0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Why don't the examples where $v_r = v_\theta = 0$ help you? Those are the same assumptions you have listed in your question... $\endgroup$ – tpg2114 Oct 3 '18 at 20:03
  • $\begingroup$ in the examples they are not zero,but the $vφ$ is zero so i have a reversed situation $\endgroup$ – Andrej Licanin Oct 3 '18 at 20:29
  • $\begingroup$ 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. $\endgroup$ – Deep Oct 4 '18 at 6:00
  • $\begingroup$ @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. $\endgroup$ – Chet Miller Oct 4 '18 at 12:02
  • $\begingroup$ @ChesterMiller I see now. $\endgroup$ – Deep Oct 5 '18 at 5:13
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.