Fluid flow around a sphere, what's the net force?

[Shmi]

Homework Statement

A sphere of radius ##a## is submerged in a fluid which is flowing in the z-hat direction. There is some associated viscosity in the fluid which will exert a force on the sphere. Use symmetry to argue that the net force will be in the z-direction. Show that it will have the form

$$ F_z = \oint \left( -P' \cos{\theta} + \sigma_{rr} \cos{\theta} - \sigma_{r \theta} \sin{\theta} \right) \; dA $$

Homework Equations

We are told earlier that it satisfies the Navier-Stokes equation

$$ - \nabla P' + \eta \nabla^2 v = 0 $$

with the boundary conditions that ## v = 0 ## at ##r=a##. Also, it satisfies ## {\bf v} = v_0 \hat{z} ## very far from the sphere.

##\sigma## is the stress tensor

Also, we earlier defined fluid force as

$$ F_i = \oint \sigma_{ij} dA_j $$

The Attempt at a Solution

So, clearly as the fluid moves around the face of the sphere, the components in y-hat cancel while the z-hat components add, and this argument applies all the way around the azimuthal coordinate to x-hat and back. So, I get that the force is only in z-hat, but I don't get the form of the integral expression. Why is there a pressure term, and what are the sin and cos terms doing next to the stress tensor components? I'm not great with tensor notation, so maybe I'm missing it. Could someone clarify what that expression means so that I can better derive it?

 

[Chestermiller]

Homework Statement

A sphere of radius ##a## is submerged in a fluid which is flowing in the z-hat direction. There is some associated viscosity in the fluid which will exert a force on the sphere. Use symmetry to argue that the net force will be in the z-direction. Show that it will have the form

$$ F_z = \oint \left( -P' \cos{\theta} + \sigma_{rr} \cos{\theta} - \sigma_{r \theta} \sin{\theta} \right) \; dA $$

Homework Equations

We are told earlier that it satisfies the Navier-Stokes equation

$$ - \nabla P' + \eta \nabla^2 v = 0 $$

with the boundary conditions that ## v = 0 ## at ##r=a##. Also, it satisfies ## {\bf v} = v_0 \hat{z} ## very far from the sphere.

##\sigma## is the stress tensor

Also, we earlier defined fluid force as

$$ F_i = \oint \sigma_{ij} dA_j $$

The Attempt at a Solution

So, clearly as the fluid moves around the face of the sphere, the components in y-hat cancel while the z-hat components add, and this argument applies all the way around the azimuthal coordinate to x-hat and back. So, I get that the force is only in z-hat, but I don't get the form of the integral expression. Why is there a pressure term, and what are the sin and cos terms doing next to the stress tensor components? I'm not great with tensor notation, so maybe I'm missing it. Could someone clarify what that expression means so that I can better derive it?

P' is the pressure distribution at the surface of the sphere and acts normal to the sphere surface, and σ_rr and σ_rθ are the viscous portions of the stress, normal and tangential to the surface, respectively. The sines and cosines give the components of these stresses in the flow direction.