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

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The net force on a submerged sphere in a fluid flowing in the z-direction is directed along the z-axis due to symmetry, with the force expressed as an integral involving pressure and stress tensor components. The integral form includes a pressure term, P', which acts normal to the sphere's surface, and viscous stress components, σrr and σrθ, which account for normal and tangential forces, respectively. The sine and cosine terms in the expression represent the directional components of these stresses in relation to the flow direction. Understanding this formulation is crucial for deriving the net force accurately. Clarification on tensor notation may be needed for those unfamiliar with the concepts.
Shmi
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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?
 
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Shmi said:

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 σrr and σ 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.
 

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