MAE 150B - 04C - Incompressible Flow over Airfoils

# MAE 150B - 04C - Incompressible Flow over Airfoils - MAE...

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MAE 150B - Chap 4 part C Incompressible Flow Over Airfoils

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Example 4.5  (pp. 334)  : NACA23012 The camber line defined as z/c = f(x/c) (a). Calculte the AOA at zero lift (b). Lift coef. At α =4 o Moment coef. About c/4 The location of center of pressure For a cambered airfoil, one of the most important parameter defining the shape is ____? It can be obtained easily from the given expression dx dz
Example 4.5  (pp. 334)  : NACA23012 (a) the zero-lift AOA can be obtained from As the integration is w.r.t. d θ , we need to transform x=(c/2)(1-cos θ ) Sub dz/dx into above equation and carry out the integration We may need to use Math. Handbook to carry out the integration ( 29 ) 61 . 4 ( 1 cos 1 0 0 0 0 θ π α d dx dz L - - = =

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Example 4.5  (pp. 334)  : NACA23012 After zero-lift AOA is known, Eq. (4.60) can be used for the lift coef. Of any other α And the result is 2 π ( α - α L=0 ) = 2 π (0.0698- 0.0191) = 0.559 ( 29 ) 60 . 4 ( 2 0 = - = L l c α π
Example 4.5  (pp. 334)  : NACA23012 (c) The quarter chord moment coeff. Is obtained from (4.64) We can get A1 and A2 from (4.51) A1 and A2 are n=1,2 respectively (what are they?) After some messy but straightforward operations, A1 and A2 are readily available ( 29 ) 64 . 4 ( 4 1 2 4 / , A A c c m - = π ) 51 . 4 ( cos 2 0 0 0 θ d n dx dz A n =

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Example 4.5  (pp. 334)  : NACA23012 Similarly, the center of pressure can be obtained by plugging A1 and A2 ( 29 ) 66 . 4 ( 1 4 2 1 - + = A A c c x l cp π
Example 4.5: NACA23012 comparison with experimental data calculated measured α L=0 -1.09 -1.1 α (4 o ) 0.559 0.55 C m,c/4 -0.0127 -0.01

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Aerodynamic Center Additional consideration (4.9) Most conventional airfoils, the aerodynamic center is
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## This note was uploaded on 10/09/2009 for the course MAE 150B taught by Professor D during the Spring '09 term at UCLA.

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MAE 150B - 04C - Incompressible Flow over Airfoils - MAE...

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