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2D axisymmetric and 3D issue

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Old   March 10, 2020, 11:58
Default 2D axisymmetric and 3D issue
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Hi all,

I have an 2D axisymmetric model (cylindrical tube) and am trying to calculate a variable through the outlet.
The thing is for reporting the resultants at outlet in an axisymmetric domain fluent integrates and shows the total results as I am selecting a surface in 3D.
But to compare CFD results with theory I want to calculate the variable using a line integral.

The question is how can I obtain line results (and not surface results) in 2D axisymmetric in Fluent?
In the other word what links 2D axisymmetric to 3D?


Any comment would be highly appreciated.
Amir
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Old   March 10, 2020, 12:09
Default 2D and 3D
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In 2D, axisymmetric problem, Fluent assumes full 2\pi region to calculate integrals.
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Old   March 11, 2020, 08:48
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Hi Vinerm,
Thanks for the reply,

It means that if I need to calculate the result at a special point on R (and not over a ring of area) I have to just divide Fluent results by 2*pi,
Is that right?

Best,
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Old   March 11, 2020, 09:12
Default 3D to 2D
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Division by 2\pi will give incorrect results since area is not a function of angle alone but of radius as well. Area is determined as

\int_\phi\int_rdrd\phi

So, any variable that is area averaged is also multiplied by the integrand and then divided by the total area, e.g., for area-weighted average Temperature, it becomes

\frac{1}{A}\int_\phi\int_r Tdrd\phi

You should be able to determine line integral from this. Since we are working with discrete mesh, it all boils down to multiplication and sum. So, you just need to multiply T at every face with the height of the face in radial direction and then divide by total radius.

Do note that this averaging would not be physically correct since each T will have weight given by the height of cell while the simulation is based on the assumption that system is axisymmetric, i.e., the outermost cell has the largest weight because of its maximum radius.
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Old   March 12, 2020, 05:12
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I got close by doing the multiplication with the radius of each face and then divide by the total radius. But still have an important difference with theory.

For instance, for the number of particle I have a huge difference between theory and fluent.
Total number of particle in my case has to be constant at on each surface at the outlet but in a line I have different number of particle which decreases from the center of pipe to the wall.

Since after integration fluent put constant values for the number of particle, doing multiplication and division by radius changes it a bite but not in a correct way as theory.


P.S. plot of theoretical results and fluent on a line
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File Type: png Untitled.png (22.1 KB, 11 views)
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Old   March 12, 2020, 07:08
Default Particles at the outlet
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I am afraid I could not understand your statement.
Quote:
Total number of particle in my case has to be constant at on each surface at the outlet but in a line I have different number of particle which decreases from the center of pipe to the wall.
Isn't theory predicting the same thing; maximum particle in the center and minimum near the walls?

And what is meant by constant values in the following statement?

Quote:
Since after integration fluent put constant values for the number of particle, doing multiplication and division by radius changes it a bite but not in a correct way as theory.
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Old   March 12, 2020, 07:45
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Sorry for not being clear,

Yes, the theory predicts as you said: ''maximum particle in the center and minimum near the wall''

But When I export the results of fluent on particle number, they are constant everywhere (same particle number at each sub area at outlet).

So when I want to determine the line integral I got wrong results (attached file).
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File Type: xlsx theory-fluent.xlsx (11.6 KB, 2 views)
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Old   March 12, 2020, 07:50
Default Line Integral
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Using line integral for axisymmetric flow would be incorrect, as mentioned earlier. It has to be area integral.
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Old   March 12, 2020, 08:07
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Yes of course, it is actually about to determine line integral from surface integral as we discussed above.

I would determine line integral from what fluent shows me to compare with my theoretical results (which are line integral).
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Old   March 12, 2020, 08:21
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In my view, you do not require an integral. The graph shows number of particles along a line, which is outlet in your case, vs. radial coordinate. Why is the integral required? You simply plot the variable along the radius.
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Old   March 13, 2020, 20:14
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A surface integral reports 1 number, not a line of numbers. Do you have 100 reports defined and are plotting this in excel? Or what?

What exactly is being reported and why are your even reporting a surface integral if you want values along a line?

If you want values along the outlet line, then just do solution data export and you get raw values that aren't integrated. If you are exporting solution this way already, then I have to ask what the heck is this theory supposed to be predicting that you are unable to match.
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