lifting line theory - virginia techdtmook/aoe5104_online/class notes/23_class_… · lifting line...
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Lifting Line Theory
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Lifting Line Theory• Applies to large aspect ratio unswept wings at small angle of attack.• Developed by Prandtl and Lanchester during the early 20th century. • Relevance
– Analytic results for simple wings– Basis of much of modern wing theory (e.g. helicopter rotor aerodynamic
analysis, extends to vortex lattice method,)– Basis of much of the qualitative understanding of induced drag and aspect
ratio
1875-19531868-1946Γ h
Vπ4Γ
=
h
Biot Savart Law:Velocity produced by a semi-infinite segment of a vortex filament
Thin-airfoil theoryCl=2π(α-αo)
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Large Aspect Ratio? Unswept?
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Out
was
h
Physics of an Unswept Wingl, Γ
y
pu<pl pu≈ pl
Downwash
Inwash
s-s
Lift varies across span
Circulation is shed (Helmholz thm)
Vortical wake
Vortical wake induces downwash on wing…
…changing angle of attack just enough to produce variation of lift across span
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Simplest Possible Model
Section A-A
αε
V∞
-w
l
ε
di
c
α=α (y) Geometric angle of attackε=ε (y) Downwash angle-w=-w (y) Downwash velocityc=c (y) Chordlengths Half spanl Lift per unit spandi Drag per unit span
Induced dragb
Total drag coeff
SVLCL 2
21
∞
=ρ
SVDC i
Di 221
∞
=ρ
Total lift coeff
A
A
Wake model Section model
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LLT – The Wake ModelΓ
y1
ydy1
Strength of vortex shed at y1=
• Assume role up of wake unimportant• Assume wake remains in a plane parallel to the free stream• Model wake using single vortex sheet starting at the quarter chord
Downwash at y due to vortex shed at y1 )(4
)(1
11
yy
dydyd
ydw y
−
Γ−=−
π
Downwash at ydue to entire wake ∫
− −
Γ
=s
s
y
yy
dydyd
yw)(4
)(1
11
π
s-s
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LLT – The Section Model
αε
V∞
-w
l
ε
di
=Γ
==∞
∞
∞ cVV
cVlCl 2
212
21 ρ
ρρ
wccV πααπ +−=Γ ∞ )( 0
Sectional lift coefficient
• Assume flow over each section 2D and determined by downwash at ¼chord, and thin airfoil theory
So
Sectional forces Γ≈ ∞Vl ρ Γ−≈Γ≈ ∞ wVdi ρερ
∫−
∞ Γ≈s
s
dyVL ρ ∫−
Γ−≈s
si dywD ρ
∫−∞∞
Γ≈=s
sL dy
SVSVLC 2
221 ρ ∫
−∞∞
Γ≈=s
s
iD dyw
SVSVDC
i 2221
2ρ
Total Forcesintegrated over span
Total Coefficients
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The Monoplane EquationWake model
Section model∫− −
Γ
=s
s
y
yy
dydyd
yw)(4
)(1
11
π
wccV πααπ +−=Γ ∞ )( 0
l, Γ
ys-s
∫−
∞ −
Γ
+−=Γs
s
y
yy
dydyd
ccV1
1
01
4)( ααπ
0 π θ
θcos/ −=sy
Substitute for θ, and express Γ as a sine series in θ
∑∞
=∞=Γ
oddnn nAsU
,1)sin(4 θ
⎥⎦⎤
⎢⎣⎡ +=− ∑
∞
=
θπθθααπ sin4
)sin(sin)(4 ,1
0 scnnA
sc
oddnn The Monoplane Eqn.
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Results
∫− −
Γ
=s
s
y
yy
dydyd
yw)(4
)(1
11
π∫−∞
Γ=s
sL dy
SVC 2 ∫
−∞
Γ=s
sD dyw
SVC
i 2
2
∑∞
=∞=Γ
oddnn nAsU
,1)sin(4 θSubstituting
into
• Lift increases with aspect ratio• For planar wings at least lift goes linearly with angle of attack and lift curve slope increases with aspect ratio (to 2π at ∞) • Drag decreases with aspect ratio and goes as the lift squared?• Downwash tends to be largest at the wing tips ?• Drag is minimum for a wing for which An=0 for n≥3.
So,
gives 1AARCL π=
)1(2
δπ
+=ARCC L
Di
∑∞
=
=oddn
n AAn,3
21)/(δ θ
θ
sin
)sin(,1∑∞
=
∞
−= oddnn nnA
Vwα
⎥⎦⎤
⎢⎣⎡ +=− ∑
∞
=
θπθθααπ sin4
)sin(sin)(4 ,1
0 scnnA
sc
oddnn
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Solution of monoplane equation
⎥⎦⎤
⎢⎣⎡ +=− ∑
∞
=
θπθθααπ sin4
)sin(sin)(4 ,1
0 scnnA
sc
oddnn
ys-s0 π θ
θcos/ −=sy
1. Decide on the number of terms N needed for the sine series for Γ2. Select N points across the half span, evenly spaced in θ3. At each point evaluate c, α, α0 and thus the NxN matrix of terms that
multiplies the An’s and the N terms on the left hand side4. Solve for the An’s by matrix division5. Evaluate CL, CDi , w(y), and Γ(y).
