Introduction to Aerospace Engineering Lecture slides Challenge the future 1 Introduction to Aerospace Engineering Aerodynamics 11&12 Prof. H. Bijl ir. N. Timmer 11 & 12. Airfoils and finite wings Anderson 5.9 – end of chapter 5 excl. 5.19 Topics lecture 11 & 12 • Pressure distributions and lift • Finite wings • Swept wings 3 Pressure coefficient Typical example Definition of pressure coefficient : p − p -Cp = ∞ Cp q∞ upper side lower side -1.0 Stagnation point: p=p t … p t-p∞=q ∞ => C p=1 4 Example 5.6 • The pressure on a point on the wing of an airplane is 7.58x10 4 N/m2. The airplane is flying with a velocity of 70 m/s at conditions associated with standard altitude of 2000m. Calculate the pressure coefficient at this point on the wing 4 2 3 2000 m: p ∞=7.95.10 N/m ρ∞=1.0066 kg/m − = p p ∞ = − C p Cp 1.50 q∞ 5 Obtaining lift from pressure distribution leading edge θ V∞ trailing edge s p ds dy θ dx = ds cos θ 6 Obtaining lift from pressure distribution TE TE Normal force per meter span: = θ − θ N ∫ pl cos ds ∫ pu cos ds LE LE c c θ = = − with ds cos dx N ∫ pl dx ∫ pu dx 0 0 NN Write dimensionless force coefficient : C = = n 1 ρ 2 2 Vc∞ qc ∞ 1 1 p − p x 1 p − p x x = l ∞ − u ∞ C = ()C −C d Cn d d n ∫ pl pu ∫ q c ∫ q c 0 ∞ 0 ∞ 0 c 7 T=Lsin α - Dcosα N=Lcos α + Dsinα L R N α T D V α = angle of attack 8 Obtaining lift from normal force coefficient =α − α =α − α L Ncos T sin cl c ncos c t sin L N T =cosα − sin α qc∞ qc ∞ qc ∞ For small angle of attack α≤5o : cos α ≈ 1, sin α ≈ 0 1 1 C≈() CCdx − () l∫ pl p u c 0 9 Example 5.11 Consider an airfoil with chord length c and the running distance x measured along the chord. The leading edge is located at x/c=0 and the trailing edge at x/c=1. The pressure coefficient variations over the upper and lower surfaces are given as x2 x C =−1 300 for 0 ≤≤ 0.1 p, u c c x x C =−+2.2277 2.2777 for 0.1 ≤≤ 1.0 p, u c c x x C =−1 0.95 for 0 ≤≤ 1.0 p, l c c Calculate the normal force coefficient. 10 Compressibility correction of the pressure coefficient Compressibility correction Cp at a certain point M 1.0 Approximate theoretical correction (valid for 0 < M < 0.7) : C C = 0,p Prandtl-Glauert Rule p 2 1− M∞ 11 Compressibility correction for lift coefficient c (C− C) dx( ) c 1pl p u 1 1 ≈o = − () Cl ()Cp C p dx ∫2 2 ∫ l u o c0 1−MM∞ 1 − ∞ c 0 1 c C≡() CCdx − () C l,0 ∫ pl p u o l c 0 C M> C = l,0 l 2 1− M ∞ α 12 Example 5.12 • Consider an NACA 4412 airfoil at an angle of attack of 4 o. If the free-stream Mach number is 0.7, what is the lift coefficient? o • - From App. D for angle of attack of 4 , c l = 0.83.However, these data were obtained at low speeds. 13 Critical Mach number and critical pressure coefficient M=0.3 Mpeak =0.44 M=0.5 Mpeak =0.71 M=0.6 Mpeak =1.00 Critical Mach number for the airfoil 14 Critical pressure coefficient Cp,min thick thick medium -1.5 medium thin -1.0 = -0.5 thin C cr,p f (M∞) M 1.0 Thicker airfoil reaches critical pressure coefficient at a lower value of M∞ 15 Critical pressure coefficient − Definition of pressure coefficient : = p p∞ = p∞ p − Cp 1 q∞ q∞ p∞ ρ = 1 ρ 2 = 1 ∞ γ 2 Dynamic pressure : q∞ 2 ∞V∞ 2 p∞V∞ γp∞ 2 = 1 V∞ γ 2 p∞ γp∞ /ρ∞ 2 γ 1 V∞ 2 2 γp∞ q = γp = p M We have found before : a = thus : ∞ 2 2 ∞ ∞ ∞ ∞ a∞ 2 ρ∞ γ p γ −1 γ −1 For isentropic flow we found : 0 = 1 + M2 p 2 γ γ − p0 γ −1 2 1 and also : = 1 + M∞ p∞ 2 Dividing these equations we find :... 