EXAMPLES

TITLE 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 INVERSE_MODELING -solution 1 2 -uncertainties .025 -balances Alkalinity 1. Br K Mg -phases H2O pre gypsum pre halite pre PHASES H2O H2O = H2O log_k 0.0 Halite NaCl = Na+ + Cl- log_k 1.582 ENDThis 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

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.

------------------------------------------- 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

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