p4-classical thin airfoil theory
TRANSCRIPT
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 ,