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Synthetic mineral dissolution
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Topic: Synthetic mineral dissolution (Read 272 times)
Hosseif
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Posts: 7
Synthetic mineral dissolution
«
on:
November 14, 2017, 12:46:46 PM »
Hi,
I'm doing a benchmark comparing our transport solver (coupled with phreeqc) with phreeqc. I've defined a synthetic mineral dissolution like AB(s)=A+B. The reactoin is a kinetic reaction and the rate relation is R=KA(1-(A*B/K_eq)). Initially AB(s) is in equilibrium with the solution (K_eq=1e-8; A=1e-4;B=1e-4). Then a solution containing A(=1e-5) and B(=1e-4) is injected from the left boundary. So I expect dissolution of the mineral. In the attached phrqc file, I'm saving mass of mineral in the first cell at first three timesteps and it's dissolving. Initially I had 0.5 mol of AB mineral and phreeqc is telling that at the first timestep I have 4.9945e-02 mol AB which means a very little dissolution. But the rate of dissolution is a bit strange to me. Because if I calculate R for the first cell in the second timestep, it would be close to:
R=KA(1-(A*B/K_eq))=(2000)(100)(1-(1e-5*1e-4)/1e-8)=1.8e4
dn=R*d_T=(1.8e4)*(0.02 seconds)=3600 mol
which means that I will dissolve 3600 mol!! but phrqc is saying I dissolve ￼￼0.450055 mol.
So could you please tell me where I am wrong?
P.S. I attached the file
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dlparkhurst
Top Contributor
Posts: 1098
Re: Synthetic mineral dissolution
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Reply #1 on:
November 14, 2017, 06:39:41 PM »
PHREEQC integrates the rate over time. You cannot have a rate that fast for the entire time step for two reasons: (1) you would consume all of the available reactant, and (2) you would be grossly supersaturated with the mineral. The rate equation goes to zero at equilibrium. If you want to compare rates, run for a very short time step, say 1e-10 seconds.
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Hosseif
Contributor
Posts: 7
Re: Synthetic mineral dissolution
«
Reply #2 on:
November 14, 2017, 10:36:31 PM »
Thanks David.
So if rate is very fast, I should get zero mol for the mineral from second time step on. But I don't get zero and I'm getting something veriy close to the initial amount of mineral.
I guess the reason I'm getting small dissolution is that I'm consuming A but at the same time I'm producing B due to dissolution. so the term
(1-A*B/K_eq) is still close to zero and basically I miscalculated the rate in my previous post.
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dlparkhurst
Top Contributor
Posts: 1098
Re: Synthetic mineral dissolution
«
Reply #3 on:
November 14, 2017, 11:19:04 PM »
Look at your output. I suspect that each cell is in equilibrium with AB at each time step. I believe you will get the same amount of reaction if you replace KINETICS with EQUILIBRIUM_PHASES containing AB
Cell 1 is continuously getting a solution advected in that is out of equilibrium, but attains equilibrium during the time step. There is mixing among cells, and during the first three shifts there is a different equilibrium in parts of the column as the infilling solution advects through.
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