5
$\begingroup$

I came across this video of Taylor-Couette Flow on YouTube. Originally I was looking for a visualization of the wavy Taylor vortices induced by the angular motion of the inner cylinder.

However, I found something strange (as you can see at the end of the video), the experimenter at the beginning injects three different dyes in a viscous liquid:

enter image description here

And then he begins to stir slowly in a specific direction until the three dyes are fully diffused into the main liquid:

enter image description here

Finally, he stopped the stirring and began to stir in the opposite direction returning the three dyes to its original spots (well sort of!):

enter image description here

So, how is this reversibility even possible? shouldn't the diffusion of dyes into the liquid be irreversible?

$\endgroup$
4
  • 3
    $\begingroup$ The experiment doesn't reverse the diffusion but only a pretty laminar shear flow. It looks impressive, but it's just a physical parlor trick. $\endgroup$
    – CuriousOne
    Sep 12, 2015 at 6:36
  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/71281/2451 $\endgroup$
    – Qmechanic
    Sep 12, 2015 at 11:54
  • $\begingroup$ The three dyes are not fully diffused into the liquid. $\endgroup$ Sep 12, 2015 at 14:25
  • $\begingroup$ My answer here identifying the experiment also contains links to lecture notes and additional videos explaining it. $\endgroup$
    – tpg2114
    Sep 12, 2015 at 15:30

3 Answers 3

9
$\begingroup$

This is what's happening in the video. I've drawn just a single drop, and for convenience I've ignored the curvature of the plates (it's harder to draw curves!):

enter image description here

It looks as if the (red in this example) ink drop is being mixed with the fluid, but actually it's just being stretched out into a thin sheet. When you turn the cylinder back again the sheet is pushed back into a drop.

$\endgroup$
1
  • $\begingroup$ This is a great explanation to understand intuitively. $\endgroup$
    – Etheryte
    Sep 12, 2015 at 17:36
6
$\begingroup$

Two typical quantities which characterize these systems are the Reynolds and Schmidt number: $$\mathrm{Re}=\frac{vL}{\nu} \qquad \mathrm{Sc}=\frac{\nu}{\mathcal{D}}$$ where $v$ and $L$ are characteristic velocity and length scales and $\nu$ and $\mathcal{D}$ are the kinematic viscosity and the diffusion coefficient (both material properties).

These characterize the system because each describes a certain regime:

  • The Reynolds number is the relative importance of convective to viscous transport; for $\mathrm{Re}\ll1$ viscosity dominates, for $\mathrm{Re}\gg1$ inertia dominates.
  • The Schmidt number is the relative importance of momentum diffusion to mass diffusion; for $\mathrm{Sc}\ll1$ mass diffusion dominates, for $\mathrm{Sc}\gg1$ momentum diffusion dominates.

Combined we can say something about the relative important of mass diffusion to convective and viscous transport. It is also typical to define a Peclet number (which is simply the product of the Reynolds and Schmidt number) which then describes the relative importance of convective transport to mass diffusion: $$\mathrm{Pe}=\mathrm{Re}\mathrm{Sc}$$

Now in the video there are some keywords which can help us estimate these value; high viscosity liquid (i.e. corn syrup) and that the system is turned slowly. This indicated that we are dealing with a very low Reynolds number (i.e. $\mathrm{Re}\ll1$) system where viscosity dominates (also known as the laminar regime).

Estimating the Schmidt number requires knowing the kinematic viscosity and diffusion coefficient. The viscosity of corn syrup is known to be around $10^{-3}$ $\left[m^2/s\right]$ (about $10^3$ times greater than water). Unfortunately, knowing the diffusion coefficient of corn syrup in corn syrup is more difficult to obtain; so let's assume a typical value of $10^{-9}\left[m^2/s\right]$ for liquids. We then estimate $\mathrm{Sc}\sim 10^6$ which indicates that momentum diffusion is much more important than mass diffusion in this system. As for the Peclet number, with the calculated values this is most likely $\mathrm{Pe}\gt1$ which indicates that convective transport is at least as important as mass diffusion but as convective transport was already less important than viscous transport this doesn't give us any new information.

Concluding this analysis, we can say that the reason why the system returns to its original state is simply because we are dealing with a laminar system in which the time scales involved are not large enough for mass diffusion to be of any importance and smear out the dyes.

$\endgroup$
2
  • $\begingroup$ I like your analysis in terms of the dimensionless numbers. You mention the Péclet number, but you do not go back to it in your analysis of the liquids. $\endgroup$
    – Bernhard
    Sep 12, 2015 at 10:32
  • 1
    $\begingroup$ @Bernhard - It was just a side note in case people are more used to using Peclet in this case. I have added a small statement. Also thanks for noticing the wrong units on the viscosity :) woops $\endgroup$
    – nluigi
    Sep 12, 2015 at 10:46
1
$\begingroup$

It's clear that the liquid in the Couette cell must have a low Reynolds number because otherwise, the injected dye would mix easily with the liquid when the rotation of the cell takes place, causing a non-reversible behavior. This is made very clear in this old but very informative video. As is explained in the video the process is not completely reversible because of diffusion of dye molecules into the fluid, so you end up with a form of the dye that's a bit more diffuse as the form of the dye you started with, but the forms do indeed don't differ much after rotating the cell a few times in one direction and then a few times in the opposite direction. Which makes one wonder how the final form of the dye with the number of rotations. I guess the final form of the dye (after the two rotations) continuously [in response to a continuously increasing (real) number of rotations] transforms into a final form that doesn't resemble the initial form anymore, meaning that non-reversibility will slowly emerge.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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