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October 29, 2007, 23:17 
Srinath Madhavan,
I figured

#21 
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Alessandro Spadoni
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Srinath Madhavan,
I figured out how to compute the drag for turbulent cases; however, I have doubts about the results obtained from liftDrag.C. I followed the instructions in the following thread: http://www.cfdonline.com/OpenFOAM_D...s/1/5067.html? The reason why I am doubful is because I have a simple naca0012 at 0 angle of attack, and OpenFOAM predicts a total drag of 0.00314, which is well below the experimental value of 0.0058 from "theory of wing sections." Later this evening, I tried the same case at 4deg angle of attack, and now I get a total drag of 0.008 and a lift coefficient of 0.05. For this case, the drag should be approximately 0.006 and the coefficient of lift shuld be about 0.5. I ran checkMesh and got no errors. Everything is incompressible, inlet V = 50 m/s chord = 1 m kepsilon model: inlet k = 0.375 inlet epsilon 0.03 Everything is reasonable, the analysis converges reasonably well, but the force coefficients are off. As Vincent suggested in http://www.cfdonline.com/OpenFOAM_D...s/1/5067.html? I will compare pressure coefficent, velocity distribution, etc. For the pressure coefficient, however, I will have to do a little coding. It is possible that my analysis is illposed, but it converges well, and both velocity and pressure seem very reasonable. Thank you, Alessandro 

December 8, 2007, 08:51 
Alessandro, I understand your

#22 
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Srinath Madhavan (a.k.a pUl)
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Alessandro, I understand your problem. However, the minute turbulence models enter the picture, I switch off the need for accurate experimental predictions in my head. A bad habit, but nevertheless quite true most of the time. Anyways, I would be interested in knowing whether you have tried the same case with Fluent or CFX and if so what dimensionless force coefficients (lift, drag etc.) do they predict?


March 17, 2008, 12:03 
Could someone verify that the

#23 
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Frank Bos
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Could someone verify that the turbulent drag calculation using liftDrag is correct......we are finding differences....anyone?
Frank
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March 20, 2008, 14:36 
Hi Frank & others,
I am som

#24 
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Srinath Madhavan (a.k.a pUl)
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Hi Frank & others,
I am somewhat confused as to why most posts in this forum suggest that modifications are required to the liftDrag utility if one has to obtain lift/drag predictions when using turbulence models. If the function of a turbulence model is to simulate (or more accurately, emulate) turbulent behavior in the fluid and if this is reasonably well done by the turbulence model in question, then turbulent behavior should automatically show up in the result (i.e. in the primitive variables u,v,w,p). Then all that remains is using these results in the integration of shear and pressure forces over the walls in question, which should directly give the correct lift and drag coefficients. Turbulent viscosity should not enter the picture at this stage as it is merely a means to get to the correct (hopefully) u,v,w and p. So when lift/drag forces are calculated, one should only use the actual density/viscosity of the fluid in question and not the turbulent viscosity? To put it in a nutshell, if the velocity and pressure fields are accurate in space (and in time for transient problems), then the liftDrag utility should give the correct forces without any modifications. I would appreciate if someone can correct me if I am wrong. 

March 20, 2008, 16:47 
Hi Srinath,
In fact, this i

#25 
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Frank Bos
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Hi Srinath,
In fact, this is exactly what I meant by my previous question. A turbulence model is used to give a better solution of u,v,w,p when the flow is turbulent. So, in fact the boundary layer becomes fuller (compared to laminar flow) which has its influence on the viscous drag, just like the laminar case. In my view, the turbulent viscosity term is not needed in the drag/lift calculation...... Anyone else?? Frank btw, I did not wrote this utility, but made some modifications such that integration in icoFoam i.e. becomes easier....
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March 25, 2008, 05:52 
Works like this:
tauw = nu

#26 
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Eugene de Villiers
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Works like this:
tauw = nu * dU/dy_wall right? Except we do not have dU/dy_wall, we have dU/dy_0 between the first node and the wall. Thus we calculate wall shear using a wall function. To keep things compact in the momentum equation we stick the wall function in the turbulent viscosity so that: tauw = nuEff_wall * dU/dy_0 So the use of turbulent viscosity to calculate the wall shear is purely an artifact used by OpenFOAM to store the influence of the wall function  nuEff_wall. 

March 25, 2008, 06:14 
Allright, so in the turbulent

#27 
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Frank Bos
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Allright, so in the turbulent case the viscous force calculation contains some turbulent contribution.....via wall function.
Then, why is there still need for a turb force contribution,present in liftDrag, derived from turbulence>R(). Frank
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March 25, 2008, 06:46 
Because (for example in kEpsil

#28 
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Eugene de Villiers
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Because (for example in kEpsilon) turbulence>R() is calculated from
2/3*I*k  nut_*2*symm(grad(U)) The wall boundary of nut_ contains the effect of the wall function. (See "#include "wallViscosity.H" line 88 kEpsilon.C) 

March 25, 2008, 07:04 
Formally and for consistency w

#29 
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Hrvoje Jasak
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Formally and for consistency we should have a correction for snGrad(U) on the wall to get the correct drag though.
Does anyone feel like changing them all? Hrv
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Hrvoje Jasak Providing commercial FOAM/OpenFOAM and CFD Consulting: http://wikki.co.uk 

September 19, 2014, 03:38 
Drag coefficient amplitude

#30 
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Hello everyone,
I need some assistance on understanding why in the drag coefficient time diagram, i am getting the correct frequency according to the Re number but the amplitude is totally different from past studies in papers. more specifically it should oscillate around 1.44 and mine oscillates around 3. Also in other cases with different mesh it drops to 0.3. although frequency is correct. Does anybody know about amplitude dependency on either numerical or physical factors that could help me understand why. I am using openfoam and i have also used another code (non commercial) that shows the same thing. Regards Konstantine 

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