I'm looking for the complete set (x,y,z component) of the Navier-Stoke equations under the Eddy Viscosity hypothesis to model turbulent fluid flow.

I found the following, but I have a really hard time believing the transition from the next to last set of equations to the last set of equations. I used the eddy viscosity hypothesis but I could not get terms to cancel to give the last set of equations. Using the eddy viscosity hypothesis: \begin{equation} - \overline{u'_i u'_j} = \nu_t \, \left( \frac{\partial \overline{u_i}}{\partial x_j} + \frac{\partial \overline{u_j}}{\partial x_i} \right) - \frac{2}{3} k \delta_{ij} = \nu_t \, \left( \frac{\partial \overline{u_i}}{\partial x_j} + \frac{\partial \overline{u_j}}{\partial x_i} \right) -\frac{1}{3} \left( \overline{u'^2} + \overline{v'^2} \right)\delta_{ij} \end{equation}

$\overline{u'^2} = -2\nu_T \frac{\partial \overline{u}}{\partial x}$, $\overline{v'^2} = -2\nu_T \frac{\partial \bar{v}}{\partial y}$ and $\overline{u'v'} = -\nu_T ( \frac{\partial \bar{u}}{\partial y} + \frac{\partial \bar{v}}{\partial x} ) = \overline{v'u'}$

For example, in two-dimensional flow, for the x-momentum equation

\begin{equation} -\frac{\partial \overline{u'^2}}{\partial x} = \frac{\partial \nu_T}{\partial x}\frac{\partial \bar{u}}{\partial x} + \nu_T \frac{\partial^2 \bar{u}}{\partial x^2} \end{equation}

\begin{equation} -\frac{\partial \overline{u'v'} }{\partial y} = \frac{\partial \nu_T}{\partial y} ( \frac{\partial \bar{u}}{\partial y} + \frac{\partial \bar{v}}{\partial x} ) + \nu_T ( \frac{\partial^2 \bar{u}}{\partial y^2} + \frac{\partial^2 \bar{v}}{\partial y \partial x} ) \end{equation}

I can't see how the sum of these two terms simplifies to give the first equation in the last set of equations on page 2 where the viscosities are added in each term.


1 Answer 1


The reference that you show, makes it overly complicated to see the steps that are taken. In fluid dynamics, it is worth to familiarize yourself with index notation. You can reduced the three sets of equations for $u$, $v$ and $w$ to a single equation for $u_i$, where $i\in{1,2,3}$. Furthermore, you sum over the repeated index (often $j$).

In non-conservative form (and divided by $\rho$, the Navier-Stokes equations can then simply be written as $$\frac{\partial u_i}{\partial t}+u_j\frac{\partial u_i}{\partial x_j}=-\frac{\partial p}{\partial x_i} + \nu \frac{\partial^2 u_i}{\partial x_j^2}$$

The continuity equation in index notation would reduce to $$\frac{\partial u_j}{\partial x_j}=0$$

Applying Reynolds averaging on the linear terms is straightforward, as the mean of the fluctuating part is always zero. Therefore, I will now only address the non-linear term.

Working out the nonlinear term using Reynolds decomposition gives

$$ u_j\frac{\partial u_i}{\partial x_j} = (\overline{u_j}+u'_j)\frac{\partial (\overline{u_i}+u'_i)}{\partial x_j}$$

which expands to four terms $$u_j\frac{\partial u_i}{\partial x_j} = \overline{u_j}\frac{\partial \overline{u_i}}{\partial x_j}+ u'_j\frac{\partial \overline{u_i}}{\partial x_j}+ \overline{u_j}\frac{\partial u'_i}{\partial x_j}+ u'_j\frac{\partial u'_i}{\partial x_j}$$

Reynolds averaging this equation, does not have an effect ong the first term.

The second term vanishes, because $\overline{u'_j}=0$ and the third because $\overline{u'_i}=0$.

The fourth term can be written in conservative form as well to give $\displaystyle\frac{\partial \overline{u'_i u'_j}}{\partial x_j}$, because $\displaystyle\frac{\partial u'_j}{\partial x_j}=0$ (from continuity). On this last term you apply the Boussinesq or eddy viscosity hypothesis.

  • $\begingroup$ I understand how he gets to the conclusion of the equations on the middle of the second page. What I don't understand is how he is able to simplify them using the Boussinesq or eddy viscosity hypothesis to the last set of equations on page 2. I have updated my original post. $\endgroup$
    – l3win
    Jun 9, 2013 at 17:32
  • $\begingroup$ The Boussinesq hypothesis is no more than substituting $\nu_\tau\frac{\partial \overline{u_i}}{\partial x_j}$ for $\overline{u'_i u'_j}$ $\endgroup$
    – Bernhard
    Jun 9, 2013 at 17:58
  • $\begingroup$ Your equations are incompressible, which simplifies it a bit. $\endgroup$
    – Bernhard
    Jun 9, 2013 at 18:09
  • $\begingroup$ only the term $\frac{\partial U_k}{\partial x_k}$ vanishes due to incompressibility? $\endgroup$
    – l3win
    Jun 9, 2013 at 18:13
  • $\begingroup$ is it not:$-\rho \overline{u'_i u'_j} = \mu_t \, \left( \frac{\partial U_i}{\partial x_j} + \frac{\partial U_j}{\partial x_i} - \frac{2}{3} \frac{\partial U_k}{\partial x_k} \delta_{ij} \right) - \frac{2}{3} \rho k \delta_{ij}$ which simplifies to $- \overline{u'_i u'_j} = \nu_t \, \left( \frac{\partial U_i}{\partial x_j} + \frac{\partial U_j}{\partial x_i} \right) - \frac{2}{3} k \delta_{ij} = \nu_t \, \left( \frac{\partial U_i}{\partial x_j} + \frac{\partial U_j}{\partial x_i} \right) -\frac{1}{3} \left( \overline{u'^2} + \overline{v'^2} \right)\delta_{ij} $ $\endgroup$
    – l3win
    Jun 9, 2013 at 18:17

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