Lifting Line Theory - Virginia Tech Line Theory AOE 5104 Advanced Aero- and Hydrodynamics Dr....

Lifting Line Theory

AOE 5104Advanced Aero- and Hydrodynamics

Dr. William Devenport andLeifur Thor Leifsson

2

Flow over a low aspect ratio wing

Bottom surface

Top surface

3

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

– Example of a phenomenological model in which elementary ideal flows are used to represent real viscous phenomena

– 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

4

Large Aspect Ratio? Unswept?

5

6

Physics of an Unswept Wing

Head, 1982, Rectangular Wing, 24o, Re=100000

Werle, 1974, NACA 0012, 12.5o, AR=4, Re=10000

Bippes, Clark Y, Rectangular Wing 9o, AR=2.4, Re=100000

7

Cessna Citation

Aircraft DataVelocity = 165 knots Wing Area = 29 m^2 Wing Span = 16 m

Mean Aerodynamic Chord = 2 m Weight = 8000 kg

Chord Reynolds Number = 1.18E7

8

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

9

Wing Nomenclature

l, Γ

y

Section A-A

αε

V∞

-w

l

ε

di

s-s

c

α=α (y) Geometric angle of attackε=ε (y) Downwash angle-w=-w (y) Downwash velocityc=c (y) ChordlengthS Planform areas Half spanb Spanc = S/b Average chordAR=b/c=b2/S Aspect ratioΛ Sweep angle of ¼ chord linel Lift per unit spandv Drag per unit span

Induced drag

bs

Area S

Total drag coeff

SVLCL 2

21

∞

=ρ

SVDC i

Di 221

∞

=ρ

Total lift coeff

A

A

10

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

1

1

dydyd

y

Γ−

dΓ

11

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 ¼ chord, and thin airfoil theory

So,

Sectional forces Γ≈ ∞Vl ρ Γ−≈Γ≈ ∞ wVdv 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

( )εααπ −− 02

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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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Solving the Monoplane Eqn.: Glauert’s method

⎥⎦⎤

⎢⎣⎡ +=− ∑

∞

=

θπθθααπ sin4

)sin(sin)(4 ,1

0 scnnA

sc

oddnn

ys-s0 π θ

θcos/ −=sy

1. Decide on the number of terms Nneeded for the sine series for Γ

2. Select N points across the half span, evenly spaced in θ

3. At each point evaluate c, α, α0 and write down the corresponding monoplane equation

4. Solve the resulting N equations for the N unknown coefficient An

5. Evaluate Γ(y) and then the integrals for w(y), L and Di, and thus the aero coefficients

∑∞

=∞=Γ

oddnn nAsU

,1)sin(4 θ

∫− −

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

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Glauert’s Method Example

⎥⎦⎤

⎢⎣⎡ +=− ∑

∞

=

θπθθααπ sin4

)sin(sin)(4 ,1

0 scnnA

sc

oddnn

ys-s0 π θ

θcos/ −=sy

∑∞

=∞=Γ

oddnn nAsU

,1)sin(4 θ

-10 -8 -6 -4 -2 02

4

6

8

10

12

∫− −

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

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Other 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

gives 1AARCL π=

)1(2

δπ

+=ARCC L

Di

∑∞

=

=oddn

n AAn,3

21)/(δ θ

θ

sin

)sin(,1∑∞

=

∞

−= oddnn nnA

Vw

• 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,

α

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The Elliptic WingSo the minimum drag occurs for a wing for which An=0 for n≥3. For this wing:

)sin(4 1 θsAU∞=Γ

1AVw

−=∞ AR

CC LDi π

2

=

)/arccos( sy−=θ14

22

1

=⎟⎠⎞

⎜⎝⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛ Γ⇒

∞ sy

sAVsince

andπααπ 10 )( AVV

c∞∞ −−

Γ=