Zero velocity divergence for incompressible flow is derived from conservation of energy equation or conservation of mass equation?

I'm a bit confused about incompressible flow definition. In many textbooks or scientific articles, they simply claim that the incompressibility condition for Navier-Stokes equation is:

$\nabla \cdot \mathbf{u} = 0$

But, nobody says explicitly how to prove that incompressible velocity field should be divergence free. Here are my findings to derive this equation from basic fundamentals of physics:

For incompressible fluid: from thermodynamics equation of state, we know that density should only depends on equilibrium potentials of pressure and temperature:

$\rho = \rho(P,T)$

If we take material derivative from this equation:

$\frac{D \rho}{D t} = (\frac{\partial \rho}{\partial P})_{T} \frac{D P}{D t} + (\frac{\partial \rho}{\partial T})_{P} \frac{D T}{D t}$

For an isothermal and incompressible fluid:

Incompressible fluid :$(\frac{\partial \rho}{\partial P})_{T} = 0$

Isothermal fluid: $\frac{D T}{D t} = 0$

So finally, these conditions will lead to:

$\frac{D \rho}{D t} = 0$

But, from mass conservation equation (continuity equation), we have:

$\frac{D \rho}{D t} = -\rho \nabla \cdot \mathbf{u}$

As a result: $\nabla \cdot \mathbf{u} = 0$

For compressible fluid: from internal energy balance equation, we know:

$\rho \frac{D e}{D t} = -\nabla \cdot \mathbf{q} + \sigma \cdot (\nabla \otimes \mathbf{u})$

Where $e$ is the internal energy of the system, which is equal to enthalpy at the constant pressure condition, $\mathbf{q}$ is the thermal heat flux, $\sigma$ is the Cauchy stress tensor, which is equal to: $\sigma = -P \mathbf{I} + \tau$, where $P$ is the pressure and $\tau$ is the deviatoric stress.

For isothermal compressible fluid: $\frac{D e}{D t} = 0$ and $\nabla \cdot \mathbf{q} = 0$.

As a result: $\sigma \cdot (\nabla \otimes \mathbf{u}) = 0$.

For a Newtonian compressible fluid, we have: $\tau = \mu (\nabla \mathbf{u} + (\nabla \mathbf{u})^{T}) + \zeta (\nabla \cdot \mathbf{u}) \mathbf{I}$.

Where $\mu$ is the shear viscosity and $\zeta$ is the bulk viscosity.

Finally, the term $\sigma \cdot (\nabla \otimes \mathbf{u})$ could be expanded as:

$\sigma \cdot (\nabla \otimes \mathbf{u}) = -P (\nabla \cdot \mathbf{u}) + \zeta (\nabla \cdot \mathbf{u})^{2} + 2 \mu S \cdot (\nabla \otimes \mathbf{u})$.

Where $S$ is the shear rate tensor, which is defined as: $S = \frac{1}{2}(\nabla \mathbf{u} + (\nabla \mathbf{u})^{T})$.

Finally, we have:

$\sigma \cdot (\nabla \otimes \mathbf{u}) = -P (\nabla \cdot \mathbf{u}) + \zeta (\nabla \cdot \mathbf{u})^{2} + 2 \mu S \cdot (\nabla \otimes \mathbf{u}) = 0$

or

$(P - \zeta (\nabla \cdot \mathbf{u})) (\nabla \cdot \mathbf{u}) = 2 \mu S \cdot (\nabla \otimes \mathbf{u})$

Now, we could argue that at low velocities (low Mach number), the viscous heat dissipation term ($2 \mu S \cdot (\nabla \otimes \mathbf{u})$) is negligble. As a result, we have:

$(P - \zeta (\nabla \cdot \mathbf{u})) (\nabla \cdot \mathbf{u}) = 0$

Finally, we should have:

$P = \zeta (\nabla \cdot \mathbf{u})$

or

$\nabla \cdot \mathbf{u} = 0$

The first equation ($P = \zeta (\nabla \cdot \mathbf{u})$) is contradictory because the thermodynamics pressure $P$ should only depends on equilibrium potentials and not kinetics variables like velocity. As a result, we have:

$\nabla \cdot \mathbf{u} = 0$

So it proves that compressible fluid could be treated as an incompressible flow, when its velocity remains small in comparison to speed of sound (low Mach number).

So my question is that why in classical fluid mechanics textbooks, always people claim the divergence free condition is a direct consequence of mass conservation?! Right now, I show that it could be derived with minimum assumptions from energy conservation equation. Any idea or suggestion is appreciated.

