Rayleigh–Plesset equation
In fluid mechanics, the Rayleigh–Plesset equation is an ordinary differential equation which governs the dynamics of a spherical bubble in an infinite body of liquid.[1][2] Its general form is usually written as
\[\frac{P_B(t)-P_\infty(t)}{\rho_L} = R\frac{d^2R}{dt^2} + \frac{3}{2}\left(\frac{dR}{dt}\right)^2 + \frac{4\nu_L}{R}\frac{dR}{dt} + \frac{2S}{\rho_LR}\]
where \[P_B(t) \] is the pressure within the bubble, assumed to be uniform \[P_\infty(t)\] is the external pressure infinitely far from the bubble \[\rho_L \] is the density of the surrounding liquid, assumed to be constant \[R(t)\] is the radius of the bubble \[\nu_L\] is the kinematic viscosity of the surrounding liquid, assumed to be constant \[S\] is the surface tension of the bubble
Provided that \(P_B(t) \) is known and \(P_\infty(t)\) is given, the Rayleigh–Plesset equation can be used to solve for the time-varying bubble radius \(R(t)\).
The Rayleigh–Plesset equation is derived from the Navier–Stokes equations under the assumption of spherical symmetry.[2] Neglecting surface tension and viscosity, the equation was first derived by John Strutt, 3rd Baron Rayleigh in 1917. The equation was first applied to traveling cavitation bubbles by Milton S. Plesset in 1949.[3]
Contents
Derivation
The Rayleigh–Plesset equation can be derived entirely from first principles using the bubble radius as the dynamic parameter.[1] Consider a spherical bubble with time-dependent radius \( R(t) \), where \( t \) is time. Assume that the bubble contains a homogeneously distributed vapor/gas with a uniform temperate \( T_B(t) \) and pressure \( P_B(t) \). Outside the bubble is an infinite domain of liquid with constant density \( \rho_L \) and dynamic viscosity \( \mu_L \). Let the temperature and pressure far from the bubble be \( T_\infty \) and \( P_\infty(t) \). The temperature \( T_\infty \) is assumed to be constant. At a radial distance \( r \) from the center of the bubble, the varying liquid properties are pressure \( P(r,t) \), temperature \( T(r,t) \), and radially outward velocity \( u(r,t) \). Note that these liquid properties are only defined outside the bubble, for \( r \ge R(t) \).
Mass conservation
By conservation of mass, the inverse-square law requires that the radially outward velocity \( u(r,t) \) must be inversely proportional to the square of the distance from the origin (the center of the bubble).[3] Therefore, letting \( F(t) \) be some function of time,
\[ u(r,t) = \frac{F(t)}{r^2} \]
In the case of zero mass transport across the bubble surface, the velocity at the interface must be
\[ u(R,t) = \frac{dR}{dt} = \frac{F(t)}{R^2} \]
which gives that
\[ F(t) = R^2dR/dt \]
In the case where mass transport occurs, the rate of mass increase inside the bubble is given by
\[ \frac{dm_V}{dt} = \rho_V\frac{dV}{dt} = \rho_V\frac{d(4\pi R^3/3)}{dt} = 4\pi\rho_VR^2\frac{dR}{dt} \]
with \( V \) being the volume of the bubble. If \( u_L \) is the velocity of the liquid relative to the bubble at \( r = R \), then the mass entering the bubble is given by
\[ \frac{dm_L}{dt} = \rho_LAu_L = \rho_L(4\pi R^2)u_L \]
with \( A \) being the surface area of the bubble. Now by conservation of mass \( dm_v/dt = dm_L/dt \), therefore \( u_L = (\rho_V/\rho_L)dR/dt \). Hence
\[ u(R,t) = \frac{dR}{dt} - u_L = \frac{dR}{dt} - \frac{\rho_V}{\rho_L}\frac{dR}{dt} = \left(1-\frac{\rho_V}{\rho_L}\right)\frac{dR}{dt}\]
Therefore
\[ F(t) = \left(1-\frac{\rho_V}{\rho_L}\right)R^2\frac{dR}{dt} \]
In many cases, the liquid density is much greater than the vapor density, \( \rho_L \gg \rho_V \), so that \( F(t) \) can be approximated by the original zero mass transfer form \( F(t) = R^2dR/dt \), so that[3]
