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Old   October 10, 2001, 11:12
Default 2/D Flow Around A Cylinder
  #1
Mark Snyder
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To get better acquainted with the FLOTRAN code, I have been using program to simulate the flow around a cylinder. The Reynolds number is about 4000, and I have selected the GIR model for turbulence (per the guidance given in the user's manual).

I have been able to obtain a transient solution with a symmetric wake. In addition, the pressure distribution around the cylinder agrees with the results obtained from the classical potential flow solution. So I have some confidence in my mesh and boundary conditions. However, I do not yet see any signs of a developing vortex trail.

This issue (i.e., the stability of a symmetric wake) has been previously discussed in this forum. However, it is not clear to me if I should let the program run and wait (how long?) for numerical roundoff to perturb the solution, or should I introduce some sort of perturbation into the flow. If I need to actually disturb the stability of the symmetric wake solution, can anyone give me some guidance as to how this should be done.

Thanks for your help.

Mark Snyder
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Old   October 10, 2001, 17:09
Default Re: 2/D Flow Around A Cylinder
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Axel Rohde
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Mark,

I am very familiar with this problem! I ran myriad tests for 2-D flows over a cylinder using my own code, an unsteady compressible Navier Stokes solver with an underlying 2nd order TVD algorithm. The keyword here is TVD as I will explain subsequently.

Think of a numerical wind tunnel being a wind tunnel that is virtually turbulence free. With 16 significant digits (double precision), your free stream turbulence intensity is about Tu = 10^-15 whether or not your code is non-dimensionalized internally. Any numerical disturbances due to round-off error will be dampened at any reasonable Reynolds number, assuming that your initial and boundary conditions are symmetric. Also, your solver has to be symmetric (predictor-corrector methods are generally asymmetric, because you start with either a forward or backward difference, which introduces the initial bias).

Now one can introduce an arbitrary amount of free stream turbulence intensity by simply modifying the far field boundary conditions through random number disturbances before each integration cycle. If I remember correctly, vortex instability will not occur unless,

Tu > 1/Re (Try also Tu^2 > 1/Re )

If you look at the non-dimensionalized Navier-Stokes equations (large terms are of order 1), then this makes perfect sense, because the factor (1/Re) appears with the viscous terms. Also, keep in mind that vortex instability takes some time to develop. In my experience, the cylinder had to travel about 50-100 diameters until periodic vortex shedding was fully developed.

If your code has a significant amount of numerical viscosity built in, you will never see vortex shedding, because your effective Reynolds number (based on physical plus numerical viscosity) will be significantly lower. By the way, my code does not have any additional "turbulence modeling" other than the random numerical fluctuations introduced at the far field boundary.

If your simulation is incompressible, the same concept should apply, assuming you have a good flow solver which does not introduce a significant amount of numerical viscosity.

I hope this helps.

Axel

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Old   October 10, 2001, 17:24
Default Re: 2/D Flow Around A Cylinder
  #3
Axel Rohde
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A quicker way to introduce vortex shedding is to have the cylinder rotating for a number of iterations (time integration steps). But once you stop the rotation, the wake will eventually become symmetric again if (1/Re) > Tu.
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Old   October 10, 2001, 18:03
Default Re: 2/D Flow Around A Cylinder
  #4
Mark Snyder
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Axel --

Thank you very much for your detailed reply. I apologize for asking what might be very basic questions, but would you explain what is the definition of Tu (turbulence intensity?) and what is the TVD algorithm?

Thanks.

Mark
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Old   October 10, 2001, 19:06
Default Re: 2/D Flow Around A Cylinder
  #5
Steve
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Hi there

I didn't think the flow past a cylinder was fully Turbulent at Re=4000 - Upstream of the separation points the boundary layer is laminar (there will be entrainment of turbulence in the separation bubbles). In any case the flow should be nothing like the symetric potential flow solution.

I think turbulence occurs at more like Re=5*10^4 (you should check this in a book). At this Re No. the Boundary layer goes fully turbulent and consequently there is a drop off in drag since the turbulent boundary layer starts to suppress separation due to its extra kinetic energy and diffusion. If you use a turbulence model at Re=4000 you will get the same phenomenon - less separation and the extra diffusion (effective viscosity) from the turbulence model could well suppress the transient wake that should occur.

You should be able to get quite good solutions with a laminar model, provided you use enough grid resolution in the wake. I've done this and you will get an unsteady wake from about Re=1000 onwards.

Check your drag coefficient - it should be around 1.2 I think. If it is lower then you must be getting less separation and the turbulence model is to blame. Also GIR is based on k-e which is notorious for supressing separation.

This is a very difficult situation to model. Not an introductionary problem.
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Old   October 10, 2001, 20:00
Default Re: 2/D Flow Around A Cylinder
  #6
Axel Rohde
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Mark,

The "statistically correct" definition of turbulence intensity can be found in any textbook on turbulence. A 'quick & dirty' CFD definition would look as follows:

Say you have a quantity q (any conserved variable), which you want to fluctuate, such that,

q = Q + q'

Q being the time average and q' the fluctuating component. And say your computer has a random function Rnd(1) which varies between 0 and 1,

0 <= Rnd(1) <= 1

Then for a given turbulence intensity, say Tu = 1/1000,

q = Q * [1 + Tu * (1 - 2 * Rnd(1))]

the value of q will vary about 0.1% around its mean value.

