Reynolds averaging and the RANS equations

Averaging the Navier–Stokes equations leaves the number of equations unchanged but adds new unknowns — introducing the closure problem that motivates all of turbulence modeling.

Opening

The Navier–Stokes equations from Chapter 3 describe every flow in principle, but in turbulence they require resolving every eddy scale, which is far too expensive for industrial computation. So in 1894 Osborne Reynolds proposed a detour: take a time average. By the end of this chapter the reader can state in one sentence why averaging adds unknowns faster than equations, and why Chapter 6 needs “turbulent viscosity models.”

Main 1 — Reynolds decomposition: splitting the signal into a mean and a fluctuation

The core idea is simple. The velocity $u_i$ at a point inside a turbulent flow fluctuates rapidly, but if we average over a long enough time, we can split it into a mean $\bar{u}_i$ (a symbol with a bar on top, read “u-bar”) and a fluctuation $u'_i$ (“u-prime”):

$$u_i = \bar{u}_i + u'_i, \qquad \overline{u'_i} = 0$$

In plain words: “average the signal, define the leftover as turbulent noise; by construction the noise averages to zero.” The second equality is not a new assumption — it is a direct consequence of the decomposition, because the average of whatever remains after subtracting the average has to be zero.

We apply the same split to every flow variable: pressure $p$, temperature $T$, concentration $c$, and so on. This is called Reynolds decomposition.

Main 2 — The new term that averaging creates

Substitute the Reynolds decomposition into the incompressible Navier–Stokes momentum equation from Chapter 3 and take the average of both sides. Averaging commutes with differentiation ($\overline{\partial f / \partial t} = \partial \bar{f} / \partial t$), so the time-derivative, pressure-gradient, and viscous terms all survive with the average sign simply pushed inside.

The problem is the convective term $u_j \partial u_i / \partial x_j$. Because it is a product of two velocities, after decomposition it becomes

$$\overline{(\bar{u}_j + u'_j)\,\frac{\partial (\bar{u}_i + u'_i)}{\partial x_j}}$$

and the cross terms that contain a single prime vanish thanks to $\overline{u'_i} = 0$. The term with two primes, $\overline{u'_j \, \partial u'_i / \partial x_j}$, remains. Using the continuity equation ($\partial u'_j / \partial x_j = 0$) we can rewrite it as a derivative of a product, $\partial \overline{u'_i u'_j} / \partial x_j$.

Collecting all the averaged pieces gives the RANS equation (Reynolds-Averaged Navier–Stokes):

$$\frac{\partial \bar{u}_i}{\partial t} + \bar{u}_j \frac{\partial \bar{u}_i}{\partial x_j} = -\frac{1}{\rho}\frac{\partial \bar{p}}{\partial x_i} + \nu \frac{\partial^2 \bar{u}_i}{\partial x_j \partial x_j} - \frac{\partial \overline{u'_i u'_j}}{\partial x_j}$$

Here $\rho$ (rho, density) and $\nu$ (nu, kinematic viscosity) are the same as in Chapter 3. The form is nearly identical to the original Navier–Stokes equation, with one new term added on the right-hand side: $-\partial \overline{u'_i u'_j} / \partial x_j$.

Main 3 — Reynolds stress and the closure problem

The quantity $\overline{u'_i u'_j}$ inside that new term is called the Reynolds stress tensor. It is named a “stress” because it transports momentum the way a viscous stress does, not because it is an actual molecular stress — it is the mean momentum exchange carried by macroscopic eddies.

The Reynolds stress is a symmetric tensor, so out of its 9 components only 6 are independent ($\overline{u'_1 u'_1}, \overline{u'_2 u'_2}, \overline{u'_3 u'_3}, \overline{u'_1 u'_2}, \overline{u'_1 u'_3}, \overline{u'_2 u'_3}$).

Now count the unknowns.

	Original Navier–Stokes	RANS
Equations	3 (momentum) + 1 (continuity) = 4	still 4
Unknowns	$u_1, u_2, u_3, p$ → 4	$\bar{u}_1, \bar{u}_2, \bar{u}_3, \bar{p}$ + 6 Reynolds stresses = 10

We started with 4 equations for 4 unknowns. After one averaging step, the same 4 equations now confront 10 unknowns. This is the closure problem — the system is not closed.

Chapter 6 covers several recipes (mixing length, k-ε, k-ω) for modeling the 6 Reynolds stresses as functions of the mean velocity field $\bar{u}_i$ alone, forcing the system shut. RANS is the workhorse of industrial CFD not because RANS is an exact equation but because that modeling step happens to work well in practice.

In Python

The Reynolds decomposition itself can be verified with a single 1D synthetic signal. After subtracting the mean, the fluctuation has mean close to zero and a positive mean-square (which corresponds to a Reynolds stress component).

import numpy as np

rng = np.random.default_rng(0)

# 1D synthetic turbulent signal: mean 1.0 + deterministic oscillation + Gaussian noise
t = np.linspace(0, 10, 1000)
u = 1.0 + 0.3 * np.sin(t) + 0.1 * rng.standard_normal(1000)

# Reynolds decomposition: split into mean and fluctuation
u_mean = np.mean(u)
u_prime = u - u_mean

# Verify
mean_of_prime = np.mean(u_prime)         # zero by construction (within float tolerance)
reynolds_stress = np.mean(u_prime ** 2)  # variance = one Reynolds stress component u'u'

print(f"Mean velocity              u_bar  = {u_mean:.6f}")
print(f"Mean of fluctuation        <u'>   = {mean_of_prime:.2e}")
print(f"Reynolds stress component  <u'u'> = {reynolds_stress:.6f}")

In the output, <u'> is on the order of $10^{-17}$ — floating-point noise, effectively zero — while <u'u'> is about 0.05, clearly positive. If this signal had been one component of a real 3D flow, that value would be exactly one diagonal entry of the Reynolds stress tensor.

To the next chapter

The closure problem has remained unsolved since 1894. Engineering chose a detour instead: rather than chasing an exact solution, settle for an approximate model that is good enough. Chapter 6: Turbulent viscosity models walks through the three canonical models — mixing length, k-ε, and k-ω — and shows what assumption each one uses to express the six Reynolds stresses as functions of the mean velocity field.