Chapter 4
Incompressible Flow Over Airfoils
Of the many problems now engaging attention,the following
are considered of immediate importance and will be
considered by the committee as rapidly as funds can be
secured for the purpose…, The evolution of the more efficient
wing sections of practical form,embodying suitable dimension
for an economical structure,with moderate travel of the
center-of-pressure and still affording a large range of angle-of-
attack combined with efficient action,
From the first annual report of the
NACA,1915
4.1 Introduction
Ludiwig Prandtl and his colleagues at
G?ttingen,Germany,showed that the
aerodynamic consideration of wings could
be split into two parts,(1) the study of
the section of a wing—an airfoil,And (2)
the modification of such airfoil properties
to account for the complete,finite wing,
Definition of airfoil
The purpose of this chapter,
Present theoretical methods for the calculation of
airfoil aerodynamic properties,
Road map of this chapter
4.2 Airfoil nomenclature
Leading edge,前缘 trailing edge,后缘
Chord line,弦线 chord length,弦长
Thickness,厚度 camber,弯度
Mean chamber line,中弧线
NACA,four digit” series airfoil
NACA2412,
The first digit,maximum camber in hundredths;
The second digit,the location of maximum camber
along the chord from leading edge
in tenth of the chord;
The last two digits,maximum thickness in hundredth
of the chord,
NACA0012,symmetrical airfoil
4.3 Airfoil characteristics(experiment)
Special definitions
lc
?
0a max,lc
0?L?
stall
lift coefficient angle of attack
lift slope Maximum lift coefficient
zero-lift attack angle
Consequence of the flow separation
It is impossible for we to calculate with inviscid
flow approximation!! max,lc
Experiment results for NACA2412
0
0 1.2??L?
6.1m a x,?lc
016?
st a l l?
Source of drag
profile drag
Aerodynamic center
acmc,
4.4 Philosophy of theoretical solutions for
low-speed flow over airfoils,
The Vortex Sheet
Schematic figure of a
point vortex
Schematic figure of a
vortex filament
Point vortex is simply a section
of a straight vortex filament
Construction of a vortex sheet
Try to remember the analogous situation for the
construction of a source sheet,
Infinite number of vortex filaments,The strength of
each vortex filament is infinitesimally small,
)(s?
Velocity induced by at point P
Definition of the strength of vortex sheet
ds?
r
dsdV
?
?
2
??
Velocity potential induced by at point P ds?
?
?
??
2
dsd ??
Difference of the superposition between the
velocity vectors and velocity potential,
The circulation around the vortex sheet is
the sum of the strengths of the elemental
vortices,
ds
b
a?
?? ?
There is a discontinuity change in the
tangential component of velocity across the
sheet,
Let to be the circulation along the dashed line,?
)( 2112 dsudnvdsudnv ??????
or dnvvdsuu )()(
2121 ?????
ds???
as
so
dnvvdsuuds )()( 2121 ?????
As the top and bottom of the dashed line approach
the vortex sheet,,become the
velocity components tangential to the vortex sheet
immediately above and below the sheet,
0?dn 21,uu
dsuuds )( 21 ???
or )( 21 uu ???
The local jump in the tangential velocity across the
vortex sheet is equal to the local sheet strength,
Philosophy of airfoil theory for inviscid,
incompressible flows,
Step 1,Replace the airfoil surface with a vortex
sheet of strength )(s?
Step 2,Find a suitable distribution of such that
the wall boundary condition can be satisfied,That is,
the combination of the free stream flow and the
vortex sheet will make the vortex sheet(the surface
of the airfoil) a streamline of the flow,
)(s?
Step 3,Calculate the circulation around the airfoil,
and then get the lift by Kutta-Joukowski theorem
??? ds? ??? ?? VL ?
Note 1,There are no general analytical solution
for an airfoil with arbitrary shape and thickness,This
should be solved numerically with suitable digital
computers,Vortex panel method (Sec,4.9)
)(s?
