In search of the perfect bubble

As discussed in a previous blog post, the frequency response of particles is dependent on the relaxation time:

\displaystyle t_0=\frac{D^2 \rho}{18\mu}

If we replaced the particle with a helium-filled (or air-filled) soap bubble, how big can we make the bubble to have the particle have the same relaxation time? Well, let’s for this purpose assume that \rho_{air}<<\rho. If that’s the case, then the effective density of the bubble is just a ratio of the shell and total volumes:

\displaystyle \rho_{eff}=\rho\frac{V_{shell}}{V_{total}}

If the bubble is a hollow shell with outer radius R and inner radius r:

\displaystyle \rho_{eff}=\rho\frac{R^3-r^3}{R^3}

Here I am seeking to find a bubble particle with the same relaxation time as the full-sphere droplet particle. Therefore, the relaxation times match:

\displaystyle t_{0,drop}=\frac{D_{drop}^2 \rho}{18\mu}=t_{0,bubble}=\frac{D_{bubble}^2 \rho_{eff}}{18\mu}

This means that the ratio between the diameters of the drop and bubble particles are:

\displaystyle \bigg(\frac{D_{bubble}}{D_{drop}}\bigg)^2 = \frac{\rho}{\rho_{eff}}=\frac{R^3}{R^3-r^3}

If we use the bubble shell thickness t, which is arguably a more useful variable:

\displaystyle \bigg(\frac{D_{bubble}}{D_{drop}}\bigg)^2 = \frac{R^3}{3R^2t-3Rt^2+t^3}

We can also express this as a function of the particle radius/thickness ratio:

\displaystyle t_r=\frac{R}{t}

Then:

\displaystyle \boxed{\bigg(\frac{D_{bubble}}{D_{drop}}\bigg)^2 = \frac{t_r^3}{3t_r^2-3t_r+1}}

I left this final formula as a ratio of the diameter of the bubble and drop squared, as the scattering cross-sections (and therefore scattering intensity) are a function of the scattering area instead of just the diameter.

Now let’s analyze the equation above. If one were to plot this function, it would look like this:

One thing that strikes immediately is the linearity, as well as the slope (1/3) at the latter part where the thickness ratio is large enough. Well, let’s physically interpret this for the Particle Image Velocimetry application: Say we have a bubble made of a water shell. I did some experiments at home and found that at ambient air conditions (20C, ~60% humidity) we get an evaporation rate of ~2-3mm/day; which is approximately ~23 to 34 nm/second. Of course, I don’t know if the evaporation rate scales down to the nanoscale that well, but it must be at least that rate otherwise I wouldn’t get the results I found by evaporating water on a dish. We also generally find that water evaporates pretty fast when we leave our wet kitchen counters for a little while; so I think this rate is not too far from reality. Further, given the spherical symmetry of the bubble and the rapid flow speeds we will look at, I can only expect the bubble to evaporate faster than what is measured in quiescent conditions.

Settling for 30nm/second of evaporation rate, if we make bubbles for use on a supersonic wind tunnel test we may afford maybe 5 to 10 seconds of lifetime between generation and the bubbles reaching the test section (by which time, we no longer need them so they can pop). This means that a bubble would have to be at least ~200-300nm thick so it can survive the run.

Let’s work with 200nm for now. If a bubble like that can be made on demand at scale with good repeatability, then the diameter ratio that such a bubble would have in comparison to a 1um oil particle would be about 1.2um (R/t=3), or a brightness ratio of (D_{bubble}/D_{drop})^2\sim 1.44.

That is not particularly great because we are producing bubbles such that we can have brighter particles, and 1.44 is not the kind of number we are after (we’re hoping for 1,000x brighter particles). Therefore, I have very little faith that bubbles can be a workable solution for tomographic PIV in high-speed flows (say, supersonic flows) that require particles to have very high frequency responses (of the order of 50kHz and more).

What do you think?

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