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Jwolf August 29, 2009 14:56

Coefficient DRAG and LIFT - Cylinder 3D

after reading some post not yet I know clearly some details to calculate these coefficient you formulate them in the CFX are:

Coeff Drag
force_x () @Tube wall/(0.5*Density*Area*Velocity^2)
Coeff Lift
force_z () @Tube wall/(0.5*Density*Area*Velocity^2)


In post law that can be calculated coefficient with the pressure forces and viscosity in wall of the cylinder which they are obtained in file.out if I I use,

force_x () @Tube wall
force_y () @Tube wall
force_z () @Tube wall

it is just like to use the forces that are obtained in .out? according to what law in Help CFX. if.

The area is the reference area and it is obtained with the projection of the area in a cross flow plane to the direction of the flow, but equal it is not to me clear.

this area as one calculates for a cylinder located in perpendicularly to the direction of the flow? in three dimensions.

Diameter 0,01 m
length 0.2 m

also law that the depth has to do for cases 3D, but what is this depth? (Dimensions cylinder)


ghorrocks August 30, 2009 07:03

Sorry but I don't understand most of what you are saying. I will have a guess, please correct me if I am wrong.

If you don't know what the force_x() function returns why don't you just calculate it in CFD-Post (using the function calculator) and find out? Then you can see for yourself whether it matches the value in the .out file (it should!)

For many applications the area is the frontal area. For a cylinder then the area becomes diameter*cylinder length. You can also calculate this using CEL using the integration functions - I will leave this as an exercise for the reader to work out the CEL expression to do this.

I don't know what you are asking about the depth.

Jwolf August 30, 2009 10:22

Glenn thanks, excuse my english… I am learning…. but you understood to me perfectly or at least you clarified my doubts.

About force_x () function needed to clarify if he is the same to use force_x () function or the forces of pressure and viscosity of the .OUT

The depth of that I speak was something that law in other commentaries of the forum. but reason why you say I am going to apply to me ( diameter*cylinder length).

Now, for coefficient Lift the results are very low, because it can pass this?

Coefficient Drag is OK comparing it with the results of the text.

ghorrocks August 31, 2009 07:42

I realise English is not the first language of many people on the forum. Your English is a lot better than my Spanish!

The force_x() should be the same as the total surface forces reported in the OUT file. But you should compare hem yourself to check this, don't just take my word for it.

I don't know what you mean by depth. Is the cylinder near a free surface?

What is generating the lift? Is the cylinder rotating or near a free surface?

Jwolf August 31, 2009 10:34

In first post this image of geometry, the outer walls of geometry (cubic) have not slip, can that influence in coeff DRAG and LIFT?

The cylinder not this rotating and the near surfaces are the outer walls.

According to I have understood lift originate the perpendicular components of the pressure forces and viscosity with respect to the flow, in the surface of the cylinder. but observing the data of the text, the values of the DRAG are correct but in the graphic one I do not obtain oscillations and the LIFT as already I said is very low.

I leave image so that they understand to me:

UP (text)
Down (my case)

It can be that I need but time of simulation for observes the behavior of the text?

Another detail the data that I have is for cylinders 2D, and being this experiment 3D, can be that the data to compare are not the best ones, but they said equal me in the university that serve like reference

ghorrocks August 31, 2009 18:32

Have you checked the issues discussed here:

cfxmar February 1, 2010 07:33


I have read all posts regarding Drag forces over a cylinder 3D and I cannot arrive to understand how to obtain the drag coefficient.
If you use this expression: force_x()@Tube wall/(0.5*Density*Area*velocity^2) you will not obtain the coefficient value since force_x = inercia force + drag force and it depends on two coeficients ( Cd and Cm for Morison forces on a cylinder).

Could anyone, please, advise me?

ghorrocks February 1, 2010 18:06

?? What are you talking about? The drag is the force in the direction of flow, the lift is the force perpendicular to the flow. The force includes both form and viscous drag components. This means the force_x/y/z functions are appropriate.

If you want to separate out the viscous drag and form drag components then that is easily done by doing a surface integral of pressure for the form drag and a surface integral of wall shear for the viscous drag.

cfxmar February 2, 2010 04:02

1 Attachment(s)
I´m talking about Morison Equation:

The Morison equation is the sum of two force components: an inertia force in phase with the local flow acceleration and a drag force proportional to the (signed) square of the instantaneous flow velocity.

The Morison equation contains two empirical hydrodynamic coefficients — an inertia coefficient and a drag coefficient — which are determined from experimental data

Please, Could you tell me how to separate both forces from the expression force_x@Location (I do a chart in CFX post to see force values as a function of time) in order to calculate Cd and Cm??

The expression I use to calculate Cd is not correct:

CDrag = (force_x()@pilote)/(998[kg \
m^-3]*area()@pilote*(areaInt(Velocity u)@pilote)^2)

since the force in x axis is the sum of Inertia and drag components, thats what i´m trying to explain in my poor english.

I don´t understand your explanation:

"If you want to separate out the viscous drag and form drag components then that is easily done by doing a surface integral of pressure for the form drag and a surface integral of wall shear for the viscous drag"

ghorrocks February 2, 2010 18:33

Why do you have the area integral of U in your CD expression? Velocity is zero at a wall.

It look to me like you will have to calculate the two terms in the Morrison eqn the same way experimentalists do. I guess they would run a range of velocities and then do a curve fit on the drag versus velocity curve for a linear and quadratic component. If you want something comparable you will need to do the same.

But as I said the viscous and form form drag components are easily calculated directly from area integrals of the relevant variables. But I doubt this will give you anything which will compare with experimental values of these amounts.

cfxmar February 4, 2010 04:51

Thank you, Glenn.
What I need is to obtain my own values of Fd and Fi to compare them with experimental results.
So, there is no way to separate both forces from force_x@location????

ghorrocks February 4, 2010 06:08

You seem to have missed my point. There is no way of separating the two components, it is just a figment of Morison's imagination anyway. If you are going to compare to experimental results you will have to extract the numbers in the same way the experimentalists did. You can't measure these numbers directly in an experiment so they must have done some data analysis to extract it. You need to do the same analysis.

The alternative is that in CFD you do know the wall shear and pressure fields. From these you can calculate friction and form drag components which should sum to the total drag (within a numerical tolerance). But this will not be much use when you compare to experimental results as the experimentalists cannot do this so you are comparing apples with oranges.

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