Note 2,Physical significance of the vortex sheet
which has been used to replace the surface of the
airfoil surface,Boundary layer is a highly viscous
region,the vorticity inside the boundary layer is finite,
Step 4,Approximation for a thin airfoil,shift the
vortex sheet from the airfoil surface to the camber
line of the airfoil,The upper and lower part of the
vortex sheet are coincide together,
This time,Find a suitable distribution of such
that the wall boundary condition can be satisfied,
That is,the combination of the free stream flow and
the vortex sheet will make the vortex sheet(camber
line of the airfoil) a streamline of the flow,
)(s?
Note 3,After the thin airfoil approximation,it is
possible to give a closed-form analytical solution of, )(s?
4.5 The Kutta Condition
For potential flows,different choice of
gives different lifting flow around circular
cylinder,And it is the same to the situation
of airfoils,
?
Two different flows around a same airfoil at
the same attack angle
The nature knows how to pick a right solution,We
need an additional condition that fixes for a given
airfoil at a given attack angle,
?
Experimental results for the development of the flow
field around an airfoil which is set into motion from an
initial state of rest,
(a)
(b)
(c)
Experimental results demonstrate that the flow is
smoothly leaving the top and bottom surface of the
airfoil at the trailing edge,
If the flow is smoothly leaving the top and bottom
surface of the airfoil at the trailing edge,then the
circulation is the value the nature adopts,
2?
The condition,that the flow is smoothly leaving the
top and bottom surface of the airfoil at the trailing
edge,which is a physical observation,is called as,
Kutta Condition
Kutta condition used in theoretical analysis
2
2
2
1 2
1
2
1 VpVp
aa ?? ???
The pressure at both the top and bottom surface
immediately adjacent to point a (trailing edge),
21 VV ?
1,For a given airfoil at a given angle of attack,
the value of circulation around the airfoil
is such that the flow leaves the trailing edge
smoothly
2,If the trailing-edge angle is finite,then the
trailing edge is a stagnation point,
3,If the trailing edge is cusped,then the
velocities leaving the top and bottom
surfaces at the trailing edge are finite and
equal in magnitude and direction,
?
At the trailing edge (TE),we have,
21)()( VVaTE ??? ??
0)()( 21 ???? VVaTE ??
For finite-angle trailing edge,
For cusped trailing edge,
0)()( 21 ???? VVaTE ??
0)( ?TE?For trailing edge,
4.6 Kelvin’s circulation theorem and the
starting vortex
Examination of Kutta condition in a detailed
way,
0??DtD
Kelvin’s Circulation
Theorem
Explanation for the generation of the
circulation around an airfoil with Kelvin’s
theorem,
From kelvin’s theorem,
0??DtD
? 021 ????
243 ?????
? 34 ????
The circulation around the airfoil is equal and opposite
to the circulation around the starting vortex,
4.7 Classical thin airfoil theory,
the symmetrical airfoil
Where are we in the road map of this chapter?
Under the assumption of thin airfoil,the
vortex is distributed along the mean camber
line,
what will be the condition for the variation
of? )(s?
The mean camber line should be a streamline of the
combined flow and Kutta condition is satisfied at the
trailing edge,i.e.,0)( ?TE?
further approximation of the placement of
the vortex sheet
what will be the condition for the variation
of? )(x?
The mean camber line should be a streamline of the
combined flow and Kutta condition is satisfied at the
trailing edge,i.e.,0)( ?c?
condition expressed by velocity components
0)(,???? swV n
:,nV?
:)(sw?
Component of the free stream velocity
normal to the camber line,
Component of the velocity induced by
vortex sheet normal to the camber line,
? ?? ?dxdzVV n ??? ??? 1,t a ns i n ?
For small angle of attack and small camber,
? ? 1t a n,1 ???? dxdz?
so that
)(,dxdzVV n ?? ?? ?
:)(sw?
:)( xw
Component of the velocity induced by
vortex sheet normal to the camber line,
Component of the velocity induced by
vortex sheet normal to the chord line,
For thin airfoil,the approximation bellow is consistent
with the thin airfoil theory
)()( xwsw ??
incremental normal velocity induced by the
vortex sheet placed on the camber line
3c o s2 ??
