An airfoil has low-speed Cl = 0.10 at α = 0° and Cl = 0.53 at α = 4°; maximum lift occurs at 16°. Use Prandtl–Glauert correction to find Cl at α = 6° and free-stream Mach 0.6, to 2 decimals.
In the pre-stall linear range, slope is $(0.53-0.10)/4=0.1075$ per degree. At 6°, incompressible Cl = 0.10+6(0.1075) = 0.745. The correction gives $C_l=C_{l,0}/\sqrt{1-M^2}=0.745/0.8=0.93125$, or 0.93.
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