# Difference between revisions of "Precursors to splashing of liquid droplets on a solid surface"

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<math>12 \mu (\rho h)_l = (\rho h^3 p_x)_x</math> | <math>12 \mu (\rho h)_l = (\rho h^3 p_x)_x</math> | ||

− | Here <math>\mu</math> is the gas viscosity, <math>\rho_l</math> is the liquid density and <math>\rho_x</math> is the gas density. Accordingly, <math>p_x</math> is the gas pressure and <math>p_l</math> the liquid pressure. | + | Here <math>\mu</math> is the gas viscosity, <math>\rho_l</math> is the liquid density and <math>\rho_x</math> is the gas density. Accordingly, <math>p_x</math> is the gas pressure and <math>p_l</math> the liquid pressure. Another equation relating the two pressures is: |

+ | |||

+ | <math>p_{ll} h_{ll} = \mathcal{H} [p_x + \sigma h_{xxx}]</math> | ||

+ | |||

+ | Where <math>\mathcal{H}</math> is a Hilbert transform. When the drop reaches a critical distance <math>H^*</math> from the wall, gas pressure rises under it and dominates surface tension and inertia.At that value, the pressure causes a dimple on the drop. Subsequently, the pressure develops two maxima as the interfacial curvature steepens rapidly. This is demonstrated in figures 1 and 2. | ||

+ | |||

+ | The authors set two parameters. One is the Stokes number <math>St = \frac{\mu}{\rho_l V R}</math>, which relates to the critical distance as <math>H^* = R St^{2/3}</math>. | ||

+ | |||

+ | The other parameter is <math>\epsilon = \frac{P_0}{(R\mu^{-1}V^7\rho_l^4)^{1/3}}</math>, which is obtained by setting equal the gas pressure gradient with the liquid deceleration. At large <math>\epsilon </math> the film thickness obeys the incompressible scaling <math>H \sim RSt^{2/3}</math>, while at small <math>\epsilon</math> compressible effects set in. | ||

+ | |||

+ | With this in mind, authors solve the set of equation using dominant balance arguments at the compressible and incompressible limit. |

## Revision as of 11:42, 18 May 2009

## Overview

**Authors:** Shreyas Mandre, Madhav Mani & Michael P. Brenner

**Source:** Physical Review Letters, Vol.102, 134502, (2009)

**Soft Matter key words:** droplets, splashing, capillary waves, surface tension, pressure, thin film

## Abstract

In this publication authors develop a theoretical model for a droplet splashing against a solid wall, which they confirm by running computer simulations. Contrary to popular belief, they stipulate that high pressure of the air film trapped between the wall and the liquid drop actually prevents the drop from contacting the wall. Instead, the droplet spreads on the thin air film and emits capillary waves.

## Soft Matter Snippet

It is interesting to take a closer look at the set of equations chosen to describe this fluid dynamics problem. The gas films deforms according to the differential equation:

<math>12 \mu (\rho h)_l = (\rho h^3 p_x)_x</math>

Here <math>\mu</math> is the gas viscosity, <math>\rho_l</math> is the liquid density and <math>\rho_x</math> is the gas density. Accordingly, <math>p_x</math> is the gas pressure and <math>p_l</math> the liquid pressure. Another equation relating the two pressures is:

<math>p_{ll} h_{ll} = \mathcal{H} [p_x + \sigma h_{xxx}]</math>

Where <math>\mathcal{H}</math> is a Hilbert transform. When the drop reaches a critical distance <math>H^*</math> from the wall, gas pressure rises under it and dominates surface tension and inertia.At that value, the pressure causes a dimple on the drop. Subsequently, the pressure develops two maxima as the interfacial curvature steepens rapidly. This is demonstrated in figures 1 and 2.

The authors set two parameters. One is the Stokes number <math>St = \frac{\mu}{\rho_l V R}</math>, which relates to the critical distance as <math>H^* = R St^{2/3}</math>.

The other parameter is <math>\epsilon = \frac{P_0}{(R\mu^{-1}V^7\rho_l^4)^{1/3}}</math>, which is obtained by setting equal the gas pressure gradient with the liquid deceleration. At large <math>\epsilon </math> the film thickness obeys the incompressible scaling <math>H \sim RSt^{2/3}</math>, while at small <math>\epsilon</math> compressible effects set in.

With this in mind, authors solve the set of equation using dominant balance arguments at the compressible and incompressible limit.