?
r
dsdV
in ??
2
0
c o s?
xxr ??
1c o s?
dxds ?
1
32
0 c o s
c o sc o s
2
1
?
???
? ? ?
??
TE
LE
in xx
dx
V
Where,if the shape of the camber line is given,
are functions of x, 321,,???
For small cambered airfoil,are small values,
that means 321,,???
1co sco sco s 321 ??? ???
? ???
TE
LE
in xx
dx
V
02
1 ?
?
1co sco sco s 321 ??? ???
? ???
TE
LE
in xx
dx
V
02
1 ?
?
Simplification introduced above is equivalent to
satisfying the boundary conditions on the x axis
instead of the mean camber line,
Conclusion,
Velocity dw at ponit x induced by the elemental
vortex segment at point ξ
)(2
)(
??
???
?
??
x
ddw
Velocity w at ponit x induced by all elemental vortex
segments at along the chord line is obtained by
integrating dw from ξ=0 to ξ=c
? ???
c
x
dxw
0 )(2
)()(
??
???
The mean camber line should be a streamline of the
combined flow and Kutta condition is satisfied at the
trailing edge,i.e.,
0)( ?c?
0)(,???? swV n
0)(,???? swV n
? )(,dxdzVV n ?? ?? ? )()( xwsw ?
?
? ?
?
??
c
x
dxw
0 )(2
)()(
??
???
?
0
)(2
)(
0
?
?
??
?
?
?
?
? ? ?
?
c
x
d
dx
dzV
??
????
?
?
?
?
?
? ??
? ?? dx
dzV
x
dc ?
?
???
? 0 )(
)(
2
1
?
?
?
?
?
?
??
?
??
dx
dz
V
x
dc
?
?
???
? 0 )(
)(
2
1
Fundamental equation of thin airfoil theory
It is simply a statement that the camber line is
a streamline of the flow,
Note,referring to the textbook(page 270),
Analysis for a symmetric airfoil
Step 1,Fundamental equation for a symmetric airfoil
?
?
???
? ?
?
??
V
x
dc
0 )(
)(
2
1 ?
0?
dx
dz
Step 2,Transform ξ into θ
)c o s1(
2
?? ?? c )c o s1(2 0???
cx
Note,ξ is a dummy variable,x is a fixed
point
?
)c o s1(
2
?? ?? c
?
??? dcd s i n
2
?
Step 3,Rewritten the fundamental equation of thin
airfoil theory in the arguments of
0,??
?
?
???
? ?
?
??
V
x
dc
0 )(
)(
2
1
? )c o s1(
2 ?? ??
c ??? dcd s i n
2? )c o s1(2 0???
cx
?
??
????
?
?
???? V
d
0 0c o sc o s
s i n)(
2
1
Note,pay attention to the limits of integral
Step 4,An rigorous solution of the equation above
for can be obtained from complex variable
analysis,but it is beyond the scope of this textbook,)(??
The solution is,
?
????
s in
)c o s1(2)( ??
?V
After the solution been given,what should
we do now? Is it a true for our problem? How
to prove it? What is the principle to be based
to prove?
Step 5,Verification for the solution
There are two conditions which the solution
must satisfy,they are wall condition and
Kutta condition,
Wall condition………
?
??
????
?
?
???? V
d
0 0c o sc o s
s i n)(
2
1
Substituting the solution into the wall
condition given above
?? ???? ?
??
??
??
?
?
??
????
? 0 00 0 c o sc o s
)c o s1(
c o sc o s
s i n)(
2
1 dVd
Now we have to prove that
?
??
??
?
? ?
?
? ?
?
??
V
dV
0 0c o sc o s
)c o s1(
Fortunately,there is a standard integral given
as
0
0
0 0 s i n
s i n
c o sc o s
c o s
?
??
??
??? ndn
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
??
?
??
??
??
?
?
?
??
??
?
?
