Cavitation: radius of a bubble for compressible flow

[JD_PM]

The liquid-vapor mass transfer (evaporation and condensation) is governed by the vapor transport equation:

$$\frac{\partial}{\partial t} (\alpha_l \rho) + \nabla \cdot (\alpha_l \rho \vec v) = \dot m^{+} + \dot m^{-}$$

In the incompressible flow case (constant density), it reduces to

$$
\frac{\partial}{\partial t} (\alpha_l) + \nabla \cdot (\alpha_l \vec v) = \frac{\dot m^{+} + \dot m^{-}}{\rho}
$$

The cavitation model of https://www.researchgate.net/profile/Guenter-Schnerr-Professor-Dr-Inghabil/publication/296196752_Physical_and_Numerical_Modeling_of_Unsteady_Cavitation_Dynamics/links/56f6b62308ae81582bf2f940/Physical-and-Numerical-Modeling-of-Unsteady-Cavitation-Dynamics.pdf aims to determine the mass transfer rate linked to the RHS term of transport equation, using the non conservative form of the continuity equation i.e.

\begin{equation}

\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho v) = 0 \Rightarrow

\end{equation}

\begin{equation}

\Rightarrow \nabla \cdot v = -\frac{1}{\rho}\Big( \frac{\partial \rho}{\partial t} + v \cdot \nabla \rho\Big) = -\frac{1}{\rho} \frac{D \rho}{D t} = \frac{\rho_l - \rho_v}{\rho} \frac{D \alpha_v}{D t}

\end{equation}

Where we used $\rho = (1 - \alpha_v ) \rho_l + \alpha_v \rho_v$

However, for compressible compressible flow

\begin{equation}

\nabla \cdot v = \frac{\rho_l - \rho_v}{\rho} \frac{D \alpha_v}{D t} + \alpha_l \frac{D \rho_v}{D t} + (1 - \alpha_l) \frac{D \rho_l}{D t}

\end{equation}

Where the vapor volume fraction is

\alpha_v = \frac{V_v}{V_l + V_v} = \frac{\frac 4 3 n_0 \pi R^3}{1 + \frac 4 3 n_0 \pi R^3}

The bubble growing and collapsing can be described by the Rayleigh Plesset equation (please see first link). Dropping the second order terms and the viscosity effects between phases in the Plesset equation, the evolution of the radius of the bubble has the expression:

$$\frac{DR}{D t} = \Big( \frac 2 3 \frac{p_b - p}{\rho_l}\Big)^{1/2}$$

It follows that the radius of a bubble is given by

$$R = \Big( \frac{3\alpha_v}{(1-\alpha_v) n_0 4 \pi} \Big)^{1/3}$$

The above radius is obtained assuming incompressible flow.

Does the same expression hold for compressible flow?

 

[pbuk]

There's a couple of typos in the above that are breaking $\LaTeX$ but in any case you might get more replies if you ask the question like this:

We can derive* the following equation for the radius of a cavitation bubble for flow of an incompressible liquid:

$$R = \Big( \frac{3\alpha_v}{(1-\alpha_v) n_0 4 \pi} \Big)^{1/3}$$

Question: does the same expression hold for compressible flow?

* Derivation:
The liquid-vapor mass transfer (evaporation and condensation) is governed by the vapor transport equation:

$$\frac{\partial}{\partial t} (\alpha_l \rho) + \nabla \cdot (\alpha_l \rho \vec v) = \dot m^{+} + \dot m^{-}$$

In the incompressible flow case (constant density), it reduces to

$$
\frac{\partial}{\partial t} (\alpha_l) + \nabla \cdot (\alpha_l \vec v) = \frac{\dot m^{+} + \dot m^{-}}{\rho}
$$

The cavitation model of https://www.researchgate.net/profile/Guenter-Schnerr-Professor-Dr-Inghabil/publication/296196752_Physical_and_Numerical_Modeling_of_Unsteady_Cavitation_Dynamics/links/56f6b62308ae81582bf2f940/Physical-and-Numerical-Modeling-of-Unsteady-Cavitation-Dynamics.pdf aims to determine the mass transfer rate linked to the RHS term of transport equation, using the non conservative form of the continuity equation i.e.

\begin{equation}

\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho v) = 0 \Rightarrow

\end{equation}

\begin{equation}

\Rightarrow \nabla \cdot v = -\frac{1}{\rho}\Big( \frac{\partial \rho}{\partial t} + v \cdot \nabla \rho\Big) = -\frac{1}{\rho} \frac{D \rho}{D t} = \frac{\rho_l - \rho_v}{\rho} \frac{D \alpha_v}{D t}

\end{equation}

Where we used $\rho = (1 - \alpha_v ) \rho_l + \alpha_v \rho_v$

However, for compressible compressible flow

\begin{equation}

\nabla \cdot v = \frac{\rho_l - \rho_v}{\rho} \frac{D \alpha_v}{D t} + \alpha_l \frac{D \rho_v}{D t} + (1 - \alpha_l) \frac{D \rho_l}{D t}

\end{equation}

Where the vapor volume fraction is

\alpha_v = \frac{V_v}{V_l + V_v} = \frac{\frac 4 3 n_0 \pi R^3}{1 + \frac 4 3 n_0 \pi R^3}

The bubble growing and collapsing can be described by the Rayleigh Plesset equation (please see first link). Dropping the second order terms and the viscosity effects between phases in the Plesset equation, the evolution of the radius of the bubble has the expression:

$$\frac{DR}{D t} = \Big( \frac 2 3 \frac{p_b - p}{\rho_l}\Big)^{1/2}$$

It follows that the radius of a bubble is given by

$$R = \Big( \frac{3\alpha_v}{(1-\alpha_v) n_0 4 \pi} \Big)^{1/3}$$