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
= αα − = αα − L Ncos T sin cl c ncos c t sin
L N T =cos αα − sin qc∞ qc ∞ qc ∞
For small angle of attack α α α ≤5o : cos ≈ 1, sin ≈ 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 cr,p 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 >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. Assume Re ≈ 5 x 10 6. the span efficiency factor is o e=e 1=0.95. If the wing is at 4 angle of attack, calculate C L and CD.
43 Swept wings
M ∞
V∞ Λ=40°
Mcr =0.7
44 Swept wings
M ∞
V∞ Λ=40° V= V cos Λ Mcr =0.7 ∞
0.7 M for swept wing = = 0.91 cr cos Λ By sweeping the wings of subsonic aircraft, the drag divergence is delayed to higher Mach numbers
45 Swept wings
Effect of wing thickness and sweepback angle on minimum wing drag coefficient
ABC t/c=9% A
CD,min t/c=6% B t/c=4% C
1.0 M 1.0 M 46 Total drag
C 2 CC= + L D D profile π Ae
CCCC= + + DDDDprofile f pressure w
47 Flaps
δ 1 2 W = L = C ρV∞S L 2
2W 2W V = = Thus : ∞ ρ and : Vstall ∞SC L ρ∞SC Lmax
The landing speed is decreased when the maximum lift coefficient is increased
48 Flaps
δ f=60° C L flaps extended δ
δ f=15°
flaps retracted δ f=0°
α
49 Flaps
Typical example
50 Flaps and slats
flaps
slats
51 Flaps
52 Flaps
53 Flaps
54