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This PDF document contains a diverse set of exercises focused on aerodynamics, a key branch of physics that studies the behavior of air and gases, especially in relation to objects in motion. The exercises provide practical examples and questions designed to deepen understanding of fundamental aerodynamic concepts. This resource is ideal for students, researchers, and engineers seeking to explore how aerodynamic forces like lift, drag, and thrust affect various objects, including airplanes, cars, and wind turbines. Topics covered include airflow around objects, pressure distribution on surfaces, and air resistance. Through these exercises, students can enhance their analytical and practical skills, improving their ability to solve complex problems in aerodynamics. By engaging with this material, they will gain valuable insights into the scientific and practical aspects of this advanced field, contributing to a stronger foundation in both theory and application.
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Thin Airfoil Theory
Charles R. O’Neill School of Mechanical and Aerospace Engineering Oklahoma State University Stillwater, OK 74078
Project One in MAE 3253 Applied Aerodynamics and Performance March 2000
The thin airfoil theory for calculation of section flight properties is reviewed. Lift and moment coefficient and center of pressure calculations are made for cambered and flapped wing sections. Effects of camber and flaps are discussed. The thin wing theory results are compared to experimental airfoil data.
surface must be zero. Thus a thin wing section at an small angle of attack can be approx- imated by v(x) U =^ α^ −^
dz dx where U is the flight velocity and dzdx is the slope of the wing section with respect to the chord line. Substituting for the downward velocity, where r = x 0 − x, yields,
1 2 πU
chord
γ(x) x 0 − x dx^ =^ α^ −^
dz dx
Additionally, the Kutta condition forces a fluid to flow off a surface tangentailly causing the vortex strength to be zero at the trailing edge. For ease of later calculation, the coordinate system is transformed to a polar system centered at the mid chord by x = 2 c (1 − cos(θ)). Thus, 1 2 πU
∫ (^) π 0
γ(θ) sin(θ) cos(θ) − cos(θ 0 ) dθ^ =^ α^ −^
dz dx A general solution of the above equation is unknown and would be in any case too complex to handle. However, an approximation can be found by assuming a distribution of γ,
γ(θ) = 2U
A 0 1 + cos(sin(θ)θ )+
n=
An sin(nθ)
Where values of A 0 through An are given by (McCormick, 1995)
A 0 = α − (^1) π
∫ (^) π 0
dz dx dθ An = π^2
∫ (^) π 0
dz dx cos(nθ)dθ Applying the Jutta Joukowski expression for lift and moment versus vortex strength yields,
Cl = 2πA 0 + πA 1 CmLE = − π 2 (A 0 + A 1 − A 2 2 )
The moment at any arbitrary point off the leading edge can be found by adding the contribution due to the lifting force and the distance from the leading edge. At the quarter
chord location, the resulting moment does not depend on the angle of attack and is given by (McCormick, 1995), Cm 0. 25 c = − π 4 (A 1 − A 2 )
Similarily, the center of pressure is given by
xcp = c 4 (1 + (^) Cπl (A 1 − A 2 ))
RESULTS AND DISCUSSION Calculations were performed and are given in Appendix A, B and C. Three profiles were analized as thin wings. These included one cambered wing consisting of constantly changing curvature along a mean chord line. One flapped and one cambered wing both consisting of straight line segments. The NACA 2412 cross section is given in Appendix A. The forward section dzdz as calculated from the mean camber line is 0.125 cos(θ) − 0 .025. The aft section has a dzdz of 0 .0555 cos(θ) − 0 .0111. The mean camber line equations are switched at θ = 1.369radians. Thus, A 0 = α − 0 .0045165. A 1 = 0.0814604 and A 2 = 0.0138724. The coefficient of lift is Cl = 2πα + 0.2275 and the moment coefficient at the quarter chord is Cm 0. 25 c = − 0 .05308. The wing section has zero lift at an angle of attack of − 2 .07 degrees. The flapped airfoil is given in Appendix B. The forward non-flapped section’s dzdx is 0.086135 and the aft flapped section has a dzdx of − 0 .3655. The wing switches from unflapped to flapped segments at θ equals 2.2377 radians. Thus, A 0 = α + 0.04381. A 1 = 0 .22591 and A 2 equals − 0 .013974. The effective coefficient of lift is Clef f = 2πα + 0. 9850 and the effective moment coefficient at the quarter chord is Cm 0. 25 cef f = − 0 .28718. The effective center of pressure is at xcp = 4 c
1 + (^2) πα^1 ef f.^14903 +0. 9850
When converted to true chord lengths and true angles of attack, Cl = 0. 9849 · 2 π + 1.5018, Cmac = − 0 .27858 and xcp = 0. 9849 c 4
1 + (^2) πα^1.^14903 +0. 9850
The cambered airfoil is given in Appendix C. The forward section’s dzdx = 0.8 and the aft section has a dzdx of − 0 .8. The wing switches profile lines at θ = 90degrees. A 0 is simply
McCormick, Barnes W., (1995) Aerodynamics, aeronautics, and flight mechanics, John Wiley & Sons, New York. Abbott, Ira H., Doenhoff, Albert E. (1959) Theory of Wing Sections, Dover Publications, New York.
Calculations: NACA 2412
From theory, the lift coefficient is Cl = 2πA 0 + πA 1 and the moment coefficient about the quarter chord is Cm 0. 25 c = − π 4 (A 1 − A 2 ). Thus, the lift and moment coefficients are
Cl = 2π(α − 0 .0045165) + 0. 0814604 π = 2πα + 0. 2275 Cm 0. 25 c = − π 4 (0. 0814604 − 0 .0138724) = − 0. 05308
The section has zero lift when Cl equals zero. Solved for alpha, the zero lift angle is
α = − 0.^22972 π = − 0. 0362 radian = − 2. 07 degrees
Thus, the angle of attack for zero lift is -2.07 degrees.
Calculations: Flapped airfoil
The effective lift and moment coefficients are
Clef f = 2π(α + 0.04381) + 0. 22591 π = 2πα + 0. 9850 Cm 0. 25 cef f = − π 4 (0. 2259158 − (− 0 .1397416)) = − 0. 28718
The location of the center of pressure with respect to the effective chord length is
xcp = 4 c(1 + (^) Cπl (A 1 − A 2 )) = 4 c
1 + (^2) παef f^1.^14903 + 0. 9850
The lift and moment coefficients and the center of pressure can be converted to the true chord length and angle of attack.
α = αef f − 4. 923 ◦ c = 0 c. 9849 ef f
Substituting the true values for the effective values and then simplifying yields,
Cl = Clef f ( cef f c ) = 0. 9849 · (2π(α + 4. 923 ◦) + 0.9850) = 0. 9849 · 2 π + 1. 5018
Cmac = Cm 0. 25 cef f · ( cef f c )^2 = − 0 .27858(0.9849)^2 = − 0. 27858 xcp = xcpef f · ( cef f c ) = 0. 9849 c 4
1 + (^2) πα^1. 14903 + 0. 9850
Calculations: Thin Cambered Airfoil