Drops, menisci, and lenses

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Revision as of 23:36, 27 September 2008 by Morrison (Talk | contribs) (Menisci shapes)

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Thickness of a large drop

Small drops are spherical segments. Large drops are flattened.

de Gennes, 2004, Fig. 2.4

To “spread” the drop requires an force per unit length:<math>\sigma _{sv}-\left( \sigma _{lv}+\sigma _{sl} \right)</math>

The hydrostatic pressure integrated over the depth of the drop is a force per unit length pushing to “spread” the drop: <math>\tilde{P}=\int\limits_{0}^{e}{\rho g\left( e-\tilde{z} \right)d\tilde{z}}=\frac{1}{2}\rho ge^{2}</math>

At equilibrium the sum of the two is zero: <math>\sigma _{sv}-\left( \sigma _{lv}+\sigma _{sl} \right)+\frac{1}{2}\rho ge^{2}=0</math>

Substituting the Young-Dupré equation: <math>\sigma _{lv}\left( 1-\cos \theta _{e} \right)=\frac{1}{2}\rho ge^{2}</math>

Re-arranging gives: <math>\text{ }e=2\kappa ^{-1}\sin \left( \frac{\theta _{e}}{2} \right)</math>

Profile of a large drop

The general ideas used to calculate the thickness are used again to calculate the profile. With a couple of modifications.

de Gennes, 2004, Fig. 2.4

The limits on the integration are changed: <math>\tilde{P}=\int\limits_{0}^{z}{\rho g\left( e-\tilde{z} \right)d\tilde{z}}=\rho g\left( ez-\frac{z^{2}}{2} \right)</math>

The “spreading” force per unit length is now: Where q<math>\theta </math> is the angle marked in the diagram: <math>\sigma _{sv}-\left( \sigma _{lv}\cos \theta +\sigma _{sl} \right)</math>

“It can be shown” from the diagram that: <math>\cos \theta =\sqrt{1+\dot{z}^{2}}</math>

This results in a differential equation for the shape consistent to the algebraic equation for the drop thickness: <math>\sigma _{lv}\left( \sqrt{1+\dot{z}^{2}}-\cos \theta _{e} \right)=\frac{1}{2}\rho g\left( 2ez-z^{2} \right)</math>

Menisci shapes

The ascending meniscus against a vertical wall (de Gennes, 2004, pp. 45f). The Laplace equation, shown on the diagram, is a differential equation that describes the shape of the meniscus. The curvature increases with height.

Pressure in a vertical meniscus.png

<math>\frac{1}{R\left( z \right)}=-\frac{{\ddot{z}}}{\left[ 1+\dot{z}^{2} \right]^{3/2}}</math>

Substituting the curvature into the Laplace equation and integrating twice gives: <math>x-x_{0}=\kappa ^{-1}\cosh ^{-1}\left( \frac{2\kappa ^{-1}}{z} \right)-2\kappa ^{-1}\left( 1-\frac{z^{2}}{4\kappa ^{-2}} \right)^{1/2}</math>

(Where x0 makes z = h at x = 0)

A correct, but not illuminating, result.