0
$\begingroup$

I want to estimate the curvature $\kappa$ of the fluid surface next to a floating sphere. The situation is static and shown here:

enter image description here

The fluid density is $\rho$, downward gravity is $g$, sphere radius is $R$, and solid-liquid-air surface tension is $\sigma$.

I know that:

1) The sphere is half submerged, so that the free surface coincides with the sphere center.

2) $\frac{\sigma}{\rho g R^2} << 1$, which implies that surface tension effects are negligible compared to gravitational effects.

I start with the Young-Laplace equation:

$$ P_i - P_o = \sigma (\frac{1}{R} + \frac{1}{R_2})$$

where one radius of curvature is the sphere radius $R$, and $\kappa = \frac{1}{R_2}$ is the curvature of the free surface.

Now, use a simple hydrostatic force balance to find $P_i$. Note that the pressures at points 1 and 2 below are the same.

enter image description here

Since we have $P_1 = P_2$, this implies that

$$P_i = P_{atm} - \rho g h$$

The outer pressure $P_o = P_{atm}$.

Substituting into Young-Laplace,

$$ -\rho g h = \sigma (\frac{1}{R} + \kappa)$$

Now, since surface tension effects are negligible, $\rho g h \sim 0$ and the final result is:

$$ \kappa \sim - \frac{\sigma}{R}$$

I'm confused by the negative sign, and I'm not sure if my $\rho g h \sim 0$ assumption is valid.

$\endgroup$
  • $\begingroup$ Near the surface of the sphere you cannot ignore surface tension effects. In your analysis, you use $R$ as the characteristic length scale for the Bond number however the region which you are considering is not characterized $R$. A more appropriate length scale may be the capillary length. $\endgroup$ – Ragnar Dec 31 '18 at 17:50
  • $\begingroup$ @PapaZulu, what is the "capillary length" here? I thought it would be the sphere circumference $2 \pi R$. Note that capillary length and capillary rise are not necessarily the same thing, correct? $\endgroup$ – Drew Dec 31 '18 at 18:49
1
$\begingroup$

I found this figure online from a book titled "Capillarity and wetting phenomena : drops, bubbles, pearls, waves" by de Gennes et al. According to the figure an estimate of the curvature near a floating sphere is given by $\kappa = 1/\ell_c$ where $\ell_c = \sqrt{\sigma/\rho g}$ is the capillary length. The capillary length is the length scale where surface tension and gravity forces are of the same order of magnitude.

enter image description here

$\endgroup$
  • $\begingroup$ That's a helpful source you cited, and I think you're right about the capillary length. However, in my problem I'm given $\frac{\sigma}{\rho g R^2} << 1$, which may not be the case in your example. I'm not sure how to use this or how to interpret it. $\endgroup$ – Drew Dec 31 '18 at 22:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.