Reynolds-Averaged Navier–Stokes Equations and Eddy Viscosity
Model
$$ \mathbf{u} = \mathbf{u} \left( t ; \mathbf{x} \right) = \left( u_{1} \left( t ; \mathbf{x} \right) , u_{2} \left( t ; \mathbf{x} \right) , u_{3} \left( t ; \mathbf{x} \right) \right) $$ In particular, suppose that the flow velocity field at time $t$ and spatial coordinates $\mathbf{x} = \left( x_{1} , x_{2} , x_{3} \right)$ in three-dimensional space is represented by the velocity vector above. Similarly, $p : \mathbb{R}^{3} \to \mathbb{R}$ represents the pressure $p = p \left( \mathbf{x} \right)$ exerted at each coordinate. Since $u_{k}$ and $p$ are hard to use for accurately capturing turbulence, we split them into a mean component and a fluctuating component as follows, and this is called the Reynolds decomposition. $$ \begin{align*} u_{k} =& \bar{u}_{k} + u_{k}^{\prime} \\ p =& \bar{p} + p^{\prime} \end{align*} $$ The equations obtained by applying the Reynolds decomposition to the Navier–Stokes equations are called the Reynolds-averaged Navier–Stokes equations (RANS). Written with the Kronecker delta $\delta_{ij}$ and Einstein notation, they read as follows. $$ {\frac{\partial \bar{u}}{\partial t}} + \rho \bar{u}_{j} {\frac{ \partial \bar{u}_{i} }{ \partial x_{j} }} = - {\frac{ \partial \bar{p} }{ \partial x_{i} }} \delta_{ij} + \mu {\frac{ \partial^{2} \bar{u}_{i} }{ \partial x_{j}^{2} }} - \rho {\frac{ \partial }{ \partial x_{j} }} \left( \overline{ u_{i}^{\prime} u_{j}^{\prime} } \right) $$ Here, the last term is called the Reynolds stress term.
Explanation
The RANS equations are averaged equations for describing turbulence, and they serve as the governing equations for the time-averaged flow velocity in the Navier–Stokes equations. The problem is that the Navier–Stokes equations, already complicated enough, become even more complicated with the extra burden of the Reynolds stress term.
Eddy Viscosity
$$ - \rho \left( \overline{ u_{i}^{\prime} u_{j}^{\prime} } \right) = \mu \left( {\frac{ \partial \bar{u}_{i} }{ \partial x_{j} }} + {\frac{ \partial \bar{u}_{j} }{ \partial x_{i} }} \right) = \mu \nabla \bar{u} $$ As an additional assumption introduced to simplify this even slightly, assuming that the Reynolds stress is proportional to the gradient of the mean flow velocity is called the eddy viscosity assumption. The proportionality constant $\mu$ in this case is called the eddy viscosity coefficient. For reference, “eddy” is a word meaning a whirl or vortex, and it is not a proper noun.
Derivation1
Assume that $\mathbf{u}$ is sufficiently differentiable, so that the order of differentiation and integration can be freely interchanged. $$ \bar{u} = \lim_{T \to \infty} {\frac{ 1 }{ T }} \int_{0}^{T} u dt $$ First, the mean is defined as above as a definite integral over time, and the means of the fluctuating components must naturally satisfy $\bar{u^{\prime}} = 0$ and $\bar{p^{\prime}} = 0$. When this averaging is regarded as a kind of operator, it is called the Reynolds operator, and by the linearity of integration the following hold. $$ \begin{align*} \bar{\bar{u}} &= \bar{u} \\ \overline{u + v} &= \bar{u} + \bar{v} \\ \overline{\bar{u} v} &= \bar{u} \bar{v} \\ \overline{\frac{\partial u}{\partial x}} &= \frac{\partial \bar{u}}{\partial x} \end{align*} $$
Since the Navier–Stokes equations assume incompressibility, we have $\nabla \cdot \mathbf{u} = 0$, and writing this in Einstein notation gives the following. $$ {\frac{ \partial u_{i} }{ \partial x_{i} }} = 0 \implies {\frac{ \partial \bar{u}_{i} }{ \partial x_{i} }} = 0 \implies {\frac{ \partial u_{i}^{\prime} }{ \partial x_{i} }} = 0 $$
Navier–Stokes equations: $$ {\frac{ \partial \mathbf{u} }{ \partial t }} + \left( \mathbf{u} \cdot \nabla \right) \mathbf{u} = - \nabla w + \nu \nabla^{2} \mathbf{u} $$
