The contradiction between dissolution degree and the amount of solute

[sama.k]

Hello;
I model a mineral (C4AF or (CaO)4. (Al2O3). (Fe2O3)) hydration kinetically. After 100 days, the degree of C4AF hydration is about 0.75, but the results show that all C4AF is dissolved. After 100 days, the amount of C4AF in moles is zero which means the degree of C4AF hydration is 1. I'm totally confused about this contradiction. Do I have some misunderstanding about this?
Here are the input data and diagrams of my model.
Any help or advice will be greatly appreciated.
Many thanks

Code: [Select]

EQUILIBRIUM_PHASES 1
    C2S       0 0
    C3A       0 0
    C3AH6     0 0
    C3FH6     0 0
    C3S       0 0
    C4AF      0 0
    CSH3T-TobH 0 0
SOLUTION 1
    temp      20
    pH        7
    pe        4
    redox     pe
    units     mmol/kgw
    density   1
    -water    0.04 # kg
REACTION 1
    C4AF       0.0174
    0.4672 moles in 1 steps
SAVE solution 1
RATES
    C4AF_RATE_HYDRATION
-start
 10 REM   PARM(1) = A,Blaine surface area of cement, m^2/kg
 20 REM   PARM(2) = A0,reference surface area of cement, m^2/kg
 30 REM   PARM(3) = E,the apparent activation energy of C4AF, j/Mmol
 40 REM   PARM(4) = R,ideal gas constant, L·Pa·K-1·mol-1
 50 REM   PARM(5) = T0,the reference temperature, K
 60 REM   M = the hydration degree
 70 a = M
 80 r1 = 0.53*(1-a)*((-1*LOG(1-a))^0.3)
 90 r2 = (((1-a)^0.67)*0.015)/(1-((1-a)^0.34))
100 r3 = (0.4*(1-a)^3.7)
110 IF(a > 0.58) THEN b = (((0.58-a)*3.333)+1)^4 ELSE b = 1
120 min = r1
130 IF min > r2 THEN min = r2
140 IF min > r3 THEN min = r3
150 a_rate = b*min*2.0736*(PARM(1)/PARM(2))*EXP((PARM(3)/PARM(4))*((1/PARM(5))-(1/TK)))
160 a_delta = a_rate*TIME/86400
170 SAVE-a_delta
-end
    C4AF
-start
10 rate = KIN("C4AF_RATE_HYDRATION")*M
20 SAVE rate * time / 86400
-end
KINETICS 1
C4AF
    -formula  (CaO)4(Al2O3)(Fe2O3)  1
    -m        0.0174
    -m0       0.0174
    -tol      1e-008
C4AF_RATE_HYDRATION
    -formula  H  0
    -m        0.001
    -m0       0.001
    -parms    413 385 34.087 8.3144 293.15
    -tol      1e-008
-steps       8640000 in 100 steps # seconds
-step_divide 1
-runge_kutta 3
-bad_step_max 5000
USER_GRAPH 2
    -headings               Time C4AF pH
    -axis_titles            "Time (days)" "C4AF  (moles)" "pH"
    -chart_title            "C4AF"
    -initial_solutions      false
    -connect_simulations    true
    -plot_concentration_vs  x
  -start
10 GRAPH_X total_time/86400
20 GRAPH_Y KIN("C4AF")
30 GRAPH_SY -LA("H+")
  -end
    -active                 true
USER_GRAPH 1
    -axis_titles            "Time (days)" "Hydration degree" ""
    -chart_title            "Hydration degree of C4AF"
    -initial_solutions      false
    -connect_simulations    true
    -plot_concentration_vs  x
  -start
10 GRAPH_X TOTAL_TIME/86400
20 GRAPH_Y KIN("C4AF_RATE_HYDRATION")
  -end
    -active                 true
END


[dlparkhurst]

Well, you are not alone; you totally confuse me too.

So, I now gather that the hydration degree is equivalent to the fraction of the mineral that has dissolved. If this is the case, then here is another stab on my part as to what you want.

As before, the C4AF_RATE_HYDRATION rate is used to integrate the total degree of hydration. In turn, the rate of change of the total degree of hydration is used to dissolve the mineral. PUT and GET are used to transfer the hydration rate (equal to dissolution rate) from the C4AF_RATE_HYDRATION rate evaluation to the C4AF rate evaluation. The amount of mineral remaining at any time should be one minus the degree of hydration (a) times the amount of mineral initially present (M0). (Earlier post mistakenly forgot the "one minus".)

If you add -cvode to your KINETICS definition and the keyword "INCREMENTAL_REACTIONS true" to the simulation, the calculation will run faster.

Code: [Select]

RATES
    C4AF_RATE_HYDRATION
-start
 10 REM   PARM(1) = A,Blaine surface area of cement, m^2/kg
 20 REM   PARM(2) = A0,reference surface area of cement, m^2/kg
 30 REM   PARM(3) = E,the apparent activation energy of C4AF, j/Mmol
 40 REM   PARM(4) = R,ideal gas constant, L·Pa·K-1·mol-1
 50 REM   PARM(5) = T0,the reference temperature, K
 60 REM   M = the hydration degree
 70 a = M
 80 r1 = 0.53*(1-a)*((-1*LOG(1-a))^0.3)
 90 r2 = (((1-a)^0.67)*0.015)/(1-((1-a)^0.34))
100 r3 = (0.4*(1-a)^3.7)
110 IF(a > 0.58) THEN b = (((0.58-a)*3.333)+1)^4 ELSE b = 1
120 min = r1
130 IF min > r2 THEN min = r2
140 IF min > r3 THEN min = r3
150 a_rate = b*min*2.0736*(PARM(1)/PARM(2))*EXP((PARM(3)/PARM(4))*((1/PARM(5))-(1/TK)))
160 a_delta = a_rate*TIME/86400
170 SAVE -a_delta
180 PUT(a_delta, 1)
-end
    C4AF
-start
10 rate = GET(1) * M0
20 SAVE rate
-end