# Does the equation of continuity hold for turbulent flows?

My textbook mainly deals with laminar flows. The book derives the equation of continuity, which states that the cross-sectional area times the velocity of a flow is always constant. But nowhere in the derivation does the textbook explicitly assumes that the flow is laminar. So, does the equation hold for turbulent flows too?

The book derives the equation of continuity, which states that the cross-sectional area times the velocity of a flow is always constant. But nowhere in the derivation does the textbook explicitly assumes that the flow is laminar. So, does the equation hold for turbulent flows too?

That is only a special case of the equation of continuity for situation where density is regarded constant and velocity is constant across the cross-section of the pipe. In general, equation of continuity is $\partial_t \rho + \nabla\cdot(\rho \mathbf v) = 0$ and holds true even for turbulent flows if the mass of the fluid is locally conserved.

Continuity is just the principle of conservation of mass in differential form.

The full continuity equation is (in index notation):

$\frac{\partial \rho}{\partial t} = -\frac{\partial }{\partial x_i}(\rho u_i)$

For example, consider an infinitesimal control volume (CV). The equation says that the local $\rho$ (inside the CV) will decrease in time if the flux divergence term $\frac{\partial }{\partial x_i}(\rho u_i)$ is positive. The flux divergence term transports quantities from one region to another in a manner that the net contribution to the system remains fixed whereby if integrated over the domain and there are no sources at the boundaries, their contribution is zero.

This is a very fundamental conservation law, if we assume incompressible flow and that the density of individual fluid particles does not change but each particle may have different densities - you arrive at the form in your book. This is derived from first principles so any flow within these parameters, including turbulent flows, no matter how disorganized, follow the same mass conservation law.

In order to have such a relation, your flow needs to be be stationary, which is never the case for turbulent flows.

The conservation of the mass gives you the local continuity equation.

$$\partial_t \rho+ \nabla . (\rho \vec{v})=0$$

For a stationary problem without sources, Ostrogradsky's theorem allows you to reach: $$\oint_S \vec{v}.d\vec{S}=0$$

But this last step is not possible for turbulent flows.

• Why is the last step not possible for turbulent flows? – Isopycnal Oscillation Mar 4 '14 at 20:07
• Because the problem is never stationary. – Loïc Henriet Mar 7 '14 at 12:47
• Are you talking about a steady flow? The continuity equation for incompressible fluids (which is OPs question) doesn't care about time because the local density of the fluid does not change (although each particle may have different densities). – Isopycnal Oscillation Mar 7 '14 at 19:36
• Hence, $D\rho/Dt = 0$, where D is the total/material derivative. Combining this with your 1st eq., it implies that $\nabla \cdot\mathbf{u} = 0$ (which is the same form for continuity as OPs question). – Isopycnal Oscillation Mar 7 '14 at 19:45