 Reynolds averaging is really pretty straightforward once you understand the rules. Each variable has an average and a perturbation or turbulent part. We need to determine the time over which we want to find the average, but after we do that, we can average all the values and then subtract the average from each individual value in order to find the perturbed or turbulent or fluctuation part of that value. The average value is, of course, the same for all the values in the average. I will use the words mean and average interchangeably for the noun meaning average, and we'll use the words like perturbation, fluctuation, and turbulent part to describe the variations of individual values about the average value. The rules are pretty simple. First, the average of a perturbed or turbulent term is zero because if it were not, then the average value would be incorrect. Second, the average of the product of a constant times a variable is just the product of the average of the constant times the average of the variable. The average of the sum of two variables is just the sum of the average of the two variables. And the average of a product of the average value of one variable and another variable is just the product of the averages of the two variables. Note that the average of a variable is just a constant. Be careful. We will soon see that the average of the product of two variables is not just the product of the average of two variables. Finally, the average of the derivative of a variable is just the derivative of the average of a variable. Let's calculate the Reynolds average of one term of the equation for kinetic energy, which is just one-half mv squared. If we divide by the air density, then we have the kinematic kinetic energy. Each term can be written as its mean and turbulent parts. Let's look only at the u term. The v and w terms can be calculated in the same way. So we multiply all the terms out, then take the Reynolds average, and apply the rules. Average values of average values are just average values. Because an average value is a constant, we get two terms of a constant times the average of a third term, which is just zero. When we are done, we see that we have two terms left. The average u squared and the perturbation term squared. You can make a simple model with a random number generator to demonstrate the average of the product of two perturbation terms is not necessarily zero. This calculation was chosen so that the averages for u and v were zero. And so the average of the product of average u and average v is zero. But the average of the perturbations is not zero.