[nima_nzm]

Hi,
I am going to remodel a CFD case in a paper to validate my turbulence model. CFD code is Fluent and the problem is 2-D flow around a circular cylinder. Re=10e6 and velocity is 12 [m/s] and Cylinder Diameter is d=9 cm. I want to capture shedding frequency of f=60 Hz, and then I chose time step of dt=0.0001.

By using Turbulent intensity as I=0.16*Re^(-1/8) = 2.84 [%] and Turbulence length scale as l=0.07*L=0.0063 [m] in CFD-online Turbulent Calculator (http://www.cfd-online.com/Tools/turbulence.php) k-e are.

k= 0.149 [m2/s2] & e=0.829 [m2/s3]

Are these values resonable?

and the most important question; due to the very low time step size, for how many seconds of flow time (Physical time) should I continue the run to see the generation and shedding of vortices or stopping the run and modifying the turbulence parameters? currently I run it for 100 seconds but it takes to much cpu time. what is your Idea?

Thanks

[FMDenaro]

Originally Posted by nima_nzm

Hi,
I am going to remodel a CFD case in a paper to validate my turbulence model. CFD code is Fluent and the problem is 2-D flow around a circular cylinder. Re=10e6 and velocity is 12 [m/s] and Cylinder Diameter is d=9 cm. I want to capture shedding frequency of f=60 Hz, and then I chose time step of dt=0.0001.

By using Turbulent intensity as I=0.16*Re^(-1/8) = 2.84 [%] and Turbulence length scale as l=0.07*L=0.0063 [m] in CFD-online Turbulent Calculator (http://www.cfd-online.com/Tools/turbulence.php) k-e are.

k= 0.149 [m2/s2] & e=0.829 [m2/s3]

Are these values resonable?

and the most important question; due to the very low time step size, for how many seconds of flow time (Physical time) should I continue the run to see the generation and shedding of vortices or stopping the run and modifying the turbulence parameters? currently I run it for 100 seconds but it takes to much cpu time. what is your Idea?

Thanks

as you can read in similar posts, the key is in the proper formulation you consider... RANS/URANS formulations, despite of the small time step you would use, cannot describe the range of frequencies you have at such high Re number. Actually RANS has only a statistical steady solution.

URANS can capture some unsteady feature but only at very low wavenumber, therefore it has no sense in using such a small time step as you wrote. You cannot capture frequencies within Pi/dt

[nima_nzm]

Originally Posted by FMDenaro

as you can read in similar posts, the key is in the proper formulation you consider... RANS/URANS formulations, despite of the small time step you would use, cannot describe the range of frequencies you have at such high Re number. Actually RANS has only a statistical steady solution.

URANS can capture some unsteady feature but only at very low wavenumber, therefore it has no sense in using such a small time step as you wrote. You cannot capture frequencies within Pi/dt

Flippo,

Experimental tests shows that the dominant frequency ( or at least one of the dominant frequencies) of the vortex shedding is in the range of [50 70] Hz in such a Reynolds number. But as you said very low wave number implies low frequency. Well I just don't know for how long should I wait to see the vortices.

[FMDenaro]

Originally Posted by nima_nzm

Flippo,

Experimental tests shows that the dominant frequency ( or at least one of the dominant frequencies) of the vortex shedding is in the range of [50 70] Hz in such a Reynolds number. But as you said very low wave number implies low frequency. Well I just don't know for how long should I wait to see the vortices.

using RANS, you'll never see developing vortex shedding eve if you run the code for infinite time... actually, you will see a (statistically) steady state with two symmetric recirculating regions behind the cylinder ...

using URANS you can try to get some unsteady component into the solution, provided that external force at low frequency is acted.

to better understand the meaning of the time-dependent solution in URANS, consider for example the motion of a piston in a cylinder: the solution at some time tn (or crank angle) is representative of the ensemble average of several realization evaluated at tn.

[flotus1]

Originally Posted by nima_nzm

Hi,
I am going to remodel a CFD case in a paper to validate my turbulence model. CFD code is Fluent and the problem is 2-D flow around a circular cylinder. Re=10e6 and velocity is 12 [m/s] and Cylinder Diameter is d=9 cm. I want to capture shedding frequency of f=60 Hz, and then I chose time step of dt=0.0001.

A better initial guess for the time step size would be the period time divided by 20.
Make sure to use a second order time-stepping.

The way you estimated the turbulent quantities at the inlet is only suited for internal flows. K and epsilon at the inlet are/should be independent of the Reynolds number of the cylinder and its diameter.
If you did not see any vortex shedding in your current setup, the high values of k might have introduced too much damping.
Typically, the values for k and epsilon at the inlet for such a flow configuration are very low and should not affect the solution. You can choose turbulence intensity values below 0.1% to start with unless you have better values from the experimental setup.

But as Filippo already pointed out, you wont gain much insight from a URANS simulation. Reducing the time step or mesh size further wont resolve new flow features like in a LES but will only reduce numerical errors.