---
title: "Cavitation"
firstLetter: "C"
publishDate: 2021-04-09
categories:
- Physics
- Fluid mechanics
- Fluid dynamics

date: 2021-04-06T19:03:36+02:00
draft: false
markup: pandoc
---

# 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](/know/concept/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:

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

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.



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