thin aerofoil theory notes

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• 8/18/2019 Thin Aerofoil Theory notes

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To fnd correct combination o elementry ows over a specifed body

1. Source panel method

2. Vortex panel method

•. It become standard aerodynamics tool in industry and a research laboratories.

•.  These are the numerical method appropriate or solutions or a computers.

 limitation to non!litin" ows

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Source sheet:

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Vortex filament:

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Vortex sheet:

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Vortex sheet over the airoil surace#

Vortex sheet over the thin airfoil surface: dsdV 2rγ = − π

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Thin Airfoil theory:

Placement of the vortex sheet for thin airfoil analysis.

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Determination of the component of freestream velocity normal to

the camber line.

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Thin \$iroil theory#

• Our purpose is to calculate the variation of (s) such

that the camber line becomes a streamline of the flow

and such that the utta condition is satisfied at the

trailin! ed!e" that is# γ \$T%& ' (. • Once we have found the particular γ \$s& that satisfies

these conditions# then the total circulation around

the airfoil is found by inte!ratin! γ \$s& from the

leadin! ed!e to the trailin! ed!e.

•  )n turn# the lift is calculated from Γ  via the utta*

+ou,ows,i theorem.

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•  The velocity at any point in the ow is the sum o the o the uniorm reestream velocity and the velocity induced by the vortex sheet.

%et V∝&n be the component o the reestreamvelocity normal to the camber line.

• 'or a thin airoil at small an"le o attac(& both are small values. )sin" the approximation that or small θ& where θ& where θ is in radians& *+uation

reduces to

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-alculation o the induced velocity at the chord line.

( )   ( )

( )

c

0

d w x

2 x

γ ξ ξ = −

π − ξ∫

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)n this section# we treat the case of a symmetric airfoil. As state in section# a symmetri

airfoil has no camber" the camber line is coincident with the chord line.

-ence# for this case# d/d ' (# and %0uation becomes

( )

( )

c

0

ddz V 0

dx 2 x ∞

γ ξ ξ  α − − = ÷

π − ξ 

∫

( )c

0

d1 dz V

2 x dx ∞

γ ξ ξ     = α − ÷π − ξ     ∫

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•  The help deal with the inte"ral in *+uations and& let us transorm into θ via the ollowin" transormation#

• Since x is a fxed point in *+uation and& it corresponds to a particular value o θ& namely& θ& such that

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Substitutin" *+uations into& and notin" that the limits o inte"rationbecomes at the leadin" ed"e /where 0 and θ0π at the trailin" ed"e /where 0c& we obtain

c d sind

2 ξ = θ θ

( ) 0

0

sind1 V

2 cos cos

π

γ θ θ θ = α

π θ − θ∫

( )   ( )1 cos 2V

sin ∞

+ θ γ θ = α

θ

( ) ( ) 0 0

0 0

sind 1 cos d1 V

2 cos cos cos cos

π π ∞γ θ θ θ + θ θθ=

π θ − θ π θ − θ∫ ∫

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0

0 0 0

sinncosnd

cos cos sin

π   π θθ θ =θ − θ∫

( )

( )

0 0 0 0 0 0

1 cos dV V d cosd

cos cos cos cos cos cos

V   0 V

π π π ∞ ∞

∞ ∞

  + θ θα α θ θ θ = + ÷π θ − θ π θ − θ θ − 

α= + π = α π

∫ ∫ ∫

( ) 0

0

sin d1 V

2 cos cos

π

∞ γ θ θ θ = α

π θ −∫

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e are now n a pos t on to ca cu ate t e t coe c ent or a t n symmetr c a r o

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e are now n a pos t on to ca cu ate t e t coe c ent or a t n# symmetr c a r o . The total circulation around the airfoil is

4sin! %0uation and# e0uation transforms to

Substitutin! %0uation into# we obtain

Substitutin! %0uation into the utta*+ou,ows,i theorem# we find that the lift per unit span is

Substitutin! %0uation into# we have

( ) c

0 dΓ = γ ξ ξ∫

( ) 0

c sind

2

π Γ = γ θ θ θ

∫  ( )

0 cV 1 cos d cV

π

∞ ∞Γ = α + θ θ = πα∫

2L' V c V∞ ∞ ∞ ∞= ρ Γ = πα ρ

l

L' c qs ∞

=

( )S c1=

( )

2

1 2

c Vc 1 Vc1

2

∞ ∞

∞ ∞

πα ρ= ρ

l

c 2= πα   ldcLift slop= 2 d

= π

α

( )   ( )1 cos 2V

sin ∞

+ θ γ θ = α

θ

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-

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-ence#

-owever# from %0uation#

5ombinin! e0uation and# we obtain

7rom %0uation# the moment coefficient about the 0uarter*chord point is

5ombinin! %0uation and# we have

' LE

m,le   2

M c

qc 2∞

πα = = −

lc

2 πα =

l m,le cc

4= −

l

m,c/4 m,le

c

c c 4= +

m,c/4 c 0

=

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Important result#

Theoretical results or a symmetric airoil#

• -l 02πα.

• %it slope 0 2π.

The center o pressure and theaerodynamic center are both located at the +uarter!chord point.

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