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What is the exact definition of an incompressible fluid flow ? Is it a flow with constant density OR Is it a flow with divergence free velocity field OR Is it a flow with Mach number less than 0.3?

As per continuity eqn,

$$\frac{\partial \rho}{\partial t}+ \nabla \cdot \rho \overrightarrow{V}=0$$

If I assume density as constant it will lead to a divergence free vector field

$\nabla \cdot \overrightarrow{V}=0$

In certain places it is defined as $\hspace{1cm}$ $Ma<0.3$ with an assumption of $\frac{\partial \rho}{\partial p} \approx 0\hspace{0.5cm}$ or $\hspace{0.5cm}\frac{\partial p}{\partial \rho} \approx \infty$

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An incompressible flow is a flow in which the volume of the fluid elements does not change over time. That means that for some sufficiently small (sometimes called infinitesimal) 3-dimensional region $\Omega$ it holds

$\frac{d}{dt} \int_\Omega 1dV = 0$.

This expression you can rewrite with Reynold's transport theorem to

$\int_\Omega \frac {\partial 1}{\partial t} dV+\int_{\partial \Omega} 1\vec{v}d \vec{S}=\int_{\partial \Omega} \vec{v}d \vec{S}=0$

with a surface element $d \vec{S}$ that points into the normal direction of the boundary region $\partial \Omega$. Gauss theorem implies

$\int_\Omega \nabla \vec{v}dV = 0$

and hence $\nabla \vec{v} = 0$.

This is the creterion of incompressibility.

Constant density (in case of a 1-component, nonreactive flow) will follow immediately from this definition.

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The continuity eq. can be rewritten in terms of the material derivative as

$$ \frac{d\ln\rho}{dt}~=~-\vec{\nabla}\cdot \vec{v}. $$

An incompressible flow means by definition that each sides of the above equation is zero.

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