lifting line theory - virginia techdevenpor/aoe5104/23. liftinglinetheory.pdf · lifting line...
TRANSCRIPT
![Page 1: Lifting Line Theory - Virginia Techdevenpor/aoe5104/23. LiftingLineTheory.pdf · Lifting Line Theory • Applies to large aspect ratiounswept wings at small angle of attack. • Developed](https://reader034.vdocuments.net/reader034/viewer/2022052516/5b905b0409d3f2e6728bd3de/html5/thumbnails/1.jpg)
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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Physics of an Unswept Wingl, Γ
y
pu<pl pu≈ pl
Downwash
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-AdiInduced dragb A
A
Wake model Section model
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LLT – The Section Model
αε
V∞w
l
ε
di
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LLT – The Wake ModelΓ
y1
ydy1
s-s
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The Monoplane EquationWake model
Section model∫− −
Γ
=s
s
y
yy
dydyd
yw)(4
)(1
11
π
0( )V c wcπ α α π∞Γ = − −
l, Γ
ys-s
1
1
01
( )4
sy
s
d dydycV cy y
π α α∞−
Γ
Γ = − −−∫
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 22
∑∞
=∞=Γ
oddnn nAsU
,1)sin(4 θSubstituting into
gives 1AARCL π= )1(2
δπ
+=ARCC L
Di
∑∞
=
=oddn
n AAn,3
21)/(δ
1,sin( )
sin
nn odd
nA nw
V
θ
θ
∞
=
∞
=∑
+=− ∑
∞
=
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
![Page 9: Lifting Line Theory - Virginia Techdevenpor/aoe5104/23. LiftingLineTheory.pdf · Lifting Line Theory • Applies to large aspect ratiounswept wings at small angle of attack. • Developed](https://reader034.vdocuments.net/reader034/viewer/2022052516/5b905b0409d3f2e6728bd3de/html5/thumbnails/9.jpg)
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));
+=− ∑
∞
=
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
0.050.1
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 Polar
Note that friction drag coefficient of 0.01 added to CDi
)(tan 0liftLD CC αα −=Curve assuming wing only generates force normal to chord
ARCC L
D π
2
= Curve for minimum drag (elliptical wing)
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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
∑∞
=∞=Γ
oddnn nAsU
,1)sin(4 θ
1,sin( )
sin
nn odd
nA nw
V
θ
θ
∞
=
∞
=∑
0( )V c wcπ α α π∞Γ = − −
1.
2.
3.
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Further results1AARCL π=
ARCC L
Di π
2
= 1w A
V∞
=
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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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Not done yet…
ARCC L
Di π
2
=2
)(2 0
+−
=AR
ARCLααπ
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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’s rescaling using LLT result to AR=5