0
0
0
0
0
0
c o sc o s
c o s
c o sc o s
c o sc o s
)c o s1(
ddV
dV
??
?
?
?
? ??? VV )0(
So,the solution satisfies the fundamental
equation of thin airfoil theory
Kutta condition… … …
?
????
s in
)c o s1(2)( ??
?V
0
02)(
?? V???
the value is undetermined
With L’Hospital’s rule(罗毕塔法则 ）
0
c o s
s i n2)( ???
? ?
???? V
So,the solution satisfies Kutta condition
0)( ???
Step 6,Calculation of the characteristics for thin
airfoils
???
c
d
0
)( ???
)c o s1(2 ?? ?? c ??? dcd s i n2? )c o s1(2 0??? cx
?
???
?
????
0
s in)(
2
dc
?
????
s in
)c o s1(2)( ??
?V
?? ? ???? cVdcV ?????
?
0
)c o s1(
?
Lift per unit span,(Kutta-joukovski)
2???? ???? VcVL ????
The lift coefficient
Sq
Lc
l
?
?
?
)1(cS ?
??
?
???
2
)1(5.0 2
2
??
??
??
cV
Vc
c l
Lift slope =
?
?
2?
d
dc l
Moment coefficient about the leading edge
Incremental lift contributed by the elemental
vortex segment dξ is, ?? ?? dVdL ?
This incremental lift creates a moment about
the leading edge )( dLdM ???
The total moment about the leading edge(per
unit span) due to the entire vortex sheet is
?? ???????
cc
dVdLM
00
)()( ??????
2
2 ??cqM
????
(Problem 4.4)
Moment coefficient
22,
????????
?? cq
M
Scq
Mc LELE
lem
as
2
2 ll corc ?? ????
so
4,
l
lem
cc ??
The moment coefficient about the quarter-
chord point is
4,4,
l
lemcm
ccc ??
04,?cmc
?
4,
l
lem
cc ??
The center of pressure is at the quarter-chord
point for a symmetric airfoil
As the moment coefficient about the quarter-
chord point of a symmetric airfoil is
independent of the attack angle,that is always
equal to zero,then,the quarter-chord point of
a symmetric airfoil is called as aerodynamic
center,
Summary,referring back to the textbook
Experiment results for a symmetric airfoil
1,??2?
lc
2,Lift slope ?2?
3,The center of pressure and the aerodynamic
center are both located at the quarter-chord
point,
Theoretical results for a symmetric airfoil
4.8 The cambered airfoil
Thin airfoil theory for a cambered airfoil is a
generalization of the method for a symmetric
airfoil,That means,the result for a
symmetrical airfoil is a special case of the
cambered airfoil,To keep the mean camber
line of a cambered airfoil be a streamline of
the flow,the condition is
?
?
?
?
?
? ??
? ?? dx
dzV
x
dc ?
?
???
? 0 )(
)(
2
1
as the value of the camber is not zero,it
makes the analysis more difficult than in the
case of symmetric airfoil,
After we use the same transform
)c o s1(
2
?? ?? c )c o s1(2 0???
cx
We obtain
?
?
?
?
?
? ??
? ?? dx
dz
V
d
?
??
????
?
?
0 0c o sc o s
s i n)(
2
1
find a solution for from the equation
above,at the same time,the solution of
must satisfy the Kutta condition
)(??
)(??
0)( ???
Step 1,Introduction of a rigorous solution
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
1
0 s in
s in
)c o s1(
2)(
n
n nAAV ?
?
?
??
At the very first,let us make a closed comparison
between the styles of the solution of symmetric
airfoil and the solution of cambered airfoil,
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
1
0 s in
s in
)c o s1(
2)(
n
n nAAV ?
?
?
??
?
????
s in
)c o s1(2)( ??
?V
Symmetric airfoil
Cambered airfoil
The form of the first term in the solution of a
cambered airfoil is nearly the same to the
solution of the symmetric airfoil,This term
can be looked as the skeleton for the
solutions for both symmetrical or cambered
airfoil,
The Fourier sine series can be looked as a
fine tuning of the solutions,so that the
camber can be taken in to account,
Step 2,To find the solution of is equal to find
the specific values of all the coefficients,
)(??
