Processes > Inverse modelling

Issue with Evaporation

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evansmanu:
Dear all,

I am in fix as to how possible I can evaporate my water at a certain percentage say 84%. The problem here is that My rainwater is affected by evaporation and would like to know the effect of evaporation on my mineral phases when performing forward modelling.

My study area has an annual rainfall of 1600 mm and a recharge of about 8% of the average annual rainfall. The rest of the rainfall goes back into the atmosphere and is also washed down into the streams as runoff.

I understand that if you want to include the evaporation phase in a model u can precipitate (H2O(g)) as shown below

EQUILIBRIUM PHASES
H2O(g) 0 0

How should the equilibrium script be if I want to evaporate my water to about 92%

Any comment will be welcomed

dlparkhurst:
There are several ways that you can produce the effects of evaporation. I am using a 1 mmol solution of NaCl, which, when removing 16% of the water, produces a concentration of 1.19 mmol/kgw.

Given that 1 kg water is 55.5 moles, the first simulation simply removes 16% of the moles of water.

Another alternative is to use MIX with a negative mixing fraction for pure water (SOLUTION 0). This calculation produces a solution with 0.84 kg water.

It may be convenient to end up with approximately 1 L of water, so the next simulation starts with a larger mass of water (~ equal to volume) of 1/0.84 and removes water so that 1 kg water remains, but concentration is again 1.19 mmol/kgw.

The last version is a bit trickier. It sets the activity of water, but in general you will not know the activity of water that corresponds to the evaporation that you want. Here, the activity of water can be calculated in one of the evaporation simulations and used in the final simulation. Otherwise, you would need to iterate on the activity of water to find the evaporation factor that you need. The activity of water approach can be useful if you have very high evaporation factors where the mass of water remaining is difficult to assess because of hydration reactions.


--- Code: ---SOLUTION 0 Pure water
END
SOLUTION 1
    Na 1
    Cl 1
END
# Remove 16% of mass of water
USE solution 1
REACTION 1
    H2O        1
    -8.88 moles
END
# Negative mix with pure water
MIX 1
    0    -0.16
    1    1
END
# Negative mix and to produce 1 kg water
MIX 1
    0    -0.19047619
    1    1.19047619
USER_PRINT
-start
10 PRINT "Log activity of water: ", LA("H2O")
-end
END
PHASES
H2O_aq
    H2O = H2O
    log_k     0
USE solution 1
EQUILIBRIUM_PHASES 1
    H2O_aq    -1.758e-05 H2O       10
END

--- End code ---

evansmanu:
Thank you very much, Parkhurst. I implemented the first one and it worked for me.
My other concern is that in your code you decided to negate the amount that you want to evaporate. I also do it this way,
REACTION 1
    H2O        -1
    8.88 moles
This is also the case in Example 4--Evaporation and Homogeneous Redox Reactions under the help menu of the PHREEQC interactive program. I hope I am not wrong.

I would also want to know how possible I can plot the sample points on a plotted activity diagram using the USER_GRAPH

Any help from the community will be gladly appreciated

Best
Evans

dlparkhurst:
The activity of water changes very little in your sample calculation. Here is a plot showing concentration increase in Na and activity change as water is incrementally removed from the solution. I am not sure what plot you have in mind, but perhaps you can modify this example.


--- Code: ---SOLUTION 0 Pure water
END
SOLUTION 1
    Na 1
    Cl 1
# Remove 16% of mass of water
REACTION 1
    H2O        1
    -8.88 moles in 10 steps
USER_GRAPH 1
    -headings               ah2o Na a(H2O)
    -axis_titles            "H2O removed, moles" "Molality" "Activity of water"
    -initial_solutions      true
    -connect_simulations    true
    -plot_concentration_vs  x
  -start
10 GRAPH_X -RXN
20 GRAPH_Y TOT("Na")
30 GRAPH_SY ACT("H2O")
  -end
    -active                 true
END

--- End code ---

evansmanu:
Thanks for your prompt reply. However, I still want to know if this reaction script is right.
REACTION 1
    H2O        -1
    8.88 moles
In your case, you negated the amount (-8.88 moles) while in my case I negated the (-1)

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