A problem from this years phy1530 problem set 2 that appears appropriate for phy454 exam prep.
Consider the steady flow between two long cylinders of radii and , , rotating about their axes with angular velocities , . Look for a solution of the form, where is a unit vector along the azimuthal direction:
- Write out the Navier-Stokes equations and find differential equations for and . You should find that these equations have relatively simple solutions, i.e.,
- Fix gthe constants and from the boundary conditions. Determine the pressure .
- Compute the friction forces that the fluid exerts on the cylinders, and compute the torque on each cylinder. Show that the total torque on the fluid is zero (as must be the case).
This is also a problem that I recall was outlined in section 2 from . Some of the instabilities that are mentioned in the text are nicely illustrated in .
We illustrate our system in figure (1).
Solution: Part 1. Navier-Stokes and resulting differential equations.
Navier-Stokes for steady state incompressible flow has the form
where the gradient has the form
Let’s first verify that the incompressible condition 3.3b is satisfied for the presumed form of the solution we seek. We have
Good. Now let’s write out the terms of the momentum conservation equation 3.3a. We’ve got
Equating and components we have two equations to solve
Expanding out our velocity equation we have
for which we’ve been told to expect that 2.2 is a solution (and it has the two integration constants we require for a solution to a homogeneous equation of this form). Let’s verify that we’ve computed the correct differential equation for the problem by trying this solution
Given the velocity, we can now determine the pressure up to a constant
Solution: Part 2. Fixing the constants.
To determine our integration constants we recall that velocity associated with a radial position in cylindrical coordinates takes the form
where is the angular velocity. The cylinder walls therefore have the velocity
so our boundary conditions (given a no-slip assumption for the fluids) are
This gives us a pair of equations to solve for and
Multipling each by and respectively gives us
Rearranging for we find
For we have
This gives us
Solution: Part 3. Friction and torque.
We can expand out the identity for the traction vector
in cylindrical coordinates and find
so we have
We want to expand the last of these
So the traction vector, our force per unit area on the fluid at the inner surface (where the normal is ), is
and our torque per unit area from the inner cylinder on the fluid is thus
Observing that our stress tensors flip sign for an inwards normal, our torque per unit area from the outer cylinder is
For the complete torque on the fluid due to a strip of width the magnitudes of the total torque from each cylinder are respectively
As expected these torques on the fluids sum to zero
Evaluating these at and respectively gives us the torques on the fluid by the cylinders, so inverting these provides the torques on the cylinders by the fluid
For the outer cylinder the total torque on a strip of width is
 D.J. Acheson. Elementary fluid dynamics. Oxford University Press, USA, 1990.
 Wikipedia. Taylor-couette flow — wikipedia, the free encyclopedia [online]. 2012. [Online; accessed 10-April-2012]. http://en.wikipedia.org/w/index.php?title=Taylor\%E2\%80\%93Couette_flow&oldid=483281707.