thin airfoil.ppt
Transcript of thin airfoil.ppt
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Chap.4
Incompressible Flow overAirfoils
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OUTLINE
Airfoil nomenclature and characteristics
The vortex sheet
The Kutta condition
Kelvins circulation theorem
Classical thin airfoil theory
The cambered airfoil
The vortex panel numerical method
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Airfoil nomenclature and characteristics
Nomenclature
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Characteristics
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The vortex sheet
Vortex sheet withstrength =(s) Velocity at Pinduced by
a small section of vortexsheet of strength ds
For velocity potential (toavoid vector addition asfor velocity)
r
dsdV
2
2
dsd
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The velocity potential at Pdue to entire vortex sheet
The circulation around thevortex sheet
The local jump in tangentialvelocity across the vortexsheet is equal to .
b
ads
2
1
b
a
ds
0,21 dnuu
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Calculate (s) such that the induced velocity field
when added to Vwill make the vortex sheet(hence the airfoil surface) a streamline of the flow.
The resulting lift is given by Kutta-Joukowskitheorem
Thin airfoil approximation
VL
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The Kutta condition
Statement of the Kutta condition
The value of around the airfoil is such that theflow leaves the trailing edge smoothly.
If the trailing edge angle is finite, then the trailingedge is a stagnationpoint.
If the trailing edge is cusped, then the velocityleaving the top and bottom surface at the trailingedge are finite and equal.
Expression in terms of
0)TE(
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Kelvins circulation theorem
Statement of Kelvins circulation theorem
The time rate of change of circulationaround aclosed curve consisting of the same fluid elementsis zero.
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Classical thin airfoil theory
Goal
To calculate (s) such that the camber linebecomes a streamline.
Kutta condition (TE)=0 is satisfied. Calculate around the airfoil.
Calculate the lift via the Kutta-Joukowski theorem.
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Approach Place the vortex sheet on
the chord line, whereasdetermine =(x) to make
camber line be astreamline.
Condition for camber linetobe a streamline
where w'(s)is thecomponent of velocitynormal to the camber line.
0)(, swV n
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Expression of V,n
For small
)(tansin1
,dx
dzVV n
)(
)()(,tansin
,dx
dzVV
xwsw
n
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Expression for w(x)
Fundamental equation ofthin airfoil theory
)()(
2
1
0 dx
dzV
x
dc
c
x
dxw
0 )(2
)()(
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For symmetric airfoil (dz/dx=0)
Fundamental equation for ()
Transformation of , xinto
Solution
Vx
dc
0
)(
2
1
)cos1(2
,)cos1(2
0 c
xc
sin
)cos1(2)( V
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Check on Kutta condition by LHospitals rule
Total circulation around the airfoil
Lift per unit span
0cos
sin2)(
V
cVdc
0)(
2
VcVL
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Lift coefficient and lift slope
Moment about leading edge and moment coefficient
2,2d
dc
cq
Lc ll
42
2
2,
2
0
lLE
lem
c
LE
c
cq
M
c
cqLdM
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Moment coefficient about quarter-chord
For symmetric airfoil, the quarter-chord pointisboth the center of pressureand the aerodynamiccenter.
0
4
4/,
,4/,
cm
llemcm
c
ccc
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The cambered airfoil
Approach
Fundamental equation
Solution
CoefficientsA0andAn
)(
coscos
sin)(
2
1
00 dx
dzV
d
1
0 sinsin
cos12)(
n
n nAAV
0
000
00 cos2
,1
dndx
dzAd
dx
dzA n
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Aerodynamic coefficients Lift coefficient and slope
Form thin airfoil theory, the lift slopeis always 2for any shape airfoil.
Thin airfoil theory also provides a means to
predict the angle of zero lift.
2,)1(cos1
20
00
d
dcd
dx
dzc ll
000
0 )1(cos1
ddx
dzL
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Moment coefficients
For cambered airfoil, the quarter-chord point isnotthe center of pressure, but still is the
theoretical location of the aerodynamic center.
)(4
)(44
124/,
21,
AAc
AAc
c
cm
llem
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The location of the center of pressure
Since
the center of pressure is not convenient fordrawing the force system. Rather, theaerodynamic center is more convenient.
The location of aerodynamic center
)(1
421 AA
c
cx
l
cp
0as lcp cx
0
4/,
0
0
0 ,where,25.0 md
dca
d
dc
a
mx
cmlac
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The vortex panel numerical method
Why to use this method
For airfoil thickness larger than 12%, or high angle ofattack, results from thin airfoil theory are notgoodenough to agree with the experimental data.
Approach
Approximate the
airfoil surface by
a series of straightpanels with strength
which is to be
determined.
j
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The velocity potential induced at P due to thej thpanel is
The total potential at P
Put P at the control point of ith panel
j
j
pjjjj
pjjxx
yyds
1tan,2
1
2
)(11
jj
pj
n
j
jn
j
j dsP
2
),(1
jj
ij
n
j
j
ii dsyx
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The normal component of the velocity is zero atthe control points, i.e.
We then have n linear algebraic equation with nunknowns.
nidsn
V
dsnVVV
VV
jj
i
ijn
j
j
i
jj
i
ijn
j
j
nin
nn
,,1,02
cos
2,coswhere
0
1
1
,
,
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Kutta condition
To impose the Kutta condition,we choose to ignore one ofthe control points.
The need to ignore one of the
control points introducessome arbitrariness in thenumerical solution.
10)TE( ii