Look at this

https://www.comsol.com/blogs/how-do-i-compute-lift-and-drag/

Dear Sourove,

Generally, the drag force is given by the following equation

Fd = 0.5 x Cd x rho x u2 x A

Fd = the drag force , Cd = the drag coefficient, rho = the mass density of the fluid, u = the flow speed of the object relative to the fluid, A = the reference area. For a sphere A = Pi x r2  

Regarding the lift force, it is given by a similar formula :

L = 0.5 x CL x rho x u2 x A

L = lift force, CL = the lift coefficient at the desired angle of attack

Note that 0.5 x rho x u2 = is the free stream fluid dynamic pressure 

With my best regards

Prof. Bachir ACHOUR

On the body-side of the fulidic domain, you have two types of forces to integrate to obtain the overall forces: pressure and friction. All you have to do is to perform the surface integration

(Cx,Cy,Cz)=[Int(-cp vec(n))dS + Int(vec(cf) \cdot vec(n))dS] / S_ref

where vec(n) is the surface normal vector, cp the pressure coefficient, vec(cf) the wall friction coefficient vector (! not the scalar friction coefficient ! vec(cf) = tensor(tau) \cdot vec(n) / q ; q=dynamic pressure )

and to probaly transform it into the aerodynamic coordinate system (I guess useless for a sphere) to obtain CL, CD.

In case you only have the pressure coefficient, the only chance to get the friction forces would be to run a 3D boundary layer code. (I assume you are not referring to CFD results but to measurements as you otherwise would get the Information out of your CFD code)

Be aware that friction forces are not equal to viscous forces. Viscous forces are the sum of the friction forces and the pressure forces due to the boundary layer and wake  displacement. The latter are included in the surface integral of the pressure coefficient.

Simply integrate pressure coefficient all along the periphery of the sphere and then resolve the integrated value along the free streamline and normal to it to get drag and lift respectively . 

 With a sphere you know the normal to the surface everywhere.   The normal force is thus easily calculated over discreetly small areas.   The normal force is then resolved into lift and drag directions you establish (up and down).  

Pll refer " WIND TUNNEL TESTING   by Alan Pope "  .......simplest method is described.

Hi,

you could also check the answers to this similar question in ResearchGate:

https://www.researchgate.net/post/How_do_I_derive_the_Lift_coefficient_data_from_the_pressure_data_gotten_from_the_simulation_wind_tunnel_tests

Regards,

Jose