We're looking at a fully developed flow along an inclined plate, the $x$ coordinate is along the plate and the $z$ coordinate is perpendicular to it.

In 2D, uniform flow I end up with the continuity equation reduced to:

$$\frac{\partial w}{\partial z} = 0$$

Now if I apply the Reynolds ensemble averaging decomposition I have :

$$\frac{\partial \bar{w}}{\partial z} + \frac{\partial w'}{\partial z} = 0$$

Now if I apply the ensemble averaging equation to that equation I end up with simply:

$$\frac{\partial \bar{w}}{\partial z} = 0$$

And injecting in the previous equation I deduce also:

$$\frac{\partial w'}{\partial z} = 0$$

Those two values are then constant along $z$. Applying bottom boundary condition I actually deduce :

$$ \bar{w} = w' = 0 $$

Now if I look at my Reynold's Stresses in my momentum equations I just have:

$ \overline{u' w'}$ in the $x$ equation and $\overline{w'^2}$ in the $z$ equation.

But knowing that $w' = 0$, all of my Reynolds stresses disappear... How is that possible ? All the material I see about it suggest that I should still have those if I want to use a turbulent model (such as Prandtl Mixing Length).

What am I missing ?


The answer is that turbulence is always a 3-dimensional phenomenon and never 2-dimensional. In other words while mean flow may be 2-dimensional, fluctuations of flow exist in all 3 dimensions. This means that if $(u,v,w)$ is the total flow velocity field, then it is to be decomposed as $(\bar{u}+u',v',w')$ because $\bar{v}=\bar{w}=0$. So Reynolds stress will be present.

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  • $\begingroup$ So a general rule is to only apply my hypothesis on mean values instead of the fluctuating part? It makes sense, I'll mark your answer as the most helpful. $\endgroup$ – Luc Rebillout Feb 10 '18 at 17:37
  • $\begingroup$ @LucRebillout By hypothesis I suppose you are referring to the assumption of 2D flow. Yes, mean flow can be 2D. Another example of 2D mean flow is shearing flow between moving plates in which mean flow is only along one direction with a gradient in the perpendicular direction i..e the mean velocity vector is $(\bar{U}(y),0,0)$. $\endgroup$ – Deep Feb 11 '18 at 4:02

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