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James Forrest February 21, 2004 19:48

Airfoil negative drag coefficent
 
Hi, I am modelling a 2D airfoil in Fluent (using segregated 2nd order solver) and as I increase the angle of attack the drag (due to pressure) is decreasing. At angles greated than 10 degrees this drag force becomes negative - i.e. producing thrust!!!!! If a turbulence model is added the pressure drag still outweighs the viscous drag so overall drag is still negative. This is obviously not correct - can anybody shed any light on this? It's quite urgent as I have to write a report on this by next week. It seems that only the upper airfoil surface is producing negative drag which is also strange. I am quite confident that the quality of the mesh is good - I'm baffled! Thanks for any help anyone can offer.

James.

Praveen February 22, 2004 00:41

Re: Airfoil negative drag coefficent
 
First question that comes to my mind is whether your equation for drag is correct. This equation will involve the surface normal; if the direction of normal is wrong then you will get the wrong sign. If the magnitude of the drag seems correct except for the sign then this is most likely the problem. Then just reverse the direction of the normal.

James Forrest February 22, 2004 07:16

Re: Airfoil negative drag coefficent
 
Thanks for your reply. I don't think this is the problem as at zero angle of attack the drag is in the correct direction. It is only once I begin to increase the angle that drag decreases, then passes through zero and becomes negative. The drag vector is set to act in the positive x direction. The thing I find strange is that, for example, at 10 degrees AoA the drag on the bottom surface is positive, but the drag on the upper surface is a larger negative value. The net sum of these two forces leads to the negative drag coefficient.

James Date February 22, 2004 13:33

Re: Airfoil negative drag coefficent
 
James

I'm sure you have got your vectors wrong somewhere! Obviously the force will be in the correct direction at zero incidence when alpha = 0 even if you have the sin's and cos's wrong!

Send me the files if you still have a problem. I assume you're either doing aeronautics or ship science at soton!

Regards James

Charles Crosby February 23, 2004 02:03

Re: Airfoil negative drag coefficent
 
James,

You did convert axial force and normal force to lift and drag via

L=Fy cos(alpha) - Fx sin(alpha) D=Fx cos(alpha) + Fy sin(alpha)

didn't you? (refer to aerodynamics 101 ....)

James Forrest February 23, 2004 03:10

Re: Airfoil negative drag coefficent
 
Well, I solved the problem last night by simply rotating the grid by the required number of degrees. However, when using the above formula on data I had obtained previously, that also works! I think the problem was that I assumed Fluent already did that calculation when displaying lift and drag forces - I overlooked the fact that forces acting in the x and y directions were not necessarily acting parallel and normal to the oncoming flow.

Thanks to all that helped out!

James.

lfc April 2, 2010 16:44

the problem could turn out to be as simple as
 
Hi there

i have spent some time now getting used to fluent it is fairly simple for the most part but the devil is in the detail, i have been under-taking a comparison between 2d and 3d aerofoil cfd results in fluent, for the 3d models i change the coordinate system pre-meshing to allow for changes in AOA but for the 2D models i just changed the flow direction accordingly in boundary conditions>>velocity,direction and magnitude. this brought about some very strange results from the forces monitors for Cl and Cd. To cut to the chase, lift is taken to be as perpendicular to the free-stream flow and drag is taken to be parallel to the flow. so for my 3d models the flow was always coming into the flow domain horizontally so i had no problems like receiving negative drag ( the chance would be a fine thing), this only occurred when i changed the flow direction in the 2d models.

simply change the force vectors in the force monitors to:
for lift
x=-sin(alpha)
y=cos(alpha)

drag
x=cos(alpha)
y=sin(alpha)


i hope this helps,

aamer April 9, 2011 07:06

HI lfc....

it is true that lift and drag components will be, the way you wrote, for an incomg flow.... but what will be done , if there is a case in which motion is in still air...

.....For instance.... If i am rotating (azimuth motion) my wing in a still air from o to pi radians, through udf (i.e no inflow velocity). Assume that the wing is given constant angle of attack of 20 degree by tilting the whole grid. Now how will the components of lift and drag be given to fluent ????


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