\[ u(r,t) = \frac{F(t)}{r^2} = \frac{R^2}{r^2}\frac{dR}{dt} \]
Momentum conservation
Assuming that the liquid is a Newtonian fluid, the Navier–Stokes equation in spherical coordinates for motion in the radial direction gives
\[ \rho_L\left(\frac{\partial u}{\partial t} + u\frac{\partial u}{\partial r}\right) = -\frac{\partial P}{\partial r} + \mu_L \left[ \frac{1}{r^2}\frac{\partial}{\partial r}\left( r^2\frac{\partial u}{\partial r}\right) - \frac{2u}{r^2}\right]\]
Substituting kinematic viscosity \( \nu_L = \mu_L/\rho_L \) and rearranging gives
\[ -\frac{1}{\rho_L}\frac{\partial P}{\partial r} = \frac{\partial u}{\partial t} + u\frac{\partial u}{\partial r} - \nu_L \left[ \frac{1}{r^2}\frac{\partial}{\partial r}\left( r^2\frac{\partial u}{\partial r}\right) - \frac{2u}{r^2}\right] \]
whereby substituting \( u(r,t) \) from mass conservation yields
\[ -\frac{1}{\rho_L}\frac{\partial P}{\partial r} = \frac{2R}{r^2}\left(\frac{dR}{dt}\right)^2 + \frac{R^2}{r^2}\frac{d^2R}{dt^2} - \frac{2R^4}{r^5}\left(\frac{dR}{dt}\right)^2 = \frac{1}{r^2}\left(2R\left(\frac{dR}{dt}\right)^2 + R^2\frac{d^2R}{dt^2}\right) - \frac{2R^4}{r^5}\left(\frac{dR}{dt}\right)^2\]
Note that the viscous terms cancel during substitution.[3] Separating variables and integrating from the bubble boundary \( r = R\) to \( r \rightarrow \infty \) gives
\[ -\frac{1}{\rho_L}\int_{P(R)}^{P_\infty} dP = \int_R^\infty \left[ \frac{1}{r^2}\left(2R\left(\frac{dR}{dt}\right)^2 + R^2\frac{d^2R}{dt^2}\right) - \frac{2R^4}{r^5}\left(\frac{dR}{dt}\right)^2 \right] dr\]
\[{ \frac{P(R) - P_\infty}{\rho_L} = \left[- \frac{1}{r}\left(2R\left(\frac{dR}{dt}\right)^2 + R^2\frac{d^2R}{dt^2}\right) + \frac{R^4}{2r^4}\left(\frac{dR}{dt}\right)^2 \right]_R^\infty = R\frac{d^2R}{dt^2} + \frac{3}{2}\left(\frac{dR}{dt}\right)^2 }\]
Boundary conditions
Let \( \sigma_{rr} \) be the normal stress in the liquid that points radially outward from the center of the bubble. In spherical coordinates, for a fluid with constant density and constant viscosity,
\[ \sigma_{rr} = -P +2\mu_L\frac{\partial u}{\partial r} \]
Therefore at some small portion of the bubble surface, the net force per unit area acting on the lamina is
\[{ \sigma_{rr}(R) + P_B - \frac{2S}{R} = -P(R) + \left.2\mu_L\frac{\partial u}{\partial r}\right|_{r=R} + P_B - \frac{2S}{R} = -P(R) + 2\mu_L\frac{\partial}{\partial r}\left( \frac{R^2}{r^2}\frac{dR}{dt} \right)_{r=R} + P_B - \frac{2S}{R} = -P(R) - \frac{4\mu_L}{R}\frac{dR}{dt} + P_B - \frac{2S}{R} }\]
where \( S \) is the surface tension.[3] If there is no mass transfer across the boundary, then this force per unit area must be zero, therefore
\( P(R) = P_B - \frac{4\mu_L}{R}\frac{dR}{dt} - \frac{2S}{R} \)
and so the result from momentum conservation becomes
\[\frac{P(R) - P_\infty}{\rho_L} = \frac{P_B - P_\infty}{\rho_L} - \frac{4\mu_L}{\rho_LR}\frac{dR}{dt} - \frac{2S}{\rho_LR} = R\frac{d^2R}{dt^2} + \frac{3}{2}\left(\frac{dR}{dt}\right)^2\]
wherby rearranging and letting \( \nu_L = \mu_L/\rho_L \) gives the Rayleigh–Plesset equation[3]
\[ \frac{P_B(t) - P_\infty(t)}{\rho_L} = R\frac{d^2R}{dt^2} + \frac{3}{2}\left(\frac{dR}{dt}\right)^2 + \frac{4\nu_L}{R}\frac{dR}{dt} + \frac{2S}{\rho_LR} \]
Using dot notation to represent derivatives with respect to time, the Rayleigh–Plesset equation can be more succinctly written as
- \[ \frac{P_B(t) - P_\infty(t)}{\rho_L} = R\ddot{R} + \frac{3}{2}(\dot{R})^2 + \frac{4\nu_L\dot{R}}{R} + \frac{2S}{\rho_LR} \]
Solutions
No analytical closed-form solution is known for the Rayleigh-Plesset equation. However, numerical solutions to any accuracy can be easily obtained. In the special case, where surface tension and viscosity are neglected, high-order analytical approximations are known [4].
References
