What is the method for deriving Stokes' law for drag around a sphere?

[fayled]

So you can solve the Stokes equation for flow around a sphere to obtain the pressure in the fluid:
p=p₀-3nuacosθ/2r²
where n is the viscosity, u is the speed of the fluid (along the z axis) far away from the sphere, a is the radius of the sphere and r,θ are the usual spherical polar coordinates.

We can also obtain an expression for the velocity field u=(u_r,u_θ,0) which I won't type here because I'm only concerned about the general principles of something.

My question is about the method for deriving Stokes' law for drag, F=6πnau. The method I have seen is to use the velocity field to work out the viscous stress tensor, which allows you to compute the stress (as a vector) on the sphere due to the fluid. This vector can then be integrated over the sphere to give the required drag force.

However, I don't understand why we can't just integrate the pressure*(-r/r) i.e multiplied by its direction of action on the surface of the sphere, the radial unit vector, and integrate this over the sphere to give the drag force. I have tried and it doesn't seem to give the same result (you don't need to perform the integral, only need notice that the stress vector and the vector obtained from the pressure are not equivalent).

Thanks for any help :)

 

[boneh3ad]

If you integrate the pressure around the surface in Stokes flow you should get zero. The pressure on the front and back should exactly cancel, if I recall. The only form of drag on a sphere in Stokes flow is viscous, and since viscous drag has nothing to do with the pressure, it shouldn't surprise you that you can't derive it from the pressure distribution.

After all, look at the pressure equation you cited. If you integrate that over $\theta = [0,\pi]$, you get zero.

 

[fayled]

If you integrate the pressure around the surface in Stokes flow you should get zero. The pressure on the front and back should exactly cancel, if I recall. The only form of drag on a sphere in Stokes flow is viscous, and since viscous drag has nothing to do with the pressure, it shouldn't surprise you that you can't derive it from the pressure distribution.

Ok - it doesn't surprise me that it's to do with the flow being viscous as opposed to inviscid, as I have used the pressure approach to find say the lift in inviscid flow. So why exactly does pressure give rise to drag and lift in inviscid flow and not viscous flow?

 

[boneh3ad]

It gives rise to drag in viscous flow, it's just that Stokes flow is a special case. The Reynolds number is so low that the flow remains attached around the sphere the entire way. In general, this is not the case. If you had a higher Reynolds number, you would end up with a wake behind the sphere with a lower pressure than in front of it, thus resulting in pressure drag.

 

[pyroknife]

If you integrate the pressure around the surface in Stokes flow you should get zero. The pressure on the front and back should exactly cancel, if I recall. The only form of drag on a sphere in Stokes flow is viscous, and since viscous drag has nothing to do with the pressure, it shouldn't surprise you that you can't derive it from the pressure distribution.

After all, look at the pressure equation you cited. If you integrate that over $\theta = [0,\pi]$, you get zero.

I think this is incorrect. The pressure drag around sphere for potential flow is zero.
For Stoke's flow you have contribution from friction (2/3) and pressure drag (1/3).
http://user.engineering.uiowa.edu/~fluids1/Creeping_flow_1.pdf