Thin Airfoil Theory

17 Pages·2007·0.2 MB·English

Thin Airfoil Theory
Airfoil Characteristics
l
y
α x
m
V
∞
2
• Lift coeffcient: C = l / 1 ρV c
l 2 ∞
• Moment coeficient:
• Center of pressure:
• Aerodynamic Center:
2
Thin Airfoil Theory – Setup
Symbols: v=V sinα+v' Assumptions:
y ∞
1. Airfoil is thin η<< c
u=V cosα+u'
2. Angles/slopes are small e.g.
∞
sinα≈ α, cosα≈ 1, slope ≈ angle
η
η t
3. Airfoil only slightly disturbs
u
η
c
free stream u', v' << V
x ∞
α
η (<0)
l
V Kutta condition? (cusped TE)
∞ Chord c
Camber η = 1 (η +η) η =η +η
c 2 u l or u c t
½ Thickness η = 1 (η −η) η =η −η
t 2 u l l c t
Non-penetration condition? Bernoulli?
3
Thin Airfoil Theory - Simplifications
Bernoulli: Assumptions:
1. Airfoil is thin η<< c
C =1−(u2 + v2)/V 2
p ∞
2. Angles/slopes are small e.g.
=1−[(V cosα+u')2 + (V sinα+ v')2]/V 2
∞ ∞ ∞ sinα≈ α, cosα≈ 1, slope ≈ angle
2u' 2v' u'2 v'2
= − cosα− sinα− − 3. Airfoil only slightly disturbs free
V V V 2 V 2
∞ ∞ ∞ ∞ stream u', v' << V
∞
“Linearized Pressure
C ≈
Coefficient”
p v=V sinα+v'
∞
u=V cosα+u'
y
∞
Kutta Condition:
V(c,0 ) =V(c,0 )
V(c,0 )
+ −
+
x
c
V(c,0 )
-
4
Thin Airfoil Theory - Simplifications
Exact:
v=V sinα+v'
Exact: y ∞
u=V cosα+u'
∞
η
dη dη v'(x,η )
η t
c + t ≈α+ u u
η
Small disturbances dx dx V c x
∞
and angles:
v'(x,η ) dη dη α η =η +η
u ≈ −α+ c + t η(<0) u c t
l
V
V dx dx η =η −η
∞
∞ l c t
v'(x,0 ) dη dη Linearized:
Airfoil thin: + ≈ −α+ c + t C (x,0 ) ≈
y
V dx dx p ±
∞
v'(x,0 ) dη dη
Likewise for − ≈ −α+ c − t y=0
+
lower surface: V dx dx
∞ x
α
v'(x,0 ) dη dη
Or: ± ≈ −α+ c ± t V∞ y=0- u(c,0 )=u(c,0 )
+ -
V dx dx
∞
5
A Source Sheet
Jump in normal velocity component
A Vortex Sheet
Jump in tangential velocity component
6
Solving for the Flow
Linearized problem: Proposed Ideal Flow Solution:
y iy
C (x,0 ) ≈ −2u(x,0 )/V
p ± ± ∞
z
v'(x,0 ) dη dη
± ≈ −α+ c ± t
y=0 V dx dx y=0
+ ∞ +
x
x
α α x dx
1
1
Source sheet
V y=0 V y=0
∞ - u(c,0+)=u(c,0-) ∞ - + vortex sheet
Complex velocity at z 1 q(x )−iγ(x ) Complex velocity at z 1 c q(x )−iγ(x )
1 1 dx W'(z) = ∫ 1 1 dx
due to element dx : 2π (z − x ) 1 due to whole sheet: 2π (x+iy − x ) 1
1
1 0 1
c