16 Critical pressure coefficient γ 1+ 1 (γ − )1 M2 γ −1 p = 2 ∞ 2 p∞ + 1 γ − 1 2 ( )1 M Substitution in Cp-equation gives : γ 1 2 γ −1 p p p 1+ (γ − )1 M∞ = ∞ − = ∞ 2 − Cp 1 2 2 1 q∞ p∞ 1 γp M 1+ 1 (γ − )1 M 2 ∞ ∞ 2 γ 1 2 γ −1 1+ (γ − )1 M∞ or : = 2 2 − Cp 1 γM2 1+ 1 (γ − )1 M2 ∞ 2 17 Critical pressure coefficient The critical pressure coefficient is found when M=1 is reached : γ 2 γ −1 2 2 + (γ −1)M∞ C = −1 p , cr 2 γ + γM∞ 1 2 cp ( M) cr 1 0 0.4 0.6 0.8 1 1.2 M 18 The intersection of this curve with the airfoil C P,min will give the value of the M cr specific for the airfoil under consideration. γ 2 1+ []()γ −1 / 2 M 2 γ −1 = cr − C ,crp 2 1 γM 1+ []()γ −1 / 2 cr Airfoil independent curve C C = p,0min M pmin 2 cr 1− M ∞ Airfoil dependent curve 19 Drag divergence 20 Drag divergence M<1 M∞<Mcr M<1 M>1 Mcr <M∞<Mdrag divergence M<1 shock wave M>1 M<1 M >M ∞ drag divergence Separated flow 21 Drag divergence Development of supercritical airfoils -Cp Cp,cr x/c 22 23 Finite Wings 24 Characteristics of finite wings tip vortex low pressure Flow around the ----- wing tip generates + + + + + tip vortices high pressure 25 26 Tip vortices Cross section of tip vortex R Vθ Induced velocity V∞ V∞ w local flow vector The trailing tip vortex causes downwash, w 27 Induced drag α i L L’ geometric angle of α attack α V∞ i α 28 Induced drag α = α i Di Lsin i L’ induced drag, D i geometric angle of α attack α V∞ i α α eff 29 Induced drag α = α i Di Lsin i L’ induced drag, D i geometric angle of α attack α V∞ i α α α α≈ α α eff Since small : sin i i D i = L i 30 Induced drag Example : elliptical lift distribution FRONT VIEW elliptical lift distribution Elliptical lift distribution results in constant downwash and therefore constant downwash, w induced angle of attack C From incompressible flow theory : α = L i π A 2 b2 C where : A = (Aspect ratio) Thus : C = L S Di πA 31 32 33 Effect of aspect ratio on induced drag High aspect ratio : Low Induced Drag DG 200 DG 500 Low aspect ratio : High Induced Drag Mirage 2000 34 Span efficiency factor C2 C = L Di π A e span efficiency factor (Oswald factor) Elliptical loading : e = 1 ; minimum induced drag Non-elliptical loading : e < 1 ; higher induced drag The total drag of the wing can now be written as : C 2 C = C + L D D p π A e Profile drag Induced drag 35 Example 5.19 • Consider a flying wing with wing area of 206 m2, an aspect ratio of 10, a span effectiveness factor of 0.95, and an NACA 4412 airfoil. The weight of the airplane is 7.5x10 5 N. If the density altitude is 3 km and the flight velocity is 100 m/s, calculate the total drag on the aircraft. 36 Lift curve slope geometric α eff angle of attack α α i V∞ The effect of a finite wing is to reduce the wing lift curve slope 37 Lift curve slope The induced angle of attack reduces α = α − α The local effective angle of attack : eff i For a wing of a general plan form we may write : C Angle is in radians! α = L e is span effectiveness factor i π 1 A e1 (theoretically different from e but in practice more or less the same) 57.3 C α = L i π For angle in degrees Ae 1 38 Lift curve slope infinite wing (effective angle of attack) dC a = l 0 dα CL a0 finite wing (geometric angle of attack) a dC a = L dα α From thin airfoil theory we find : = π a 0 2 39 Lift curve slope Effective local angle of attack αeff = α − αi For an arbitrary loading distribution wing we find : dC α = C L L = a Integrate : i π α − α 0 Ae 1 d( i ) C C = a (α−α )+const, C = a α − L + const L 0 i L 0 π Ae 1 a α const C = 0 + L a a 1+ 0 1+ 0 π π Ae 1 Ae 1 40 Lift curve slope dC a Differentiating this equation results in : L = 0 dα a 1 + 0 π Ae 1 Example : dC Infinite wing : A= ∞ then L = 2π dα dC L Finite wings : A=12 (Fokker 50) then = 2π⋅0.857 = 1.71 π dα dC A=5 then L = 2π⋅0.714 =1.43 π dα 41 dC a dC a =L = 0 a = l α a 0 dα d 1+ 0 π Ae 1 dC a L = 0 a per degree α 57.3 a 0 d 1+ 0 π Ae 1 42 Example 5.21 • Consider a wing with an aspect ratio of 10 and NACA 23012 airfoil section.