Categories: Fluid dynamics, Fluid mechanics, Physics.

Rayleigh-Plesset equation

In fluid dynamics, the Rayleigh-Plesset equation describes how the radius of a spherical bubble evolves in time inside an incompressible liquid. Notably, it leads to cavitation.

Consider the main Navier-Stokes equations for the velocity field $$\va{v}$$:

\begin{aligned} \frac{\mathrm{D} \va{v}}{\mathrm{D} t} = \pdv{\va{v}}{t} + (\va{v} \cdot \nabla) \va{v} = - \frac{\nabla p}{\rho} + \nu \nabla^2 \va{v} \end{aligned}

We make the ansatz $$\va{v} = v(r, t) \vu{e}_r$$, where $$\vu{e}_r$$ is the basis vector; in other words, we demand that the only spatial variation of the flow is in $$r$$. The above equation then becomes:

\begin{aligned} \pdv{v}{t} + v \pdv{v}{r} = - \frac{1}{\rho} \pdv{p}{r} + \nu \bigg( \frac{1}{r^2} \pdv{r} \Big( r^2 \pdv{v}{r} \Big) - \frac{2}{r^2} v \bigg) \end{aligned}

Meanwhile, the incompressibility condition in spherical coordinates yields:

\begin{aligned} \nabla \cdot \va{v} = \frac{1}{r^2} \pdv{(r^2 v)}{r} = 0 \end{aligned}

This is only satisfied if $$r^2 v$$ is constant with respect to $$r$$, leading us to a solution $$v(r)$$ given by:

\begin{aligned} v(r) = \frac{C(t)}{r^2} \end{aligned}

Where $$C(t)$$ is an unknown function that does not depend on $$r$$. We then insert this result in the main Navier-Stokes equation, and isolate it for $$\pdv*{p}{r}$$, yielding:

\begin{aligned} \pdv{p}{r} = - \rho \bigg( \frac{1}{r^2} C' - \frac{2}{r^5} C^2 - \nu \Big( \frac{2}{r^4} C - \frac{2}{r^4} C \Big) \bigg) = - \rho \bigg( \frac{1}{r^2} C' - \frac{2}{r^5} C^2 \bigg) \end{aligned}

Integrating this with respect to $$r$$ yields the following expression for $$p$$, where $$p_\infty(t)$$ is the (possibly time-dependent) pressure at $$r = \infty$$:

\begin{aligned} p(r) = p_\infty + \rho \bigg( \frac{1}{r} C' - \frac{1}{2 r^4} C^2 \bigg) \end{aligned}

From the definition of viscosity, we know that the normal stress $$\sigma_{rr}$$ in the liquid is given by:

\begin{aligned} \sigma_{rr}(r) = - p(r) + 2 \rho \nu \pdv{v(r)}{r} \end{aligned}

We now consider a spherical bubble with radius $$R(t)$$ and interior pressure $$P(t)$$ along its surface. Since we know the liquid pressure $$p(r)$$, we can find $$P$$ from $$\sigma_{rr}(r)$$. Furthermore, to include the effects of surface tension, we simply add the Young-Laplace law to $$P$$:

\begin{aligned} P = - \sigma_{rr}(R) + \alpha \frac{2}{R} = p(R) - 2 \rho \nu \Big( \frac{-2}{R^3} C \Big) + \alpha \frac{2}{R} \end{aligned}

We isolate this for $$p(R)$$, and equate it to our expression for $$p(r)$$ at the surface $$r\!=\!R$$:

\begin{aligned} P - \rho \nu \frac{4}{R^3} C - \alpha \frac{2}{R} = p_\infty + \rho \bigg( \frac{1}{R} C' - \frac{1}{2 R^4} C^2 \bigg) \end{aligned}

Isolating for $$P$$, and inserting the fact that $$R'(t) = v(t)$$, such that $$C = r^2 v = R^2 R'$$, yields:

\begin{aligned} P &= p_\infty + \rho \bigg( \frac{1}{R} \dv{(R^2 R')}{t} - \frac{1}{2 R^4} (R^2 R')^2 + \nu \frac{4}{R^3} (R^2 R') \bigg) + \alpha \frac{2}{R} \\ &= p_\infty + \rho \bigg( 2 (R')^2 + R R'' - \frac{1}{2} (R')^2 + \nu \frac{4}{R} R' \bigg) + \alpha \frac{2}{R} \end{aligned}

Rearranging this and defining $$\Delta p \equiv P - p_\infty$$ leads to the Rayleigh-Plesset equation:

\begin{aligned} \boxed{ \frac{\Delta p}{\rho} = R \dv{R}{t} + \frac{3}{2} \bigg( \dv{R}{t} \bigg)^2 + \nu \frac{4}{R} \dv{R}{t} + \frac{\alpha}{\rho} \frac{2}{R} } \end{aligned}

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

© Marcus R.A. Newman, a.k.a. "Prefetch". Available under CC BY-SA 4.0.