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Time-dependent flow past a cylinder at Re=100

Posted 21 set 2010, 06:10 GMT-4 12 Replies

Iman Lashgari
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I am new in Comsol! I am working on Comsol/4.0.
I got convergence in that case. However, even for long total time and short time step, my result is somehow far from numerical result.
For example, I got CD (Drag coefficient) about 1.05, while in numerical report, it is given about 1.34.

This is my set up detail:
Domain : 30*d from each side
Cylinder diameter(d) = 0.2 m
Laminar inflow with average velocity= 1m/s
zero pressure no viscous outflow
Symmetry boundary condition on top and bottom walls.


12 Replies Last Post 27 set 2010, 03:39 GMT-4
Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)
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Posted: 1 decade ago 21 set 2010, 10:50 GMT-4
Hi

have you checked that your model is mesh independent ? and that you are using 100% the same physics hypothesis for the fluid

--
Good luck
Ivar
Hi have you checked that your model is mesh independent ? and that you are using 100% the same physics hypothesis for the fluid -- Good luck Ivar
Danial Tz
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Posted: 1 decade ago 21 set 2010, 14:18 GMT-4

I am new in Comsol! I am working on Comsol/4.0.
I got convergence in that case. However, even for long total time and short time step, my result is somehow far from numerical result.
For example, I got CD (Drag coefficient) about 1.05, while in numerical report, it is given about 1.34.

This is my set up detail:
Domain : 30*d from each side
Cylinder diameter(d) = 0.2 m
Laminar inflow with average velocity= 1m/s
zero pressure no viscous outflow
Symmetry boundary condition on top and bottom walls.


Hopefully you are not using water properties, because that would make Re=200,000. You can upload your model so we can check.

Edit: One the second thought since you have Cd already 1, shouldn't be Re problem anymore. Check the tolerances to see if they are small enough, use less artificial diffusion if you can (means finer mesh), and finally check if your boundary conditions are exactly the same as your benchmark (i.e. symmetry on the wall, which usually in benchmarks it is no-slip)
[QUOTE] I am new in Comsol! I am working on Comsol/4.0. I got convergence in that case. However, even for long total time and short time step, my result is somehow far from numerical result. For example, I got CD (Drag coefficient) about 1.05, while in numerical report, it is given about 1.34. This is my set up detail: Domain : 30*d from each side Cylinder diameter(d) = 0.2 m Laminar inflow with average velocity= 1m/s zero pressure no viscous outflow Symmetry boundary condition on top and bottom walls. [/QUOTE] Hopefully you are not using water properties, because that would make Re=200,000. You can upload your model so we can check. Edit: One the second thought since you have Cd already 1, shouldn't be Re problem anymore. Check the tolerances to see if they are small enough, use less artificial diffusion if you can (means finer mesh), and finally check if your boundary conditions are exactly the same as your benchmark (i.e. symmetry on the wall, which usually in benchmarks it is no-slip)
Iman Lashgari
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Posted: 1 decade ago 22 set 2010, 04:01 GMT-4

Hi

I ran the program with finer mesh yesterday, So I found that my result is mesh independent.
Right now I am running the case with "normal inflow velocity equal to 1m/s" instead of "Laminar inflow average velocity equal to 1m/s", according to the reference paper.

I set up total time= 300s with time step=0.05s, So It will take time. Then I will inform you about the result

Regards
Iman
Hi I ran the program with finer mesh yesterday, So I found that my result is mesh independent. Right now I am running the case with "normal inflow velocity equal to 1m/s" instead of "Laminar inflow average velocity equal to 1m/s", according to the reference paper. I set up total time= 300s with time step=0.05s, So It will take time. Then I will inform you about the result Regards Iman
Danial Tz
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Posted: 1 decade ago 22 set 2010, 04:11 GMT-4

Hi

I ran the program with finer mesh yesterday, So I found that my result is mesh independent.
Right now I am running the case with "normal inflow velocity equal to 1m/s" instead of "Laminar inflow average velocity equal to 1m/s", according to the reference paper.

I set up total time= 300s with time step=0.05s, So It will take time. Then I will inform you about the result

Regards
Iman


Salam Iman,

Well, that's a difference already. Normal equal to 1 m/s means you have 1m/s on all your boundary, but inflow average velocity means that you will have 0 on both ends and probably 1.5 m/s in middle - where cylinder is. I would suggest use laminar entry option and set that to 1.

One more thing, this is the case if upper and lower walls are no-slip condition. If they are symmetry (slip) then both average and normal inflows are the same, 1 m/s.

So, once again, comes to the boundary conditions. Are they identical?

Also, how are you calculating the drag coefficient? do you use reacf() operator or lagrange multiplier?
[QUOTE] Hi I ran the program with finer mesh yesterday, So I found that my result is mesh independent. Right now I am running the case with "normal inflow velocity equal to 1m/s" instead of "Laminar inflow average velocity equal to 1m/s", according to the reference paper. I set up total time= 300s with time step=0.05s, So It will take time. Then I will inform you about the result Regards Iman [/QUOTE] Salam Iman, Well, that's a difference already. Normal equal to 1 m/s means you have 1m/s on all your boundary, but inflow average velocity means that you will have 0 on both ends and probably 1.5 m/s in middle - where cylinder is. I would suggest use laminar entry option and set that to 1. One more thing, this is the case if upper and lower walls are no-slip condition. If they are symmetry (slip) then both average and normal inflows are the same, 1 m/s. So, once again, comes to the boundary conditions. Are they identical? Also, how are you calculating the drag coefficient? do you use reacf() operator or lagrange multiplier?
Iman Lashgari
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Posted: 1 decade ago 22 set 2010, 04:58 GMT-4
Salam Danial ;)

Thanks for your answer!

