Posts for the month of June 2008

Contact Angle Metrics

Given the simulations from here, we can predict the triple point motion and contact angles.

The following figure show the triple point position against time normalized by the initial radius of the drop.

source:/trunk/reactiveWetting/contact.png@latest

The triple point position is given by

$\vec{r}_T$

where

$P_T\left(\vec{r}_T \right) = \underset{\vec{r}}{\min}\,P_T\left(\vec{r}\right)$

and

$P_T\left( \vec{r}; \phi_T, X_{2T}, \rho_T \right) = 1 + \frac{ \left(\phi \left( \vec{r} \right) - \phi_T\right)^2 }{ \underset{\vec{r}}{\max}\,\left(\phi \left( \vec{r} \right) - \phi_T \right)^2} + \frac{ \left(X_2 \left( \vec{r} \right) - X_{2T}\right)^2 }{ \underset{\vec{r}}{\max}\,\left(X_2 \left( \vec{r} \right) - X_{2T} \right)^2} + \frac{ \left(\rho \left( \vec{r} \right) - \rho_T\right)^2 }{ \underset{\vec{r}}{\max}\,\left(\rho \left( \vec{r} \right) - \rho_T \right)^2}$

and

$\phi_T = \frac{\phi_l + \phi_v + \phi_s}{3}$

$X_{2T} = \frac{X_{2l} + X_{2v} + X2_{2s}}{3}$

$\rho_T = \frac{\rho_l + \rho_v + \rho_s}{3}$

The roughness of the curves in the figure is due to the zero order interpolation of $P_T$. Currently, the center of the cell with the minimum value is deemed to be the triple point. Higher order interpolation should improve the jagged nature of the curves. Need at least second order interpolation.

This movie shows the four points that are chosen to determine the triple point angles. The position of the four points seems reasonable.

The solid vapor interface point, $\vec{r}_{sv}$ such that,

$P_{sv}\left(\vec{r}_{sv} \right) = \underset{\vec{r}}{\min}\,P_{sv}\left(\vec{r}; \phi_{sv}, X_{2sv}, \rho_{sv} \right) C\left (\vec{r} ; \vec{r}_T \right)$

where

$C \left( \vec{r} ; \vec{r}_T \right) = 1 + 10 \left(\frac{|\vec{r} - \vec{r}_T| - R_0}{R_0}\right)^2$

with

$\phi_{sv} = \frac{\phi_v + \phi_s}{2}$

$X_{2sv} = \frac{X_{2v} + X2_{2s}}{2}$

$\rho_{sv} = \frac{\rho_v + \rho_s}{2}$

The radius of the circle $R_0$ is chosen to be $20 \Delta x$. We'll need to see how contact angle varies with this parameter.

Contact Angles

The links in the table below show contact angles versus time using the method described above.

ID $R$ $\mu_L$ Figure
2 $5\times10^{-7}$ $10^4$ figure
4 $1\times10^{-6}$ $10^4$ figure
6 $1\times10^{-6}$ $10^3$ figure
7 $5\times10^{-7}$ $10^2$ figure
8 $1\times10^{-6}$ $10^2$ figure
9 $5\times10^{-7}$ $10^3$ figure
10 $2\times10^{-6}$ $10^4$ figure
12 $2\times10^{-6}$ $10^3$ figure

Thoughts:

  • The initial contact angles in these figures seem to be somewhat wrong. The solid contact angle should be $\pi$ while the other angles should be $\pi / 2$.
  • Initial contact angle should improve with bigger droplets and this isn't happening.
  • The liquid and vapor angles both seem to drift away from the equilibrium values.
  • The curves are extremely jagged.
  • The surface tensions used to calculate the equilibrium contact angle could be for a different thermo. Need to double check.
  • Even though the system is not in equilibrium the contact angles should probably be in equilibrium.

What next?

  • Figure out why the initial angles are wrong.
  • Plot the equilibrium solution against that the last frame in the simulation to figure out how close to equilibrium.
  • Use higher order interpolation for the contact points to improve the smoothness.
  • Making $R_0$ smaller may improve the results.
  • Test the equilibrium contact angle in the case where the solid has the same viscosity as the liquid, so that the system reaches equilibrium. Probably need to run that case again.
  • run a full sphere on a solid surface that doesn't melt (correct equilibrium concentrations). Use this to compare with cox stuff.
  • vary $R_0$ for one frame of a simulation to see how the angle changes.

