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Abhishek JAin (ajain)
New member
Username: ajain

Post Number: 1
Registered: 08-2004
Posted on Thursday, August 19, 2004 - 08:52 am:   

Dear Sir
I have been working with FlexPDE for last one month now. I am solving a solute transfer problem in a thin tube with advection and diffusion within the tube along with adsorption taking place at the tube wall. It means that in the boundary condition of the main equation, i have the equation of adsorption at the wall.

I solve for C variable as given in the attached file. There are two coefficients involved in the equation: a)Peclet No. which is defined in the diffusion equation and b)kappa which is the coefficient of mass transfer in the adsoption equation defined at the boundary.

I fail to get a solution to my problem unless I keep my errlim >=0.01 which is not desired.
Another strange thing that happens is if I multiply my equation on both sides by a constant, say 0.5, I get a different solution altogether although the problem is still the same. I fail to understand if my solution is inaccurate or there is a numerical error.

I have also observed that if I decrease the error limit, I can get a solution only if I have a course mesh? Also, I want to know how can the threshold values can make a difference to the solution? Can u plz explain?
application/octet-streamAdvection-Diffusion-Reaction of solute in a tube
test1.pde (1.6 k)
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Robert G. Nelson (rgnelson)
Moderator
Username: rgnelson

Post Number: 212
Registered: 06-2003
Posted on Thursday, August 19, 2004 - 01:09 pm:   

1.
In the first place, you MUST MUST MUST include some graphical output while the problem runs! You cannot simply close your eyes and wait for the end to see what happened. Add monitoring contours of all variables at 10 cycle intervals until you are satisfied that you have set the problem up correctly. (See User Guide | Basic Usage | Problem Setup Guidelines).

2.
The Natural boundary condition defines the value of the surface terms generated by integrating the second order terms of your equation by parts. If you multiply the equation by 2, you must multiply the natural boundary condition by 2 as well. (See User Guide | Addressing More Difficult Problems | Natural Boundary Conditions.)

3.
Ds is not properly a variable. It is merely dt(C). I assume that you did it this way in order to use it as a boundary condition on C. But I don't believe this boundary condition. The flux depends on its own rate of change? I would have thought that the adsorption depended on concentration, not rate of change of concentration. Try the problem first without this BC and without Ds.
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Abhishek JAin (ajain)
New member
Username: ajain

Post Number: 2
Registered: 08-2004
Posted on Friday, August 20, 2004 - 04:50 am:   

Dear Sir
Thank you very much for your prompt response.
I wish to add further to my doubts

1) Regarding the formulation of the problem of adsorption, I am considering equilibrium adsorption and hence, at the boundary, I have the b.c. as given...I am therefore attaching a document describing the problem statement for your perusal. Kindly refer.

2) The problem of having different solutions to the same problem originated from the fact that I was getting different solutions on keeping the constant Peclet on either side of the equation. To my confusion, I also get a different solution if I keep the term dc/dt on the LHS and bring the other terms on the RHS....I have put these remarks in detail in the attached document.
However, I do not encounter such dissimilarities if I do not have any adsorption, i.e. I have no such thing as Ds.

3)Regarding your concern to include graphical output, I intentionally took them out of the file that I sent to you in order to reduce to your confusion, if it may arise.

Therefore, I will be thankful to you if you can let me know what am I doing wrong in my formulation.

Best Regards
Abhishek
application/mswordDescription of the problem statement
FlexPDE_problem_description.doc (32.8 k)

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Robert G. Nelson (rgnelson)
Moderator
Username: rgnelson

Post Number: 214
Registered: 06-2003
Posted on Friday, August 20, 2004 - 03:15 pm:   

1.
As far as the boundary condition with the time derivative in it, just because you wrote it in a .DOC file doesn't make me believe it. Your argument seems to be that since you want a steady state solution, you must have a time derivative in the boundary flux? Steady-state solutions are those that don't have any time derivatives. If its supposed to be zero, take it out.

2.
In point 2 of my previous post, I used the example of multiplying an equation by 2. This example applies for other numbers as well, like 3,4, pi, and most importantly, -1. If you swap sides of your equation, you have effectively multiplied by -1, so you must multiply natural boundary conditions by the same -1. I think you are not comprehending the meaning of the natural boundary condition. Re-read the pertinent sections of the User Guide.

3.
You hid all the activitiy so I wouldn't be confused? Do you hide symptoms from your doctor so he won't be confused? ("You're the doctor, You figure it out!".)

4.
The problem takes forever to run because the time derivative in the boundary condition is unstable.
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Abhishek JAin (ajain)
Junior Member
Username: ajain

Post Number: 3
Registered: 08-2004
Posted on Monday, August 23, 2004 - 04:30 am:   

Dear Sir
Thank you again for your kind response.

1) I indeed have understood the meaning of natural(C). However, as a user, I must express my opinion that natural(C) should provide only n.grad(C), which is regardless of the way the PDE has been formulated. A little careless reader can easily get confused over it as you have given inforamtion about it in the Technical Notes of the User Guide. In the Chapter Basic Usage|Setting the Boundary Condition, you have mentioned, "The NATURAL boundary condition specifies flux at the boundary of the domain". Although the statement is correct, but a reader can associate it with the Neumann B.C. which is only under a particular case of Laplace Equation.

2) Regarding the problem I am solving, I think I have not been able to communicate to you properly. I do not wish to imply a steady-state solution. The problem has been well published and solved by many hydrogeologists.
kindly refer to a revised document.

I hope I can know why the B.C. becomes unstable in this case.
Regards
Abhishek
application/mswordProblem_Description(Revised)
FlexPDE_problem_description.doc (35.3 k)
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Robert G. Nelson (rgnelson)
Moderator
Username: rgnelson

Post Number: 216
Registered: 06-2003
Posted on Monday, August 23, 2004 - 03:33 pm:   

1.
The use of the term "Natural Boundary Condition" in FlexPDE conforms to standard usage in variational calculus. The statement you reference in the User Guide is immediately followed by this text:
[Note: The precise meaning of the NATURAL boundary condition depends on the PDE for which the boundary condition is being specified. Details are discussed in the Chapter "Natural Boundary Conditions." ]
The link takes you to a more extensive discussion of the topic, which appeals to the Divergence theorem as a model for understanding the meaning. It seams to me the presentation is quite clear.
We have used the word DNORMAL to mean the normal derivative boundary condition.

2.
Ok, I see the logic, although it seems somewhat circular to me. Nevertheless, we are left with the problem that FlexPDE is unstable with time derivatives in the boundary fluxes. Whether this is a bug or something more fundamental, I will have to discover by investigation.
There may, however, be other ways to address this.
Your "s" is in fact analogous to "c" in the adsorbing material.
If you add a layer of adsorbing material to your model, and define c=s/k/d, where d is the thickness of the layer, then the interface flux (1/Pe)dr(c) is -alpha*JUMP(c), conforming to the model of contact resistance introduced in FlexPDE version 4.1.
Perhaps you can address the problem this way.

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