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Turbulent external flow around cylinder

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Old   April 30, 2013, 16:18
Default Turbulent external flow around cylinder
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Steve Mcharg
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Folks,

I have been studying external flow around a cylinder at Re = 50000 and 100000

I have been using k-e model and looking at pressure distribution.. I have taken it that this is turbulent flow... Is this correct?

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Old   April 30, 2013, 16:35
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That range of Re is more likely laminar or transitional. The separation is likely to be laminar, while the wake is turbulent.
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Old   April 30, 2013, 16:41
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Thanks.. am I right to use a k-e turbulence model for this flow? The more work I do the mire I realise how complex this type of flow can be.

On my pressure coefficient graphs I am seeing separation at around 120 degrees

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Old   April 30, 2013, 20:57
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For something so simple, this flow can be challenging to model. A significant portion of the drag is coming from pressure drag, therefore the turbulent wake must be modeled correctly. This usually requires something like DES. The cylinder must be modeled as 3D so the vortices can bend, twist, and get all mangled up. The more 2Dish the turbulence model is, i.e. higher stiffness due to high eddy viscosity, the higher the pressure drag will get and the less accurate it will be.
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Old   May 1, 2013, 00:43
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You can give transition model a try... I have recently used transition model to predict the drag and lift on wind turbine airfoil and results were within 1-2% error band for drag.
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Old   May 1, 2013, 02:06
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Thanks Martin and far. I have got results from my k-e model which seem pretty good when compared to other data.... The cp value drops to around 2 AT 90 degrees and then separation ooccurs at 120 degrees.... For basic predictions does this sound okay?
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Old   May 1, 2013, 04:42
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Don't know what it should be for the k-e model. But, the separation point (angle) seems to be too high. At these Re numbers, I believe, a SA model will give separation at about 80 degrees. What is your cd? It sounds like your cd will be low, as if the Re was greater than 400K.
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Old   May 1, 2013, 05:06
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Ahh I see. 80 degrees would be laminar separation and produce a bigger wake.

Is the CD found by the difference between 1 and the base coefficient (where the graph plateaus) if so it is about 1.5 or 1.6....

Cheers for the help Martin
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Old   May 1, 2013, 11:57
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Cd is calculated by dividing the force in the x direction by the dynamic pressure and the base area (diameter*length). In the real world, the drag will be unsteady so it is an averaged value. At the Reynolds numbers you mentioned, the cd is about 1.2. At around Re=4e5 the cd will drop down to about 0.3 (drag crisis). That's full all out turbulent. The turbulence of the wake has worked it's way forward to the cylinder.

I'm not curtain what you mean by "Is the CD found by the difference between 1 and the base coefficient" I think my answer may be no. Off the top of my head, I'm not sure what the forward face pressure integrates to.

The separation point on the cylinder will depend on both on the boundary layer turbulence (i.e. transition) and the character of the vorticity behind the cylinder.

In regards to vortex characteristics, at low Reynolds numbers (100s) the vorticity is laminar and large. This is the Karman vortex street wake. It is basically 2D. As the Re increases the 2D nature will break down to 3D. This is for Re < 300K. Then the drag crisis is reached around 300K to 400K.

The SA turbulence model (without rotation correction) creates so much eddy viscosity in the wake that the solution is steady and symmetric. Thus two steady vortices exist behind the cylinder. The drag will be over predicted. Therefore a DDES (delayed detached eddy simulation) model is used which cuts down on the eddy viscosity in the wake to capture the unsteady wake but retains eddy viscosity next to the body to get the boundary layer.

I'm not sure how the k-e model does for a cylinder.

A paper I googled:
http://www.csc.kth.se/~jhoffman/arch...nDESCC1999.pdf

Also, googleing images for "pressure distribution around the circumference cylinder" will give you an idea of how correct your pressure plot is.
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