Do vortex generators (VGs) increase Re of a wing?

I understand that VGs re-energize the boundary layer on a wing.

I understand it turns laminar flow into attached turbulent flow. Doesn't that mean the Re increases substantially?

If so, wouldn't that increase it's cl/cd, which means lower drag in the 600-900k Re range?

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    $\begingroup$ @Fred, although most people here understand what you meant, it is better to explain the key term you are using. VG is not such a universal abbreviation everyone knows immediately. I took the liberty to edit your text slightly. $\endgroup$
    – Zeus
    Jul 23, 2019 at 1:01
  • $\begingroup$ You completely misunderstood what the Reynolds number is and what vortex generators do. Re depends on flow speed, viscosity and length, nothing more. And it cannot be dialled up or down, except by changing those parameters. $\endgroup$ Aug 24, 2019 at 6:07
  • $\begingroup$ lol: I guess you're right. I thought wrongly, that Re was also indirectly a relationship between laminar and turbulent flow. $\endgroup$
    – Fred
    Aug 24, 2019 at 11:41

2 Answers 2


Reynolds number depends on the free-stream speed, the chord length, fluid density and viscosity. So vortex generator does not modify the Reynolds number. However, it trips the flow aft of the device into turbulent flow. Turbulent flow, in fact, has higher drag than a laminar flow (at the same Re). Therefore, on top of the additional skin friction drag introduced by the VGs, the aerodynamic surface would also have higher drag.

  • $\begingroup$ Again, the correct answer gets fewer votes. +1 $\endgroup$ Aug 24, 2019 at 6:04

The Reynolds number is a construct that measures the relative magnitude of viscous and inertial forces, and it is defined in a way that is convenient. It is not a measurable physical quantity, so the answer will depend on how you define it.

Note that its expression: $$ \mathrm{Re} = \frac{\rho u L}{\mu} = \frac{u L}{\nu} $$

includes a length term $L$ that is arbitrarily chosen as a characteristic length of the problem in question.

This means that one can define a Reynolds number for the free flow over a wing ($L=L_{wing}=c$) and a totally different one for the boundary layer ($L=L_{layer}$). In practice the boundary layer $\mathrm{Re}_{layer}$ and the wing $\mathrm{Re}_{wing}$ will be closely linked as turbulence from the free flow seeps to the boundary layer, but they will not be the same.

For example, the boundary layer near the forward stagnation point will be extremely low and the layer will be laminar, even if the prevailing free flow conditions would indicate turbulence. Of course in such a situation the layer is bound to transition very soon unless the airfoil is specifically designed to prevent this, like the NACA 6-series.

Layer manipulation devices like trip strips and vortex generators muddy the picture further still by altering the conditions in the boundary layer:

A trip strip will force a turbulent transition by means of a local spike in surface roughness, with the intention of using a turbulent boundary layer´s more favourable separation properties.

zig zag tape or trip strip

Vortex generators, on the other hand, mix the flow from the free stream out of the layer into it, increasing its energy, delaying separation. They also increase the $\mathrm{Re}_{layer}$ because they are also a source of roughness and their modus operandi directly increases the local vorticity. While it is conceivable that at very low Reynolds a set of vortex generators would operate fully in laminar mode and not transition the layer, this is an academic case, and in the real world you will likely have a turbulent layer downstream of them.

vortex generators on a wing

Now, to tackle you question more directly: if you want to increase the effective Reynolds of the airfoil for the purpose of layer attachment, you can obtain some success with vortex generators, and probably even more with a trip strip, but it do not expect either method to take the value over that of the entire wing.

  • $\begingroup$ The definition of the Reynolds number does not depend on vorticity. The correct answer to the question is no – there is no way that VGs can influence the Reynolds number, effective or otherwise. $\endgroup$ Aug 24, 2019 at 6:02
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    $\begingroup$ @PeterKämpf I´m afraid your comment that my answer is wrong only had the effect of getting it more upvotes. Could you elaborate? This topic is on the edge of my practical knowledge and I'd appreciate a deeper explanation :) $\endgroup$ Aug 24, 2019 at 19:50
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    $\begingroup$ You say the answer depends. It doesn't. There is only one definition for the Reynolds number and Re grows along the flow path until it reaches what is quoted for airfoils at the trailing edge. Tripping the boundary layer doesn't change Re (and doesn't muddy any waters), and VGs certainly don't increase Re (but boundary layer thickness). Lastly, there is no special effective Reynolds number. Just the Reynolds number. However, the other information in your answer (which is the majority) is correct. I just pointed out the wrong parts. $\endgroup$ Aug 25, 2019 at 6:14

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