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flow over a cylinder 2D

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Old   December 18, 2009, 11:04
Unhappy flow over a cylinder 2D
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Arash
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Dear all,

I'm new with fluent, and I'm trying to investigate flow over a cylinder.

it works perfectly in low Reynolds number, but I can not see the vortex shedding in higher Reynolds number (!!!!

I enclosed a photo of my gambit file in this message.

It would be kind of you if you help me in this matter

Thanks a lot in advance
Cheers
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File Type: jpg cylinder.jpg (99.3 KB, 66 views)
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Old   December 18, 2009, 14:18
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finally,I solved it, But still something confusing me:

First time, I tried to solve it like Non-dimensional value so that, I put air density equal to 1 and velocity equal to 1 and also the Diameter of cylinder equal to 1. the only thing that i change each time was viscosity to get my favorite Re. But i couldn't get the right answer in high Re.

Second time, I change everything to the default value that is density:1.225 and etc.
then i got the correct answer for all of the Re number.

anybody could tell why does this happen?
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Old   December 18, 2009, 16:35
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karine
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hi
which model are u using???
what do u mean in right answer????
for gigh Re, coherent structures still exist, but u cant see it clearly. U must use a FFT transform to see it.
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Old   December 18, 2009, 16:44
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Hey
I have done this before and I got good results. First of all, about the domain dimensions: there must be a free space of at least 10D long behind and in front of the cylinder where D is the cylinder's diameter. The same rule applies to the height of the domain. I mean 10D above the cylinder and 10D below. By the way, you don't have to model the whole cylinder. a half-cylinder with symmetry lines on the lower edge will do the work.
Second of all: use real properties since it is a physical problem. it is better to keep the real properties and change the velocity in order to change the ReD. Remember that for 15<ReD<40 you should get two laminar vortexes behind the cylinder. For ReD above 100 the flow is turbulent and laminar viscosity model may not be a good choice.
And finally, You are an Iranian. right? I am too. Nice to meet you.
Have fun and Good luck.

Last edited by behrang2009; December 18, 2009 at 17:20.
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Old   December 18, 2009, 16:54
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what are u talking about???????????????????????????????
u are iranian, i am not , and u know nothing in CFD
what the hell does the mach number do in an incompressible flow???
How are u planning to simulate VK street with a symetrical boundary????
Laminar model is not a good choice??? Do u know what is a laminar model???????

When u know nothing dont post ure comments.
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Old   December 18, 2009, 17:13
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Dear CFD user,
Thanks for being polite. I answered Arash's question. Not yours. He must be Iranian. Anyway, there must be a misunderstanding here. I know that low-Reynolds number flows are practically incompressible. For Ma>0.3 compressibility is serious. Not for less than that. Maybe I shouldn't have said that which led to this stiff opposition (!) from you. I have done it using symmetry and I got good results. I calculated Cd and NuD and they are consistent with empirical correlations. For turbulent flow you are correct. Symmetry might not do good for that case. If you are interested, give me your email and I will send you the mesh, the case and the figures and the report on this. the work was done for 1<ReD<50.
Have a good one and Good luck
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Old   December 18, 2009, 17:26
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Dear Behrang,
in a scientific forum we must not talk a lot about nationality no???
U have done it with a symetrical condition and it worked well because u are using a RANS model right?? with a RANS model u cant really model the vortex schedding, however AVERAGE quantities can be well simulated.
Anyway my friend, the laminar model is a DNS, wich means that it works for laminar AND for turbuelnt flows if ure mesh is sufficiently small.

1<Re<50 is however a laminar flow. Vortex schedding occurs for Re=48.5. Laminar model will gib=ve u best results, but it is however better to use is in a 3D geometry.
Regards
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Old   December 18, 2009, 17:41
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Hi guys,

Thanks for your response.

@CFDUSER: you said "with a RANS model u cant really model the vortex shedding" so does it mean that i always use the laminar model? even for Re=3x10^3 ???
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Old   December 18, 2009, 17:48
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laminar model = DNS=the best solution
the problem is when Re become high, u will need more cells and a smaller time step...
If u are only interested by mean quantities, u can use RANS models but u will never see a vortex schedding with it since it will always average ure result

(Never use RANS for simulating unsteady symetrical geometry)
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Old   December 18, 2009, 17:50
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Oh! Sorry
I did not see the word 'Shedding'. I misread the whole thing. By the way, I used the laminar steady model cause I was looking for something else.
Bye
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Old   December 18, 2009, 17:50
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Thanks a lot CFDuser. I got it
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Old   December 18, 2009, 18:14
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Sorry friend, but one more time u do a big mistake. U did not verify ure flow convergence. LAMINAR STEADY model will NEVER converge in a vortex schedding simulation.
(Residuals are NEVER enough to judge convergence)
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Old   December 18, 2009, 18:24
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yes, you are right because of that i change it to "e-epsilon" model...after that i could not see vortex shedding, it converge but i cannot see the vortex shedding, BUT when use "laminar" model it doesnt converge but i can see vortex shedding--> Exactly the same that you told me :-)
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Old   December 18, 2009, 18:30
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Again there's a giant misunderstanding. When I did that I was not looking for the vortex shedding. When I saw Arash's post, I did not pay attention to the word 'Shedding'. I thought he was just after the vortexes behind the cylinder and velocity vectors and contours. That's when I misread the whole thing. With the laminar model you can get laminar vortexes which appear for 15<ReD<40. For ReD>40 the laminar model may not be reliable. Vortex Shedding is a different story. It's inherently unsteady flow while I used the steady model. I was not after that. That's why I said 'Sorry' in my previous post.
Have a good one.
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Old   December 18, 2009, 18:40
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no problem my friend
no need to say sorry
but u still do a mistake: laminar model is always the best. The only problem is that it will need very much cells and small time steps.
Laminar model is the direct simulation of the NS equations.
We use RANS just become the cost of laminar (DNS) model become prohibitive for high Re, in terms of compuatational time (since we need very fine mesh and small time steps)
Regards
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Old   December 18, 2009, 18:54
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It's obvious that laminar model is the solution for NS equations since NS eqn's are based on the notion that the fluid is Newtonian. But Mesh Generation, Run time, and sometimes round-off errors are prohibitive factors.
When I said 'may not be reliable' I meant 'if you use the same mesh'.
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Old   December 18, 2009, 19:05
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yes
if the mesh is symetric, RANS models will keep a symetrical flow.
If u are interested in mean quantities, use Low-Re Rans models. If u really want unsteady behavior, u must use laminar or LES.....or URANS but after that u induce a small asymetry in ure geometry
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Old   March 3, 2010, 18:07
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Behrang Asgharian
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Hi!
I think you know the answer to the following question of mine:
I am modeling oscillating flow which is naturally unsteady. I need to integrate some field functions over one period T. In what menu can I find such an option?
Thanks Good luck
Behrang
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Old   March 4, 2010, 19:42
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Hi
u must monitor this data and write it to file
(solve-->monitors--->surface)
Then u can treat it on Matlab or Excel
This is the best way
If u want to do it on Fluent, u can do data sampling but u must know ure period a priori
regards
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Old   March 4, 2010, 20:09
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Hi
I have the data in .out files. the excel or Matlab solution seems to be good. I know the period. What Menu contains such options?
Thanks
Behrang
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