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Example 12.-- Inverse Modeling with Evaporation

Evaporation is handled in the same manner as other heterogeneous reactions for inverse modeling. To model evaporation (or dilution) it is necessary to include a phase with the composition H2O. The important concept in modeling evaporation is the water mole-balance equation that is included in every inverse problem formulation (see Equations and Numerical Method for Inverse Modeling). The moles of water in the initial solutions times their mixing fractions plus water gained or lost by dissolution or precipitation of phases plus water gained or lost through redox reactions must equal the moles of water in the final solution. The equation is approximate because it does not include the moles of water gained or lost in homogeneous hydrolysis and complexation reactions.

Table 21. Input data set for example 12

Example 12.--Inverse modeling of Black Sea water evaporation
SOLUTION 1  Black Sea water
        units   mg/L
        density 1.014
        pH      8.0     # estimated
        Ca      233
        Mg      679
        Na      5820
        K       193
        S(6)    1460
        Cl      10340
        Br      35
SOLUTION 2  Composition during halite precipitation
        units   mg/L
        density 1.271
        pH      5.0     # estimated
        Ca      0.0 
        Mg      50500
        Na      55200
        K       15800
        S(6)    76200
        Cl      187900
        Br      2670
        -solution 1 2
        -uncertainties .025
                Alkalinity 1.
                H2O     pre
                gypsum  pre
                halite  pre
        H2O = H2O
        log_k   0.0
        NaCl = Na+ + Cl- 
        log_k   1.582
This example uses data for the evaporation of Black Sea water that is presented in Carpenter (1978). Two analyses are selected, the initial Black Sea water and a water composition during the stage of evaporation in which halite precipitates. The hypothesis is that evaporation and precipitation of gypsum and halite are sufficient to account for the changes in water composition of all of the major ions and bromide. The input data set (table 21) contains the solution compositions in the SOLUTION keyword data blocks.

The INVERSE_MODELING keyword defines the inverse model for this example. Solution 2, the solution during halite precipitation, is to be made from solution 1, Black Sea water. Uncertainties of 2.5 percent are applied to all data. Water, gypsum, and halite are specified to be the potential reactants (-phases). Each of these phases must precipitate, that is, must be removed from the aqueous phase in any valid inverse model.

By default, mole-balance equations for water, alkalinity, and electrons are included in the inverse formulation. In addition, mole-balance equations are included by default for all elements in the specified phases. In this case, calcium, sulfur, sodium, and chloride mole-balance equations are included by the default. The -balances identifier is used to specify additional mole-balance equations for bromide, magnesium, and potassium and to change the uncertainty on alkalinity to 100 percent. In the absence of alkalinity data, the calculated alkalinity of these solutions is controlled entirely by the choice of pH. No pH values were given and thus the alkalinities are unknown. For reasonable values of pH, alkalinity is a minor contributor to charge balance and no alkalinity is contributed by the reactive phases. Thus, setting the uncertainties to 100 percent allows the alkalinity balance equation effectively to be ignored.

Only one model is found in the inverse calculation. This model indicates that Black Sea water (solution 1) must be concentrated 62 fold to produce solution 2, as shown by the fractions of the two solutions in the inverse-model output (table 22). Thus approximately 62 kg of water in Black Sea water is reduced to 1 kg of water in solution 2. Halite precipitates (13.7 mol) and gypsum precipitates (.35 mol) during the evaporation process. Note that these numbers of moles are relative to 62 kg of water. To find the loss per kilogram of water in Black Sea water, it is necessary to divide by the mixing fraction of solution 1. The result is that 54.6 mol of water, 0.0056 mol of gypsum, and 0.22 mol of halite have been removed per kilogram of water. (This calculation could be accomplished by making solution 1 from solution 2, taking care to reverse the constraints on minerals from precipitation to dissolution.) All other ions are conservative within the 2.5-percent uncertainty that was specified. The inverse modeling shows that evaporation and halite and gypsum precipitation are sufficient to account for all of the changes in major ion composition between the two solutions.

Table 22. Selected output for example 12

Beginning of inverse modeling calculations.

Solution 1: Black Sea water
     Alkalinity      5.280e-06  +   0.000e+00  =   5.280e-06
             Br      4.320e-04  +   0.000e+00  =   4.320e-04
             Ca      5.733e-03  +  -1.249e-04  =   5.608e-03
             Cl      2.876e-01  +   7.646e-04  =   2.884e-01
           H(0)      0.000e+00  +   0.000e+00  =   0.000e+00
              K      4.868e-03  +   1.015e-04  =   4.969e-03
             Mg      2.754e-02  +  -6.886e-04  =   2.685e-02
             Na      2.497e-01  +   0.000e+00  =   2.497e-01
           O(0)      0.000e+00  +   0.000e+00  =   0.000e+00
          S(-2)      0.000e+00  +   0.000e+00  =   0.000e+00
           S(6)      1.499e-02  +   3.747e-04  =   1.536e-02

Solution 2: Composition during halite precipitation
     Alkalinity      1.770e-04  +   1.523e-04  =   3.294e-04
             Br      2.629e-02  +   6.556e-04  =   2.695e-02
             Ca      0.000e+00  +   0.000e+00  =   0.000e+00
             Cl      4.170e+00  +   1.042e-01  =   4.274e+00
           H(0)      0.000e+00  +   0.000e+00  =   0.000e+00
              K      3.179e-01  +  -7.948e-03  =   3.100e-01
             Mg      1.634e+00  +   4.086e-02  =   1.675e+00
             Na      1.889e+00  +  -3.092e-02  =   1.858e+00
           O(0)      0.000e+00  +   0.000e+00  =   0.000e+00
          S(-2)      0.000e+00  +   0.000e+00  =   0.000e+00
           S(6)      6.241e-01  +  -1.560e-02  =   6.085e-01

Solution fractions:                   Minimum        Maximum
   Solution   1      6.238e+01      0.000e+00      0.000e+00
   Solution   2      1.000e+00      0.000e+00      0.000e+00

Phase mole transfers:                 Minimum        Maximum
            H2O     -3.406e+03      0.000e+00      0.000e+00   H2O
         Gypsum     -3.498e-01      0.000e+00      0.000e+00   CaSO4:2H2O
         Halite     -1.372e+01      0.000e+00      0.000e+00   NaCl

Redox mole transfers:    

Sum of residuals:                                     2.443e+02
Maximum fractional error in element concentration:    8.605e-01

Model contains minimum number of phases.

Summary of inverse modeling:

	Number of models found: 1
	Number of minimal models found: 1
	Number of infeasible sets of phases saved: 4
	Number of calls to cl1: 8

Table 21. Input data set for example 12
Table 22. Selected output for example 12

User's Guide to PHREEQC - 07 MAY 96
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