Porosity Changes With Mineral Dissolution/Precipitation

[davidblevy]

I am a bit perplexed by the comparison of results from two sensitivity models where three brines are equilibrated with a set of minerals and run at different T/P (using PHREEQC.dat). The difference in the two models is the set of Equilibrium_Phases. The models produce different calculated porosity results as might be expected, but the amounts of minerals precipitated/dissolved are identical, even though the Equilibrium_Phases are different. Any explanation would be appreciated. I have inserted one code, the second code would be identical except it would use these phases instead:

EQUILIBRIUM_PHASES 1 #LOW Carbonate
Quartz      0.0   654.10 dissolve only
Dolomite      0.0   213.13 dissolve only
Siderite      0.0   82.24
Pyrite      0.0   39.7 dissolve only
Calcite      0.0   11.9
CO2(g)      2.216 #log PCO2 (164.5 atm)

Code: [Select]

#High Carb Mineralogy

PRINT
-reset false
-saturation_indices true
-headings true

SOLUTION 1 Knox Dolomite #Well F25CP001
units mg/L
pH 6.6
Alkalinity 118 as HCO3
Br 133
Ca 2570
Cl 43900 charge
Mg 1100
K 935
Na 22200
Sr 310

EQUILIBRIUM_PHASES 0
Quartz 0.0 10

SAVE Solution 1
END

SOLUTION 2 #High TDS/High Ca (Well I24AP003A)
units mg/L
pH 6.2
Alkalinity 22 as HCO3
Br 292
Ca 20100
Cl 114000 charge
Mg 4420
K 3000
Na 41800
S(6) 289

EQUILIBRIUM_PHASES 0
Quartz 0.0 10

SAVE Solution 2
END

SOLUTION 3 #Low TDS/High SO4 (Well I17C0159)
units mg/L
pH 7.8
Alkalinity 192 as HCO3
Ca 476
Cl 7800 charge
Mg 329
K 135
Na 4630
S(6) 1640

EQUILIBRIUM_PHASES 0
Quartz 0.0 10

SAVE Solution 3
END

EQUILIBRIUM_PHASES 1 #HIGH Carbonate
Quartz 0.0 495.53 dissolve only
Dolomite 0.0 290.63 dissolve only
Siderite 0.0 61.68
Pyrite 0.0 99.26 dissolve only
Calcite 0.0 11.9
CO2(g) 2.216 #log PCO2 (164.5 atm)

REACTION_TEMPERATURE 1 #LOW T/P
49.4 #121 F

REACTION_PRESSURE  1 #LOW T/P
130.7 #1921 PSI

REACTION_TEMPERATURE 2 #HIGH T/P
61.1 #142 F

REACTION_PRESSURE  2 #HIGH T/P
213.4 #3136 PSI

END

SELECTED_OUTPUT
-file c:/output/Maquoketa_Sensitivity_Model_1.dat
-reset false

USER_PUNCH
-headings Step Pressure pH Delta_Quartz Delta_Dolomite Delta_Siderite Delta_Pyrite Delta_Calcite Porosity
-start

10 PUNCH STEP_NO
20 PUNCH PRESSURE
30 PUNCH -LA("H+")
40 PUNCH EQUI_DELTA("Quartz")
50 PUNCH EQUI_DELTA("Dolomite")
60 PUNCH EQUI_DELTA("Siderite")
70 PUNCH EQUI_DELTA("Pyrite")
80 PUNCH EQUI_DELTA("Calcite")
110 REM Calculate Porosity
120 qtz = EQUI("Quartz")
130 dol = EQUI("Dolomite")
140 sid = EQUI("Siderite")
150 pyr = EQUI("Pyrite")
160 cal = EQUI("Calcite")
190 ill = 40.33
200 qtz_V = (qtz * 22.67)
210 dol_V = (dol * 64.5)
220 sid_V = (sid * 29.2)
230 pyr_V = (pyr * 23.48)
240 cal_V = (cal * 36.9)
270 ill_V = (ill * 141.48)
280 Vtot = 1000 + qtz_V + dol_V + sid_V + pyr_V + cal_V + ill_V
290 por = 1-(qtz_V + dol_V + sid_V + pyr_V + cal_V + ill_V)/Vtot
300 PUNCH por
END

#ROW 1

USE SOLUTION 1
USE REACTION_TEMPERATURE 1
USE REACTION_PRESSURE 1
USE EQUILIBRIUM_PHASES 1
END

USE SOLUTION 1
USE REACTION_TEMPERATURE 2
USE REACTION_PRESSURE 2
USE EQUILIBRIUM_PHASES 1
END

USE SOLUTION 2
USE REACTION_TEMPERATURE 1
USE REACTION_PRESSURE 1
USE EQUILIBRIUM_PHASES 1
END

USE SOLUTION 2
USE REACTION_TEMPERATURE 2
USE REACTION_PRESSURE 2
USE EQUILIBRIUM_PHASES 1
END

USE SOLUTION 3
USE REACTION_TEMPERATURE 1
USE REACTION_PRESSURE 1
USE EQUILIBRIUM_PHASES 1
END

USE SOLUTION 3
USE REACTION_TEMPERATURE 2
USE REACTION_PRESSURE 2
USE EQUILIBRIUM_PHASES 1
END



[dlparkhurst]

In both cases, you have sufficient moles of minerals to reach equilibrium. The same amount of each mineral reacts to produce equilibrium in each case. The only thing different in the two simulations is the amount of each mineral initially, and the amount remaining after reaction.

[davidblevy]

I'm still struggling a little. I used the same initial porosity with the bulk density to calculate the moles of minerals in both cases. Because everything is based on the same initial porosity, then it seems that if the calculated equilibrium porosity is different, shouldn't the moles of minerals dissolved/precipitated be different? Thank you.

[dlparkhurst]

No.

If the solubility of halite is 6 mol/kgw and you have 3 mol of Halite, it will all dissolve and you will be undersaturated with halite.

If you have 6 moles of halite, it will all dissolve and you will have equilibrium with halite.

If you have 10 moles of halite, 6 will dissolve, and you will be left with 4 moles of halite.

The amount of halite dissolving in case 2 and 3 is exactly the same.

[dlparkhurst]

The amount of water in the two calculations is the same, as defined in SOLUTION.

[davidblevy]

Thank you for that perspective!