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Lets say we have a fully developed turbulent pipe flow. We consider the flow in cylindrical coordinates (x aligns in the streamwise direction). So $\bar{u_x}(r)$, we decomposition the flow as follow $u_r=u_r^{'}(x,r,\theta)$, $u_r=u_r^{'}(x,r,\theta)$ and $u_x=\bar{u_x}+u_x^{'}(x,r,'\theta)$.

The continuity equation, when the flow in incompressible, as follow: $\frac{\partial u_x}{\partial x} + \frac{1}{r}\frac{\partial ru_r}{\partial r} + \frac{\partial u_\theta}{\partial \theta}=0$. We decomposition it and take the average, but can we say: $\overline{\frac{1}{r}\frac{\partial ru_r}{\partial r}}=\frac{1}{r}\frac{\partial r\overline{u_r}}{\partial r}=0$? Because averaging is integrating just over time so all the other coordinates can be left out?

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    $\begingroup$ Consider to spell out acronyms. $\endgroup$ – Qmechanic Apr 14 '18 at 12:25
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    $\begingroup$ Yes, that's right. But, you left out a 1/r in from of the theta component. $\endgroup$ – Chet Miller Apr 14 '18 at 12:55
  • $\begingroup$ @ChesterMiller that looks like an answer. $\endgroup$ – Kyle Kanos Apr 14 '18 at 14:32
  • $\begingroup$ @chestermiller You are correct. I forget the 1/r for the theta. But thanks for the conformation $\endgroup$ – user1200276 Apr 14 '18 at 14:37

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