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.

 

[rezeew]

for the interesting problem! The initial drag force on the sphere can be calculated using the Stokes' drag equation, which is given by Fd = 6πμaU, where μ is the viscosity of the fluid, a is the radius of the sphere, and U is the velocity of the fluid at the surface of the sphere.

In this case, since the flow is laminar and Re<<1, the shear stress at the surface of the sphere can be calculated using the Hagen-Poiseuille equation, which you have already included in your post. This equation takes into account the non-uniform velocity profile and the resulting shear stress can be used to calculate the drag force.

As for the effect of the linear shear profile on the drag of the sphere, it will depend on the orientation of the sphere in the flow. If the sphere is aligned with the flow direction, the shear stress will not have an effect on the drag force. However, if the sphere is at an angle to the flow, the shear stress will contribute to the overall drag force.

Regarding the potential spinning of the sphere, this will also depend on the orientation and shape of the sphere. If the sphere is perfectly spherical and aligned with the flow, it is unlikely to spin. However, if the sphere is not perfectly spherical or at an angle to the flow, the varying velocity of the fluid may cause it to spin.

I hope this helps in solving your problem! Good luck!