1
$\begingroup$

A Newtonian fluid of constant density $\rho$ is in a vertical cylinder of radius R with the cylinder rotating about its axis at angular velocity $\omega$. Find the shape of the free surface at steady state.Consider the cylindrical coordinate system for analysis. Consider the pressure (P) to be a function of two coordinate system r and z. Refer to the figure below for more details.

enter image description here

I used Navier Stoke's equation in the angular direction since the principal motion is in ${\theta}$ direction.

$$\begin{aligned} &\theta \text { -component: }\\ &\rho\left(\frac{\partial u_{\theta}}{\partial t}+u_{r} \frac{\partial u_{\theta}}{\partial r}+\frac{u_{\theta}}{r} \frac{\partial u_{\theta}}{\partial \theta}+\frac{u_{r} u_{\theta}}{r}+u_{z} \frac{\partial u_{\theta}}{\partial z}\right)\\ &=-\frac{1}{r} \frac{\partial P}{\partial \theta}+\rho g_{\theta}+\mu\left[\frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{\partial u_{\theta}}{\partial r}\right)-\frac{u_{\theta}}{r^{2}}+\frac{1}{r^{2}} \frac{\partial^{2} u_{\theta}}{\partial \theta^{2}}+\frac{2}{r^{2}} \frac{\partial u_{r}}{\partial \theta}+\frac{\partial^{2} u_{\theta}}{\partial z^{2}}\right] \end{aligned}$$

Reducing it with assumptions I get

\begin{aligned} 0=\frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{\partial u_{\theta}}{\partial r}\right)-\frac{u_{\theta}}{r^{2}} \end{aligned}

\begin{aligned} c=\left(r \frac{\partial u_{\theta}}{\partial r}\right)-\int\frac{u_{\theta}}{r} dr \end{aligned}

Now it's clear that in order to solve it I have to integrate it twice, i.e., consider $u_{\theta}$ not a function of $r$. But physics bites me from inside. Can someone provide me a logic to this? Another thing is why $u_{\theta} = r \omega$ i.e., ( $sin (\theta) = 1 $) not valid here?

$\endgroup$
9
  • $\begingroup$ Is $ u_{\theta}$ the theta component of velocity? $\endgroup$ – Buraian Oct 26 '20 at 18:03
  • $\begingroup$ yes, ingforum.haninge.kth.se/armin/FLUID/EXER/NAVIER_STOKES_EQ.pdf $\endgroup$ – Vishesh Mangla Oct 26 '20 at 18:04
  • $\begingroup$ I have a feeling that your diagram is inaccurate because usually it looks like a parabola cavity once it starts rotating $\endgroup$ – Buraian Oct 26 '20 at 18:06
  • $\begingroup$ I am not dismissing your problem but am suggesting edits for clarity's sake $\endgroup$ – Buraian Oct 26 '20 at 18:08
  • $\begingroup$ Talking about the diagram, well how will you know that the profile is parabolic without actually performing the expt?(I know how to prove it parabolic). $\endgroup$ – Vishesh Mangla Oct 26 '20 at 18:11
1
$\begingroup$

Using Navier-Stokes is taking a steamroller to crack a nut.

Your fluid is roting as a solid body so $v_\theta =r\omega$. You can find the shape of the surface by observing that in the fluid is stationary in the rotating frame, and in that frame the potential energy is $$ V(r,z)=\rho g z- \frac 12 \rho \omega^2 r^2, $$ a sum of the gravitational and centrifugal potentials. The surface must be an equipotential, so $$ z(r)= \frac 1{2g} \omega^2 r^2, $$ which is a parabola of revolution.

Your original problem askes you to use the pressure. Euler (divided by $\rho$) tells us that $$ \frac{\partial {\bf v}}{\partial t}+ \left({\bf v}\cdot \nabla \right) {\bf v}= - \frac 1 \rho \nabla P- {\bf g} $$ Now ${\bf v}=(-\omega y, \omega x, 0)$ so $$ \left({\bf v}\cdot \nabla \right) {\bf v}= -\frac 12 \omega^2 \nabla (x^2+y^2) $$ so $$ \nabla\left(\frac P\rho - \frac 12\omega^2 r^2 +gz\right)=0. $$ Thus again $P$ is constant on $$ z= \frac 1{2g}\omega^2 r^2. $$

$\endgroup$
7
  • $\begingroup$ thanks for the answer but if you see it's a homework problem. I have to follow the instructions. On the other hand in the description, I have written about why $v_{\theta} = r \omega$ is wrong. I know how to do it by Newton's ways. Also an answer which I do not require is here too physics.stackexchange.com/questions/88340/… $\endgroup$ – Vishesh Mangla Oct 26 '20 at 19:09
  • $\begingroup$ If $v$ were not $\rho \omega$ you are not rotating a s rigid body. Viscosity will come into play until it is rotating rigidly. $\endgroup$ – mike stone Oct 26 '20 at 19:11
  • $\begingroup$ that means that I have reduced the equation incorrectly. Thanks. Please also add this too your answer $\endgroup$ – Vishesh Mangla Oct 26 '20 at 19:12
  • $\begingroup$ I would not have answered in so detailed a manner if I known it was an hw problem. But anyway: what is the "$\sin \theta$ tha ask about? $\endgroup$ – mike stone Oct 26 '20 at 19:13
  • 1
    $\begingroup$ There is Coriolis in the rotating frame, but the fluid is not moving in that frame so there is only centrifugal. I wrote Euler in an inertial lab frame, so neither Coriolis nor centrigulal --- just Euler's equation with its convective derivative. It's the same physics, but differently descrubed in different frames. $\endgroup$ – mike stone Oct 26 '20 at 19:23

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.