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To obtain the time average of an unsteady term like $\frac{\partial u_{i}}{\partial t}$ by definition we perform the following:

\begin{align} \overline{\frac{\partial u_{i}}{\partial t}} &= \frac{1}{T}\int_{t}^{t+T} \frac{\partial }{\partial t}(U_i + {u}'_i)\, dt \\ & = \frac{U_i(x,t + T) - U_i(x,t)}{T} + \frac{u'_i(x,t + T) - u'_i(x,t)}{T} \end{align}

where $U_i$ is the mean value of velocity in $x$-direction and $u'_i$ is the fluctuating part.

My question is that why this term $\frac{u'_i(x,t + T) - u'_i(x,t)}{T}$ equals zero so that $$ \overline{\frac{\partial u_{i}}{\partial t}} = \frac{U_i(x,t + T) - U_i(x,t)}{T} = \frac{\partial U_{i}}{\partial t}$$

Somehow the reason is because $T$ effectively approaches $\infty$ on the time scale of the turbulent fluctuations so that it equals zero, but why isn't that the case for the first term?

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  • $\begingroup$ I would use ensemble average instead of time average for all purposes, then this relation is exact. $\endgroup$ – Yrogirg Nov 16 '17 at 17:19
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It's simply a definition of the properties of a fluctuating component. We require that, on large enough time scales, the fluctuating component averages out to zero: $$ \frac{1}{T}\int_t^{t+T}u'(t)\,dt=0 $$ so that you are left with just the bulk flow term, $U$, that contributes to the mean velocity, $\bar{u}$.

I think this is more easily seen visually than mathematically, though. Basically your velocity vs time plot looks like this1,

home-made turbulent velocity plot

As you can see in the plot above, if you look at very short time scales (e.g., $T\sim3$), the turbulence can dominate the value of the velocity. However, over a large enough chunk of time (e.g., $T\sim20$), the effect of turbulence is pretty small and the mean velocity, $\bar{u}$, is simply the bulk velocity, $U$.


1 This was generated using R by taking up <- rnorm(n=100, mean=0, sd=1) and u <- 8 adding them for the plot: plot(0, xlim=c(0,100), ylim=c(0,16); lines(u+up). So $u'$ is a normal distribution with mean of zero and a standard deviation of 1.

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This is a very good question, which illustrates that Reynolds averaging is a very special form of averaging. Indeed the Reynolds averaging procedure assumes three properties of the averaging operator:

  1. Linearity: Let $a,b$ be constants and $f,g$ observables $\overline{af+bg} = a \overline{f}+ b \overline{g}$.
  2. Commutes with derivatives: $\overline{\frac{\partial f}{\partial s}} =\frac{\partial\overline{ f}}{\partial s} $, for $s=x,y,z$ or $t$
  3. Factorising property: $\overline{f\overline{g}} = \overline{f}\overline{g}$.

As discussed here, these properties are not satisfied, in an exact sense, by many common averaging procedures. For example the running average you suggest doesn't satisfy either 3. or 2., but as Kyle said correctly, in practice, when handling actual data, one simply chooses the smoothing period long enough and one may also use a weighting function to minimise the problem. An example of an averaging operator which satisfies 1.-3. is the zonal average around a latitude circle of Earth which is often considered in studying the general circulation of the atmosphere (the basic state in this case is zonally symmetric).

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