Edition:

Proof of negligible viscous dissipation heat rate:

Full internal energy balance equation:

$\rho \frac{\partial e}{\partial t} + \rho \mathbf{u} \cdot \nabla e = -\nabla \cdot \mathbf{q} + \sigma \cdot (\nabla \otimes \mathbf{u})$

The internal energy will be equal to enthalpy at the constant pressure. As a result we have:

$e = C_{p} \Delta T$

Where $C_{p}$ is the constant pressure specific heat capacity and $\Delta T$ is the temperature difference from reference point. Also by assuming the Fourier heat transfer law, we have:

$\mathbf{q} = -k \nabla T$

Where $k$ is the heat conductivity.

The internal energy equation could be rewritten as:

$\rho C_{p} \frac{\partial T}{\partial t} + \rho C_{p} \mathbf{u} \cdot \nabla T = k \nabla^{2} T + \sigma \cdot (\nabla \otimes \mathbf{u})$

If we put the expansion of $\sigma \cdot (\nabla \otimes \mathbf{u})$ for a Newtonian compressible fluid, finally we will find:

$\rho C_{p} \frac{\partial T}{\partial t} + \rho C_{p} \mathbf{u} \cdot \nabla T = k \nabla^{2} T -P (\nabla \cdot \mathbf{u}) + \zeta (\nabla \cdot \mathbf{u})^{2} + 2 \mu S \cdot (\nabla \otimes \mathbf{u})$

This equation could be nondimensionalized by taking:

$\theta = \frac{\Delta T}{\Delta T_{0}}$, $t^{'} = \frac{t}{t_{0}}$, $\mathbf{u}^{'} = \frac{\mathbf{u}}{u_{0}}$, $\nabla^{'} = \epsilon \nabla$, $P^{'} = \frac{P}{P_{0}}$, $S^{'} = \frac{\epsilon S}{u_{0}}$

So the above equation could be rewritten as:

$\frac{\rho C_{p} \Delta T_{0}}{t_{0}} \frac{\partial \theta}{\partial t^{'}} + \frac{\rho C_{p} u_{0} \Delta T_{0}}{\epsilon} \mathbf{u}^{'} \cdot \nabla^{'} \theta = \frac{k \Delta T_{0}}{\epsilon^{2}} {\nabla^{'}}^{2} \theta - \frac{P_{0} u_{0}}{\epsilon} P^{'} (\nabla^{'} \cdot \mathbf{u}^{'}) + \frac{\zeta u_{0}^{2}}{\epsilon^{2}} (\nabla^{'} \cdot \mathbf{u}^{'})^{2} + \frac{2 \mu u_{0}^{2}}{\epsilon^{2}} S^{'} \cdot (\nabla^{'} \otimes \mathbf{u}^{'})$

Finally, by taking $\alpha = \frac{k}{\rho C_{p}}$ and its nondimensionalized form $\alpha^{'} = \frac{\alpha t_{0}}{\epsilon^{2}}$, we have:

$\frac{1}{\alpha^{'}} \frac{\partial \theta}{\partial t^{'}} + Pe \mathbf{u}^{'} \cdot \nabla^{'} \theta = {\nabla^{'}}^{2} \theta - \frac{P_{0} u_{0} \epsilon}{k \Delta T_{0}} P^{'} (\nabla^{'} \cdot \mathbf{u}^{'}) + Br_{bulk} (\nabla^{'} \cdot \mathbf{u}^{'})^{2} + 2Br_{shear} S^{'} \cdot (\nabla^{'} \otimes \mathbf{u}^{'})$

Where Peclet, bulk Brinkman, and shear Brinkman numbers are defined as:

$Pe = \frac{u_{0} \epsilon}{\alpha}$

$Br_{bulk} = \frac{\zeta u_{0}^{2}}{k \Delta T_{0}}$

$Br_{shear} = \frac{\mu u_{0}^{2}}{k \Delta T_{0}}$

Finally, for an isothermal fluid: $\theta = \theta_{0} = const.$ and we will have:

$2Br_{shear} S^{'} \cdot (\nabla^{'} \otimes \mathbf{u}^{'}) = (\frac{P_{0} u_{0} \epsilon}{k \Delta T_{0}} P^{'} - Br_{bulk} (\nabla^{'} \cdot \mathbf{u}^{'})) (\nabla^{'} \cdot \mathbf{u}^{'})$

For low Mach number Brinkman numbers (both shear and bulk) are negligible. In fact, Brinkman number should be at least in order of $O(1)$ to consider viscous dissipation heat rate in internal energy equation. For conventional fluids at low Mach number regime Brinkman number is in the order of $O(10^{-3})$, which is negligible.

