# Spin-up time for fluid in an open cylinder with no endwalls?

I'm interested in how quickly (if at all) the fluid in an open cylinder would reach solid-body rotation if the cylinder is suspended vertically in an unbounded fluid.

I've found plenty of work that considers spin-up in a cylinder with a single end-wall, but can't seem to track down any information on whether the end-wall is essential for driving the flow. Even a pointer to the correct keywords to search for would help.

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Is it supposed to be simple laminar flow? If yes, you can just consider a single-end-wall cylinder of infinite length. – leftaroundabout Aug 17 '11 at 1:36
My ultimate aim is to look at turbulent situations, but any initial steps via laminar flow are useful. I'm probably being a bit dense, but does the infinite length case imply that the end wall does impact, but is just negligible compared to the surface area of the side wall? Thanks – R_usr Aug 17 '11 at 14:48
Oh, presumably the unbounded external fluid is also considered irrelevant? i.e. assume there should be no leakage into or out of the cylinder? – R_usr Aug 17 '11 at 14:49
Im not sure if I fully understand the setup ... But I think stationary solid body rotation should be attained after all of the turbulent kinetic energy is dissipated by the molecular viscosity of the medium. – Dilaton Aug 17 '11 at 19:36
I suppose, viscous fluid flow' is the key word. For laminar flow, back-of-the envelope estimate of the spin-up time is t=A*R^2/nu, where R is the radius of the cylinder (or the small radius of the torus), nu is the kinematic viscosity [cm^2 s^-1], A is a coefficient of order 1. To calculate A, one needs to solve the Navier-Stokes equations. Gravity is unimportant, and the end-wall matters only if the cylinder is short. Without the end-wall, the problem is actually easier, because it is 1-D. If this sounds like what you were looking for, I can expand. – drlemon Aug 18 '11 at 20:57

This problem can get as complicated as you like, but let me show you how it can be approached in the most simple case. The Navier-Stokes equations describe the flow:

$\rho\left(\displaystyle\frac{\partial {\bf u}}{\partial t}+({\bf u}\cdot\nabla){\bf u}\right)=-\nabla p + \mu \nabla^2 {\bf u}$. $\qquad$ (1)

Assume that the motion of the fluid is slow enough so that the non-linear term $({\bf u}\cdot \nabla){\bf u}$ is negligible. This will be true as long as viscosity forces are large compared to inertial forces from acceleration. Ugh, now that it is a linear equation, something can be done analytically.

Consider the problem in cylindrical coordinates with $z$, $r$ and $\phi$. Let us assume $\partial/\partial z=0$ and $\partial/\partial \phi=0$ for all quantities due to the axial symmetry of the problem. Then the continuity equation will offer additional simplifications. In cylindrical coordinates, the continuity equation reads

$\displaystyle\frac{\partial \rho}{\partial t}+ \frac{1}{r}\displaystyle\frac{\partial(\rho r u_r)}{\partial r}+ \frac{1}{r}\frac{\partial\rho u_{\phi}}{\partial \phi} + \frac{\partial (\rho u_z)}{\partial z}=0$

With $\rho=\mathrm{const}$ (incompressibility assumption), the first term is zero, and last two terms are equal to zero due to the axial symmetry assumption. The remainder yields $u_r=0$. Hooray! Let us choose the problem so that $u_z=0$, too (true if the cylinder is not moving with respect to the fluid in $z$-direction) Then we are left with just $u_{\phi}$, and let us denote it $u\equiv u_{\phi}$.

If the cylinder is suspended vertically, i.e., the axis of the cylinder is along the gravity force, then the pressure gradient $\nabla p$ can also be neglected. This is because the only component of $\nabla p$ is in z-direction due to gravity, and we already assumed $\partial {\bf u}/\partial z$=0. And there may be no $\phi$-component of $\nabla p$, because it would violate our $\partial /{\partial \phi}=0$ assumption.

All of the above leaves us with the only equation for $u_\phi\equiv u$:

$\displaystyle\frac{\partial u}{\partial t} = \mu \frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial u}{\partial r}\right)$. $\qquad$ (2)

To understand where it came from, look up the explicit form of equation (1) in cylindrical coordinates at this Wikipedia page, take the $\phi$-component of it and use the assumptions outlined below to get rid of $\partial/\partial z$ and $\partial/\partial \phi$ terms, and of the non-linear and pressure terms.

The cylinder is spinning, which makes the boundary conditions for equation (2)

$u(r=0, t)=0$ and $u(r=R, t)=R\omega_0$,

where $R$ is the radius and $\omega_0$ is the angular frequency of the cylinder. To understand the second one, look at the well-known solution of a similar problem at this Wikipedia page or this paper referenced in the Wiki page. The initial conditions are

$u(r, t=0)=0$.

Equation (2) is a very well-known equation. It is the diffusion or the heat equation in disguise. Well, no surprise about it, because viscosity in is nothing more than diffusion of momentum. At least, this is how it is expressed in the viscosity term of equation (1). The solution is readily obtained using separation of variables and Fourier analysis. I will not solve it here, because it is a well understood mathematical problem, with which graduate students are routinely tortured with great success. This Wikipedia page describes it.

And the back-of-the envelope estimate I mentioned in my comment comes from dimensional analysis of equation (2). If the typical velocity scale of the problem is $U$, spatial scale is $R$ and time scale is $T$, then (2) can be approximated as

$\displaystyle\frac{U}{T} \approx \mu \frac{1}{R} \frac{1}{R} R \frac{U}{R}$,

or

$T \approx R^2/\mu$,

which is a crude estimate of your spin-up time.

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For your rotating torus, this may not be much help, because you may have to consider the pressure gradient. But this paper might shed some light pre.aps.org/abstract/PRE/v60/i5/p6192_1 They do some interesting trick, integrating with respect to z. – drlemon Aug 25 '11 at 20:35
Thanks for a thought-provoking, humorous and enjoyable answer. It's going to take some time for me to fully digest, but I now appear to have something to get me started. – R_usr Aug 25 '11 at 22:07
You are welcome. – drlemon Aug 26 '11 at 1:52