Floating and pendant drops

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Floating due to rigid surface tension

The surface tension of a fluid is a true tension that shows up in different phenomena. Water bugs, for example, can float supported by the surface tension. When a paper clip or a bug is floating, the water surface is distorted. Since the water surface acts as a tense membrane, each portion of the perimeter pulls the clip upwards and prevents if from sinking. Therefore what determines whether an object can be supported by surface tension is the perimeter of the distorted surface. This is why the surface tension is measured in force per unit length. In particular, surface tension of water = 0.0004 pounds/inch. This looks like a small number but is enough to support some objects. For a paper clip, the perimeter is around a couple of inches, which means that the water is capable of supporting a weight of about 0.0008 pounds, which is roughly the weight of a paper clip. If soap is to be added, the surface tension decreases three times, and is now not enough to support the weight of the clip and it sinks. Caution: Do not confuse the floating due to surface tension with the floating due to buoyancy. Ships and wood float because they are less dense than water. A piece of wood will float regardless of its shape. On the other hand, if we change the shape of a paper clip and make it spherical, the perimeter of the deformed surface will be smaller and the surface tension will not be enough to keep it floating.

Liquid-liquid-air interfaces

Neumann’s construction for a liquid drop (2) on a liquid substrate (1):

Neumann's Triangle, Morrison, 2002, Fig. 6.6


<math>\frac{\sigma _{1}}{\sin \theta _{1}}=\frac{\sigma _{2}}{\sin \theta _{2}}=\frac{\sigma _{12}}{\sin \theta _{12}}</math>





Small drops

When gravity is negligible (small drops) the drop is spherical and Neumann’s vector relation is satisfied:

<math>\vec{\sigma }_{A}+\vec{\sigma }_{B}+\vec{\sigma }_{AB}=\vec{0}</math>


de Gennes, 2004, Fig. 2.19a

Which gives the capillary lengths:

<math>\begin{align}
 & \kappa _{A}^{-1}=\sqrt{\frac{\sigma _{A}}{\rho _{A}g}} \\ 
& \kappa _{B}^{-1}=\sqrt{\frac{\sigma _{B}}{\rho _{B}g}} \\ 
& \kappa _{AB}^{-1}=\sqrt{\frac{\sigma _{AB}}{\left( \rho _{A}-\rho _{B} \right)g}} \\ 

\end{align}</math>







Large drops

Gravity flattens the drop.

de Gennes, 2004, Fig. 2.19b

The “area” of the drop” is the volume divided by the thickness: <math>\frac{\Omega }{e}</math>

The surface energy is: <math>\left( \sigma _{B}-\sigma _{A}-\sigma _{AB} \right)\cdot \frac{\Omega }{\varepsilon }=S\cdot \frac{\Omega }{\varepsilon }</math>

The total energy also depends on thickness, e: <math>F\left( e \right)=\left[ \frac{1}{2}\rho _{A}g\left( e-{e}' \right)^{2}+\frac{1}{2}\left( \rho _{B}-\rho _{A} \right)g{e}'^{2}-S \right]\frac{\Omega }{e}</math>

The balance of hydrostatic pressures gives: <math>\rho _{A}ge=\rho _{B}g{e}'</math>

Which gives: <math>F\left( e \right)=\left[ \frac{1}{2}\tilde{\rho }ge^{2}-S \right]\frac{\Omega }{e}</math>

with <math>\tilde{\rho }=\frac{\rho _{A}}{\rho _{B}}\left( \rho _{B}-\rho _{A} \right)</math>

Minimizing F(e) at constant volume, Ω, gives: <math>\frac{1}{2}\tilde{\rho }ge_{c}^{2}=-S</math>

Which looks like the solution for the sessile drop except for the density. (Equation 2.8 in de Gennes)





Pendant drops

A drop adopts its shape based on two effects: surface tension that strives towards minimizing energy in a spherical form, and force of gravity that distorts that spherical shape. There are several methods currently used to measure surface tension: The Pendant Drop Method and the more commonly used Sessile Drop Method. We’ll discuss the former.


The main idea of the pendant drop method is that we let a drop dangle from the end of a capillary tube, taking the shape shown in the figure below (Fig. 2.20). Pressures balancing out in the drop are Laplacian and hydrostatic. Taking <math>C</math> to be curvature of the surface of the drop, <math>\sigma</math> surface tension, and <math>\rho</math> density of liquid, we obtain the following:

<math>\sigma C=\rho g z\,\!</math>

de Gennes, 2004, Fig. 2.20

We can express curvature using cylindrical coordinate system in the following manner:

<math>C=-\frac{r_{zz}}{(1+r_z^2)^{3/2}}+\frac{1}{r(1+{r_z}^2)^{1/2}}</math>

where <math>r_z=\frac{dr}{dz}</math> and <math>r_{zz}=\frac{d^2 r}{dz^2}</math>

Using the two aforementioned formulas, we can solve the problem numerically by treating surface tension as an adjustable parameter, and slightly changing its value until our results agree with experiments.

