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If airfoils at low speed are able to generate a higher coefficient of lift, why is this not reflected in airfoil polars?

I understand airfoils generate a higher coefficient of lift at low Re numbers.

So I entered the NACA2312 airfoil in javafoil and generated a lift polar for the airfoil from Re 5,000 to 10 million. There was no increase in lift at lower Re numbers...

Why not?

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    $\begingroup$ Where do you see that airfoil generates higher lift at lower Re? $\endgroup$
    – JZYL
    Aug 23 '19 at 17:11
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No, they don't.

The Reynolds number is the ratio of inertial to viscous forces in a fluid. That means viscosity has proportionally more effect at lower Reynolds numbers. The boundary layer is proportionally thicker, friction steals more energy from the flow and lift is lower.

This answer contains a graph with a collection of empirical data, as does this answer. Here is another answer which covers both attached and separated flow. As you can see, Reynolds number effects mainly affect attached flow, and in a way that reduces lift for the same angle of attack; the more so, the closer you come to the stall angle of attack.

Just to save you from following all the links: See below for a plot of the venerable NACA 4412 from Abbott and Doenhoffs collection of airfoil data (picture source):

NACA 4412 lift curve and drag polar

Note that the lift coefficient is plotted for Reynolds numbers R of 3, 6 and 9 million.

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  • $\begingroup$ Above you state: The Reynolds number is the ratio of inertial to viscous forces in a fluid. That means viscosity has proportionally more effect at lower Reynolds numbers. The boundary layer is proportionally thicker, friction steals more energy from the flow and lift is lower. i.e. which states at lower speeds, lift is lower. $\endgroup$
    – Fred
    Aug 24 '19 at 11:39
  • $\begingroup$ in the question of wether a wing can achieve a cLma of 6.8, you state: "Another practical limit is the maximum Mach number in the suction peak of all the airfoil elements. Once this reaches 1.58, no lift growth could be observed in experiments. This translates to a maximum for the product of Mach squared and pressure coefficient of -1.0. In other words: You need to fly very slowly in order to achieve high values of c𝑙𝑚𝑎𝑥". $\endgroup$
    – Fred
    Aug 24 '19 at 11:47
  • $\begingroup$ That's where this question originates: You need to fly very slowly ( i.e. low Re number) to achieve high values of climax. At what Re are we talking about? i.e. Will a wing at Re 200k produce more lift than at Re 600k or Re 1 million? . $\endgroup$
    – Fred
    Aug 24 '19 at 11:51
  • $\begingroup$ @Fred: I understand – you expanded the validity of that statement too much. Slowly means first low Mach. At the same Mach you should increase Re for more lift. Going from 200 K (model aircraft) to 600 K (glider wingtip, slow speed) to 1 million (glider root, slow speed) will increase lift each time. And if you progress further, lift will increase even more (see diagram). $\endgroup$
    – Peter Kämpf
    Aug 24 '19 at 11:55

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