Failed to calculate when water inputted in solution is unsuitable

[guobingbing212]

Dear David:
I used the following codes to calculate the concrete pore solution, including Equilibrium-Phases and Surface. Sometime it may be failed to calculate when the water inputted in solution is unsuitable, e.g., when the water inputted is from 0.02 kg to 0.06 kg. But it is successful to calculate in the other cases, e.g., when the water inputted is 0.01 kg or 0.08 kg.
Could you tell me the reasons? Thank you very much.

TITLE Pore Solution in concrete
# Define Double Electric Layer of Pore solid
Surface_Master_Species
      [mySurf1]    [mySurf1]OH
Surface_Species
   [mySurf1]OH = [mySurf1]OH
   log_k   0.0
   [mySurf1]OH + OH- = [mySurf1]O- + H2O
   log_k   -12.7
   [mySurf1]OH + Ca+2 = [mySurf1]OCa+ + H+
   log_k   -9.4
   [mySurf1]OH + Cl- = [mySurf1]OHCl-
   log_k   -0.35
   [mySurf1]OH + Na+ = [mySurf1]ONa + H+
   log_k   -13.6
   [mySurf1]OH + K+ = [mySurf1]OK + H+
   log_k   -13.6
# Define Dissolution and Deposition of Cement Hydrates
Phases
CSH
Ca1.6SiO3.6:2.58H2O + 3.2H+ = 1.6Ca+2 + 2.18H2O + H4SiO4
   log_k   28.0
    delta_h -133.31 kcal
CH
Ca(OH)2 + 2H+ = Ca+2 + 2H2O
   log_k   22.81
   delta_h -130.11
AFm
Ca6Al2(SO4)3(OH)12:26H2O + 12H+ = 2Al+3 + 6Ca+2 + 38H2O + 3SO4-2
   log_k   57.73
   delta_h -389.36
AFt
Ca4Al2(SO4)(OH)12:6H2O + 12H+ = 2Al+3 + 4Ca+2 + SO4-2 + 18H2O
   log_k   73.68
   delta_h -553.08
Fs
Ca4Al2Cl2O6:10H2O + 12H+ = 2Al+3 + 4Ca+2 + 2Cl- + 16H2O
   log_k   74.93
   delta_h -486.20

#Pore Solution of Concrete
Solution 1
      units   mmol/L
      pH      13.23
      temp    25.0
       Ca      4.8
      Na      88.9
      K       74.8
      Cl      1.4
       S(6)    0.2   as SO4
        -water  0.06 # kg
Equilibrium_Phases 1
   CSH 0   1.176
   CH  0   1.215
   AFm 0   0.015
   AFt 0   0.018
   FS   0   0.000
SURFACE 1
    -equilibrate with solution 1
    [mySurf1]OH CSH             equilibrium_phase 0.7656      95700
    -donnan debye_lengths 1 limit_ddl 0.8
End

[dlparkhurst]

PHREEQC is going to work best with 1 kg water. You could consider scaling your reactants so that there is 1 kgw.  (Although I made a quick adjustment and that did not seem to help.)

I think the numerical problem is with the -donnan calculation including the debye length. If I increase the number of Debye lengths to 2 or limit the ddl to 0.5, then I get a solution to the problem with 0.02 kgw. Also, if I fix the thickness of the double layer rather than using debye lengths, it seems to work. It is possible there is no numerical solution with some sets of parameters, but it is also possible that the numerical method simply fails under some conditions. I guess I would try to adjust the parameters slightly to get results. The thickness of the diffuse layer is not very well defined, especially in small pores. Both a fixed thickness and the Debye approach are purely assumptions to try to make the problem tractable.

[guobingbing212]

Okey, and I have a try. Thank you for your prompt reply very much.