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*To*: David L Parkhurst <dlpark@xxxxxxxx>*Subject*: Re: PHREEQC problem*From*: xiaomin Mao <xmao@xxxxxxxx>*Date*: Tue, 3 Jun 2003 12:05:58 +0100*References*: <OFECEB172F.29DD6BA8-ON87256D34.007CA3E5@usgs.gov>

David, Thank you very much! It can work very well now. Yes, the .0001(org_cuex1 + org_cuex2) = kin(org_xcu)" represents equilibrium. In fact it's only part of the Nica-Donnan model, as at first I think it is too difficult to model it using phreeqc. Now maybe I should continue to explore how to perform the whole model. I am further convinced that PHREEQC can do anything you want. :-) cheers, xiaomin David L Parkhurst wrote: > > Sorry I didn't explain the process clearly. I assume the sorption of Cu > by > organicmatter site (Org_x) as follow (see Table 6.6, Werner Stumm, 1995, > the > book: Aquatic Chemistry): > Cu+2 + Org_xH2 = Org_xCu + 2H+ > > > To the formula of Org_xh, I mean H is 0 (zero), not H O. Maybe as you > said, > it's not necessary to mention this kinetic expression as it has the same > rate > like Org_xcu. > > Sorry, my bad. I am a little slow, but I understand your input file now. I > don't have the latest Stumm and Morgan, but I did some investigation into > your problem. I am assuming the ".0001(org_cuex1 + org_cuex2) = > kin(org_xcu)" represents equilibrium; if not, then I what follows would > have to be modified. > > Here are a few comments: > > (1) First you are using kinetics to represent equilibrium and PHREEQC is > not particularly good for this kind problem. PHREEQC is not good for > handling stiff differential equations, an implicit integration method would > work better than the explicit Runge-Kutta routine that is currently used. > However, I think PHREEQC can beat the problem to death, although slowly. > Better yet would be to include the sorption model as a type of equilibrium > for PHREEQC, which would probably be hundreds of times faster. > > (2) I'm not sure of the database you are using, but if it has both Cu(1) > and Cu(2), you could be running into some redox problems. You should define > the Cu in solution as "Cu(2)", otherwise, you may be removing Cu(1) and > effectively replacing it with one H+ and one H(0). The H(0) will cause some > unwanted redox reactions. With wateq4f.dat, if you define copper as Cu in > your initial solution, a significant fraction is Cu(1), so be careful that > the Cu starts out in the right redox state [Cu(2)]. > > (3) Aside from the redox-state issue, I think the fundamental problem is > integrating the rate expression. You have a dissolved Cutot [TOT("Cu")] and > sorbed Cukin [KIN(org_cux]. You calculate the sorbed Cu from the Cutot (and > H+), compare it to the Cukin, and calculate a rate from the difference > divided by 10. The rate is sufficient to ensure equilibrium in about 10 > seconds; however, phreeqc attempts to extrapolate the rate for times up to > the time step (30000 seconds). There tends to be overshoot and undershoot > if the rate is extrapolated for two reason: (1) the long extrapolation > relative to the time to equilibrium, and (2) poor estimate of equilibrium. > The time scale of 10 seconds for the rate probably limits the internal time > step for the integration to about 1 second. thus there are a lot of > internal iterations to integrate the rate expression in each cell. > > As for the poor estimation of equilibrium, the original rate expression > does not account for the change in solution and sorbed concentrations to > reach equilibrium. Let equilibrium be represented by Cutot,eq = Cutot + > dM/TOT("water") and Cukin,eq = Cukin - dM for some dM, where dM is moles. > You have equilibrium as dM = f(CuTot, H+) - Cukin, where CuTot and CuKin > are fixed at the values at the beginning of the integration step. I think > this leads to the possibility of oscillation, where one integration step > overshoots and the next over-corrects. > > I've written a little routine that estimates dM and then calculates the > rate as a fraction of dM. I've used dM/10000 as discussed below. Regardless > of the factor used, the integration appears to be robust, only more or less > slow. > > (4) Your dispersivity is large relative to the cell size and column length. > This (and -correct_disp) requires 9 "mixruns" to simulate a time step. As > you know, a smaller dispersivity would run faster. With as much dispersion > as you have, it would be considerably faster and equally accurate (I think) > to run the problem with phast; it would not require the 9 "mixruns" to > account for the dispersion. > > (5) Finally, a rate that