From 45af77f068aaa57c052cd861412d53beecbe5e3b Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Fri, 9 Apr 2021 20:44:44 +0200
Subject: Expand knowledge base

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+---
+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.
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