kind of troubles that I'm sure they will be easily solved for everyone not like me :-) I have to reproduce a batch experiment. I have to simulate the dissolution of several minerals with a kinetic reaction. My problem is that for the first hours (4 h) of the experiment my constant rate of dissolution has a certain value while for the last hours (20) it has to be modeled with another constant rate value. My problem is that I'm not able to reproduce these two steps in a row (I mean with a unique input file) and create a unique user graph able to consider the total amount of ions dissolved along the all experiment of 24 h. I hope that someone could help me, I'm stuck on this problem for weeks and I don't see a solution. If something is not clear let me know. Thanks very much!!

Selene

[dlparkhurst]

The following simple examples dissolves quartz at a faster rate for 4 hours and a slower rate for 20 hours.

Code: [Select]

RATES
    Qtz
-start
10 if TOTAL_TIME > 60*60*4 then goto 100
20 rate = 2e-5/3600 # mol/s
30 goto 200
100 REM after 4 hours
110 rate = 1e-5/3600 # mol/s
200 moles = rate * TIME
210 SAVE moles
-end
END
SOLUTION
END
INCREMENTAL_REACTIONS
KINETICS
Qtz
-formula SiO2 1
-step 86400 in 48 steps
END
USER_GRAPH 1
    -headings               Hours Si
    -axis_titles            "Time, in hours" "Si, mmol/kgw" ""
    -initial_solutions      false
    -connect_simulations    true
    -plot_concentration_vs  x
  -start
10 graph_x TOTAL_TIME/3600
20 graph_y TOT("Si")*1000
  -end
    -active                 true
USE solution 1
USE kinetics 1
END


[Selene Eliana Patani]

Great!! Thank you very much!!!

Selene