0A nA
Substituting the solution into the fundamental
equation,
?
?
?
?
?
? ??
? ?? dx
dz
V
d
?
??
????
?
?
0 0c o sc o s
s i n)(
2
1
???
?
???
?
??? ?
?
?
?
1
0 s ins in
)c o s1(2)(
n
n nAAV ??
???
dx
dz
dnAdA
n
n
??
?
?
?
?
? ??
?
?
?
??
???
???
??
?
??
1
0
0
0
0
0
c o sc o s
s ins in1
c o sc o s
)c o s1(1
0
0
0 0 s i n
s i n
c o sc o s
c o s
?
??
??
??? ndn
?
??
0
0 0
0
c o sc o s
)c o s1(1
A
dA
?
?
?? ?
??
??
?
from
We can have
and as
0
0 0
c o s
c o sc o s
s i ns i n
??
??
????
n
dn
??
??
then
dx
dz
dnAdA
n
n
??
?
?
?
?
? ??
?
?
?
??
???
???
??
?
??
1
0
0
0
0
0
c o sc o s
s ins in1
c o sc o s
)c o s1(1
Reduced to
dx
dz
nAA
n
n ??? ?
?
?
??
1
00 c o s
or
?
?
?
???
1
00 c o s)(
n
n nAAdx
dz
??
It is the transformed version of the
fundamental equation of thin airfoil theory
Note,the equation
?
?
?
?
?
? ??
? ?? dx
dzV
x
dc ?
?
???
? 0 )(
)(
2
1
is evaluated at a given point x along the chord
hence the equation
?
?
?
???
1
00 c o s)(
n
n nAAdx
dz
??
is also evaluated at a given point x along the
chord,here,and correspond to the
same point x along the chord,
dxdz 0?
is a function of, And dxdz
0? )c o s1(2 0???
cx
)( 0?fdxdz ?
Step 3,Investigation of Fourier cosine series
expansion
The general form of the Fourier cosine series
expansion representing a function of f(θ)
over an interval 0≤ θ ≤π is given by
?
?
?
??
1
0 c o s)(
n
n nBBf ??
If an integration is taken on the both side over
the interval 0≤ θ ≤π,then
?????
???
dnBdBdf
n
n? ???
?
?
??
1
00
0
0
c o s)(
?
0)( 0
0
??? Bdf ???
?
?
??
?
?
?? 00 )(1 dfB
If we multiplies cos(θ) on the both sides,and
an integration is taken on the both side over
the interval 0≤ θ ≤π,then
?????
???
??
?
dnBdB
df
n
n? ??
?
?
?
??
1
00
0
0
c o sc o sc o s
c o s)(
?
??????
??
dBdf ?? ??
0
1
0
c o sc o s0c o s)(
)(0c o sc o s
0
nmifdnm ??? ???
?
)(
2
c o sc o s
0
nmifdnm ??? ????
?
??????
??
dBdf ?? ??
0
1
0
c o sc o s0c o s)(
?
???
?
?
?? 01 c o s)(2 dfB
In the same way,we can prove
???
?
?
?? 0 c o s)(2 dnfB n
Conclusion,if a function of f(θ) over an
interval 0≤ θ ≤π is given by
?
?
?
??
1
0 c o s)(
n
n nBBf ??
The coefficients and should be
0B nB
??
?
?
?? 00 )(1 dfB
???
?
?
?? 0 c o s)(2 dnfB n
Step 4,Solution of and
0A nA
as
?
?
?
????
1
000 c o s)()(
n
n nAAfdx
dz
???
With the use of the results for the investigation
of Fourier series expansion,we can get a direct
expression of and
0A nA
0
0
0
1 ?
?
?
?
??? ddxdzA
or
0
0
0
1 ?
?
?
?
??? ddxdzA
and
0
0
0c o s
2 ??
