1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
|
---
title: "Cavitation"
date: 2021-04-09
categories:
- Physics
- Fluid mechanics
- Fluid dynamics
layout: "concept"
---
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:
$$\begin{aligned}
R \dvn{2}{R}{t} + \frac{3}{2} \bigg( \dv{R}{t} \bigg)^2
= - \frac{p_\infty}{\rho}
\end{aligned}$$
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.
|
