PhreeqcUsers Discussion Forum

Processes => Oxidation and reduction equilibria => Topic started by: Drew_2006 on April 01, 2016, 02:25:21 PM

Title: Nitrification/Denitrification redox
Post by: Drew_2006 on April 01, 2016, 02:25:21 PM
Hello!
I'm a master's student from Catalonia, and I'm trying to make model the redox reactions that occur for nitrification and denitrification. When I modeled in BATCH format and both reactions operate separately. But when I try that reactions go together to see the route of the nitrogen specie it doesn't work. I've try to induce all the oxygen runs out to start reducing nitrogen but does not work either.
I hope you can help me.

Thank you very much!

xoxox

SOLUTION_MASTER_SPECIES
Doc             Doc                 0.0         Doc          30
Nitrat       Nitrat        0.0         Nitrat              62
Amp           Amp                   0.0           Amp          17.0

SOLUTION_SPECIES
Doc = Doc;           log_k 0
Nitrat = Nitrat;            log_k 0
Amp = Amp;                   log_k 0

SOLUTION 1
units          mol/kgw
pH               7.02
temp          22
N(3)                  3.50e-3
N(5)                  3.50e-2
Amp                   3.50e-3
Doc              3.50e-3
O(0)                 3.50e-1
Ca             5.892821523e-3
Alkalinity       7.91e-3 as HCO3-
P             1.77e-3 as PO4-3
S             1.35e-3 as SO4-2
Mg            1.38e-3
K             2.41e-4
Na            2.25e-3
Cl            1.43e-3

INCREMENTAL_REACTIONS true

KINETICS
   Amp_degradation
         -formula Amp 1  NH4+ -1
         -steps     0 150*0.05

   Doc_degradation
         -formula Doc 1    Nitrat 0.8 CH2O -1
         -steps     0 150*0.05


RATES
   Amp_degradation
         -start
   10 Biomass = 1e-4
   20 Ks =  1.855555E-03
   30 Ka =  7.277990E-04
   40 qm =  65.2151
   50 Monod_Amp = (tot"Amp") / (Ks + tot"Amp")
   60 Monod_Doc = (tot"O(0)") / (Ka + tot"O(0)")
   70 rate_Doc =  - qm * Monod_Amp * Monod_Doc * Biomass
   80 if tot("O(0)") < 1e-8 then goto 90
   90 moles_Doc = rate_Doc * time
   100 SAVE moles_Doc
            -end

   Doc_degradation
         -start
   10 Biomass = 1e-4
   20 Ks =  1.4e-06
   30 Ka =  2.277990e-06
   40 qm =  53.9475
   50 Ki =  1e-8
   50 Monod_Nitrat = (tot("N(5)") / (Ks + tot("N(5)")))
   80 Monod_Doc = (tot("Doc") / (Ka + tot("Doc")))
   100 rate_Doc =  - qm * Monod_Nitrat * Monod_Doc * Biomass
   110 moles_Doc = rate_Doc * time
   120 SAVE moles_Doc
         -end

USER_GRAPH
      -headings Time Amp Oxigen Nitrat Doc DIC pH totcarbon
       -initial_solutions false
       -axis_scale x_axis 0 6 0.5
        -axis_scale y_axis 0 15 0.5
      -axis_scale SY_axis 0 100 10
         -axis_titles "Time [days]" "Concentration [mmol/L]"
      -chart_title "Experiment BATCH de Degradació de Amoni"
       -plot_concentration_vs time
        -start

   10    GRAPH_X TOTAL_TIME
   20    GRAPH_Y   tot("N(5)")*1000 tot("Doc")*1000 tot("C(4)")*1000
   25   GRAPH_Y   tot("Amp")*1000 tot("O(0)")*1000 tot("N(5)")*1000
   30  totalcarbon = (tot("C(4)")+tot("C(-4)")+tot("Doc"))
   40  GRAPH_Y - la("H+")
   300 GRAPH_Y totalcarbon
        -end

END

Title: Re: Nitrification/Denitrification redox
Post by: dlparkhurst on April 04, 2016, 05:06:48 PM
Here is input to handle the nitrogen system kinetically. By using the Amm.dat database, ammonia is defined as the element Amm, with species Amm and AmmH+.

The nitrogen system contains N(0), N(3), and N(5). By the SOLUTION_SPECIES definition below, the aqueous species N2 is made negligibly small. So N has just NO2- and NO3- species, and NO2- is usually negligible.

Amm.dat has Ntg gas defined, which is used to simulate N2(aq) and N2(g).

Thus nitrification converts Amm to NH3, and denitrification converts N2 to Ntg. The RATES data block defines the kinetic rate expressions, and KINETICS defines the stoichiometry of the reaction. For nitrification, thermodynamics will determine the electron acceptor, but it will be oxygen (the reaction stops in the absence of dissolve oxygen). Denitrification is inhibited by dissolved oxygen. For denitrification, consumption of organic matter (CH2O) is assumed, and specified in the KINETICS formula.


SOLUTION_SPECIES
2 NO3- + 12 H+ + 10 e- = N2 + 6 H2O
   #-log_k   207.08
        -log_k        0

RATES
  Nitrification
-start
10 Amm = MOL("AmmH+")
#20 if (Amm <= 0) then goto 200
30 p_Oxg = SR("O2(g)")
40 moles = 3.10e-10*Amm/(100.10e-6+Amm)*(p_Oxg/(15.10e-6+p_Oxg))*TIME
200 SAVE moles
-end
  Denitrification
-start
10 NO3 = MOL("NO3-")
#20 if (NO3 <= 0) then goto 200
25 p_Oxg = SR("O2(g)")
30 moles = 2.22e-9*NO3*(1 - p_Oxg/(15.10e-6+p_Oxg))*TIME
200 SAVE moles
-end
END
KINETICS  1-19
#-cvode
Nitrification
    -formula Amm -1.0 NH3 1.0
    -m 4.60
    -tol 1e-8
-step_divide 1
-runge_kutta 3
Denitrification
    -formula N2 -1 CH2O +1 Ntg 1
    -m 24.4
    -tol 1e-8
-step_divide 1
-runge_kutta 3
END
Title: Re: Nitrification/Denitrification redox
Post by: MichaelZ20 on December 05, 2021, 06:32:51 PM
Hi David!
In the PHREEQC manual, in KINETICS explanation, it is written: "The product of the coefficient times the moles of reaction progress gives the mole transfer for formula relative to the aqueous solution".
The denitrification reaction can be written as 4NO3- + 5CH2O + 4H+ = 2N2 + 7H2O + 5CO2.
In the KINETICS block of the above example, you wrote:
-formula N2 -1 CH2O +1 Ntg 1, however by the manual it should be
-formula N2 -2 CH2O +5 Ntg 2.
I have checked that the results are the same for the two cases.
Can you please comment on this?
Thank you in advance for your reply.
Title: Re: Nitrification/Denitrification redox
Post by: dlparkhurst on December 06, 2021, 03:43:42 AM
You caught me. My stoichiometry is wrong. Yours is correct if you are not keeping track of dissolved organic carbon or the organic carbon source is not dissolved.

Keeping track of dissolved organic carbon, the -formula in https://phreeqcusers.org/index.php/topic,1847.msg6408.html#msg6408 is correct, but the delta for C is omitted. Here is the complete justification:

4NO3- + 5Doc + 4H+ = 2Ntg + 5CO2 + 7H2O

delta N = -4
delta Ntg = 2
delta Doc = -5
delta C = 5
delta O = 5
delta H = 10

or -formula N -4 Ntg 2 Doc -5 CH2O 5