# Difference between revisions of "Floating and pendant drops"

## Liquid-liquid-air interfaces

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

$\frac{\sigma _{1}}{\sin \theta _{1}}=\frac{\sigma _{2}}{\sin \theta _{2}}=\frac{\sigma _{12}}{\sin \theta _{12}}$

## Small drops

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

$\vec{\sigma }_{A}+\vec{\sigma }_{B}+\vec{\sigma }_{AB}=\vec{0}$

Which gives the capillary lengths:

\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}

## Large drops

Gravity flattens the drop.

The “area” of the drop” is the volume divided by the thickness: $\frac{\Omega }{e}$

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

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

The balance of hydrostatic pressures gives: $\rho _{A}ge=\rho _{B}g{e}'$

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

with $\tilde{\rho }=\frac{\rho _{A}}{\rho _{B}}\left( \rho _{B}-\rho _{A} \right)$

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

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 $C$ to be curvature of the surface of the drop, $\sigma$ surface tension, and $\rho$ density of liquid, we obtain the following:

$\sigma C=\rho g z\,\!$

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

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

where $r_z=\frac{dr}{dz}$ and $r_{zz}=\frac{d^2 r}{dz^2}$

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.

Another way you can determine the value of $\sigma$ 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 $2\pi R \sigma$ where $R$ is the inner radius of the capillary. By measuring the weight of the fallen drop, one can theoretically compute the value of $\sigma$. 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:

$\frac{4}{3}R_g^3 \rho \pi g= 2 \pi \sigma R$

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

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

In practice, we often have that only certain percentage $\alpha$ 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:

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