p1174 - corresponding side of the first, we find β + π /...

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11.74. CHAPTER 11, PROBLEM 74 119 11.74 Chapter 11, Problem 74 Problem: A thin, cambered airfoil has a lift coefficient of 0.34 at zero angle of attack and 1.02 at 6 o . Estimate the maximum thickness, 2 T max /c , and maximum camber, C max /c , (in percent) for this airfoil. HINT: To simplify your solution, use the fact that sin 6 tan 6 6 for small 6 . Solution: Using the formula for Kutta-Joukowski airfoils, we can estimate the lift coef- ficient as C L =2 π w 1+0 . 77 T max c W sin( α + β ) , β =tan 1 w 2 C max c W So, approximating sin( α + β ) ( α + β ) , the given values tell us that 0 . 34 = 2 π w 1+0 . 77 T max c W β ( α =0 o ) 1 . 02 = 2 π w 1+0 . 77 T max c Ww β + π 30 W ( α =6 o ) whe reweusethefac ttha t 6 o = π
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Unformatted text preview: corresponding side of the first, we find β + π / 30 β = 1 . 02 . 34 = 3 = ⇒ β + π 30 = 3 β = ⇒ β = 0 . 0524 = 3 o Substituting this value of β back into the equation for α = 0 , we can solve for T max /c , i.e., 1 + 0 . 77 T max c = . 34 2 πβ = 1 . 0327 = ⇒ T max c = 0 . 0425 Therefore, the maximum thickness of this airfoil is 2 T max c = 8 . 5% Also, the maximum camber is given by β = tan − 1 w 2 C max c W = ⇒ C max c = 1 2 tan β = 0 . 026 Therefore, the maximum camber for this airfoil is C max c = 2 . 6%...
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This note was uploaded on 02/09/2012 for the course AME 309 taught by Professor Phares during the Spring '06 term at USC.

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