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Jean October 4, 2002 16:05

drag coefficient
I am using SA and LES to simulate flow pass a sphere (both Fluent 5 and Fluent 6). I found that drag coefficients are not right. But

( drag coefficient ) / (density * U-infin. *U-infin.)

is very close to the experimental data. Is it possible that Fluent has a bug in the drag coefficient calculation?


Glenn October 7, 2002 09:03

Re: drag coefficient
The drag coefficient is calculated using the values in: report > reference values So its possible that your reference area is set wrong.

Jean October 7, 2002 13:33

Re: drag coefficient
Dear Glenn:

Your message is not clear for me. I didn't set the reference area. I got the drag coefficient from:

Solve - > Monitors - > Force - > drag coefficient

and that value matches with

Report - > Forces - > total coefficient for vector (1,0,0).


-- Jean

Jean October 7, 2002 14:25

Re: drag coefficient
Hi Glenn,

I understand what you mean now. In Report -> Referance Values, the Area = 1 m^2 and it should be Pi*D^2. Since D=1, so it will be Cd_new=Cd/Pi.

The value is still not right!


-- Jean

Rex October 8, 2002 03:19

Re: drag coefficient
Guys, Sorry to interject, but I have a converged flow field with no objects in it and was wondering if I can report the drag coefficient on an imaginary sphere where I would set its properties (i.e. density, diameter)?


Glenn October 8, 2002 05:48

Re: drag coefficient
Im not too sure about this, but I think that you can only calculate the drag on a 'wall' boundary condition. However maybe someone knows different????

Chetan Kadakia October 9, 2002 19:00

Re: drag coefficient
Rex, what do you mean by an imaginary sphere? How do you model a sphere as being imaginary?

Rex October 10, 2002 03:12

Re: drag coefficient
I mean that if there was a sphere in the flow field (there isn't one really), could you report, for instance, the drag force being experienced at that point?


Chetan Kadakia October 11, 2002 20:23

Re: drag coefficient
If you imagine its there, then the model will not realize it and it wont give you anything. But if you incorporate it in your model, you can set the reference values, and monitor the drag coefficient.

Your answer is yes. Report->reference values: set the frontal area, the velocity, density, etc Solve->monitors->force: select drag and the sphere wall

bob October 30, 2002 05:39

Re: drag coefficient
Hey Rex, surely the flow would be perturbed by the sphere, so changing its characteristics and hence the Drag ? unless you can some how take into account the effect the sphere has upon the flow ? I guess though if the shere were very small then its influence on the flow would be negligable, but then so would its drag ??? What is the interest of calculating the drag this way round ?? bob

Laika October 30, 2002 08:50

Re: drag coefficient
That's an interesting study.

What's is the status of your project? For who are you doing this?

For wich values of Re are you calculating? How is your mesh?

are you willing to share some results with us?

greetings, Laika, still orbiting

Rex October 30, 2002 09:48

Re: drag coefficient
The so called mechanistic models for particulate transport I am using in conjunction with Fluent flow data assume that the spheres do not effect the characteristices of the flow field (velocity profile, etc.), so how the particles modify the flow is not an issue I'm looking at. I just want to know the forces being experienced over a certain (pre-defined by me) area are.

Jean October 31, 2002 15:51

Re: drag coefficient
Dear Glenn,

As you have mentioned that the drag coefficient is calculated using the values in: report > reference values. I have found that in my case, the problem is not reference area but reference velocity. For both pressure coefficient and drag coefficient, correction should be made by divided (V_inf * V_inf).

Many thanks,

-- Jean

Jean October 31, 2002 15:55

Re: drag coefficient
Hi Laika,

Thanks for the email. I am willing to share some my results with you. I will email you later.

-- Jean

Laika November 4, 2002 04:49

Re: drag coefficient

Laika, still orbiting

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