Aerospace Engineering DISCUSSION

Wind tunnel test of a 1:10 scale wing: why can't I match the Reynolds number by going faster?

Started by fonseca Reynolds numberwind tunnel scalingdynamic similarityscale model testingMach number
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Latest activity · 30 Sep 2026

Wind tunnel test of a 1:10 scale wing: why can't I match the Reynolds number by going faster?

fonseca Aerospace Engineering Forum
#1

For a student project I want to test a 1:10 scale model of a wing that has a 1.5 m chord and flies at 50 m/s at full scale. Our tunnel reaches about 50 m/s. I was told the lift and drag coefficients from the model only transfer to the full-size wing if the Reynolds number matches.

What speed would the model need, why is that not practical, and how do people get useful data from small models anyway?

Community replies 4

Re: Wind tunnel test of a 1:10 scale wing: why can't I match the Reynolds number by going faster?

#2

Reynolds number is Re = V × c / ν, with ν the kinematic viscosity of air, about 1.46 × 10^-5 m²/s at sea level and 15 °C. Full scale: 50 × 1.5 / (1.46 × 10^-5) = about 5.1 million. The model has a 0.15 m chord, so in the same air it would need ten times the speed, 500 m/s, to match.

That is about Mach 1.5. The flow around the model would be supersonic with shock waves, while the real wing flies at about Mach 0.15. Matching Re that way destroys the similarity you were trying to achieve. In your tunnel at 50 m/s the model is at a Reynolds number of about 510,000.

Re: Wind tunnel test of a 1:10 scale wing: why can't I match the Reynolds number by going faster?

#3

The coefficients are the reason scale testing works at all. CL = L / (½ρV²S) removes the effects of size, speed and density, so what is left depends only on shape, angle of attack and the similarity parameters, mainly Reynolds number and Mach number. If those match, the coefficients match, and you scale forces back up with the full-size ½ρV²S.

At low speed, below about Mach 0.3, compressibility is negligible and Reynolds number is the parameter that matters. For transonic and supersonic testing the Mach number has to be matched first.

Re: Wind tunnel test of a 1:10 scale wing: why can't I match the Reynolds number by going faster?

#4

What changes with Re is the boundary layer. At a few hundred thousand, much of the model's boundary layer is laminar, which separates easily; laminar separation bubbles form, maximum lift is lower and drag is different from the full-scale wing, where the boundary layer becomes turbulent much earlier along the chord. The lift-curve slope in the attached-flow range transfers reasonably well; CLmax, stall behaviour and drag do not.

A standard remedy is a trip strip, a thin band of roughness near the leading edge that forces transition on the model where it would occur at full scale.

Re: Wind tunnel test of a 1:10 scale wing: why can't I match the Reynolds number by going faster?

#5

Since Re = ρVc/µ, facilities that need a high Reynolds number change the fluid rather than the speed. Pressurised tunnels raise the density; cryogenic tunnels cool the gas, which raises density and lowers viscosity at once, and lowers the speed of sound too, so Mach number and Reynolds number can be set more independently. Bigger tunnels with bigger models do the rest.

None of that is available for a student project, so the realistic approach is the largest model the test section allows without excessive blockage, trip strips, and an honest statement of the test Reynolds number alongside the results.

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