# Stokes flow between 2 spheres

The problem is given like this:

Given two rotating spheres with constant angular velocity $$\Omega_1$$ and $$\Omega_2$$ around the vertical axis and pressure on the borders of spheres is $$p_1$$ and $$p_2$$ ,filled with fluid between them, using Navier-Stokes equations, find the physical velocity components in spherical coordinates (no gravity, no slip condition).

Assume ($${v_r} = 0,\quad {v_\theta } = 0,\quad {v_\varphi } = f(r)\sin > \theta$$)

It's a Stokes flow problem, and I need to find the function $$f(r)$$. The equations I derived look like this:

\begin{align} \frac{{\partial p}}{{\partial r}} &= {\rho _0}\sin^{2}\left( \theta \right)\frac{{{f^2}\left( r \right)}}{r}\tag{1}\\ \frac{{\partial p}}{{\partial \theta }} &= 0\tag{2}\\ \frac{{\partial p}}{{\partial \varphi }} &= \, - \mu \sin\left( \theta \right)\frac{{\left( { - 2f\left( r \right) + {r^2}f''\left( r \right)} \right)}}{r}\tag{3} \end{align}

I have no idea how to solve this. Is there any extra equations I can use to find $$f(r)$$? When I searched for Stokes flow around a sphere, I keep finding examples where the $${v_r},{v_\theta }$$ is something given in form of a steam function but $${v_\varphi }$$ is zero.

• Why don't the examples where $v_r = v_\theta = 0$ help you? Those are the same assumptions you have listed in your question... – tpg2114 Oct 3 '18 at 20:03
• in the examples they are not zero,but the $vφ$ is zero so i have a reversed situation – Andrej Licanin Oct 3 '18 at 20:29
• Could you provide a schematic drawing for the problem? From your description, it's difficult to see how the spheres are located w.r.t. each other. – Deep Oct 4 '18 at 6:00
• @Deep You have 2 concentric spheres of different radii rotating about a common axis passing through their centers. There is a viscous liquid filling the space between the spheres. – Chet Miller Oct 4 '18 at 12:02
• @ChesterMiller I see now. – Deep Oct 5 '18 at 5:13

I don't confirm your 2nd and 3rd equations. I get:

$$\frac{\partial p}{\partial \theta}=\rho f^2\sin{\theta}\cos{\theta}$$ $$\frac{\partial p}{\partial \phi}=\frac{\mu \sin^2{\theta}}{r}\left[\frac{\partial}{\partial r}\left(r^2\frac{\partial f}{\partial r}\right)-2f\right]$$ By symmetry, p is not a function of $$\phi$$. Therefore, $$\frac{d}{d r}\left(r^2\frac{d f}{d r}\right)-2f=0$$

This is a Stokes (low Reynolds number) flow, and the radial- and latitudinal equations contain only inertial terms. So they can be neglected. The only relevant equation is the one in the $$\phi$$ direction.