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?

  • $\begingroup$ When just looking at the math: Say that $\rho$ is constant. If the density is constant for all positions and all time then you have a blob in space (filling all the space where it it's position is defined) not moving. It is not changing shape, there are however certain types of ways the matter can flow. The derivative being $0$ means that you could have a $\rho(\vec{r})$ not constant, that just doesn't change wrt time. So it doesn't move but certain flow inside is still allowed. The density can be a time independent non constant function of position. $\endgroup$ – ty. Mar 3 at 19:03

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.

  • $\begingroup$ But without speaking about compressibility effects. Why $\dfrac{D\rho}{Dt} =0$ does not imply $\rho = \text{cst}$ everywhere at any time $\endgroup$ – IamNotaMathematician Mar 3 at 19:22
  • $\begingroup$ Could you please give me an example when $\dfrac{D\rho}{Dt} = 0$ and the desnity is not constant everywhere? $\endgroup$ – IamNotaMathematician Mar 3 at 19:25
  • 1
    $\begingroup$ @IamNotaMathematician Done. These examples come from inspecting the equation. I encourage you to understand what the terms mean in the equations you write out and to think about what happens under various situations (like steady, or no velocity, or spatial gradients exist, etc). Everything will fall out of the equations. $\endgroup$ – tpg2114 Mar 3 at 19:33
  • $\begingroup$ In the last example, I imagine that as follow: $\vec u$ is a vector and $\nabla \rho$ (gradient) is a vector.The scalar product = 0 implies that they are perpendicular that the direction of density change is perpendicular to flow velocity. $\endgroup$ – IamNotaMathematician Mar 3 at 20:00
  • $\begingroup$ @IamNotaMathematician I'd have to think about whether such a flow could exist and be stable. Remember, this is just the continuity equation, a real flow will have to satisfy the momentum equation also. A flow with velocity perpendicular to density gradient will end up generating baroclinic torque which will try to align the velocity and density gradients. So I suppose mathematically looking at this one equation, yes, flow perpendicular to density gradient is a possible solution. But I don't think it would hold for a real flow. $\endgroup$ – tpg2114 Mar 3 at 20:07

The general definition of an incompressible flow is $\frac{D\rho }{Dt}=0$ : the density of a fluid particle does not change along its path.

For example, if $\overrightarrow{v}=v(x)\overrightarrow{{{e}_{x}}}$ and $\rho =\rho (y)$ : the path lines are horizontal lines and on such a line, the density does not change.

The condition $\rho =cst$ is a particular case ("incompressible fluid" rather than "incompressible flow").

But frequently, one mean $\rho =cst$ when one speak of incompressible fluid !

Sorry for my poor english !


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