I have an intriguing case of numerical dispersion that I would like to share and perhaps, get a reason for that.
I am running some models in Hydrus and I included 2 solutes in my model, being one reactive and one conservative, in order to calculate travel times. The reactive solute results were ok, but my Conservative solute showed very high and, for some models, very low concentrations, where they are not supposed to be. Sometimes even negative concentrations appeared.
I tried to implement many changes according to tips available in pcprogress forum and procedures described in user and technical manuals. The changes were selecting different space weighting scheme, stability criterion, mesh (by reducing target element size, reducing from 2 to 1,5 the ratio of neighboring mesh), changing soil specific parameters (dispersion and bulk density). Nothing worked...
Then I decided to change only the space weighting scheme and include one more solute, conservative as well (identical). I still have the same numerical dispersion in my first conservative solute, however not for my second conservative solute!! So, now I can analyze my results considering my reactive solute and my second conservative solute!
Why does it happen? What might be the explanation?
Thank you,
Natalia
Using a Second Conservative Solute to avoid numerical dispersion
Re: Using a Second Conservative Solute to avoid numerical dispersion
I have trouble believing that this can happen? Can you send me the project where you observe this behavior?
Note that different solutes, if considered, affect (or at least should) each other only by affecting the numerical time step (for all of them), which is always adjusted to the least favorable solute, i.e., to the conservative solute.
J.
Note that different solutes, if considered, affect (or at least should) each other only by affecting the numerical time step (for all of them), which is always adjusted to the least favorable solute, i.e., to the conservative solute.
J.
Re: Using a Second Conservative Solute to avoid numerical dispersion
Dear Mr. Jirka,
Thank you for your attention.
I am sending the model, but I had to delete the results in order to attach the file here. But it runs fast, in a couple minutes.
I am also sending the figures showing the plumes of the reactive (concentration 1) and conservative solutes (concentration 2 and 3).
Thank you,
Natalia
Thank you for your attention.
I am sending the model, but I had to delete the results in order to attach the file here. But it runs fast, in a couple minutes.
I am also sending the figures showing the plumes of the reactive (concentration 1) and conservative solutes (concentration 2 and 3).
Thank you,
Natalia
 Attachments

 Concentration3.jpg (239.19 KiB) Viewed 816 times

 Concentration2.jpg (243.72 KiB) Viewed 816 times

 Concentration1.jpg (231.86 KiB) Viewed 816 times

 HydrusOutput12.zip
 (426.73 KiB) Downloaded 37 times
Re: Using a Second Conservative Solute to avoid numerical dispersion
Natalia,
1. First of all, I do not see any oscillations in any of the three solutes.
2. Second, you say that you have three solutes while the last two are conservative. That’s not the case. The first solute has firstorder degradation. The second solute has zeroorder production (as a result of the degradation of the third solute). And only the third solute is a conservative solute. One would not expect numerical oscillations for the first two nonconservative solutes. However, I do not see any oscillation for the third solute either, likely since you were using the numerical scheme with the artificial dispersion.
Thus, I’m really not sure what you say the problem is.
J.
1. First of all, I do not see any oscillations in any of the three solutes.
2. Second, you say that you have three solutes while the last two are conservative. That’s not the case. The first solute has firstorder degradation. The second solute has zeroorder production (as a result of the degradation of the third solute). And only the third solute is a conservative solute. One would not expect numerical oscillations for the first two nonconservative solutes. However, I do not see any oscillation for the third solute either, likely since you were using the numerical scheme with the artificial dispersion.
Thus, I’m really not sure what you say the problem is.
J.
Re: Using a Second Conservative Solute to avoid numerical dispersion
Mr. Jirka,
Thank you for your feedback, it means a lot. I have the same understanding in relation to your item 1, but I have doubts regarding item 2, which is exactly touching the problem.
If I delete the third solute and run the model again, the second solute will present the same results as before. Based on that, I think we cannot consider the second solute has zeroorder production as a result of the degradation of the third solute, because the third solute is not in the system anymore.
Besides that, the second and the third solute have the same reaction parameters in the model (all values are zero). That is why I am considering that both are the same and conservative solutes. Then, when I observed the value of 195mmol/m³ in the second solute, I thought that I was looking at an oscillation behavior (as the concentration applied in the well is 100 mmol/m³).
Thank you,
Natalia
Thank you for your feedback, it means a lot. I have the same understanding in relation to your item 1, but I have doubts regarding item 2, which is exactly touching the problem.
If I delete the third solute and run the model again, the second solute will present the same results as before. Based on that, I think we cannot consider the second solute has zeroorder production as a result of the degradation of the third solute, because the third solute is not in the system anymore.
Besides that, the second and the third solute have the same reaction parameters in the model (all values are zero). That is why I am considering that both are the same and conservative solutes. Then, when I observed the value of 195mmol/m³ in the second solute, I thought that I was looking at an oscillation behavior (as the concentration applied in the well is 100 mmol/m³).
Thank you,
Natalia
Re: Using a Second Conservative Solute to avoid numerical dispersion
There was a mistake in my previous answer. This sentence should read:
“The second solute has zeroorder production (as a result of the degradation of the first solute).”
J.
“The second solute has zeroorder production (as a result of the degradation of the first solute).”
J.