Similar, but slightly different equation can be acquired from considering the difference in density between fluids at interface:

<math>\sigma = Dr g Ro^2/b </math>

where <math>\sigma</math>=surface tension Dr = difference in density between fluids at interface g = gravitational constant Ro = radius of drop curvature at apex b = shape factor

b, the shape factor can be defined through the Young-Laplace equation expressed as 3 dimensionless first order equations as shown in the figure below.

Example.jpg [1]

Another way you can determine the value of <math>\sigma</math> is an experiment many of us have done in their early physics classes. A pendant drop on the capillary can break loose when the force of gravity exceeds capillary force of <math>2\pi R \sigma</math> where <math>R</math> is the inner radius of the capillary. By measuring the weight of the fallen drop, one can theoretically compute the value of <math>\sigma</math>. As straightforward as it seems, the experiment runs into numerous complications due to the nature of drop separation. Still, remarkably, it gives us well calibrated drops whose radius can be calculated in the following manner:

<math>\frac{4}{3}R_g^3 \rho \pi g= 2 \pi \sigma R</math>

<math>R_g=(\frac{3}{2} \frac{\sigma R}{\rho g} )^{1/3}</math>

<math>R_g=(\frac{3}{2} \kappa^{-2} R)^{1/3}</math>

In practice, we often have that only certain percentage <math>\alpha</math> of the drop falls of. This value can be looked up and is usually around 60% depending on the type of liquid. Taking this into account, we can write:

<math>R_g=(\frac{3}{2 \alpha} \kappa^{-2} R)^{1/3}</math>

de Gennes, 2004, Fig. 2.21

Throbbing Oil Drop Phenomena

Last year a phenomena that has been witnessed for centuries was solved at MIT by Professors Roman Stocker and John Bush. The basic idea is that if you mix a surfacant in oil and put a drop of it on a surface of water the oil drop will expand and contract due to evaporation induced changes in surface tension.

"Think of the oil detergent drop as a small lens with a rounded bottom. The surfactant in the drop moves to the bottom surface of the lens, where it interacts with the water to decrease the surface tension where oil meets water. This change in tension increases the forces pulling on the outer edges of the drop, causing the drop to expand.

The center of the drop is deeper than the edges, so more surfactant settles there, reducing the surface tension correspondingly. This causes the oil and surfactant near the outer edges of the drop to circulate. This circulation creates a shear (think of it as two velocities going in opposite directions), which generates very tiny waves rolling outward toward the edge. When these waves reach the edge, they cause small droplets to erupt and escape onto the water surface outside the drop. Videomicroscopy - essentially, attaching a video camera to a microscope - was critical in observing this step in the process. Those droplets of oil and surfactant disperse on the water and decrease the surface tension of the water surface, so the drop contracts.

As the surfactant evaporates, the surface tension of the water increases again, and the system is reset. Forces pull at the outer edges of the lens, and the cyclical process begins again."

A video of the phenomena is available on this link as well as the rest of the release article. Their work was published in the July 25, 2007 Journal of Fluid Mechanics. [2]

Any idea why the interface ruptures in the manner it does and sprays tiny oil droplets into the water phase? if the surfacant is evaporating it would have to be on the surface and when it evaporates that part of the surface would shrink back. If a surfactant molecule evaportated it would have that large of an effect on the surface of the drop?

Can anyone think of ideas for how this motion could be useful? Some kinda of micropump that stops pumping when it runs out of surfactant or is controled by its need for air?

Amazing Dynamic Surface Effects

Drops on surfaces can have some very unusual behavior when subject to non-equilibrium forces. Here are two very interesting and fun experiments conducted and published in recent years:

1) The Leidenfrost effect [3] applied to super-heated non-uniform surfaces can actually cause water droplets to go against gravity when the surface is tilted. Results of such experiments were published in 2006. [4] This shows how interactions between heat transfer, liquid-vapor transition and surface geometry can lead to a very cool effect, where a droplet never comes into contact with the surface it's "sitting on" and keeps moving in a counter-intuitive direction (against gravity).

2) In an even more counter-intuitive fashion, a droplet of liquid will not coalesce with a bath of the same liquid when falling on it, as long as the bath is accelerated in the right direction and at the right amplitude at the expected moment of contact. This leads to very interesting examples of what people call "bouncing droplets" or even "dancing droplets". The full story on this phenomenon can be read in Couder et al.'s 2005 work in PRL [5], and shows how the finite time it takes for a thin film of air under a drop to leave creates forces large enough to lift off the drop from the surface of a liquid bath. If you can find the movies of these bouncing and orbiting droplets, please post the links.


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