?
?
?? dndxdzA n
What shall we keep in mind? and what we have
to think about at this moment?
Please back to our textbook at page 276,
after we get the solution of,then we
are ready to obtain expressions for the
aerodynamic coefficients for a cambered
airfoil
)(??
Step 1,Calculation of the total circulation of the
vortex sheet
?? ???
?
???????
00
s in)(
2
)( dcd
c
???
?
???
?
??? ?
?
?
?
1
0 s ins in
)c o s1(2)(
n
n nAAV ??
???
?
?
?
?
?
?
?
?
???? ? ? ?
?
?
?
? ?
?????
0
1
0
0 s ins in)c o s1( dnAdAcV
n
n
?
?
?
?
?
?
??
?
?
10
12
s ins in
)c o s1(
0
0
nfo r
nfo r
dn
d
?
???
???
?
?
??
?
??
? ???
? 10 2 AAcV
??
Step 2,After the total circulation being evaluated,
the lift per unit span is
??
?
??
? ?????
???? 10
2
2
AAcVVL ????
Step 3,The lift coefficient is
? ?102 2
)1(5.0
AA
cV
Lc
l ??
?
?
??
?
?
000
1 ?
??
???? d
dx
dzA
00 0c o s
2 ??
?
??? dn
dx
dzA
n
?
?
?
?
?
?
??? ? 0
0
0 )1( c o s
1
2 ??
?
??
?
d
dx
dz
c l
?
?
2??
d
dcs l o p eL i f t l
?? 2?ddc l
From thin airfoil theory for any shape airfoil
Step 4,Definition of zero lift angle
)( 0??? Lll
d
dcc ??
?
)(2 0??? Llc ???
?
?
?
?
?
?
??? ? 0
0
0 )1( c o s
1
2 ??
?
??
?
d
dx
dz
c l
)(2 0??? Llc ???
Comparison between two expressions of the lift
coefficient
We have
0
0
00 )1( c o s
1 ??
?
?
?
? ???? ddxdzL
For symmetric airfoil
00 ??L?
?
?
?
?
?
? ????
22
2
10,
AAAc
lem
?
Step 5,Moment coefficient
? ?102 AAc l ?? ??
??
?
??
? ???? )(
44 21,
AAcc llem ?
For symmetric airfoil
4,llem cc ??
Moment coefficient about the quarter-chord point
)(
4 124,
AAc cm ?? ?
It is independent to the attack angle,it depends on
the shape of the camber line of the airfoil,
Thus,the quarter-chord point is the theoretical
location of the aerodynamic center for a cambered
airfoil
Step 6,Location of the pressure center
l
lemLE
cp c
cc
L
M
x,??
?
?
??
??
?
??
? ???? )(
44 21,
AAcc llem ??
?
?
?
?
?
?
??? )(1
4 21
AA
c
c
x
l
cp
?
4.9 Lifting flows over arbitrary bodies,the
vortex panel numerical method
Advantages for thin airfoil theory,and its
limit for applications,
Method suited to calculated the
aerodynamic characteristics of bodies of
arbitrary shape,thickness and orientation,
the vortex panel method
Comparison between the vortex panel
method and source panel method,lifting
body and nonlifting body,
Philosophy of the vortex panel method
Covering the body surface with a vortex sheet,
Find the strength distribution to make the surface a
streamline of the flow,
Approximate the vortex sheet by a series of straight
panels,
Let the vortex strength per unit length be
constant over a given panel,
)(s?
If the number of vortex panels is n,there will be n
unknowns to be solved,that is
Solve the n unknowns,such that the body surface
becomes a streamline of the flow and that the Kutta
condition is satisfied,
n????,,,,321 ??
Algorithm of the vortex panel method
Velocity potential introduced at P due to the jth panel
j
j
jpjj ds???? ???? 2
1
j
j
jpjj ds???? ???? 2
1
where is a constant over the jth panel
j?
j
j
pj xx
yy
?
?
? ? 1t a n?
and
Velocity potential introduced at P due to all the panels
? ??