The boundary conditions are identical.
Right now these are symmetry on top and bottom, normal inflow velocity for inlet and pressure no viscous equal to zero for outlet.

For Drag Coefficient I used "Total stress in x-dir /( 0.5 * density * Inflow velocity^2 * cylinder diameter)"

Regards
Iman


Salam Danial ;) Thanks for your answer! The boundary conditions are identical. Right now these are symmetry on top and bottom, normal inflow velocity for inlet and pressure no viscous equal to zero for outlet. For Drag Coefficient I used "Total stress in x-dir /( 0.5 * density * Inflow velocity^2 * cylinder diameter)" Regards Iman
Danial Tz
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Posted: 1 decade ago 22 set 2010, 05:24 GMT-4

Salam Danial ;)

Thanks for your answer!

The boundary conditions are identical.
Right now these are symmetry on top and bottom, normal inflow velocity for inlet and pressure no viscous equal to zero for outlet.

For Drag Coefficient I used "Total stress in x-dir /( 0.5 * density * Inflow velocity^2 * cylinder diameter)"

Regards
Iman


I would recommend you to use reacf() operator, as it is more accurate. You use it the same as T_xx

so use reacf(u) for drag force, and reacf(v) for lift.

[QUOTE] Salam Danial ;) Thanks for your answer! The boundary conditions are identical. Right now these are symmetry on top and bottom, normal inflow velocity for inlet and pressure no viscous equal to zero for outlet. For Drag Coefficient I used "Total stress in x-dir /( 0.5 * density * Inflow velocity^2 * cylinder diameter)" Regards Iman [/QUOTE] I would recommend you to use reacf() operator, as it is more accurate. You use it the same as T_xx so use reacf(u) for drag force, and reacf(v) for lift.
Iman Lashgari
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Posted: 1 decade ago 23 set 2010, 03:54 GMT-4
Hi

I got the attached results for lift and drag coefficient using "reacf()" function.
Please take a look at them.
I guess I should decrease the time step further.

Regards
Iman


Hi I got the attached results for lift and drag coefficient using "reacf()" function. Please take a look at them. I guess I should decrease the time step further. Regards Iman
Danial Tz
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Posted: 1 decade ago 23 set 2010, 05:50 GMT-4
You can calculate the CFL number and follow it.

Also, you can calculate your required time.step roughly too. Assume a Strouhal number of say 0.15 and calculate the frequency. Then period. And finally your timestep should be like 1/20th of a period for enough accuracy.
You can calculate the CFL number and follow it. Also, you can calculate your required time.step roughly too. Assume a Strouhal number of say 0.15 and calculate the frequency. Then period. And finally your timestep should be like 1/20th of a period for enough accuracy.
Iman Lashgari
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Posted: 1 decade ago 23 set 2010, 06:59 GMT-4
Hi

I did that. I can also use Lift coefficient graph to find frequency.
So the time step is less than 1/20th of the period for (0.05s)!
I am running to case right now:
1- Using even smaller time step (0.025s)
2- Using finer mesh (almost two times in number of cells)

The CD graph is so erratic as you see in attachment. I don't know why?
Another problem is that the average Lift coefficient is not exactly zero. it seems the field is not perfectly symmetric, but as I checked It is totally symmetric .

Regards
Iman
Hi I did that. I can also use Lift coefficient graph to find frequency. So the time step is less than 1/20th of the period for (0.05s)! I am running to case right now: 1- Using even smaller time step (0.025s) 2- Using finer mesh (almost two times in number of cells) The CD graph is so erratic as you see in attachment. I don't know why? Another problem is that the average Lift coefficient is not exactly zero. it seems the field is not perfectly symmetric, but as I checked It is totally symmetric . Regards Iman
Danial Tz
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Posted: 1 decade ago 23 set 2010, 13:50 GMT-4
It will be easier to identify if you upload your model and link to paper. This problem is easy to setup but difficult to fine tune for a specific accuracy.
It will be easier to identify if you upload your model and link to paper. This problem is easy to setup but difficult to fine tune for a specific accuracy.
Iman Lashgari
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Posted: 1 decade ago 27 set 2010, 03:19 GMT-4
Hi
Sorry, I answer u a bit late.
As I decreased the time step and used "reacf()" for calculating drag coefficient, I got an acceptable result in this case. So the problem is solved :)

Thanks for helping me very much.
I hope I can get your advise for my next simulation as well.

Regards
Iman
Hi Sorry, I answer u a bit late. As I decreased the time step and used "reacf()" for calculating drag coefficient, I got an acceptable result in this case. So the problem is solved :) Thanks for helping me very much. I hope I can get your advise for my next simulation as well. Regards Iman
Danial Tz
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Posted: 1 decade ago 27 set 2010, 03:39 GMT-4

Hi
Sorry, I answer u a bit late.
As I decreased the time step and used "reacf()" for calculating drag coefficient, I got an acceptable result in this case. So the problem is solved :)

Thanks for helping me very much.
I hope I can get your advise for my next simulation as well.

Regards
Iman


eyval ;)
[QUOTE] Hi Sorry, I answer u a bit late. As I decreased the time step and used "reacf()" for calculating drag coefficient, I got an acceptable result in this case. So the problem is solved :) Thanks for helping me very much. I hope I can get your advise for my next simulation as well. Regards Iman [/QUOTE] eyval ;)

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