8/18/2019 Thin Aerofoil Theory notes

1/25

8/18/2019 Thin Aerofoil Theory notes

2/25

8/18/2019 Thin Aerofoil Theory notes

3/25

8/18/2019 Thin Aerofoil Theory notes

4/25

8/18/2019 Thin Aerofoil Theory notes

5/25

 To fnd correct combination o elementry ows overa specifed body

1. Source panel method

2. Vortex panel method

•. It become standard aerodynamics tool in industryand a research laboratories.

•.  These are the numerical method appropriate orsolutions or a computers.

limitation to non!litin" ows

8/18/2019 Thin Aerofoil Theory notes

6/25

Source sheet:

8/18/2019 Thin Aerofoil Theory notes

7/25

 Vortex filament:

8/18/2019 Thin Aerofoil Theory notes

8/25

Vortex sheet:

8/18/2019 Thin Aerofoil Theory notes

9/25

Vortex sheet over the airoilsurace#

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

8/18/2019 Thin Aerofoil Theory notes

10/25

Thin Airfoil theory:

Placement of the vortex sheet for thin airfoil analysis.

8/18/2019 Thin Aerofoil Theory notes

11/25

Determination of the component of freestream velocity normal to

the camber line.

8/18/2019 Thin Aerofoil Theory notes

12/25

 Thin $iroil 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.

8/18/2019 Thin Aerofoil Theory notes

13/25

•  The velocity at any point in the ow is the sum othe o the uniorm reestream velocity and thevelocity induced by the vortex sheet.

•

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

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

reduces to

8/18/2019 Thin Aerofoil Theory notes

14/25

 -alculation o the induced velocity at the chordline.

( )  ( )

( )

c

0

dw x

2 x

γ ξ ξ= −

π − ξ∫ 

8/18/2019 Thin Aerofoil Theory notes

15/25

)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

ddzV 0

dx 2 x∞

γ ξ ξ  α − − = ÷

π − ξ  

  ∫ 

( )c

0

d1 dzV

2 x dx∞

γ ξ ξ    = α − ÷π − ξ    ∫ 

8/18/2019 Thin Aerofoil Theory notes

16/25

•  The help deal with the inte"ral in *+uations and&let us transorm into θ via the ollowin"transormation#

 

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

 

8/18/2019 Thin Aerofoil Theory notes

17/25

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

cd sind

2ξ = θ θ

( )0

0

sind1V

2 cos cos

π

∞

γ θ θ θ= α

π θ − θ∫ 

( )  ( )1 cos2V

sin∞

+ θγ 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∫ ∫ 

8/18/2019 Thin Aerofoil Theory notes

18/25

0

00 0

sinncosnd

cos cos sin

π   π θθ θ=θ − θ∫ 

( )

( )

0 0 00 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 = απ

∫ ∫ ∫ 

( )0

0

sin d1V

2 cos cos

π

∞γ θ θ θ = α

π θ −∫ 

8/18/2019 Thin Aerofoil Theory notes

19/25

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

8/18/2019 Thin Aerofoil Theory notes

20/25

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

0dΓ = γ ξ ξ∫ 

( )0

csind

2

πΓ = γ θ θ θ

∫ ( )

0cV 1 cos d cV

π

∞ ∞Γ = α + θ θ = πα∫ 

2L' V c V∞ ∞ ∞ ∞= ρ Γ = πα ρ

l

L'cqs∞

=

( )S c1=

( )

2

12

c Vc1Vc1

2

∞ ∞

∞ ∞

πα ρ=ρ

l

c 2= πα   ldcLift slop= 2d

= π

α

( )  ( )1 cos2V

sin∞

+ θγ θ = α

θ

8/18/2019 Thin Aerofoil Theory notes

21/25

-

8/18/2019 Thin Aerofoil Theory notes

22/25

-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

Mc

qc 2∞

πα= = −

lc

2πα =

lm,le cc

4= −

l

m,c/4 m,le

c

c c 4= +

m,c/4c 0

=

8/18/2019 Thin Aerofoil Theory notes

23/25

Important result#

 Theoretical results or a symmetricairoil#

• -l 02πα.

• %it slope 0 2π.

•

 The center o pressure and theaerodynamic center are both locatedat the +uarter!chord point.

8/18/2019 Thin Aerofoil Theory notes

24/25

8/18/2019 Thin Aerofoil Theory notes

25/25