In a fluid mechanics course I found that an incompressible fluid flow means literally:
$$\rho = \text{constant} \quad \forall \vec r,\, \forall t$$
Where $\vec r = (x, y, z)$
In my understanding, this means literally that the fluid density is uniform? (Am I wrong?)
In the other hand we can find also that an incompressible fluid means:
$$\dfrac{D\rho}{Dt} = 0$$ which does not necessarily mean that the fluid density is uniform.
What's wrong here?
Best Answer
The definition of incompressible is often unclear and changes depending on which community uses it. So let's look at some common definitions:
Constant density
This means the density is constant everywhere in space and time. So: $$ \frac{D\rho}{Dt} = \frac{\partial \rho}{\partial t} + \vec{u}\cdot\nabla{\rho} = 0 $$ Because density is constant everywhere in space and time, the temporal derivative is zero, and the spatial gradient is zero.
Low Mach Number
This shows up when the flow velocity is relatively low and so all pressure changes are hydrodynamic (due to velocity motion) rather than thermodynamic. The effect of this is that $\partial \rho / \partial p = 0$. In other words, the small changes in pressure due to flow velocity changes do not change the density. This has a secondary effect -- the speed of sound in the fluid is $\partial p/\partial \rho = \infty$ in this instance. So there is an infinite speed of sound, which makes the equations elliptic in nature.
Although we assume density is independent of pressure, it is possible for density to change due to changes in temperature or composition if the flow is chemically reacting. This means:
$$ \frac{D\rho}{Dt} \neq 0 $$ because $\rho$ is a function of temperature and composition. If, however, the flow is not reacting or multi-component, you will also get the same equation as the constant density case:
$$ \frac{D\rho}{Dt} = 0 $$
Therefore, incompressible can mean constant density, or it can mean low Mach number, depending on the community and the application. I prefer to be explicit in the difference because I work in the reacting flow world where it matters. But many in the non-reacting flow communities just use incompressible to mean constant density.
Example of non-constant density
Since it was asked for an example where the material derivative is zero but density is not constant, here goes:
$$ \frac{D\rho}{Dt} = \frac{\partial \rho}{\partial t} + \vec{u}\cdot\nabla \rho = 0 $$
Rearrange this:
$$ \frac{\partial \rho}{\partial t} = -\vec{u}\cdot\nabla\rho $$
gives a flow where $\rho \neq \text{const.}$ yet $D\rho/Dt = 0$. It has to be an unsteady flow.
Is there another example of steady flow? In steady flow, the time derivative is zero, so you have:
$$ \vec{u}\cdot\nabla\rho = 0 $$
If velocity is not zero, $\vec{u} \neq 0$, then we have $\nabla \rho = 0$ and so any moving, steady flow without body forces (gravity) or temperature/composition differences must have constant density.
If velocity is zero, you can have a gradient in density without any issues. Think of a column of the atmosphere for example -- density is higher at the bottom than the top due to gravity, and there is no velocity. So again, $D\rho/Dt = 0$ but density is not constant everywhere. The challenge here of course is that the continuity equation is not sufficient to describe the situation since it becomes $0 = 0$. You would have to include the momentum equation to incorporate the gravity forces.