is of the same scale as the time step will run > faster (less stiff). I've used a rate that should be close to equilibrium > in 10000 seconds. I experimented a little with faster rates and the > accuracy seemed sufficient with the slower (dM/10000) rate. With this rate, > I was able to run the entire simulation in about 6 minutes on an early > pentium IV. > > David > > DATABASE ../database/wateq4f.dat > Rates > Org_xcu > -start > 10 Cu_conc0=Tot("Cu") > 20 H_conc=ACT("H+") > 30 sorbed0 = M > 35 moles_cu = cu_conc0 * TOT("water") + sorbed0 > #40 print "cell, cu, sorbed, moles_cu: ", cell_no, cu_conc0, sorbed0, moles_cu, time, total_time > 50 if (moles_cu <= 1e-14) then goto 350 > 60 if (Cu_conc0 > 1e-14) then x = log10(Cu_conc0/2) else x = log10(.01*moles_cu/TOT("water")) > #80 print "10^x, 10^x_cu, 10^x_tot", 10^x, Cu_conc0/2, .01*moles_cu/TOT("water") > 130 REM loop to calculate equilibrium > 140 gosub 500 > 150 fx1 = fx > 160 REM calculate df(x)/dx > 165 dx = 1e-7 > 170 x = x + dx > 180 gosub 500 > 185 fx2 = fx > 190 dfx = (fx - fx1)/dx > #200 x = x - dx - fx1/dfx > 210 if (ABS(fx1/dfx) < 1) then x = x - dx - fx1/dfx else x = x - dx - (fx1/dfx)/ABS(fx1/dfx) > #220 print "cu_est, sorbed_est, resid, delta, mass_err", 10^x, sorbed, fx1, ABS(fx1/dfx), (10^x*TOT("water") + sorbed) - moles_cu, dfx > 230 if ABS(fx1) > 1e-14 then goto 130 > > #0.0001*Cu_conc/(Cu_conc+H_conc)*parm(1) > #35 rate1=-(0.0001*(Org_xcue1+Org_xcue2)-Kin("Org_xcu"))/10 > #335 rate1 = -(sorbed-kin("Org_xcu"))/10 # diff from equilibrium / 10 > 335 rate1 = -(sorbed-kin("Org_xcu"))/10000 # diff from equilibrium / 10000 > # 0.0001--based on the quantity of organic matter > 337 put(rate1,1) > 340 moles=rate1 * time > 345 if ABS(moles) < 1e-14 then moles = 0 > 350 save moles > 360 goto 1000 > > # calculates sorbed_Cu from cu_conc, H_conc > 500 REM x is log10 of Cu concentration > 510 Cu_conc = 10^x > 520 sorbed = sorbed0 - TOT("water")*(Cu_conc - Cu_conc0) > 530 Org_xcue1=4.722*(1.8197*Cu_conc)^0.53/((1.8197*Cu_conc)^0.53+(218.78*H_conc)^0.66) \ > *((1.8197*Cu_conc)^0.53+(218.78*H_conc)^0.66)^0.59/(1+((1.8197*Cu_conc)^0.53+(218.78*H_conc)^0.66)^0.59) > 540 Org_xcue2=0.8811*(181970086*Cu_conc)^0.36/((181970086*Cu_conc)^0.36+(398107171*H_conc)^0.76) \ > *((181970086*Cu_conc)^0.36+(398107171*H_conc)^0.76)^0.7/(1+((181970086*Cu_conc)^0.36+(398107171*H_conc)^0.76)^0.7) > 550 fx = .0001*(org_xcue1 + org_xcue2) - sorbed > 560 return > > 1000 REM end of program > -end > Org_xh > -start > 35 rate1=get(1) > 40 moles=-rate1*time > 50 save moles > -end > end > > SOLUTION 0 CCA concentration > units mol/kgw > pH 7.0 charge > pe 4 > -water 0.03375 > Na 1 > Cl 1 > > SOLUTION 1 CCA concentration > units mol/kgw > pH 4.0 > pe 4 > -water 0.03375 > > Cu(2) 0.123562 > Na 1 > Cl 1.3 charge > > SOLUTION 2-16 CCA concentration > units mol/kgw > pH 4.0 charge > pe 4 > -water 0.03375 > Cu 0.00 > Na 1 > Cl 1 > > kinetics 1-16 > Org_xcu > -runge_k 6 > -Formula H -2.0 Cu +1 > -m0 0.00 > -m 0.00 > -tol 1e-10 > #-Parms 0.1 > Org_xh > -Formula H 0 > -m0 0.1 > -Parms 0.1 > > TRANSPORT > -cells 16 > -shifts 33 > -time_step 30857 #10.5d/28*86400s/d > -flow_direction forward > -lengths 16*5e-3 > -boundary_conditions flux flux > -dispersivities 16*0.012 > -correct_disp true > -diffusion_coefficient 0.0000 > -punch_cells 16 > -punch_frequency 1 > -warnings false > > SELECTED_OUTPUT > -file mao.7.sel > -selected_out true > -user_punch true > -reset false > > USER_PUNCH > -headings time cell Cu pH Org_xcu Tot_Cu fx > -start > 10 Cu_conc=Tot("Cu(2)") > 20 H_conc=ACT("H+") > 30 Org_xcue1=4.722*(1.8197*Cu_conc)^0.53/((1.8197*Cu_conc)^0.53+(218.78*H_conc)^0.66) \ > *((1.8197*Cu_conc)^0.53+(218.78*H_conc)^0.66)^0.59/(1+((1.8197*Cu_conc)^0.53+(218.78*H_conc)^0.66)^0.59) > 40 Org_xcue2=0.8811*(181970086*Cu_conc)^0.36/((181970086*Cu_conc)^0.36+(398107171*H_conc)^0.76) \ > *((181970086*Cu_conc)^0.36+(398107171*H_conc)^0.76)^0.7/(1+((181970086*Cu_conc)^0.36+(398107171*H_conc)^0.76)^0.7) > 50 fx = .0001*(org_xcue1 + org_xcue2) - Kin("Org_xcu") > 100 PUNCH total_time, cell_no, ToT("Cu"), -LM("H+"),Kin("Org_xcu"), Cu_conc*TOT("water") + KIN("Org_xcu"), fx > -end > > USER_Graph > -start > 10 PUNCH ToT("Cu"), -LM("H+"),Kin("Org_xcu"),Kin("Org_xh") > -end > > END > > David Parkhurst (dlpark@xxxxxxxx) > U.S. Geological Survey > Box 25046, MS 413 > Denver Federal Center > Denver, CO 80225 > > Project web page: https://wwwbrr.cr.usgs.gov/projects/GWC_coupled

**Follow-Ups**:**Re: PHREEQC problem***From:*David L Parkhurst

**References**:**Re: PHREEQC problem***From:*David L Parkhurst

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