Categories: Fluid dynamics, Fluid mechanics, Physics.

# Cavitation

In a liquid, cavitation is the spontaneous appearance of bubbles, occurring when the pressure in a part of the liquid drops below its vapour pressure, e.g. due to the fast movements. When such a bubble is subjected to a higher pressure by the surrounding liquid, it quickly implodes.

To model this case, we use the simple form of the Rayleigh-Plesset equation for an inviscid liquid without surface tension. Note that the RP equation assumes incompressibility.

We assume that the whole liquid is at a constant pressure $$p_\infty$$, and the bubble is empty, such that the interface pressure $$P = 0$$, meaning $$\Delta p = - p_\infty$$. At first, the radius is stationary $$R'(0) = 0$$, and given by a constant $$R(0) = a$$. The simple Rayleigh-Plesset equation is then:

\begin{aligned} R \dv{R}{t} + \frac{3}{2} \bigg( \dv{R}{t} \bigg)^2 = - \frac{p_\infty}{\rho} \end{aligned}

To solve it, we multiply both sides by $$R^2 R'$$ and rewrite it in the following way:

\begin{aligned} - 2 \frac{p_\infty}{\rho} R^2 R' &= 2 R^3 R' R'' + 3 R^2 (R')^3 \\ - \frac{2 p_\infty}{3 \rho} \dv{t} \Big( R^3 \Big) &= \dv{t} \Big( R^3 (R')^2 \Big) \end{aligned}

It is then straightforward to integrate both sides with respect to time $$\tau$$, from $$0$$ to $$t$$:

\begin{aligned} - \frac{2 p_\infty}{3 \rho} \int_0^t \dv{\tau} \Big( R^3 \Big) \dd{\tau} &= \int_0^t \dv{\tau} \Big( R^3 (R')^2 \Big) \dd{\tau} \\ - \frac{2 p_\infty}{3 \rho} \Big[ R^3 \Big]_0^t &= \Big[ R^3 (R')^2 \Big]_0^t \\ - \frac{2 p_\infty}{3 \rho} \Big( R^3(t) - a^3 \Big) &= \Big( R^3(t) \: \big(R'(t)\big)^2 \Big) \end{aligned}

Rearranging this equation yields the following expression for the derivative $$R'$$:

\begin{aligned} (R')^2 = \frac{2 p_\infty}{3 \rho} \Big( \frac{a^3}{R^3} - 1 \Big) \end{aligned}

This equation is nasty to integrate. The trick is to invert $$R(t)$$ into $$t(R)$$, and, because we are only interested in collapse, we just need to consider the case $$R' < 0$$. The time of a given radius $$R$$ is then as follows, where we are using slightly sloppy notation:

\begin{aligned} t = \int_0^t \dd{\tau} = - \int_{a}^{R} \frac{\dd{R}}{R'} = \int_{R}^{a} \frac{\dd{R}}{R'} \end{aligned}

The minus comes from the constraint that $$R' < 0$$, but $$t \ge 0$$. We insert the expression for $$R'$$:

\begin{aligned} t = \sqrt{\frac{3 \rho}{2 p_\infty}} \int_{R}^{a} \Bigg( \sqrt{ \frac{a^3}{R^3} - 1 } \Bigg)^{-1} \dd{R} = \sqrt{\frac{3 \rho a^2}{2 p_\infty}} \int_{R/a}^{1} \frac{1}{\sqrt{x^{-3} - 1}} \dd{x} \end{aligned}

This integral needs to be looked up, and involves the hypergeometric function $${}_2 F_1$$. However, we only care about collapse, which is when $$R = 0$$. The time $$t_0$$ at which this occurs is:

\begin{aligned} t_0 = \sqrt{\frac{3 \rho a^2}{2 p_\infty}} \sqrt{\frac{3 \pi}{2}} \frac{\Gamma(5/6)}{\Gamma(1/3)} \approx \sqrt{\frac{3 \rho a^2}{2 p_\infty}} 0.915 \:\mathrm{s} \end{aligned}

With our assumptions, a bubble will always collapse. However, unsurprisingly, reality turns out to be more complicated: as $$R \to 0$$, the interface velocity $$R' \to \infty$$. By looking at the derivation of the Rayleigh-Plesset equation, it can be shown that the pressure just outside the bubble diverges due to $$R'$$. This drastically changes the liquid’s properties, and breaks our assumptions.

1. B. Lautrup, Physics of continuous matter: exotic and everyday phenomena in the macroscopic world, 2nd edition, CRC Press.