Navier-Stokes - Reynolds decomposition of energy equation

Question

I am trying to apply the Reynolds decomposition to the Navier-Stokes equations for incompressible flows. At the moment I am doing that for the energy equation following the book Viscous Fluid Flow by Frank M. White (3rd edition).

For incompressible flows with constant properties and no heat source the energy equation reads:

$\rho c_p\dfrac{\mathrm{D}T}{\mathrm{D}t}=k\nabla^2T+\Phi$,

where $\Phi$ is the dissipation function (given for a newtonian fluid):

$\Phi=\mu\left[2\left(\dfrac{\partial u}{\partial x}\right)^2+2\left(\dfrac{\partial v}{\partial y}\right)^2+2\left(\dfrac{\partial w}{\partial z}\right)^2+\left(\dfrac{\partial u}{\partial y}+\dfrac{\partial v}{\partial x}\right)^2+\left(\dfrac{\partial u}{\partial z}+\dfrac{\partial w}{\partial x}\right)^2+\left(\dfrac{\partial v}{\partial z}+\dfrac{\partial w}{\partial y}\right)^2\right]$.

The term with $\lambda$ is not included in $\Phi$ because of the incompressible flow hypothesis.

Applying the Reynolds decomposition ($a=\overline{a}+a^{\prime}$ ) and averaging the energy equation, the result should be:

$\rho c_p\dfrac{\mathrm{D}\overline{T}}{\mathrm{D}t}=-\dfrac{\partial}{\partial x_i}\left(q_i\right)+\overline{\Phi}$,

where

$q_i=-k\dfrac{\partial \overline{T}}{\partial x_i}+\rho c_p\overline{u_i^{\prime}T^{\prime}}$,

and

$\overline{\Phi}=\dfrac{\mu}{2}\overline{\left(\dfrac{\partial \overline{u_i}}{\partial x_j} + \dfrac{\partial u_i^{\prime}}{\partial x_j} + \dfrac{\partial \overline{u_j}}{\partial x_i} + \dfrac{\partial u_j^{\prime}}{\partial x_i}\right)^2}$.

My question is: how should the Einstein summation convention be expanded in the last expression? I want to double check my derivation and I am not sure about how to deal with that.

EDIT

Just to clarify what I am looking for, I would like to understand how to expand the following expression in Einstein summation convention form:

$\dfrac{\partial \overline{u_i}}{\partial x_j} + \dfrac{\partial u_i^{\prime}}{\partial x_j} + \dfrac{\partial \overline{u_j}}{\partial x_i} + \dfrac{\partial u_j^{\prime}}{\partial x_i}$

Answer 1

Using ISBN 978-2-85428-483-6, it is more obvious to use the symmetrical tensor: $s_{ij} = \frac{1}{2} \cdot ( \frac{\partial u_i}{x_j} + \frac{\partial u_j}{x_i})$ and $S_{ij} = \frac{1}{2} \cdot ( \frac{\partial U_i}{x_j} + \frac{\partial U_j}{x_i})$ and $\overline{S}_{ij} = \frac{1}{2} \cdot ( \frac{\partial \overline{U}_i}{x_j} + \frac{\partial \overline{U}_j}{x_i})$, we have $\overline{\Phi} = \overline{\phi + \varphi}$ with $\phi = 2 \mu \overline{S}_{ij} \overline{S}_{ij}$ and $\varphi = 2\mu_{ij} \mu_{ij}$ so using your second expression: $\varphi=\mu\left[2\left(\dfrac{\partial \overline{u}}{\partial x}\right)^2+2\left(\dfrac{\partial \overline{v}}{\partial y}\right)^2+2\left(\dfrac{\partial \overline{w}}{\partial x}\right)^2+\left(\dfrac{\partial \overline{u}}{\partial y}+\dfrac{\partial \overline{v}}{\partial x}\right)^2+\left(\dfrac{\partial \overline{u}}{\partial z}+\dfrac{\partial \overline{w}}{\partial x}\right)^2+\left(\dfrac{\partial \overline{v}}{\partial z}+\dfrac{\partial \overline{w}}{\partial y}\right)^2\right]$. The same applies for $\phi = \mu\left[2\left(\dfrac{\partial \overline{U}}{\partial x}\right)^2+2\left(\dfrac{\partial \overline{V}}{\partial y}\right)^2+2\left(\dfrac{\partial \overline{W}}{\partial x}\right)^2+\left(\dfrac{\partial \overline{U}}{\partial y}+\dfrac{\partial \overline{V}}{\partial x}\right)^2+\left(\dfrac{\partial \overline{U}}{\partial z}+\dfrac{\partial \overline{W}}{\partial x}\right)^2+\left(\dfrac{\partial \overline{V}}{\partial z}+\dfrac{\partial \overline{W}}{\partial y}\right)^2\right]$