Conditions for flow incompressibility The condition for flow incompressibility is usually stated as:
$$
\frac{D\rho}{Dt}=0
$$
This sort of vaguely makes sense intuitively. However it seems more natural to me to understand incompressibility as "density doesn't increase/decrease when pressure increases/decreases", which can be formulated mathematically as:
$$
\frac{\partial\rho}{\partial p}=0
$$
Are the two formulations equivalent? Is there a way to show it? Or maybe there is a way of understanding the first equation?
 A: They are related, but it is a bit more complicated than that. There is no perfect incompressibility in real world, it is an aproximation. The rationale to relate the two would be the conection between time evolution of the fluid and the pressure. By euler and navier-stokes equation, the 'engine' that drives the fluid is pressure, therefore, the 'usual' thermodynamic pressure can't assure incompressibility ($\nabla \cdot {\bf u} = 0$) exactly. 
From a thermodynamical point of view, what regulates the behaviour of density with the pressure is the 'bulk modulus' or 'Young modulus', defined as 
$Y = \left[\frac{1}{\rho}\frac{\partial \rho}{\partial p}\right]^{-1}$
Which is a 'scale of pressure' in a certain sense. So, the idea is not to ask if $\partial \rho/\partial p = 0$ but wether 
$\frac{p_0}{Y} = p_0\frac{1}{\rho}\frac{\partial \rho}{\partial p} \approx 0$
Where $p_0$ is a typical pressure in your fluid system. Y is the pressure where 'the fluid start to suffer compression'. In a pictorial view, $\partial \rho/\partial p = 0$ would be equivalent to infinite bulk modulus $Y$.
Well, how does this relate to fluid dynamic evolution?
The continuity equation for density goes as:
$\frac{\partial \rho}{\partial t} + ({\bf u}\cdot \nabla)\rho = \frac{D\rho}{Dt} = - \rho (\nabla \cdot {\bf u})$
If you believe that that the fluid have a simple equation of state of the form $p = p(\rho)$ bijectively (which is resoanble for liquid water on normal conditions, for example), you can invert this relation and write $\rho = \rho(p)$, and now comes the conection with the previous explation. 
$\frac{D\rho}{Dt} = \frac{\partial \rho}{\partial p}\frac{Dp}{Dt} = - \rho (\nabla \cdot {\bf u})$
Reorganizing all terms we get:
$\frac{1}{\rho} \frac{\partial \rho}{\partial p}\frac{Dp}{Dt}  = Y^{-1} \frac{Dp}{Dt}= - \nabla \cdot {\bf u}$
Therefore, the divergence of velocity, which is the usual 'cinematic' sensor for compressibility, is proportional to the Bulk modulus of the fluids corresponding thermodynamic equation of state. 
Now, lets imagine that you have a large $Y$, like $10^9\ {\rm Pa}$ large (because they are). If you have $\nabla \cdot {\bf u}$ anywhere near a large value, like $0.1 \ s^{-1}$, it would be necessary huge pressure time variations to cope with such a large divergence, which only occurs you also have huge density variations. In mostly daily cases, you have pressure diferences of the order of $10^0 - 10^3\ {\rm Pa}$, also ocurring over a few seconds. The corresponding $\nabla \cdot {\bf u}$ are around $10^{-9} - 10^{-6}\ s^{-1}$.
So, this is the overall conection between 'stiffness' of the fluid, in the sense of the bulk modulus, and it's 'cinematic' incompressibility condition. My order of maginitude estimates may be a bit off, but at least they should be close enought to give a decent feel of what is going on.
A: The simplest definition of incompressibility in my opinion is to express the invariance of volume with regards to pressure:
$$\frac{\partial V}{\partial p} \approx 0$$
As $\rho=\frac{m}{V} \implies V=m \rho$ with $m=\text{constant}$, then:
$$\frac{\partial (m\rho)}{\partial p} \approx 0$$
$$m\frac{\partial \rho}{\partial p} \approx 0$$
$$\implies \frac{\partial \rho}{\partial p} \approx 0$$
