Processes > Surface Complexation
Surface complexation model_Effect of dosage on uranium removal
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Francis Adu:
Dear All,
I would be glad if anyone could share his/her thought on my model for me. I am modeling the removal of uranium by iron, silicon, and aluminum hydroxides. The main objective is to study the effect of varying the adsorbent dosage on the removal at a fixed pH of 8. In the model attached; I am only varying the dosage from 0.05-0.2g. However, when I change the mass, there is no appreciable change in the removal as I am recording also 100% removal at all the different dosages. I would be glad if anyone could check my codes or suggest any reason why significant differences are not recorded whenever there is a change in the dosage. I have attached the codes below for your reference;
PHASES 1
fix_pH
H+ = H+
log_k 0
SURFACE_MASTER_SPECIES 1
Surf_al Surf_alOH
Surf_si Surf_siOH
Surf_fe Surf_feOH
SURFACE_SPECIES 1
Surf_alOH = Surf_alOH
log_k 0.0
-no_check
Surf_alOH + H+ = Surf_alOH2+
log_k 12.3
Surf_alOH = Surf_alO- + H+
log_k -13.16
Surf_alOH + UO2+2 = Surf_alOUO2+ + H+
log_k 1.4
Surf_alOH + 2UO2+2 + CO3-2 + 3H2O = Surf_alO(UO2)2CO3(OH)3-2 + 4H+
log_k 0.99
Surf_siOH = Surf_siOH
log_k 0.0
Surf_siOH + H+ = Surf_siOH2+
log_k -0.95
Surf_siOH = Surf_siO- + H+
log_k -6.90
Surf_siOH + UO2+2 = Surf_siOUO2+ + H+
log_k 0.99
Surf_siOH + UO2OH+ = Surf_siOUO2OH + H+
log_k 1.0
Surf_siOH + UO2(CO3)3-4 = Surf_siOHUO2(CO3)3-4
log_k 8.0
Surf_siOH + UO2(OH)3- = Surf_siOHUO2(OH)3-
log_k 6.90
Surf_feOH = Surf_feOH
log_k 0.0
Surf_feOH +H+ = Surf_feOH2+
log_k 7.47
Surf_feOH = Surf_feO- + H+
log_k -9.51
Surf_feOH + UO2+2 = Surf_feOUO2+ + H+
log_k 5.20
Surf_feOH + UO2+2 + CO3-2 = Surf_feOHUO2CO3
log_k 15.95
SURFACE 1
Surf_siOH 2.773e-04 500 0.05
Surf_alOH 9.243e-06
Surf_feOH 8.596e-04
SOLUTION 1
temp 20
pH 8
pe 4
redox pe
units mmol/l
density 1
C(4) 1
Cl 10
Na 11
U 250 ug/L
-water 0.1 # kg
END
Thank you
dlparkhurst:
I'm not sure how you were adjusting the surface parameters. If you changed only the number of grams, then the number of sorption sites would not vary; only the surface area would change. The surface area would most affect the potential factor, which is related to the amount of work needed to bring a charged ion from solution to a surface site.
Regardless of the surface area, the number of sites is sufficient to remove almost all of the uranium. Surf_fe is the most important surface site for sorbing the uranium.
Another way to define the number of surface sites is by use of the -sites DENSITY option, which allows the number of sites to depend on the number of grams that is defined. The input is sites per nanometer squared. I have adjusted the SURFACE definition to use -site DENSITY to produce the same number of sites in your definition for 0.05 g. I repeated the calculation with fewer and fewer grams, which corresponded to fewer and fewer sites. The amount of U sorbed is plotted against the total number of Surf sites. It requires about 1e-4 fewer sites before uranium is not completely sorbed. You seemed to expect a variation as the number of sites increased from 0.05 g. You need to look at the log Ks of the sorption reactions if you want sorption to be less complete.
--- Code: ---PHASES 1
fix_pH
H+ = H+
log_k 0
SURFACE_MASTER_SPECIES 1
Surf_al Surf_alOH
Surf_si Surf_siOH
Surf_fe Surf_feOH
SURFACE_SPECIES 1
Surf_alOH = Surf_alOH
log_k 0.0
-no_check
Surf_alOH + H+ = Surf_alOH2+
log_k 12.3
Surf_alOH = Surf_alO- + H+
log_k -13.16
Surf_alOH + UO2+2 = Surf_alOUO2+ + H+
log_k 1.4
Surf_alOH + 2UO2+2 + CO3-2 + 3H2O = Surf_alO(UO2)2CO3(OH)3-2 + 4H+
log_k 0.99
Surf_siOH = Surf_siOH
log_k 0.0
Surf_siOH + H+ = Surf_siOH2+
log_k -0.95
Surf_siOH = Surf_siO- + H+
log_k -6.90
Surf_siOH + UO2+2 = Surf_siOUO2+ + H+
log_k 0.99
Surf_siOH + UO2OH+ = Surf_siOUO2OH + H+
log_k 1.0
Surf_siOH + UO2(CO3)3-4 = Surf_siOHUO2(CO3)3-4
log_k 8.0
Surf_siOH + UO2(OH)3- = Surf_siOHUO2(OH)3-
log_k 6.90
Surf_feOH = Surf_feOH
log_k 0.0
Surf_feOH +H+ = Surf_feOH2+
log_k 7.47
Surf_feOH = Surf_feO- + H+
log_k -9.51
Surf_feOH + UO2+2 = Surf_feOUO2+ + H+
log_k 5.20
Surf_feOH + UO2+2 + CO3-2 = Surf_feOHUO2CO3
log_k 15.95
SOLUTION 1
temp 20
pH 8
pe 4
redox pe
units mmol/l
density 1
C(4) 1
Cl 10
Na 11
U 250 ug/L
-water 0.1 # kg
END
#SURFACE 1
#Surf_siOH 2.773e-04 500 0.05
#Surf_alOH 9.243e-06
#Surf_feOH 8.596e-04
END
USER_GRAPH 1
-axis_titles "Surf sites, moles" "U sorbed, moles" ""
-axis_scale x_axis auto auto auto auto log
-axis_scale y_axis auto auto auto auto log
-initial_solutions false
-connect_simulations true
-plot_concentration_vs x
-start
10 GRAPH_X TOTMOL("Surf_al") + TOTMOL("Surf_fe") + TOTMOL("Surf_si")
20 GRAPH_Y SURF("U","Surf")
-end
-active true
USE solution 1
SURFACE 1
-sites DENSITY
Surf_siOH 6.6803 500 0.05
Surf_alOH 0.22267
Surf_feOH 20.708
END
USE solution 1
SURFACE 1
-sites DENSITY
Surf_siOH 6.6803 500 0.0005
Surf_alOH 0.22267
Surf_feOH 20.708
END
USE solution 1
SURFACE 1
-sites DENSITY
Surf_siOH 6.6803 500 0.000005
Surf_alOH 0.22267
Surf_feOH 20.708
END
USE solution 1
SURFACE 1
-sites DENSITY
Surf_siOH 6.6803 500 0.0000005
Surf_alOH 0.22267
Surf_feOH 20.708
END
USE solution 1
SURFACE 1
-sites DENSITY
Surf_siOH 6.6803 500 0.00000005
Surf_alOH 0.22267
Surf_feOH 20.708
END
USE solution 1
SURFACE 1
-sites DENSITY
Surf_siOH 6.6803 500 0.000000005
Surf_alOH 0.22267
Surf_feOH 20.708
END
--- End code ---
Francis Adu:
Dear Parkhurst,
Thank you very much for your time in answering my question. The explanation has explicitly clarified a lot of things for me.
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