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 $\hspace{0.5cm}\frac{\partial p}{\partial \rho} \approx \infty$


2 Answers 2


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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.