p4-classical thin airfoil theory

Classical Thin Airfoil Classical Thin Airfoil TheoryTheorySymmetric Airfoil

Symmetric AirfoilSymmetric Airfoil

Chamber line, z = z (x)

w

Chord lineco

V x

z

Chamber line, z = z (x)

sw

Chord lineco

V x

z

s

s xw

Symmetric AirfoilSymmetric Airfoil

Chamber line, z = z (x)

V

P

dxdz1tan

dxdz1tan

o90 nV ,

Chamber line: stream line

0 swV n,

dxdzVV n

1tansin,

(4.12)

(4.13)

small are and attack, of angel smallfor 1

dxdztan

dx

dzVV n , toreduces (4.13)equation (4.14)

Symmetric AirfoilSymmetric Airfoil

wd

s

xwsw

Thin airfoil, chamber line close to chord line

(4.15)

Velocity at point x induced by elemental vortex

xddw

2(4.16)

c

xdxw

0 2

Subst. To eq. (4.12)

(4.17)

020

c

xd

dxdzV

dxdzV

xdc

021

(4.18)Fundamental equation of thin Airfoil theory

Symmetric AirfoilSymmetric Airfoil

The central problem of thin airfoil theory is to solve eq. 4.18 for vortex strength, subject to the Kutta condition

Vxd

dxdz

c

021

0 line, chordlinechamber airfoil, symmetricfor

(4.19)

Exact expression for inviscid, incompressible flow over a flat plate

dcd

cx

c

o

2

12

12

into transform

sin

cos

cos

(4.20)

(4.21)

(4.22)

Substitution into eq. 4.19

(4.23)

Vd

o0

21

coscossin

Symmetric AirfoilSymmetric Airfoil

Vd

o0

21

coscossin

(4.23)

(4.24)

00

1 21

(4.23) into (4.24) eq.on substituti

oo

dVdcoscos

coscoscos

sin

(4.25)

oo

ndn

sinsin

coscoscos

0

(4.26)

VV

ddV

dV

oo

o

0

1

(4.25) eq. of side handright (4.26), eq. using

0 0

0

coscoscos

coscos

coscoscos

(4.27)

(4.23) eq. osolution t theindeed is

(4.24) Eq. (4.24). eq. toidentical is

21

(4.25) into (4.27) eq.on substituti

0

Vd

ocoscossin

sin

cos

12

issolution the

V

Symmetric AirfoilSymmetric Airfoil

condition Kutta the

satisfies also (4.24) eq. thus

02

(4.24) eq.on rule HospitalL' using002

(4.24) eq. TE, at the

(4.24) 12

cossin

,sincos

V

V

V

spanunit per Lift theorem,

Joukowski-Kutta into (4.30) eq. subst.

)0(4.3 1

(4.29) into (4.24) eq.on substituti

(4.29) sin 2

(4.22) eq. and (4.20) eq. using

(4.28)

airfoil aroundn circulatio Total

0

0

0

cVdV

dc

dc

cos

(4.32)

coeficientlift

(4.31) 2

SqLc

VcVL

l

(4.33) 2

121

1 where

2

2

cV

Vcc

cS

l

(4.34) 2slopeLift

ddcl

Symmetric AirfoilSymmetric Airfoil

d

dd

dL

LE

(4.35)

LE about theMomen

00

cc

LE dVdLM

dLdM

dVdL

dd

(4.37) 2

1 where

coeficientmoment

(4.36) 2

(4.22) and (4.20) eq. using

2

2

cqM

c

cSScq

Mc

cqM

LELEm

LELEm

LE

,

,(4.39)

4

(4.38) and (4.37) eq.

(4.38) 2

(4.33) eq. from however,

lLEm

l

cc

c

,

Symmetric AirfoilSymmetric Airfoil

(4.41) 0

(4.40) and (4.39) eq.

(4.40) 4

(1.22) eq. point,

chord-quarterabout coeficientmoment

4

4

cm

lLEmcm

c

ccc

,

,,

center of pressure

(4.24) sin

cos

12 V

(4.33) 2lc

(4.34) 2slopeLift

ddcl

(4.41) 04 cmc ,