Processes > Reactive transport modelling
Water cleaning by air stripping
MichaelZ20:
Thank you, David!
MichaelZ20:
Hi David!
I tried to reproduce the experimental data of ammonia removal (https://doi.org/10.1016/j.cej.2024.153715) by using the proposed approach.
For achieving the final experimental percent of ammonia removal the simulation gives about 3 L of the passing air, whereas in the experiment -120 L.
Could you please look through the attached input file? Do I omit something? Thank you in advance.
--- Code: ---DATABASE c:\phreeqc\database\minteq.v4.dat
TITLE
SOLUTION_MASTER_SPECIES
Oxg Oxg 0 Oxg 32
N NO3- 0 14.0067 14.0067
N(-3) NH4+ 0 14.0067
N(0) N2 0 14.0067
Phenol Phenol 0 Phenol 94.11
SOLUTION_SPECIES
Oxg = Oxg
log_k 0
Phenol = Phenol
Log_k 0
NH4+ = NH3 + H+
log_k -9.252
delta_h 12.48 kcal
-analytic 0.6322 -0.001225 -2835.76
-gamma 2.5 0
2 NO3- + 12 H+ + 10 e- = N2 + 6 H2O
log_k 207.08
delta_h -312.13 kcal
NO3- + 10 H+ + 8 e- = NH4+ + 3 H2O
log_k 119.077
delta_h -187.055 kcal
PHASES
Oxg(g)
O2 + 4 H+ + 4 e- = 2 H2O
log_k 83.0894
delta_h -571.66 kJ
N2(g)
N2 = N2
log_k -3.26
delta_h -1.358 kcal
NH3(g)
NH3 = NH3
log_k 1.77
delta_h -8.17 kcal
Phenol(g)
Phenol = Phenol
-log_k -3.45
-delta_h 96.4 KJ
Fix_pH
H+ = H+
log_k 0
GAS_PHASE 0
-fixed_volume
-pressure 1
-volume 0.2 # L in headspace
-temperature 25
N2(g) 0.78 # a portion of N2 in atmospheric air
Oxg(g) 0.242 # a portion of O2 in atmospheric air
CO2(g) 0.00044 # a portion of CO2 in atmospheric air
NH3(g) 1e-6 # a portion of NH3 in atmospheric air
Phenol(g) 1e-6 # a portion of Phenol in atmospheric air
END
SOLUTION 0 Initial water
units mg/l
Density 1 calculate
pH 7.7
temp 25
N(-3) 5470 # Total ammonia nitrogen: 4500 mg/l
Acetate 4620
Phenol 1640
Cl 10 charge
water 1
END
MIX 1
0 0.1 # 100 mL
SAVE SOLUTION 0
SELECTED_OUTPUT; -reset false; -file NH3_gas.txt
USER_PUNCH
-start
10 for i = 0 to 120 step 1
20 punch "USE SOLUTION 0" + eol$
25 punch "EQUILIBRIUM_PHASES" + eol$
27 punch " Fix_pH -9.3 NaOH 10" + eol$
30 punch "USE GAS_PHASE 0" + eol$
40 punch "REACTION_TEMPERATURE 85" + eol$
45 punch "SAVE SOLUTION 0" + eol$
50 punch "END" + eol$
60 next i
-end
END
PRINT; -reset false; -selected_out false
USER_GRAPH 1
-axis_titles "The total volume of air passed, L" "Ammonium mass removal, %"
-start
10 GRAPH_X GAS_VM * (GAS("NH3(g)") + GAS("N2(g)") + GAS("Oxg(g)") + GAS("CO2(g)") + GAS("H2O(g)") + GAS("Phenol(g)")) * (SIM_NO - 3)
20 GRAPH_Y ((0.3917 - TOT("N(-3)")) / 0.3917) * 100
-end
INCLUDE$ NH3_gas.txt
END
--- End code ---
dlparkhurst:
Mainly, I think you need to consider disequilibrium in the nitrogen system. I would use Amm.dat with the extra gases added, although you really don't need the other gases; if pH is fixed, the main reaction is simply between ammonia and the head space.
Here is a revision of your script. It takes about 120 L of introduced air to remove virtually all of the ammonia, assuming it reacts independently of all other aqueous species and gases. I redefined the gases to avoid Peng-Robinson calculations, which is reasonable at low pressures.
--- Code: ---#DATABASE c:\phreeqc\database\minteq.v4.dat
TITLE
SOLUTION_MASTER_SPECIES
Phenol Phenol 0 Phenol 94.11
Acetate Acetate- 1 59.045 59.045
SOLUTION_SPECIES
Phenol = Phenol
log_k 0
Acetate- = Acetate-
log_k 0
Acetate- + H+ = H(Acetate)
log_k 4.757
delta_h 0.41 kJ
-gamma 0 0
PHASES
Amm(g)
Amm = Amm
log_k 1.7966
-analytical_expression -18.758 0.0003367 2511.3 4.8619 39.192 0
Oxg(g)
Oxg = Oxg
-analytical_expression -7.5001 0.0078981 0 0 200270 0
N2(g)
N2 = N2
log_k -3.1864
-analytical_expression -58.453 0.001818 3199 17.909 -27460 0
CO2(g)
CO2 = CO2
log_k -1.468
delta_h -4.776 kcal
-analytical_expression 10.5624 -0.023547 -3972.8 0 587460 1.9194e-05
H2O(g)
H2O = H2O
log_k 1.506
delta_h -44.03 kJ
-analytical_expression -16.5066 -0.0020013 2710.7 3.7646 0 2.24e-06
Phenol(g)
Phenol = Phenol
log_k -3.45
delta_h 96.4 kJ
Fix_pH
H+ = H+
log_k 0
GAS_PHASE 0
-fixed_volume
-pressure 1
-volume 0.2
-temperature 25
Amm(g) 1e-06
CO2(g) 0.00044
N2(g) 0.78
Oxg(g) 0.242
Phenol(g) 1e-06
END
SOLUTION 0 Initial water
temp 25
pH 7.7
pe 4
redox pe
units mg/l
density 1
Acetate 4620
Amm 5470 as N
Cl 10 charge
Phenol 1640
-water 1 # kg
END
MIX 1
0 0.1 # 100 mL
SAVE SOLUTION 0
SELECTED_OUTPUT; -reset false; -file NH3_gas.txt
USER_PUNCH 1
-headings
-start
10 FOR i = 0 TO 5*120 STEP 1
20 PUNCH "USE SOLUTION 0"+EOL$
30 PUNCH "EQUILIBRIUM_PHASES"+EOL$
40 PUNCH " Fix_pH -9.3 NaOH 10"+EOL$
50 PUNCH "USE GAS_PHASE 0"+EOL$
60 PUNCH "REACTION_TEMPERATURE 1"+EOL$
70 PUNCH "85"+EOL$
80 PUNCH "SAVE SOLUTION 0"+EOL$
90 PUNCH "END"+EOL$
100 NEXT i
-end
END
SELECTED_OUTPUT 1
-active false
END
USER_GRAPH 1
-headings vol %_removal Amm
-axis_titles "The total volume of air introduced, L" "Ammonium mass removal, %" "Amm, mol/kgw"
-initial_solutions false
-connect_simulations true
-plot_concentration_vs x
-start
10 stp = GET(0)+1
20 PUT(stp, 0)
30 GRAPH_X stp * 0.2
40 GRAPH_Y((0.3917-TOT("Amm"))/0.3917)*100
50 GRAPH_SY TOT("Amm")
-end
-active true
INCLUDE$ NH3_gas.txt
END
--- End code ---
MichaelZ20:
Thank you, David!
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