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Unformatted text preview: ) What would the lift be on a full‐scale 1.00m by 5.00m wing flying at the same angle of attack through the atmosphere at the same velocity and same lift coefficient? e) What are the chord‐based Reynolds numbers for the model wing and the full‐scale wing? f) What should be the velocity in the wind tunnel to achieve similarity with respect to Reynolds number if the full‐scale velocity is 50 m/s? The pressure difference will be ρgh for the manometer. ρwater is 1000 kg/m3. g = 9.81 m/s2 h = 0.020m So ΔP = 196 kg‐m2/m3‐s2 = 196 Pa L = clΔP = (0.10m)(0.500m)(196N/m2) = 9.8N q∞ = (0.5)(1.2250kg/m3)(50m/s)2 = 1500 Pa CL = L/ q∞‐S = (9.8N)/{(1500Pa)(0.10m)(0.50m)} = 0.13 Lf‐s = CL q∞‐S = (0.13)(1500 Pa)(1.00m)(5.00m) =...
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