Let us write one row of the Navier–Stokes equations in Einstein notation. $$ \frac{\partial u_{i}}{\partial t} + u_{j} \frac{\partial u_{i}}{\partial x_{j}} = - \frac{1}{\rho} \frac{\partial p}{\partial x_{i}} + \nu \frac{\partial^{2} u_{i}}{\partial x_{j}^{2}} $$ From now on, we will repeat the process of applying the Reynolds decomposition to each term and taking the mean. It is easy to understand by looking at the first term on the left-hand side as an example. $$ \begin{align*} \overline{\frac{\partial u_{i}}{\partial t}} =& \overline{\frac{\partial \bar{u}_{i} + u_{i}^{\prime}}{\partial t}} \\ =& \overline{\frac{\partial \bar{u}_{i}}{\partial t}} + \overline{\frac{\partial u_{i}^{\prime}}{\partial t}} \\ =& \frac{\partial \bar{u}_{i}}{\partial t} + \frac{\partial \bar{u_{i}^{\prime}}}{\partial t} \\ =& \frac{\partial \bar{u}_{i}}{\partial t} + \frac{\partial 0}{\partial t} \\ =& \frac{\partial \bar{u}_{i}}{\partial t} \end{align*} $$ From here on it will not be quite this detailed, but we will still repeat the process as carefully as possible. The key is to focus on how the growing number of terms turn into $0$ or get multiplied. The second term on the left-hand side is the nonlinear term, for which we use a trick based on the chain rule. $$ \begin{align*} & \overline{u_{j} \frac{\partial u_{i}}{\partial x_{j}}} \\ =& \overline{\frac{\partial u_{j} u_{i}}{\partial x_{j}}} - \overline{{\frac{\partial u_{j}}{\partial x_{j}} u_{i}}} \\ =& \overline{\frac{\partial \left( \bar{u}_{j} \bar{u}_{i} + \bar{u}_{j} u_{i}^{\prime} + u_{j}^{\prime} \bar{u}_{i} + u_{j}^{\prime} u_{i}^{\prime} \right)}{\partial x_{j}}} - \overline{{0 \cdot u_{i}}} \\ =& \overline{\frac{\partial \bar{u}_{j} \bar{u}_{i}}{\partial x_{j}}} + \overline{\frac{\partial \left( \bar{u}_{j} u_{i}^{\prime} + u_{j}^{\prime} \bar{u}_{i} \right)}{\partial x_{j}}} + \overline{\frac{\partial u_{j}^{\prime} u_{i}^{\prime} }{\partial x_{j}}} \\ =& \frac{\partial \bar{u}_{j} \bar{u}_{i}}{\partial x_{j}} + \frac{\partial \left( \bar{u}_{j} \bar{u}_{i}^{\prime} + \bar{u}_{j}^{\prime} \bar{u}_{i} \right)}{\partial x_{j}} + \frac{\partial \overline{u_{j}^{\prime} u_{i}^{\prime} } }{\partial x_{j}} \\ =& \bar{u}_{j} \frac{\partial \bar{u}_{i}}{\partial x_{j}} + \frac{\partial \overline{u_{j}^{\prime} u_{i}^{\prime} } }{\partial x_{j}} \end{align*} $$ The first term on the right-hand side can simply be passed over, apart from the attached Kronecker delta. For the second term, let us set $\nu$ aside for convenience and examine it. $$ \begin{align*} & \overline{\frac{\partial^{2} u_{i}}{\partial x_{j}^{2}}} \\ =& \overline{\frac{\partial^{2} \left( \bar{u}_{i} + u_{i}^{\prime} \right)}{\partial x_{j}^{2}}} \\ =& \overline{\frac{\partial^{2} \bar{u}_{i}}{\partial x_{j}^{2}}} + \overline{\frac{\partial^{2} u_{i}^{\prime}}{\partial x_{j}^{2}}} \\ =& \frac{\partial^{2} \bar{u}_{i}}{\partial x_{j}^{2}} + \frac{\partial^{2} \bar{u_{i}^{\prime}}}{\partial x_{j}^{2}} \\ =& \frac{\partial^{2} \bar{u}_{i}}{\partial x_{j}^{2}} + \frac{\partial^{2} 0}{\partial x_{j}^{2}} \\ =& \frac{\partial^{2} \bar{u}_{i}}{\partial x_{j}^{2}} \end{align*} $$ Let us substitute all the results so far. $$ \frac{\partial \bar{u}_{i}}{\partial t} + \bar{u}_{j} \frac{\partial \bar{u}_{i}}{\partial x_{j}} + \frac{\partial \overline{u_{j}^{\prime} u_{i}^{\prime} } }{\partial x_{j}} = - \frac{1}{\rho} \frac{\partial \bar{p}}{\partial x_{i}} \delta_{ij} + \nu \frac{\partial^{2} \bar{u}_{i}}{\partial x_{j}^{2}} $$ Moving the Reynolds stress term, currently the last term on the left-hand side, to the other side yields the Reynolds-averaged Navier–Stokes equations.
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