p4-classical thin airfoil theory
TRANSCRIPT
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Classical Thin Airfoil Classical Thin Airfoil TheoryTheorySymmetric Airfoil
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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
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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)
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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
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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
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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
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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
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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
,
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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 ,