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May 8, 2006, 11:45 
Does anybody know how the lift

#1 
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Srinath Madhavan (a.k.a pUl)
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Does anybody know how the lift coefficient is calculated using the lift/drag utility in OpenFOAM 1.2. To my knowledge, the equation should be in the following lines. When using icoFoam, how does the utility know what the density (rho) is?
C_L = F_lift / (A_ref * 0.5 * rho * U_ref^2) 

May 8, 2006, 15:29 
Hi pUI
the liftDrag utility

#2 
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kumar
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Hi pUI
the liftDrag utility is for solvers where pressure is pressure/density . you can check the dimensions of pressure in icoFoam to see what dimensions it is using and then apply a correction to the output of icoFoam regards kumar 

May 9, 2006, 04:17 
Hi all,
I also did a calcul

#3 
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Anja Stretz
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Hi all,
I also did a calculation with icoFoam and the results for p are of the dimensio [m2/s2]. Now I want to use FoamToTecplot. Does that change the dimension of the pressure? Or do I have to do that myself? But therefor I have to know with which density icoFoam is calculating?! Thanks Anja 

May 9, 2006, 05:03 
Hi Anja
foamToTecplot does

#4 
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kumar
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Hi Anja
foamToTecplot does not change the dimensions. so you will have to do the conversion yourself. since icoFoam is solving p/rho , all fluid densities are same for icoFoam . i am not sure if viscosity is used by icoFoam . but if it is then you will have to change the viscosity to what you want regards kumar 

May 9, 2006, 05:18 
Hi kumar,
so I just need to

#5 
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Anja Stretz
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Hi kumar,
so I just need to know with which density icoFoam is calculating?? Where can I find that value?? Thanks Anja 

May 9, 2006, 05:25 
There is no density: it is con

#6 
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Hrvoje Jasak
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There is no density: it is constant and the whole equation has been divided through by the density. This is why you get different units for nu and p, i.e. kinematic viscosity and kinematic pressure.
If you really have to have density, it is equal to 1. Hrv
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May 9, 2006, 05:27 
icoFoam does not calculate a d

#7 
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Eugene de Villiers
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icoFoam does not calculate a density, it calculates p/rho, using a specified value of nu. You should choose rho based on the kinematic viscosity (mu) and rho you used to calculate dynamic viscosity nu.


May 9, 2006, 05:28 
But we never did input a rho t

#8 
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Srinath Madhavan (a.k.a pUl)
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But we never did input a rho to start with?


May 9, 2006, 05:31 
Does that also mean that the w

#9 
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Srinath Madhavan (a.k.a pUl)
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Does that also mean that the when the liftDrag utility extracts the forces on walls, it assumes a density of unity? So if I wish to use the definition as mentioned in the beginning of this thread, I should account for rho based on the nu I input say in the Aref term?


May 9, 2006, 05:45 
My mistake its the other way r

#10 
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Eugene de Villiers
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My mistake its the other way round. nu is of course kinematic visc not dynamic.
You dont need to put in rho for icoFoam becuase you are solving p* = p/rho. Not p. If you want to calculate lift and drag forces you have to pass rho into the calculation somehow. Otherwise you will end up calculating F/rho. 

May 9, 2006, 05:48 
Actually F_Lift/rho is just wh

#11 
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Srinath Madhavan (a.k.a pUl)
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Actually F_Lift/rho is just what I need. However, since rho can be taken as unity as Hrv menttioned, I can modify A_ref to account for rho by using A_ref_eff = A_ref * rho. The 'rho' here would be based on the actual density I had in mind when I decided nu. Would this be ok?


May 9, 2006, 05:54 
So there is the equation:
kin

#12 
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Anja Stretz
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So there is the equation:
kinematic viscosity mu = dynamic viscosity nu / rho In the file "transportation Properties" you can write a value for: nu [0 2 1 0 0 0 0] certainValue (shouldn't this be mu then??) But, as OpenFoam uses rho=1kg/m³ the values (apart from the dimensions) for mu and nu are the same?? 

May 9, 2006, 05:56 
Anja,
It should be the othe

#13 
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Srinath Madhavan (a.k.a pUl)
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Anja,
It should be the other way around: kinematic viscosity nu = dynamic viscosity mu / rho 

May 9, 2006, 05:58 
Yes, I just did not see the ot

#14 
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Anja Stretz
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Yes, I just did not see the other post, as long as I was writing mine.
But what about my last sentence before? 

May 9, 2006, 06:13 
I guess 'mu' does not have muc

#15 
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Srinath Madhavan (a.k.a pUl)
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I guess 'mu' does not have much meaning in icoFoam. Case in point, since both sides of the transport equation are divided by rho, p* = p/rho as Eugene mentioned. So the pressure Openfoam calculates should be actually 'p/rho'. As a result, when we want to extract the pressure we should multiply the 'p' output with rho (this rho should be what you intended it to be, for instance, consider flow of water. rho = 1000 and mu = 0.001 => nu = 0.001/1000 = 0.000001 [all in SI units]). This is only for the grad(p) term. the 'mu grad(grad(v))' term in the original NS equation is now 'nu grad(grad(v))' as it should be when it is divided by 'rho'.
The problem now is trying to follow 'rho'. If 'rho' needs to be passed for the calculation of lift coefficient, then one should multiply the pressure 'p' obtained from OpenFoam with 'rho' (which is 1000) to get the actual pressure. However, to calculate C_L using: C_L = F_lift / (A_ref * 0.5 * rho * U_ref^2) one need not bother about calculating the actual pressure because the pressure itself is 'p/rho' and we can be certain that rho is 1000 here since we fixed nu to be 10^(6). So the expression for C_L becomes: C_L = F_lift / (A_ref * 0.5 * U_ref^2) Hrv, Eugene: Does this sound OK? 

May 9, 2006, 06:17 
Yup, it's good.
Hrv

#16 
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Hrvoje Jasak
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Yup, it's good.
Hrv
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