Lime Addition for Fluoride Removal

[sreuter]

I am modeling a process unit that adds lime (Ca(OH)2) to precipitate CaF2. Goal is fluoride removal. I would like to include in my model a way to evaluate the lime's solubility. As pH increases, I assume it will be more and more difficult for the lime to dissolve due to the common ion effect/La Chatlier's principle. To address this, I changed the database to minteq.v4.dat and ran the below model. The output was nonsensical (pH at first step lime dose was ~15). The model without lime added in Equilibrium_Phases makes a lot more sense, but I do not think it accounts for reduced solubility of lime at elevated pH. Any ideas on what I am doing wrong?

TITLE Precipitating fluoride as CaF2

#Adding water quality results from analytical testing.
SOLUTION 1 Spill Pond
    temp      15
    pH        8.7
    pe        4
    redox     pe
    units     mg/l
    density   1
    Ca        16.2
    Cl        20
    F         195
    K         10
    Mg        10
    Na        0.59 charge
    S(6)      10
    Alkalinity 1040
    -water    1 # kg

SAVE solution 1
END   

#Adding lime to Spill Pond
USE solution 1
REACTION 1
   Ca(OH)2   1
   0.1 in 500 steps

Equilibrium_PHASES 1
   Calcite   0.0
   Fluorite   0.0
   Lime      0.0

SELECTED_OUTPUT
   -file   output_Lime_Addition
   -pH   true
   -alkalinity   true
   -totals   Ca   F
   -equilibrium_phases   Calcite   Fluorite   Lime
END

[dlparkhurst]

In EQUILIBRIUM_PHASES, you set the target saturation index to be zero (equilibrium), but you did not set the number of moles of each phase. By default, there are 10 moles of each phase available to react, so, the REACTION really had no effect. At each point in the reaction, the equilibrium phases reacted to equilibrium, producing the same solution composition.

pH 15 sounds a bit high. I have added Portlandite from the wateq4f.dat database. Fluorite starts to precipitate immediately, followed by calcite, and ultimately portlandite. Maximum pH is about 13.

Code: [Select]

PHASES
Portlandite         539
        Ca(OH)2 + 2H+ = Ca+2 + 2H2O
        log_k           22.8
        delta_h -31.0 kcal
END
TITLE Precipitating fluoride as CaF2
#Adding water quality results from analytical testing.
SOLUTION 1 Spill Pond
    temp      15
    pH        8.7
    pe        4
    redox     pe
    units     mg/l
    density   1
    Ca        16.2
    Cl        20
    F         195
    K         10
    Mg        10
    Na        0.59 charge
    S(6)      10
    Alkalinity 1040
    -water    1 # kg
END   
#Adding lime to Spill Pond
USE solution 1
REACTION 1
   Ca(OH)2   1
   0.1 in 500 steps
Equilibrium_PHASES 1
   Calcite   0.0 0
   Fluorite   0.0 0
   Portlandite      0.0 0
USER_GRAPH 1
    -headings               rxn Fluorite Calcite Portlandite pH
    -axis_titles            "Ca(OH)2 added, moles" "Moles" "pH"
    -initial_solutions      false
    -connect_simulations    true
    -plot_concentration_vs  x
  -start
10 GRAPH_X RXN
20 GRAPH_Y EQUI("Fluorite"), EQUI("Calcite"), EQUI("Portlandite")
30 GRAPH_SY -LA("H+")
  -end
END


[davidblevy]

Also, are you intending to model a closed system? Because if you were to treat it as an open system by maintaining PCO2(g) at 10^-3.5, the pH will be lower.

[dlparkhurst]

If you do add CO2(g) -3.4 to the equilibrium phases, the pH does remain lower, but you also do not lower the fluoride concentration as much--only to about 60 mg/L compared to less than 4 mg/L if it is considered a closed system.

In theory, addition of gypsum requires lesser amounts than Ca(OH)2 to lower the fluoride concentration. However, the pH should decrease rather than increase. It assumes the gypsum dissolves quickly enough to effectively remove fluoride.