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I would like to understand how to calculate the Reynolds number in the test chamber of a wind tunnel.

It is known that the Reynolds number for a pipe is:

$$Re = \frac{UL}{\nu}$$

where

$U$ is the fluid velocity.

$L$ is the characteristic length (usually the diameter of the pipe).

$\nu$ is the kinematic viscosity.

Analogously, the test chamber of a wind tunnel could be considered a rectangular pipe, and $L$ the section length.

The problem arises when we want to calculate the $U$ for which the turbulent flow appears. Considering that the turbulent flow arises at $Re=4000$, the air kinematic viscosiy ($\nu = 10^{-5}$) and a relatively small section length of a test chamber ($L = 200$ mm), we obtain that turbulence appears if $U > 0.2$ m/s (a very slow velocity).

My question is: ¿How is it therefore possible that in large wind tunnels at high speeds free streamlines can be obtained, as in the following image? ¿Shouldn't everything be mixed?

Example:

link

fog tracer in a wind tunnel

My theory: the honeycomb grids and the difusors used in the wind tunnels reduce the turbulence, which would be theoretically recovered far away from them. Therefore, the Reynolds number makes not sense in the test chamber due to the distance from these elements. I'm wrong?

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  • $\begingroup$ I was of the opinion that turbulent flow can occur from about an Re of 1700 upwards... $\endgroup$
    – user207455
    Commented Jul 24, 2019 at 12:16
  • $\begingroup$ The Reynolds number exceeding the turbulence threshold doesn't mean that the entire cross-section is turbulent though, right? If memory serves me, that would only mean the boundary layer is turbulent. Presumably a good wind chamber would be set up such that the boundary layer is too small to interact with the test piece. $\endgroup$
    – JMac
    Commented Jul 24, 2019 at 12:37
  • $\begingroup$ Shouldn't you be using the hydraulic diameter, not the length, to calculate the Reynolds number? After all, for flow in a pipe, we use the diameter, not the length. Also, who says that, for a wind tunnel, the flow is supposed to be laminar? $\endgroup$ Commented Jul 24, 2019 at 14:16
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    $\begingroup$ @ChetMiller Yeah, I think he is calculating the Re for flow across a flat plate; which only really tells him if a turbulent layer would be expected at the edges of the windtunnel, whereas hydraulic diameter would tell him if the entire flow profile in the tunnel is turbulent (if I'm remembering my fluids correctly at all). $\endgroup$
    – JMac
    Commented Jul 24, 2019 at 15:04
  • $\begingroup$ @JMac thanks for your answers. Yeah, Reynolds number is related to the boundary layer length. The good wind chamber depends on the setups that you want to perform in it. $\endgroup$
    – JuanMi
    Commented Jul 25, 2019 at 13:23

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The flow behind the screen with the grid has a specific structure unlike the flow in the pipe - see fig.1. First, the flow is uniform across the channel, which is important for aerodynamic experiments. Secondly, the transition to turbulence occurs in the boundary layer at large Reynolds numbers, and the core of the flow remains unperturbed. All this allows to realize the laminar flow in the wind tunnel even at a flow velocity of 10-25 m / s. Figure 1 shows the magnitude of velocity (left), the velocity profile at the outlet (in the center) and the longitudinal component of velocity in a laminar flow in a channel behind the grid. Figure 1

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  • $\begingroup$ Thank you so much @AlexTrounev. It is all more clear for me now. As could be expected, the grid is key to change the flow structure and let to perform aerodynamic experiments in a proper way. Reynolds number is, as I applied it, a too simpler way to characterize this type of flow. $\endgroup$
    – JuanMi
    Commented Jul 29, 2019 at 6:25
  • $\begingroup$ @JuanMi You're welcome! The Reynolds number in this flow is determined by the length of the boundary layer on the wall. The transition to turbulence can be tightened up to $Re=5*10^5$. $\endgroup$ Commented Jul 29, 2019 at 7:46
  • $\begingroup$ Thanks for the information @AlexTrounev. Now I know how I have to analyse the flow in wind tunnels. $\endgroup$
    – JuanMi
    Commented Aug 1, 2019 at 10:55

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