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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