Analytical solutions for two-phase

Laplace’s law

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The difference between the pressure inside the droplet \(P_{in}\) and the pressure outside \(P_{out}\) is equal to

(692)\[\underbrace{P_{in}-P_{out}}_{\Delta P}=\frac{\sigma}{R}\]

where \(\sigma\) is the surface tension and \(R\) is the droplet radius.

Analytical solution of double-Poiseuille flow

(693)\[\begin{split}u_{x}(y)=\begin{cases} \frac{Gh^{2}}{2\eta_{A}}\left[-\left(\frac{y}{h}\right)^{2}-\frac{y}{h}\left(\frac{\eta_{A}-\eta_{B}}{\eta_{A}+\eta_{B}}\right)+\frac{2\eta_{A}}{\eta_{A}+\eta_{B}}\right] & \mbox{if }0\leq y\leq h\\ \frac{Gh^{2}}{2\eta_{B}}\left[-\left(\frac{y}{h}\right)^{2}-\frac{y}{h}\left(\frac{\eta_{A}-\eta_{B}}{\eta_{A}+\eta_{B}}\right)+\frac{2\eta_{B}}{\eta_{A}+\eta_{B}}\right] & \mbox{if }-h\leq y\leq0 \end{cases}\end{split}\]
(694)\[G=\frac{u_{c}}{h^{2}}(\eta_{A}+\eta_{B})\,\,\mbox{and}\,\,u_{c}=5\times10^{-5}\]

Analytical solution of Prosperetti for capillary wave

The example of such a study is given by the test case of capillary wave. For that test case, an analytical solution exists [1]. The objective is to study the influence of mesh on the solution accuracy. The amplitude is

(695)\[a(t)=\frac{4(1-4\beta)\nu^{2}k^{4}}{8(1-4\beta)\nu^{2}k^{4}+\omega_{0}^{2}}a_{0}\text{erfc}(\nu k^{2}t)^{1/2}+\sum_{i=1}^{4}\frac{z_{i}}{Z_{i}}\left(\frac{\omega_{0}^{2}a_{0}}{z_{i}^{2}-\nu k^{2}}-u_{0}\right)\times\exp[(z_{i}^{2}-\nu k^{2})t]\text{erfc}(z_{i}t^{1/2})\]

where \(\nu\) is the kinematic viscosity which is identical for both fluids, \(k\) is the wave number which is related to the wavelength \(\lambda\) by \(k=2\pi/\lambda\). The coefficient \(\beta\) is defined by the liquid \(\rho_l\) and gas \(\rho_g\) densities:

(696)\[\beta=\frac{\rho_{l}\rho_{g}}{(\rho_{l}+\rho_{g})^{2}}\]

\(\omega_0\) is the inviscid natural frequency given by

(697)\[\omega_{0}=\frac{\rho_{l}-\rho_{g}}{\rho_{l}+\rho_{g}}gk+\frac{\sigma k^{3}}{\rho_{l}+\rho_{g}}\]

where \(g\) is the gravity. In the following verifications \(g=0\). The \(z_i\)’s are the four roots of the algebric equation

(698)\[z^{4}-4\beta(k^{2}\nu)^{1/2}z^{3}+2(1-6\beta)k^{2}\nu z^{2}+4(1-3\beta)(k^{2}\nu)^{3/2}z+(1-4\beta)\nu^{2}k^{4}+\omega_{0}^{2}=0\]

and

(699)\[Z_{1}=(z_{2}-z_{1})(z_{3}-z_{1})(z_{4}-z_{1})\]

Coalescence of two equal cylinders

The analytical solution is written in [2]. The contour is given by:

(700)\[x(\theta)=R_{0}\left[(1-m^{2})(1+m^{2})^{-1/2}(1-2m\cos2\theta+m^{2})^{-1}\right](1+m)\cos\theta\]
(701)\[y(\theta)=R_{0}\left[(1-m^{2})(1+m^{2})^{-1/2}(1+2m\cos2\theta+m^{2})^{-1}\right](1-m)\sin\theta\]

where \(R_0\) is the radius of both droplets. In Eqs (700) and (701) the parameter \(m\) can be related to physical properties of surface tension \(\sigma\), dynamic viscosity \(\eta\) and time \(t\) by

(702)\[\frac{\sigma t}{\eta R_{0}}=\frac{\pi}{4}\int_{m^{2}}^{1}\left[\mu(1+\mu)^{1/2}K(\mu)\right]^{-1}d\mu\]

where the function \(K(\mu)\) is defined by

(703)\[\begin{split}K(\mu)&=\int_{0}^{1}\left[(1-x^{2})(1-\mu x^{2})\right]^{-1/2}\\ &=\int_{0}^{\pi/2}(1-\mu\sin^{2}\theta)^{-1/2}d\theta\end{split}\]

References

Section author: Alain Cartalade