### 11.5 Here’s How Reynolds Did Averaging

For any variable, the observed value can be written as a sum of the mean value and a turbulent value. The turbulent part is sometimes called the fluctuating part or the perturbation. For example, we can write the zonal component of the velocity as:

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

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

The average can be a spatial average or a temporal average. Mathematically, we can write those averages using integration if we have continuous data or by summation if we have discrete data. Here are the formulas for computing the mean of the zonal component of the velocity:

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

where the integrals are over specific time periods or distances and the summations are for *N* discrete measurements. 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. It is important to remember that the mean values defined in the above equation are not necessarily constant in space or time.

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

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

These rules should make sense to you. For example, consider the first rule, which says that the average of the turbulent value is zero. That makes sense because the turbulent value is *defined* as the departure from the mean. 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 the product 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:

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