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:

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

References

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

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