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I'm trying to develop a very basic scaling law/unit analysis for viscous droplet formation, and I'd like to get some rough numerical values of the Reynolds number to play with. To be specific, I'm looking at the behavior of the smaller of the two droplets shown in the picture below (experimental setup shown sideways):

Experiment setup (sideways)

The darker fluid is a glycerol/water mixture and the lighter fluid is mineral oil - so it's a viscous fluid dropping into another (dissimilar) viscous fluid. I'm trying to understand how the viscosity of the dark fluid effects the size of the small droplet - the large droplet pretty much remains the same size, but the small one gets smaller for lower viscosities.

Since I'm trying to work only with dimensionless parameters, I'd like to work with something like Reynolds number instead of viscosity. I know that $\text{Re}=\frac{\rho \mathbf{v} L}{\mu}$ is the "standard" formula, but I want to make sure this applies here, and if so, that I'm using the correct values for the parameters. I can calculate $\bf{v}$ from my high speed video, and I can calculate $\rho$ and $\mu$ for both fluids using a formula, but I'm wondering:

  • Do I need to use a ratio of the two densities and viscosities, or do I just use one? Would I use the viscosity of the stationary fluid or the moving fluid?
  • Does the width of the channel make sense for the characteristic length $L$, or should I go with something more related to the droplet?

Feel free to suggest a book or online resource if this doesn't have a simple answer. I have to admit I'm very hazy on the "physicist's" viewpoint on fluids (I'm a math grad student). Thanks in advance!

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My recommendation is that, rather than shopping for ready made dimensionless numbers, you list all the variables you think affect the outcome, and see how many dimensionless numbers you can build out of them. If there is a velocity, a viscosity, a density and a length, the Reynolds number will show up naturally, but there is a chance your problem will be better represented by another number. See – Jaime Dec 4 '12 at 17:08
I would also be surprised if surface tension doesn't play a role in this, so if you still want to shop for dimensionless numbers, make sure to check these out:,, and – Jaime Dec 4 '12 at 17:13
Is there a reasonable way to calculate surface tension here (either in theory or empirically)? Something with contact angles maybe? – icurays1 Dec 4 '12 at 18:01
Yes there is, and you already have a suitable test set up: or$file/… – Jaime Dec 4 '12 at 18:09
@Jaime Marangoni number is certainly not necessary here, as surface tension will be constant. However, the others may be relevant, but most of them are somehow related. – Bernhard Dec 7 '12 at 11:59

3 Answers 3

I'm not sure I fully understand your problem, as I only see the one droplet. However, I agree 100% with Jaime. You will undoubtedly form a Reynolds number from applying Buckingham's Pi Theorem, but that doesn't necessarily mean it dominates the physics of your problem.

I will add, however, that if you have an exact match of all your criteria of similarity, then the choice of length scale is irrelevant, since every length will have the same scale factor.

Your droplet looks pretty spherical, which implies constant surface tension (and low Re).

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I'm not sure that the roundness of the droplets implies only constant surface tension. I think rather that it implies that surface tension dominates over inertial or buoyancy forces (cf. Weber and Bond numbers) – Dai Apr 25 at 12:02

What is your objective in calculating a characteristic Reynolds number of the droplet formation? If you believe you can rule out inertia from this (which may be, droplet pinching happens in many regimes, including visous forces/capillary forces balance), then pick the characteristic velocity, density, etc., so that the result will be largest: if this upper estimate is still small ($<10^{-1}$ say), you may approach the problem as a Stokes problem.

Other nondimensional groups have been defined and are useful for droplet pinching, e.g. Weber number and Ohnesorge number.

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Just seen the question is 2012... although other answer is 2 month old... Well, will leave the answer anyhow... – Joce Apr 3 '14 at 20:31

Calculating Reynolds number for a droplet is not reasonable.

It should be noted that Reynolds number provides only some distant correlation between Turbulence, viscosity and flow velocity. There ie. is no such Reynolds Number which absolutely definates the flow to "Turbulent". It should be noted that the Laminar flow conditions can be hold up to Re > 150 000, and there actually isn't any upper limit for Laminar flow. (ie. Ven Te Chow, Open Channel Hydraulics) There is a good old video about the issue here; They say there the same; over Re> 100 000 laminar flow is possible. (~8 min 25 s) So there is no Causality between velocity and Turbulence. It's just a correlation which implies only to typical cases. And thus Reynolds number is actually quite meaningless. Look this video, and note at 8 min, how small details, like adding a funnel, is more important than the plain velocity.

Reynolds number is usable there where it's correlations to reality are well studied and known; like pipeflows.

If we go back to your original question; "Very basic Scaling law", I propose you simple use Froude scaling law. If you are ought to get something usable from the lab scale to real scale. Ie. the book from Hubert Chanson "Hydraulics of Open Channel flow" provides full information about this.

The error caused by wrong Reynolds number in Turbine lab tests are ie. typically only few %-units. And the error is never negative; ie lab provides 92% efficiency and if Reynolds scaling is taken account it could provide 93.5 % in real scale.

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Further, if you have a droplet of fluid in a fluid, there is surely a dividing surface between these fluids having some kind of own surface tension. And all internal velocities inside the droplet might be zero, when compared to this surface. Analogy could be found from a ballbearing; what would be the reynolds number of a single ball of a ballbearing? I think it wont make any difference if it's installed in a wheel of ferrari going 300 km/h or if it's lying in a store. There is no internal deformation and the viscous forces inside are rather meaningless. – JokelaTurbine Jun 23 at 9:43

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