As a result, we should have:

$\frac{P_{0} u_{0} \epsilon}{k \Delta T_{0}} P^{'} = Br_{bulk} (\nabla^{'} \cdot \mathbf{u}^{'})$

or

$\nabla^{'} \cdot \mathbf{u}^{'} = 0$

Again, we could argue that the first equation ($\frac{P_{0} u_{0} \epsilon}{k \Delta T_{0}} P^{'} = Br_{bulk} (\nabla^{'} \cdot \mathbf{u}^{'})$) could be not true because the thermodynamics pressure should only depend on equilibrium potential and not kinetics variables (e.g. velocity). As a result finally we will find:

$\nabla^{'} \cdot \mathbf{u}^{'} = 0$

or in its dimensional form:

$\nabla \cdot \mathbf{u} = 0$

If we look at the mass conservation equation in an Eulerian framework (because it's easier), we have:

$$\frac{\partial \rho}{\partial t} + \nabla \cdot \rho \vec{u} = 0$$

Where, if density is constant, obviously $\partial \rho / \partial t = 0$ and then we can factor the $\rho$ out of the derivative and divide, which gives:

$$\nabla \cdot \vec{u} = 0$$

In other words, this is exactly true and no approximations or assumptions are made, other than the fact that density is constant.

On the other hand, when you derive the same expression via other means, like you tried to do with the momentum equation, you have to introduce many assumptions and approximations. You assumed it was low Mach number. You assumed viscous heating was negligible.

But that isn't required to be the case if density is constant. You could have a fast, constant density flow, with significant viscous heating (mathematically at least). So the conservation of mass is the simplest, least restrictive way to say that for a constant density flow, $\nabla \cdot \vec{u} = 0$. Other ways are more restrictive and less direct.

In other words, using conservation of mass means you say "Assuming a constant density fluid..." whereas assuming low-Mach and negligible viscous heating, you are saying "Density can be shown as constant for a flow that is...," which are two different statements that both end up giving $\nabla \cdot \vec{u} = 0$ for very different reasons.

The other thing that is worth pointing out, because it comes up often, is that incompressible fluid is a vague term. It can mean either constant density or low-Mach. The former means density never changes. The latter means the flow is relatively low speed but allows density to change as a function of temperature, but not pressure. You get different equations and different behavior depending on which form of "incompressible" you want to look at.

• But, it turns out that constant density (strictly constant density means: $\frac{\partial \rho}{\partial t} = 0$ and $\nabla \rho =0$) is just a toy model which does not have physical meaning. In fact, incompressible flow does not require to have a constant density (cref [en.wikipedia.org/wiki/Incompressible_flow]). – Mehrdad Yousefi Aug 8 '18 at 14:43
• @MehrdadYousefi Yes, that's exactly right (and why I made the point to clarify the differences between constant density incompressible and low-Mach incompressible). Ultimately, even low-Mach is a mathematical "toy" because a real flow doesn't suddenly decide it's low-Mach or not. The compressible equations are always true, anything else is to make our lives easier when analyzing them. Whether it has physical meaning or not depends on whether our assumptions used to make life easier are valid somewhere. – tpg2114 Aug 8 '18 at 14:49
• Incompressible, linearized, potential flow, for example, is perhaps the ultimate mathematical toy problem. But still describes a few things well enough to be useful. – tpg2114 Aug 8 '18 at 14:50
• And, pedantically, even the compressible Navier-Stokes equations are "unphysical toy models" of the real equations, which you would have to track every collision between every molecule exactly. But it holds for certain classes of flows well enough to be useful. – tpg2114 Aug 8 '18 at 14:51
• I agree. I just pointed to low Mach number flows because I'm working on an application of calculating pressure at low Mach number regime from CFD simulations and I had a really hard time to wrap my head around how an incompressible flow with negligble density fluctuation (your constant density assumption) could have spatial and temporal pressure gradient?! – Mehrdad Yousefi Aug 8 '18 at 14:52

Physical meaning of divergence is rate of change of control volume per unit volume. If density is not changing then rate of charge of control volume will be zero this is directly from conservation of mass.