1 q(x )−iγ(x )
Complex velocity at
W'(x,0 ) = ∫ 1 1 dx
K
y=0 due to sheet: ± 2π (x−x ) 1
± 0 1
So what? v'(x,0 ) ⎧W'(x,0 )⎫ 1 c γ(x ) q(x)
± = −Im⎨ ± ⎬ = ∫ 1 dx ±
1
V V 2πV (x− x ) 2V
⎩ ⎭
∞ ∞ ∞ 0 1 ∞
N.B. Karamcheti defines the vortex u'(x,0 ) ⎧W'(x,0 )⎫ 1 c q(x ) γ(x)
± = Re⎨ ± ⎬ = ∫ 1 dx m 7
sheet strength with the opposite sign 1
V V 2πV (x − x ) 2V
⎩ ⎭
∞ ∞ ∞ 0 1 ∞
Solving for the Flow
Linearized problem: Proposed Ideal Flow Solution:
y iy
C (x,0 ) ≈ −2u(x,0 )/V v'(x,0 ) 1 c γ(x ) q(x)
p ± ± ∞ ± = ∫ 1 dx ±
V 2πV (x− x ) 1 2V
v'(x,0 ) dη dη
∞ ∞ 0 1 ∞
± ≈ −α+ c ± t
y=0 V dx dx y=0 u'(x,0 ) 1 c q(x ) γ(x)
+ ∞ + ± = ∫ 1 dx m
V 2πV (x− x ) 1 2V
∞ ∞ 0 1 ∞
x
x
α α x dx
1
1
Source sheet
V y=0 V y=0
∞ - u(c,0+)=u(c,0-) ∞ - + vortex sheet
Kutta u'(c,0 ) u'(c,0 ) 1 c q(x ) γ(c) 1 c q(x ) γ(c)
+ = − so ∫ 1 dx − = ∫ 1 dx + so γ(c) = 0
condition: V V 2πV (c− x ) 1 2V 2πV (c− x ) 1 2V
∞ ∞ ∞ 0 1 ∞ ∞ 0 1 ∞
c
Non- dη dη 1 iγ(x ) q(x)
−α+ c ± t = ∫ 1 dx ±
1
penetration: dx dx 2πV (x− x ) 2V
∞ 0 1 ∞
c
1 q(x ) γ(x)
Pressure: C (x,0 ) = − ∫ 1 dx ±
p ± 1
πV (x − x ) V
∞ 0 1 ∞
Solution
Pressure Difference:
order?
8
Need q(x)?
General Algebraic Solution
Non-penetration
c
dη dη 1 iγ(x ) q(x)
−α+ c + t = ∫ 1 dx +
1
dx dx 2πV (x− x ) 2V
c
∞ 0 1 ∞ dη 1 iγ(x )
add −α+ c (x) = ∫ 1 dx
1
dx 2πV (x− x )
c
dη dη 1 iγ(x ) q(x) ∞ 0 1
−α+ c − t = ∫ 1 dx −
1
dx dx 2πV (x− x ) 2V
∞ 0 1 ∞ +Kutta γ(c) = 0
dη
How to solve for γ(x) w/o specifying c ( x ) ?
dx
•Write γ as
x/c = 1 (1+cosθ)
2
•Solve integral for each term in
x /c = 1 (1+cosθ)
1 2 1
dη
•Relate to for c
dx
0 c
α x
V π 0
∞ θ
9
General Algebraic Solution
Fourier Series Solution gives:
x/c = 1 (1+cosθ)
1−cosθ ∞ 2
l
γ(θ)/V = −(2α− B ) + 2∑B sin(nθ)
∞ o sinθ n m ΔC
O
n=1 p
0 c
2 π dη α x
where: B = ∫ c (θ)cos(nθ)dθ
n π dx V π 0
0 ∞ θ
and we have that ΔC (θ) = 2γ(θ)/V
p ∞
l
C = =
l 1 ρV 2c
2 ∞
Substitute for γ(θ) and evaluate: C = 2πα−π(B + B )
l 0 1
c 1 0
m −1 x x
C = O = ∫ xΔpdx = −∫ ΔC (x)d = − 1 ∫ ΔC (θ)(1+ cosθ)sinθdθ
mO 1 ρV 2c2 1 ρV 2c2 c p c 4 p
2 ∞ 2 ∞ 0 0 π
Substitute for γ(θ) and evaluate: C = − 1πα+ 1π(B + 2B + B )
mO 2 4 0 1 2
= − 1 C + 1π(B + B ) 10
4 l 4 1 2