Compressible fluid equation

Question

We know the continuity equation of a continuum (in this case I want to discuss fluids, equation reference):

$$\frac{\partial \rho}{\partial t} + \nabla . (\rho u) = 0$$

where $\rho$ is the mass density of the fluid, $u$ is its velocity.

If it had a constant density, then equation becomes:

$$ \nabla . u = 0$$

My confusion is about another equation that concerns incompressible fluid, and somehow I think it is related to the continuity equation because it has the velocity divergence term. I found the equation in the research paper. The equation is:

$$\gamma \dot{p} + \nabla . u = 0$$

where $\gamma$ is the fluid compressibility, $\dot{p}$ is the time derivative of the pressure, and $u$ is its velocity.

I tried to look for a derivation, but I couldn't find until now.

Answer 1

Let's write density $\rho$ as a function of pressure $P$ and entropy $s$, \begin{equation} \rho(P, s) \ , \end{equation} being every thermodynamic quantity a field, function of space and time.

Thus, it's possible to write differential, time derivative, and gradient of the density as a derivative of composite function, as \begin{equation} d \rho = \left(\dfrac{\partial \rho}{\partial P}\right)_s d P + \left(\dfrac{\partial \rho}{\partial s}\right)_P d s \\ \partial_t \rho = \left(\dfrac{\partial \rho}{\partial P}\right)_s \partial_t P + \left(\dfrac{\partial \rho}{\partial s}\right)_P \partial_t s \\ \nabla \rho = \left(\dfrac{\partial \rho}{\partial P}\right)_s \nabla P + \left(\dfrac{\partial \rho}{\partial s}\right)_P \nabla s \\ \end{equation} Recalling the definition of isentropic compressibility \begin{equation} \beta := \dfrac{1}{\rho}\left(\dfrac{\partial \rho}{\partial P} \right)_s \ , \end{equation} and inserting in the continuity equation, recasted in the convective form (introducing the material derivative $D_t \_ = \partial_t \_ + \mathbf{u} \cdot \nabla \_$) \begin{equation} \begin{aligned} 0 & = \partial_t \rho + \nabla \cdot (\rho \mathbf{u}) = \\ & = \partial_t \rho + \mathbf{u} \cdot \nabla \rho + \rho \nabla \cdot \mathbf{u} = \\ & = D_t \rho + \rho \nabla \cdot \mathbf{u} \ , \end{aligned} \end{equation} we get \begin{equation} 0 = \left(\dfrac{\partial \rho}{\partial P}\right)_s D_t P + \left(\dfrac{\partial \rho}{\partial s}\right)_P D_t s + \rho \nabla \cdot \mathbf{u} \ . \end{equation} Assuming constant and uniform entropy, $D_t s \equiv 0$, and dividing by $\rho \ne 0$, we get \begin{equation} \begin{aligned} 0 & = \dfrac{1}{\rho} \left(\dfrac{\partial \rho}{\partial P}\right)_s D_t P + \nabla \cdot \mathbf{u} = \\ & = \beta D_t P + \nabla \cdot \mathbf{u} \end{aligned} \end{equation}