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∞p_\infty, and the bubble is empty, such that the interface pressure P=0P = 0, meaning Δp=−p∞\Delta p = - p_\infty. At first, the radius is stationary R′(0)=0R'(0) = 0, and given by a constant R(0)=aR(0) = a. The simple Rayleigh-Plesset equation is then:

Rd2Rdt2+32(dRdt)2=−p∞ρ\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 R2R′R^2 R' and rewrite it in the following way:

−2p∞ρR2R′=2R3R′R′′+3R2(R′)3−2p∞3ρddt(R3)=ddt(R3(R′)2)\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 τ\tau, from 00 to tt:

−2p∞3ρ∫0tddτ(R3)dτ=∫0tddτ(R3(R′)2)dτ−2p∞3ρ[R3]0t=[R3(R′)2]0t−2p∞3ρ(R3(t)−a3)=(R3(t) (R′(t))2)\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′R':

(R′)2=2p∞3ρ(a3R3−1)\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)R(t) into t(R)t(R), and, because we are only interested in collapse, we just need to consider the case R′<0R' < 0. The time of a given radius RR is then as follows, where we are using slightly sloppy notation:

t=∫0tdτ=−∫aRdRR′=∫RadRR′\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′<0R' < 0, but t≥0t \ge 0. We insert the expression for R′R':

t=3ρ2p∞∫Ra(a3R3−1)−1dR=3ρa22p∞∫R/a11x−3−1dx\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 2F1{}_2 F_1. However, we only care about collapse, which is when R=0R = 0. The time t0t_0 at which this occurs is:

t0=3ρa22p∞3π2Γ(5/6)Γ(1/3)≈3ρa22p∞0.915 s\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→0R \to 0, the interface velocity R′→∞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′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.