# Navier-Stokes - Reynolds decomposition of energy equation

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:

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

## 1 Answer

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 $$\phi = \mu\left[2\left(\dfrac{\partial \overline{U}}{\partial x}\right)^2+2\left(\dfrac{\partial \overline{V}}{\partial y}\right)^2+2\left(\dfrac{\partial \overline{W}}{\partial x}\right)^2+\left(\dfrac{\partial \overline{U}}{\partial y}+\dfrac{\partial \overline{V}}{\partial x}\right)^2+\left(\dfrac{\partial \overline{U}}{\partial z}+\dfrac{\partial \overline{W}}{\partial x}\right)^2+\left(\dfrac{\partial \overline{V}}{\partial z}+\dfrac{\partial \overline{W}}{\partial y}\right)^2\right]$$

• +1 Better use the same notation as in the question. – Deep Feb 19 at 4:19
• I am not sure I understand the notation you used. I tried to look up the book you mentioned but unfortunately I could not find any resource available. Why you distinguish between $u$ and $U$? I have edited my question to clarify what I am looking for. – Francesco Feb 19 at 11:38
• $\dfrac{\partial \overline{u_i}}{\partial x_j} = \dfrac{\partial \overline{u}}{\partial x} + \dfrac{\partial \overline{u}}{\partial y} + \dfrac{\partial \overline{u}}{\partial z} + \dfrac{\partial \overline{v}}{\partial x} + \dfrac{\partial \overline{v}}{\partial y} + \dfrac{\partial \overline{v}}{\partial z} + \dfrac{\partial \overline{w}}{\partial x} + \dfrac{\partial \overline{w}}{\partial y} + \dfrac{\partial \overline{w}}{\partial z}$ Notice how components are summed twice. Same applies for $u'$. – PackSciences Feb 19 at 11:46
• Thanks for showing the expansion of $\dfrac{\partial \overline{u_i}}{\partial x_j}$. However what about $\dfrac{\partial \overline{u_j}}{\partial x_i}$? Does it expand in the same way? If so, is it true that $\dfrac{\partial \overline{u_i}}{\partial x_j}+\dfrac{\partial u^{\prime}_i}{\partial x_j}+\dfrac{\partial \overline{u_j}}{\partial x_i}+\dfrac{\partial u^{\prime}_j}{\partial x_i}=2\left(\dfrac{\partial \overline{u_i}}{\partial x_j}+\dfrac{\partial u^{\prime}_i}{\partial x_j}\right)$? – Francesco Feb 19 at 12:05
• No, it is a tensorial operation. What you have to square it the Tensor S described previously. So $\Phi = \sum_i \sum_j \frac{1}{2} \cdot ( \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}) = \frac{1}{2} \cdot ( \frac{\partial u}{\partial x} + \frac{\partial u}{\partial x} + \frac{\partial v}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} + \frac{\partial w}{\partial x} + \dots)$ – PackSciences Feb 19 at 12:23