How Is Drag Calculated on a Sphere in a Non-Uniform Flow Channel?

[Red_CCF]

Hi

This isn't a homework question, just a problem I thought up of but would require some implementation I don't really know how to do.

Homework Statement

If I have a microfluidic channel with laminar flow and a parabolic velocity profile as predicted in the Hagen-Poiseuille flow, and I insert into the channel a sphere of diameter 1/16th of that of the channel (can be anything really just something that's small but not negligible compared to the diameter), how would I calculate the initial drag force on the sphere assuming that Re<<1 such that there is no wake/flow separation?

The main problem here is that the velocity profile or uinf is not uniform, which means that all of the Cd data available is pretty much useless as they all assume uniform incoming flow.

I'm also wondering how the linear shear profile of the flow is reflected in the drag of the sphere since for a uniform incoming flow, there is no inherent shear stress in the flow. Also, would the varying incoming velocity cause the sphere to spin as well?

Homework Equations

Re = UD/v, v = 1/4μ * (r^2-R^2) * dp/dz


Thanks

 

[rude man]

Sorry you're not getting any responses.

About all I can do is suggest that, by symmetry, there would not be spin imparted to your sphere. The only asymmetry would be gravity, so if you ignore that, I see no asymmetry in the fluid flow past the sphere, complex though it be.

 

[Red_CCF]

Sorry you're not getting any responses.

About all I can do is suggest that, by symmetry, there would not be spin imparted to your sphere. The only asymmetry would be gravity, so if you ignore that, I see no asymmetry in the fluid flow past the sphere, complex though it be.

Hi

I believe that to be true if the sphere was placed at the center of the pipe, but if I place it say a quarter diameter from the wall of the pipe, the dv/dr would not be zero and if the flow velocity past the sphere is different intuitively I would think the shear stress would also be different, hence why I think there would be a spin. Is this deduction correct?

Thanks

 

[rude man]

Hi

I believe that to be true if the sphere was placed at the center of the pipe, but if I place it say a quarter diameter from the wall of the pipe, the dv/dr would not be zero and if the flow velocity past the sphere is different intuitively I would think the shear stress would also be different, hence why I think there would be a spin. Is this deduction correct?

Thanks

Afraid you're right. I assumed center position for the sphere.

I'd try to find some good simulation software! Sorry I'm not of more help.

I also agree there would in the off-center case be spin unless viscosity = 0.