P1170 - β is β = tan − 1(0 20 = 11 31 o Then the lift coefficient is C L = 2 π[1 0 77(0 06 sin(6 o 11 31 o = 1 96

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11.70. CHAPTER 11, PROBLEM 70 101 11.70 Chapter 11, Problem 70 Using the formula for Kutta-Joukowski airfoils, we can estimate the lift coefficient 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, for the given data, the angle
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Unformatted text preview: β is β = tan − 1 (0 . 20) = 11 . 31 o Then, the lift coefficient is C L = 2 π [1 + 0 . 77(0 . 06)] sin (6 o + 11 . 31 o ) = 1 . 96...
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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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