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Old   November 16, 2003, 18:43
Default Calculating Drag
Ajay Rao
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The following code I am using to calculate the Drag is way off ! Could any one point out the mistake . Thanks in advance -Ajay



c=F_C0(face,tf); /* gets the adjacent cell */

t=THREAD_T0(tf);/* gets the adjacent cell thread */

/* doing this so that i can use C_DUDX(c,t) */ /* F_DUDX notpresentto get derivative at wall */

sigmaxy=C_MU_L(c,t)*(C_DUDY(c,t)+ C_DVDX(c,t));

sigmaxz=C_MU_L(c,t)*(C_DUDZ(c,t)+ C_DWDX(c,t));

sigmaxx=C_MU_L(c,t)*2*(C_DUDX(c,t)) - C_P(c,t) ;


sigmayz=C_MU_L(c,t)*(C_DVDZ(c,t)+ C_DWDY(c,t));

sigmayy=C_MU_L(c,t)*2*(C_DVDY(c,t))-C_P(c,t) ;



sigmazz=C_MU_L(c,t)*2*(C_DWDZ(c,t))- C_P(c,t);


/*compute the force vector at each cell centroid */

/* area vector points inwards hence a= -a */






printf(" Drag force = %2.20e ",force_vector[0]);

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Old   November 17, 2003, 02:15
Default Re: Calculating Drag
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I think the values used to compute the force on the wall should be the ones located on the face centroid.
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Old   November 17, 2003, 09:13
Default Re: Calculating Drag
Ajay Rao
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Ya True , But fluent does not give access to the derivatives at the wall . They have functions like C_DUDX which gives the derivative at the cell centroid. But they dont have any functions like F_DUDX to give the derivative at the wall . Any idea how to go around this ?

Regards, Ajay
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Old   November 18, 2003, 10:19
Default Re: Calculating Drag
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You should fix if you want to name your stress tensor "sigma" or "tau"...
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Old   November 19, 2003, 11:55
Default Re: Calculating Drag
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hi, Ajay

I will attend a Fluent User Conference tomorrow and try to find some help from engineers.

A possible way I am considering is to find the stresses by interpolating from the values on the centroid.

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Old   November 19, 2003, 20:51
Default Re: Calculating Drag
Ajay Rao
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It would be nice to hear what the FLUENT ppl have to say , please keep me posted , Thanks a lot -Ajay
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Old   November 20, 2003, 09:30
Default Re: Calculating Drag
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hehe, the one I asked said nothing. it seems we have to solve this problem by ourselves.

let us inform the other when new ideas come into mind. good luck
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Old   November 30, 2009, 01:28
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Hi Ajay,

I have been facing the same trouble. I have to give a traction free condition at the outlet of a pipe with a porous layer surrounding it. The condition applies to the entire outlet. Its, -p+(mew)du/dx=0. Since I do not know how to apply a traction free condition, I am trying to specify a pressure profile at the outlet in terms of (mew)du/dx. However, the same problem persists: derivative is calculated considering cell centroid and pressure profile is hooked at face centroids. If you find a solution to this please let me know. Thank you so much.
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Old   February 15, 2010, 09:15
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Hi guys, i'm tring to calculate the drag force coefficient in a 3-d case and the fluid is behind a cylinder.
the fluid is:
density = 800 kg/m^3
dinamic viscosity = 0.005 kg/ms
velocity = 8 m/s

the cylinder is:
diameter = 13 mm
length = 300 mm
stiffness modulus = 200000 MPa
poisson coefficient = 0.3

to calculate the drag coefficient i push: solve/monitors/force/drag coefficient.....
the direction of the vector of force is in parallel to the flow flowing

i set the reference value:
Area = 0.013*0.3 that is the product for diameter and the length of the cylinder,
Length = 0.013 that is the diameter
then i set other values: viscosity, velocity, density of fluid as i wrote upforward.

i put 90 nodes around the circonference of the cylinder, and 25 along to the length of the cylinder.

the algorytm used is LES, because this is a 3d case
I used PISO
and second order upwind for momentum...

the Cd value is incresing but after 0.02 seconds it is still to 0.7.

How can i do, please help me.
Thank to all

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