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s=2.8; %Half span (distances normalized on root chord)alpha=5*pi/180; %5 degrees angle of attackalpha0=-5.4*pi/180; %Zero lift AoA=-5.4 deg. for Clark YN=20; %N=20 points across half spanth=[1:N]'/N*pi/2; %Column vector of theta's y=-cos(th)*s; %Spanwise positionc=ones(size(th)); %Rectangular wing, so c = c_r everywheren=1:2:2*N-1; %Row vector of odd indices
res=pi*c/4/s.*(alpha-alpha0).*sin(th); %N by 1 result vectorcoef=sin(th*n).*(pi*c*n/4/s+repmat(sin(th),1,N)); %N by N coefficient matrixa=coef\res; %N by 1 solution vector
gamma=4*sin(th*n)*a; %Normalized on uinf and sw=-(sin(th*n)*(a.*n'))./sin(th);AR=2*s/mean(c);CL=AR*pi*a(1);CDi=CL^2/pi/AR*(1+n(2:end)*(a(2:end).^2/a(1).^2));
1. Decide on the number of terms N needed for the sine series for Γ2. Select N points across the half span, evenly spaced in θ3. At each point evaluate c, α, α0 and thus the NxN matrix of terms that
multiplies the An’s and the N terms on the left hand side4. Solve for the An’s by matrix division5. Evaluate CL, CDi , w(y), and Γ(y).
⎥⎦⎤
⎢⎣⎡ +=− ∑
∞
=
θπθθααπ sin4
)sin(sin)(4 ,1
0 scnnA
sc
oddnn
1AARCL π=)1(
2
δπ
+=ARCC L
Di
∑∞
=∞=Γ
oddnn nAsU
,1
)sin(4 θ
llt.m
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Example
-1 -0.9 -0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1 00
0.05
0.1
0.15
0.2
Γ/V∞
s
CL=0.80783, CDi=0.038738
-1 -0.9 -0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1 0-0.2
-0.15
-0.1
-0.05
0
y/s
-w/V
∞
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.05
0.1
x/c
y/c αo≈-5.4o
Determine aerodynamic characteristics of our rectangular Clark Y wing
Our AR=5.6 Rectangular Clark Y Wing
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Drag PolarARCC L
D π
2
= Curve for minimum drag (elliptical wing)
Note that friction drag coefficient of 0.01 added to CDi
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If we pretend wing is elliptical…
-1 -0.9 -0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1 00
0.05
0.1
0.15
0.2
Γ/V∞
s
CL=0.80783, CDi=0.038738
-1 -0.9 -0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1 0-0.2
-0.15
-0.1
-0.05
0
y/s
-w/V
∞
041.0
856.02
)(2
2
0
==
=+−
=
ARCC
ARARC
LD
L
i π
ααπ
AR=5.6, α=5o, α0=-5.6o
Thus, an elliptical lift distribution can often be a good approximation!
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The Elliptic WingThe minimum drag occurs for a wing for which An=0 for n≥3. For this wing:
( )sy /cos −=θ
14
22
1
=⎟⎠⎞
⎜⎝⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛ Γ⇒
∞ sy
sAV
)sin(4)sin(4 1,1
θθ sAUnAsUoddn
n ∞
∞
=∞ ==Γ ∑
1,1
sin
)sin(A
nnA
Vw oddn
n
−=−=∑∞
=
∞ θ
θ
wccV πααπ +−=Γ ∞ )( 0
πααπ 10 )( AVVc
∞∞ −−Γ
=⇒• If the wing is untwisted, the chordlength is proportional to circulation and thus also has an elliptical form
• Lift distribution has an elliptical shape.
• Downwash velocity is constant across span
1.
2.
3.
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Spitfire
Note that the chordlengths are all lined up along the quarter chord line so the actual wing shape is not an ellipse
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Further results1AARCL π=
ARCC L
Di π
2
= 1AVw
−=∞
But what is A1?
sAVssAV r
r1
22
1
4104 ∞
∞
=Γ⇒=⎟⎠⎞
⎜⎝⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛ Γ
πααπ 10 )( AVVc r
r∞∞ −−
Γ=
Planform area of elliptic wing is rscS π21=
Now
and
Substituting and solving for A1 gives )2/()(2 01 +−= ARA αα
And thus 2
)(22
)(2 00
+−
−=+−
=∞ ARV
wAR
ARCLααααπ
…confirming our earlier presumption about aspect ratio effects on CL
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Not done yet…
ARCC L
Di π
2
=
Consider two elliptical wings with the same section but different AR producing the same lift coefficient:
2)(2 0
+−
=AR
ARCLααπ
⎟⎟⎠
⎞⎜⎜⎝
⎛−=−⇒
+=−
+=−
BA
LBA
B
BLB
A
ALA
ARARC
ARARC
ARARC
11
2)2(
2)2(
00
παα
παα
παα
Similarly, we can show the two drag coefficients are related as:
⎟⎟⎠
⎞⎜⎜⎝
⎛−=−
BA
LDiBDiA ARAR
CCC 112
π
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Geometrically Similar WingsThese results work quite well even for non-elliptical wings:
⎟⎟⎠
⎞⎜⎜⎝
⎛−=−
BA
LBA ARAR
C 11π
αα ⎟⎟⎠
⎞⎜⎜⎝
⎛−=−
BA
LDD ARAR
CCCiBiA
112
π
Prandtl’s Classic Rectangular Wing Data for Different Aspect Ratios
Prandtl’srescaling using LLT result to AR=5