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Phreeqc seems to halve first kinetic step
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Topic: Phreeqc seems to halve first kinetic step (Read 2467 times)
peterwadeuk
Top Contributor
Posts: 52
Phreeqc seems to halve first kinetic step
«
on:
20/04/18 09:44 »
Greetings good persons.
In the transport-kinetic script below I attempt to introduce a solution in steps of 30 days, to a 2-cell column.
One of my difficulties is that I seem to be receiving output from Phreeqc at periods of 15, 45, 75, 105 days. The start time is not zero or 30 days.
Grateful for any advice.
All the best,
Peter
========================================
TITLE Rebuild kinetic model of Mike's experiments
PHASES
Plagioclase
Ca0.5Na0.5Al1.5Si2.5O8 + 6H+ + 2H2O = 1.5Al+3 + 0.5Ca+2 + 2.5H4SiO4 + 0.5Na+
log_k 914.2 # log_k 14.2 # Change to ensure reactivity
delta_h -12.467 kJ
Diopside
CaMgSi2O6 + 4H+ + 2H2O = Ca+2 + 2H4SiO4 + Mg+2
log_k 919.894 # log_k 19.894 # Change to ensure reactivity
delta_h -32.348 kcal
SOLUTION 0 Pore waters in absence of basalt
temp 25
pH 8.156
pe 12.471
redox pe
units mol/l
density 1
Al 3.251E-07
Ba 5.471E-06
C 9.689E-04
Ca 7.100E-04
Cl 1.630E-04
F 5.841E-05
Fe 2.560E-08
K 2.010E-05
Li 3.360E-07
Mg 6.131E-05
Mn 4.011E-08
N 4.647E-04
Na 1.800E-04
S 5.081E-05
Si 5.389E-05
Sr 9.761E-07
Zn 2.310E-06
-water 1 # kg
SOLUTION 1-2 Pore waters in absence of basalt
temp 25
pH 8.156
pe 12.471
redox pe
units mol/l
density 1
Al 3.251E-07
Ba 5.471E-06
C 9.689E-04
Ca 7.100E-04
Cl 1.630E-04
F 5.841E-05
Fe 2.560E-08
K 2.010E-05
Li 0
Mg 6.131E-05
Mn 4.011E-08
N 4.647E-04
Na 1.800E-04
S 5.081E-05
Si 5.389E-05
Sr 9.761E-07
Zn 2.310E-06
-water 1 # kg
RATES
Plagioclase
-start
10 moles = 0
20 IF(M <= 0 OR SI("Plagioclase")) >= 0 THEN GOTO 220
25 rock_area = PARM(1)
30 area_fract = PARM(2)
35 a0 = rock_area * area_fract
40 v = PARM(3)
50 dif_temp = 1/TK-1/298
60 k_acid = 10^-7.87
70 k_neut = 10^-10.91
80 k_base = 10^-15.57
90 eapp_acid = 42.1
100 eapp_neut = 45.2
110 eapp_base = 71
120 n_acid = 0.63
130 n_base = -0.57
150 hplus = MOL("H+")
160 r_acid = k_acid*EXP((-eapp_acid/8.314e-3)*dif_temp)*(hplus^n_acid)
170 r_neut = k_neut*EXP((-eapp_neut/8.314e-3)*dif_temp)*(hplus^n_neut)
180 r_base = k_base*EXP((-eapp_base/8.314e-3)*dif_temp)*(hplus^n_base)
190 r_all = r_acid+r_neut+r_base
200 rate = (a0/v)*(M/M0)^0.67*r_all*(1-SR("Plagioclase"))
205 print "r_acid, r_neut, r_base, SR-plag, rate, time = ", r_acid, r_neut, r_base, SR("Plagioclase"), rate, time
210 moles = rate*TIME
220 SAVE moles
-end
Diopside
-start
10 moles = 0
20 IF(M <= 0 OR SI("Diopside")) >= 0 THEN GOTO 220
25 rock_area = PARM(1)
30 area_fract = PARM(2)
35 a0 = rock_area * area_fract
40 v = PARM(3)
50 dif_temp = 1/TK-1/298
60 k_acid = 10^-6.36
70 k_neut = 10^-11.11
80 k_base = 0
90 eapp_acid = 96.1
100 eapp_neut = 40.6
110 eapp_base = 0
120 n_acid = 0.71
130 n_base = 0
150 hplus = MOL("H+")
160 r_acid = k_acid*EXP((-eapp_acid/8.314e-3)*dif_temp)*(hplus^n_acid)
170 r_neut = k_neut*EXP((-eapp_neut/8.314e-3)*dif_temp)*(hplus^n_neut)
180 r_base = k_base*EXP((-eapp_base/8.314e-3)*dif_temp)*(hplus^n_base)
190 r_all = r_acid+r_neut+r_base
200 rate = (a0/v)*(M/M0)^0.67*r_all*(1-SR("Diopside"))
210 moles = rate*TIME
220 SAVE moles
-end
KINETICS 1
Plagioclase
-formula Ca0.5Na0.5Al1.5Si2.5O8Cs0.001 1
-m 0.2789
-m0 0.2789
-parms 100.9 0.4661 1.131
-tol 1e-08
Diopside
-formula CaMgSi2O6Rb0.001 1
-m 0.0621
-m0 0.0621
-parms 100.9 0.0783 1.131
-tol 1e-08
-steps 1
-step_divide 1
-runge_kutta 3
-bad_step_max 500
SELECTED_OUTPUT 1
-file 180419_1333_MikeExpts1_kinetic.tsv
-totals Na Ca Mg C Li Cs Rb
-equilibrium_phases Calcite Kaolinite CO2(g)
-kinetic_reactants Plagioclase Diopside
TRANSPORT
-cells 2
-shifts 4
-time_step 2592000 # seconds
-lengths 2*0.25
-diffusion_coefficient 0
-thermal_diffusion 2 0
-print_cells 1-2
-punch_cells 1-2
EQUILIBRIUM_PHASES 1
Calcite 0 10
Kaolinite 3 0
CO2(g) -2.39 10
EQUILIBRIUM_PHASES 2
Calcite 0 0
Kaolinite 3 0
CO2(g) -3.39 10
Logged
dlparkhurst
Global Moderator
Posts: 3949
Re: Phreeqc seems to halve first kinetic step
«
Reply #1 on:
20/04/18 16:45 »
I believe the calculation is accounting for the location of the node. The nodes are at half the length of the cell, so it is accounting for when a concentration front would reach the node.
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peterwadeuk
Top Contributor
Posts: 52
Re: Phreeqc seems to halve first kinetic step
«
Reply #2 on:
20/04/18 18:15 »
Thank you David
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