Jobs

Two new jobs

ID PID machine status $R$ ($\mu$m) movie Angles
25 13638.0 cluster stopped 0.4 movie
26 13639.0 cluster R 0.8 movie figure
27 13740.0 cluster R 0.35 movie figure
28 12539 poole R 1.8 movie figure
29 14230.0 cluster R 0.35 movie figure

Various plots for the simulations above,

Issues:

  • The $Ca$ number seems to be larger than Cox's theory predicts.
  • The drop off is slower than Cox's theory predicts.
  • Cox's theory is for high $Pe$ number. Need to run some simulations in this regime.
  • Need to improve data sampling. This is relatively easy to do by refining the grid and interpolating the base fields.
  • axi-symmetric may also be required

Peclet Number

A rough estimate for the Peclet Number $Pe$ is given by

$Pe = \frac{X m V_{\text{tp}} \rho}{\bar{M} R}$

If we take an interface length for $X$ of $~10^{-7}$. At early times the Peclet number is 100 in the solid and 0.01 in the liquid. This indicates that if we give the solid value of $\bar{M}$ to the fluid then the Peclet number will be large everywhere and will hopefully allow a better fit to Cox's theory. Simulation 29 in the table above is running with $\bar{M}_{f}=10^{-11}$

Jobs with lower Peclet number

ID PID machine status $R$ ($\mu$m) $\bar{M}_f$ $\mu_f$ $\Delta t_{\text{max}}$ movie Angles
29 14230.0 cluster unstable 0.35 $1 \times 10^{-11}$ $1 \times 10^{4}$ $1 \times 10^{-7}$ movie figure
30 14256.0 cluster finished 0.35 $1 \times 10^{-8}$ $1 \times 10^{4}$ $1 \times 10^{-4}$ movie figure
31 14257.0 cluster finished 0.35 $1 \times 10^{-9}$ $1 \times 10^{4}$ $1 \times 10^{-4}$ movie figure
32 14258.0 cluster unstable 0.35 $1 \times 10^{-10}$ $1 \times 10^{4}$ $1 \times 10^{-4}$ movie figure
33 14280.0 cluster R 0.35 0 $1 \times 10^{4}$ $1 \times 10^{-4}$ movie figure
34 14296.0 cluster R 0.35 0 $1 \times 10^{-3}$ $1 \times 10^{-4}$ movie figure

Higher order interpolation

It works!

High Peclet number

Can't seem to run the code with a Peclet number of about 10 and above. This is certainly connected with other issues. The figure below shows the Cox theory fit for decreasing $\bar{M}$ corresponding to increasing Peclet number. As the Peclet number is increasing, the curvature of the simulation points is increasing. This seems to show it is getting closer to the theory, though it is hard to tell. It is probable that as the Peclet number passes through 1 there will be a shift from one spreading regime to another.

Mbar=0

Far out. By solving for $\rho_1$ and $\rho_2$ separately and with corrections the system can be solved when $\bar{M}=0$. It's interesting that the system remains stable. The movie for this simulations is movie. I can't use this for Cox's theory as the system does not seem to movie towards the equilibrium contact angles. What are the equilibrium contact angles in the case when $\bar{M}=0$. It's also interesting that the pressure remains higher in the fluid state.

Low viscosity

The next step is try the $\bar{M}=0$ system with a much lower viscosity in the fluid. Running a job with $\mu_f=1 \times 10^{-3}$. Working on this issue made me think about modeling the system by solving for $\rho_1$ and $\rho_2$ instead. Simulation 35 is my attempt at this issue. I added RHS flux back in to each of these equations and did the variational derivative. Unfortunately this wasn't really successful at low viscosity. Same sort of issues as before.

  • Posted: 2008-06-06 16:54 (Updated: 2009-04-29 17:46)
  • Author: wd15
  • Categories: (none)
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