How to simulate surface tension? - Chemistry Stack Exchange most recent 30 from chemistry.stackexchange.com 2019-09-21T04:43:27Z https://chemistry.stackexchange.com/feeds/question/32044 https://creativecommons.org/licenses/by-sa/4.0/rdf https://chemistry.stackexchange.com/q/32044 11 How to simulate surface tension? RaaziR https://chemistry.stackexchange.com/users/16549 2015-05-27T01:48:53Z 2016-06-14T22:01:34Z <p>I am trying to create a water drop simulation for measuring hydrophobicity of surface. I don't know how to simulate the contact angle which is related by younges equation to the surface tensions for each of the solid-liquid-gas interfaces. I have come across some models for molecular dynamics, which for chemistry novice like me (I'm a EE) are bit advanced. I am looking for simpler method in which I could simulate the interfacial tensions by some sort of derived mathematical model which would be simple enough to act as an objective function to an evolutionary optimizer.</p> https://chemistry.stackexchange.com/questions/32044/-/53742#53742 2 Answer by Lucademicus for How to simulate surface tension? Lucademicus https://chemistry.stackexchange.com/users/22838 2016-06-14T22:01:34Z 2016-06-14T22:01:34Z <p>If I understand your question correctly, you want to relate a droplet's geometry to it's environment. The curvature of a liquid droplet is described by the Laplace equation of capillarity, also known as the Young-Laplace equation. This is valid for both a sessile (sitting on a surface) and a pendant (hanging) droplet. I am assuming a static experiment in which the droplet is axisymmetric, with $z$ defined as the axis of symmetry. Such methods of solving for interfacial tension and contact angles are called Axisymmetric Drop Shape Analysis (ADSA) methods. ADSA methods are based on fitting the shape of an experimental droplet to the Laplace equation. This differential equation has been numerically solved in 1883, but thanks to computers and camera's this method is easier and more accurately to apply.</p> <p>The Laplace equation is given by:</p> <p>$$\gamma\left(\frac{1}{R_1}+\frac{1}{R_2}\right)=\Delta P\label{eq:laplace}\tag{1}$$</p> <p>in which $\gamma$ is the surface tension and $\Delta P$ is the pressure difference over the interface. After some work, this may be written as a system of differential equations. $R_1$ and $R_2$ are the principal radii of curvature. $R_1$ is called the meridional curvature, which turns in the plane of the screen you are reading from. Perpendicular to this is the azimuthal curvature $R_2$.</p> <p>The meridional curvature is defined as:</p> <p>$$\frac{1}{R_1}=\frac{\partial \phi}{\partial s}$$ The azimuthal curvature is defined as: $$\frac{1}{R_2}=\frac{\sin{\phi}}{x}$$</p> <p>`In the absence of external forces, other than gravity, the pressure difference is a linear function of the elevation' (Del Rio and Neumann, 1996):</p> <p>$$\Delta P= \Delta P_0+\left(\Delta\rho\right)gz\label{eq:rio}\tag{2}$$ $\Delta P_0$ is the pressure at the origin, at which the droplet is not perturbed and $R_1=R_2=R_0$. Using \eqref{eq:laplace} this leads to the following expression: $$\Delta P_0 =\gamma\left(\frac{2}{R_0}\right)\label{eq:p0}$$ Combining \eqref{eq:laplace} and \eqref{eq:rio} and filling in the expression for $\Delta P_0$ and the curvatures leads to:</p> <p>\begin{align} \gamma\left(\frac{\partial \phi}{\partial s}+\frac{\sin{\phi}}{x}\right)&amp;=\frac{2\gamma}{R_0}+\left(\Delta\rho\right)gz\\ \frac{\partial\phi}{\partial s}&amp;=\frac{2}{R_0}+\frac{\left(\Delta\rho\right)gz}{\gamma}-\frac{\sin{\phi}}{x} \end{align} Now the coordinates are transformed to dimensionless coordinates: \begin{align} \bar{x}=\frac{x}{R_0}, \bar{z}=\frac{z}{R_0}, \bar{s}=\frac{s}{R_0} \end{align}</p> <p>This changes the differential: \begin{align} \frac{\partial\phi}{\partial s}=\frac{\partial\phi}{\partial\left(R_0\bar{s}\right)}=\frac{\partial\phi}{R_0\partial\bar s} \end{align} Multiplying both sides with $R_0$ gives: \begin{align} \frac{\partial\phi}{\bar s}&amp;=2+\frac{\left(\Delta\rho\right)gR_0^2}{\gamma}\bar z-\frac{\sin\phi}{\bar x}\label{eq:phi} \end{align}</p> <p>Furthermore the following geometric interpretation leads to the remaining differential equations. The arc length $s$ follows the curvature of the interface of the drop. $\phi$ is the tangential contact angle, so a change in $x$ over a change in $s$ is equal to the cosine of $\phi$:</p> <p>\begin{align} \frac{\partial x}{\partial s} &amp;= \frac{\partial \bar{x}}{\partial \bar{s}} = \cos(\phi)\label{eq:cos} \end{align} Similarly, a change in $z$ over a change in $s$ is equal to the sine of $\phi$: \begin{align} \frac{\partial z}{\partial s} &amp;= \frac{\partial \bar{z}}{\partial \bar{s}} =\sin(\phi)\label{eq:sin} \end{align} The same holds for the dimensionless coordinates, resulting in the given system of differential equations.</p> <p>Solving the curvature of the droplet is an initial value problem, with the initial values being $x_0$, $z_0$ and $\phi_0$. You can feed this to your numerical differential equation solver, e.g. Matlab's ODE45 function. With this method you will be able to solve the contact angle and surface tension from a given curvature, or draw a curvature for a given contact angle and surface tension.</p> <p>Suggested reading on which I have based my answer:</p> <p>Y. Rotenberg, L. Boruvka, and A. W. Neumann, “Determination of surface tension and contact angle from the shapes of axisymmetric fluid interfaces,” Journal of Colloid and Interface Science, vol. 93, pp. 169–183, May 1983.</p> <p>Rı́o, O. I. del, and A. W. Neumann. “Axisymmetric Drop Shape Analysis: Computational Methods for the Measurement of Interfacial Properties from the Shape and Dimensions of Pendant and Sessile Drops.” Journal of Colloid and Interface Science 196.2 (1997): 136–147. ScienceDirect. Web.</p>