5. Pendant drop of maximum volume

In an effort to determine the maximum value a droplet can form without detaching from the substrate, we will consider an axisymmetric droplet forming on the underside of a ceiling, as illustrated in Figure 5. Here z is the distance perpendicular to the substrate, x is the distance from the centerline of the drop parallel to the substrate, and h is the distance from the substrate to the free surface of the drop. g is the acceleration of gravity, and the contact angle with the substrate z ¼ 0 is given by θc. The pressure jump across the free surface due to surface tension is given by σκ. At equilibrium this must equal the hydrostatic pressure due to gravity, or

$$
\sigma \mathbf{x} - \rho \mathbf{g} \mathbf{z} = \text{constant}.
$$

If we nondimensionalize the problem by scaling z, x, and h with the capillary length, lc <sup>¼</sup> ffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>σ</sup>=ð Þ <sup>ρ</sup><sup>g</sup> <sup>p</sup> , then the drop shape is defined by the following differential equation:

$$\frac{\varkappa\_{xx}}{\left(\left(\mathbbm{1}+\varkappa\_x^2\right)^{3/2}\right)} - \frac{1}{\varkappa\left(\mathbbm{1}+\varkappa\_x^2\right)^{1/2}} + \mathbbm{z} = \varkappa\_0. \tag{27}$$

Here κ<sup>0</sup> is the mean curvature at x ¼ 0 and z ¼ h.

Using a standard Runge–Kutta method, we can numerically integrate Eq. (27) assuming θ<sup>c</sup> ¼ 0. We find that the maximum volume is given by Vmax≈19:0l 3 <sup>c</sup> and the maximum height is given by hmax≈2:2lc. For a curved, rotating substrate, the body force g is replaced by a centrifugal force that is a function the rotation rate ω,

Figure 5. Profile view of axisymmetric pendant drop.

radius from the axis of rotation r sð Þ, and the inclination angle of the substrate θð Þs . For this case lc is modified to

$$l\_c = \sqrt{\frac{\sigma}{\left(\rho r \sin \theta w^2\right)}}.$$

If h sð Þ , t is greater than 2:2lc, we assume that the drop has detached from the substrate and name this time as tmax.
