For any variable, the observed value can be written as a sum of the mean value and a turbulent value:

$$u=\overline{u}+{u}^{\prime}$$

where $\overline{u}$ is the mean, or average value, and ${u}^{\prime}$ is the turbulent part.

The average can be obtained by averaging *u* over time or over space or even by doing a number of samples and averaging over the samples.

$$\text{temporalaveraging:}\overline{u}=\frac{{\displaystyle \int u(t)dt}}{{\displaystyle \int dt}}=\frac{{\displaystyle \sum _{i=0}^{N-1}u({t}_{i})}}{N};\text{\hspace{1em}}\text{spatialaveraging}\text{\hspace{1em}}\overline{u}=\frac{{\displaystyle \int u(x)dx}}{{\displaystyle \int dx}}=\frac{{\displaystyle \sum _{i=0}^{N-1}u({x}_{i})}}{N}$$

If the turbulence does not change with time and is homogeneous (i.e., the same in all directions and for all time), then these averages equal each other.

In Lesson 10, we developed the equation of motion without really considering the short-term and small-scale variations, except to say that they led to turbulent drag, which acts to resist the mean flow in the upper boundary layer. Now we want to think about how to correctly capture the dynamic effects of turbulent motion. What we want to do is to write down the equations of motion that you learned in Lesson 10; substitute mean and turbulent parts for the variables such as *u*, *v*, and *w*; average over all the terms; and then see if we can sort out the terms to create an equation for the mean wind and an equation for the turbulent wind. This type of averaging is called **Reynolds averaging**.

But first we need to learn the rules for averaging.

### Rules of Averaging

$$\begin{array}{l}c\text{isconstant;}u\text{and}v\text{arevariables}\hfill \\ \overline{{u}^{\prime}}=0\hfill \\ \overline{c\text{\hspace{0.17em}}u}=c\text{\hspace{0.17em}}\overline{u}\hfill \\ \overline{u+v}=\overline{u}+\overline{v}\hfill \\ \overline{\left(\overline{u}\text{\hspace{0.17em}}v\right)}=\overline{u}\text{\hspace{0.17em}}\overline{v}\hfill \\ \overline{\left(\frac{\partial u}{\partial t}\right)}=\frac{\partial \overline{u}}{\partial t}\hfill \end{array}$$

Now let’s apply these rules to a variable with a mean and a turbulent part. For example, consider the product *uv*, which is just the advection of horizontal wind in one direction by the horizontal wind in the other direction.

So, using the rules:

$$\begin{array}{l}\overline{u\text{\hspace{0.17em}}v}=\overline{(\overline{u}+u\text{'})(\overline{v}+v\text{'})}\hfill \\ \text{}=\overline{\overline{u}\text{}\overline{v}}\text{\hspace{0.17em}}+\overline{\overline{u}v\text{'}}+\overline{u\text{'}\overline{v}}+\overline{u\text{'}v\text{'}}\hfill \\ \text{and}\hfill \\ \overline{\overline{u}\overline{v}}=\overline{u}\text{}\overline{v}\hfill \\ \overline{\overline{u}\text{\hspace{0.17em}}v\text{'}}=\overline{u}\text{\hspace{0.17em}}\overline{v\text{'}}=\overline{u}\cdot 0=0\hfill \\ \overline{u\text{'}\overline{v}}=\overline{u\text{'}}\overline{\overline{v}}=0\cdot \overline{v}=0\hfill \\ \text{so}\hfill \\ \overline{u\text{\hspace{0.17em}}v}=\overline{u}\text{}\overline{v}\text{\hspace{0.17em}}+\overline{u\text{'}v\text{'}}\hfill \end{array}$$

This second term, a product of two turbulent terms is *not necessarily zero*! Whereas the average of one turbulent term is zero by definition, the average of two turbulent terms is not necessarily zero.

The following video (3.11) further describes Reynolds averaging:

### Example

Consider two random numbers varying between –0.5 and 0.5, called *u* and *v*. The figure below shows *u*, *v*, the mean of *uv* and the mean of *u’v’*. Of course, the mean of *u’v’* might be zero, but is it not necessarily zero, as shown in this figure.

The same thinking applies to *u*^{2}.

$$\overline{u{\text{\hspace{0.17em}}}^{2}}=\overline{(\overline{u}+u\text{'})(\overline{u}+u\text{'})}=...=\overline{u}{\text{\hspace{0.05em}}}^{2}\text{\hspace{0.17em}}+\overline{u{\text{'}}^{2}}$$

Remember your statistics and the concept of variance:

$${\sigma}_{u}{}^{2}=\frac{1}{N-1}{{\displaystyle \sum _{i=0}^{N-1}\left({u}_{i}-\overline{u}\right)}}^{2}\approx \frac{1}{N}{{\displaystyle \sum _{i=0}^{N-1}\left({u}_{i}-\overline{u}\right)}}^{2}=\frac{1}{N}{{\displaystyle \sum _{i=0}^{N-1}\left({u}_{i}\text{'}\right)}}^{2}=\overline{u{\text{'}}^{2}}$$

So, the variance is the same as the mean value for the square of the turbulent part of the variable.

The covariance of *u* and *v* is given by the equation:

$$\text{covariance}(u,v)=\overline{u\text{'}v\text{'}}$$

The standard deviation is just:

$${\sigma}_{u}={\left(\overline{u{\text{'}}^{2}}\right)}^{1/2}$$

We can get a better sense of how large this variance is by dividing by the mean value:

$$\text{I=}\frac{{\sigma}_{u}}{\overline{u}}$$