TVD stands for "Total Variation Diminishing". There are entire books written on this subject. For a quick review of this concept, let me refer you to my dissertation and a recently published AIAA paper,

www.cfd4pc.com/papers.htm

In essence, a TVD scheme will not "blow up", while at the same time it has a minimal amount of numerical viscosity, which can even be controlled through a certain parameter epsilon (I set epsilon = 0 in my cylinder simulation, since entropy is sufficiently enforced via physical viscosity).

You can make any scheme TVD by adding enough artificial or numerical viscosity. This is what a lot of "turbulence models" are about, and some commercial codes claim that their smeared and overpredicted boundary layers are simply "turbulent". They essentially destroyed the solution through some complex averaging technique, and then "fudged" the answer by correlating experimental data to closure constants that give "perfect" results in a few benchmark cases.

Axel
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Old   October 10, 2001, 21:53
Default Re: 2/D Flow Around A Cylinder
  #7
Guus Jacobs
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Hi I think the word transient might need some clarification here. Do you mean transient in the sense of transient phase between laminar and turbulent flow or a non-steady calculation? In case of the former: I am not familiar with the GIR turbulence model, but using any turbulence model in the region up to Re 4000 will probably damp the waves that cause instability, which prevents vortex shedding. In case it means non-steady: due to the non-linearity of the equation of NS you will eventually see a low frequency vortex at high enough Re numbers, but there cannot be attributed much physical meaning to it.

Hope this helps.
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Old   October 12, 2001, 10:54
Default Re: 2/D Flow Around A Cylinder
  #8
Mark Snyder
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Axel --

Thanks for the expanation.

Mark
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Old   October 12, 2001, 11:08
Default Re: 2/D Flow Around A Cylinder
  #9
Mark Snyder
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Steve --

I'm not sure where your value of Re =50,000 comes from. I did check two good fluid mechanics/heat transfer textbooks, and they both "state" that the von Karman vortex street is formed at Reynolds numbers on the order of 100 - 300. At higher values of Re, say greater than 1000, the pictures in the texts show a wake with turbulent eddies. So, I'm now wondering if I should actually reduce the value of the Reynolds number to see if my FLOTRAN model can actually reproduce the vortex trail.

Thanks.

Mark
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Old   October 12, 2001, 11:12
Default Re: 2/D Flow Around A Cylinder
  #10
Mark Snyder
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Guus --

By transient, I meant that I was marching out the soulution in time.

Mark
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Old   October 12, 2001, 12:32
Default Re: 2/D Flow Around A Cylinder
  #11
Guus Jacobs
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Hi. A von Karman vortex street cannot really be labeled turbulent. It is a transient flow in the sense that it represents the a transitional phase between laminar and turbulent flow. It is essentially 2D, which already gives you a hint that it is not a turbulent flow.

Averaging a transitional flow around a cylinder leads to a velocity field with two stable vortices behind the cylinder. Most turbulence models actually calculate an average flow field, which is one explanation why you are seeing no Karman street. This is basically the same as stating that the turbulent viscosity damps the spurious waves that cause the instability.

To calculate a von Karman street you need to perform Direct Numerical Simulation, which implies solving the NS eq. without using models.

For a review on the vortex dynamics behind a cylinder I would strongly recommend the following paper;

C.H.K. Williamson; Vortex dynamics in the cylinder wake; Ann. Rev. Fluid. Mech., 28:477-539, 1996.

Guus

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Old   October 12, 2001, 12:56
Default Re: 2/D Flow Around A Cylinder
  #12
Mark Snyder
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Thank you very much for the information. It looks like a trip to library and some reading is called for!

Mark
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Old   October 12, 2001, 12:58
Default Re: 2/D Flow Around A Cylinder
  #13
Axel Rohde
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My computation was in essence a Direct Numerical Simulation, although greatly underresolved within the wake. As Guus pointed out correctly, a 2-D simulation can never be truly turbulent. But a 2-D von Karman vortex street is the closest thing to real turbulence. To me it was pretty exciting to see them develop by simply introducing tiny disturbances into the free stream/far field boundary and letting the solution march in time long enough.
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Old   October 13, 2001, 19:08
Default Re: 2/D Flow Around A Cylinder
  #14
steve
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Mark,

At these low Reynolds numbers the boundary layer is definately Laminar. Transition to turbulence does not occur until Re number over 10000. Just because eddies are present does not mean the flow is turbulent. Yuo should be getting laminar separation. To resolve any turbulence in the wake you will need (as Axel suggested) to use DNS. However modelling the boundary layer is much more important and so solving the raw Navier-Stokes equations with high enough grid resolution should give a resonable solution. At the very least it will give the correct flow regime.
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Old   October 13, 2001, 20:53
Default Re: 2/D Flow Around A Cylinder
  #15
John C. Chien
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(1). 2-D Flow over a cylinder has been used as a test case for a long time. There are numerous published papers and reports world-wide. (2). If you are having problems with this case, then it is obviously that you are using the wrong code, or using the code wrongly. (3). It is a good idea to use a finite-difference formulation, if you are interested in getting the right answer. If you are interested in getting approximate solution, then try finite-volume formulation. (4). If you still think that there is a bright future for the finite-element formulation, then ask the vendor to show you the results first, I would be interested in knowing the results also.
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