Reynolds-averaged Navier–Stokes equations

Remarks

The derivation writes the convective term as $\partial_j (u_i u_j)$, which is allowed because $\partial_j u_j = 0$, and uses $\overline{u_i u_j} = \bar u_i \bar u_j + \overline{u_i' u_j'}$. The extra term is the divergence of the Reynolds stress $-\rho \overline{u_i' u_j'}$ divided by $\rho$. There are four equations for the ten unknowns $\bar u_i$, $\bar p$ and the six independent components of $\overline{u_i' u_j'}$, so the system is not closed.

Example: for statistically steady, fully developed flow in a channel $0 < x_2 < 2h$ with mean velocity $\bar u_1(x_2)$, the equations reduce to $$0 = -\frac{1}{\rho} \frac{d \bar p}{d x_1} + \frac{d}{d x_2} \left( \nu \frac{d \bar u_1}{d x_2} - \overline{u_1' u_2'} \right),$$ so the total shear stress varies linearly across the channel. Non-example: replacing $\overline{u_i u_j}$ by $\bar u_i \bar u_j$, that is, dropping the Reynolds stress, gives the Navier–Stokes equations for $\bar u$, which are not the averaged equations of a turbulent flow.