??
????
n
j
j
j
pj
j
n
j
j dsP
11
2
)( ?
?
?
??
Put P at the control point of the ith panel
Put P at the control point of the ith panel,that means
point P is located at,then ),(
ii yx
ji
ji
ij xx
yy
?
?
? ? 1t a n?
and
? ?
?
??
n
j
j
j
ij
j
ii dsyx
1
2
),( ?
?
?
?
At the control points,the normal component of the
velocity is zero,that is
0,??? nn VV
iin VnVV ?c o s,??? ???
??
? ?
?
?
?
??
n
j j
j
i
ijj
n dsnV
1
2
?
?
?
0
2
c o s
1
?
?
?
? ? ?
?
?
n
j
j
j
i
ijj
i ds
n
V
?
?
?
?
j
j i
ij
ji dsn? ?
?
?
?
,J 0
2
c o s
1
,?? ?
?
?
n
j
ji
j
iV J
?
?
?
0
2
c o s
1
,?? ?
?
?
n
j
ji
j
iV J
?
?
?
It represents n equations with n unknowns
Kutta condition should be satisfied at the trailing edge,
0)( ?TE?
?
01 ?? ?ii ?? ? 1??? ii ??
Incompressible Flow Over Airfoils
Of the many problems now engaging attention,the following
are considered of immediate importance and will be
considered by the committee as rapidly as funds can be
secured for the purpose…, The evolution of the more efficient
wing sections of practical form,embodying suitable dimension
for an economical structure,with moderate travel of the
center-of-pressure and still affording a large range of angle-of-
attack combined with efficient action,
From the first annual report of the
NACA,1915
4.1 Introduction
Ludiwig Prandtl and his colleagues at
G?ttingen,Germany,showed that the
aerodynamic consideration of wings could
be split into two parts,(1) the study of
the section of a wing—an airfoil,And (2)
the modification of such airfoil properties
to account for the complete,finite wing,
Definition of airfoil
The purpose of this chapter,
Present theoretical methods for the calculation of
airfoil aerodynamic properties,
Road map of this chapter
4.2 Airfoil nomenclature
Leading edge,前缘 trailing edge,后缘
Chord line,弦线 chord length,弦长
Thickness,厚度 camber,弯度
Mean chamber line,中弧线
NACA,four digit” series airfoil
NACA2412,
The first digit,maximum camber in hundredths;
The second digit,the location of maximum camber
along the chord from leading edge
in tenth of the chord;
The last two digits,maximum thickness in hundredth
of the chord,
NACA0012,symmetrical airfoil
4.3 Airfoil characteristics(experiment)
Special definitions
lc
?
0a max,lc
0?L?
stall
lift coefficient angle of attack
lift slope Maximum lift coefficient
zero-lift attack angle
Consequence of the flow separation
It is impossible for we to calculate with inviscid
flow approximation!! max,lc
Experiment results for NACA2412
0
0 1.2??L?
6.1m a x,?lc
016?
st a l l?
Source of drag
profile drag
Aerodynamic center
acmc,
4.4 Philosophy of theoretical solutions for
low-speed flow over airfoils,
The Vortex Sheet
Schematic figure of a
point vortex
Schematic figure of a
vortex filament
Point vortex is simply a section
of a straight vortex filament
Construction of a vortex sheet
Try to remember the analogous situation for the
construction of a source sheet,
Infinite number of vortex filaments,The strength of
each vortex filament is infinitesimally small,
)(s?
Velocity induced by at point P
Definition of the strength of vortex sheet
ds?
r
dsdV
?
?
2
??
Velocity potential induced by at point P ds?
?
?
??
2
dsd ??
Difference of the superposition between the
velocity vectors and velocity potential,
The circulation around the vortex sheet is
the sum of the strengths of the elemental
vortices,
ds
b
a?
?? ?
There is a discontinuity change in the
tangential component of velocity across the
sheet,
Let to be the circulation along the dashed line,?
)( 2112 dsudnvdsudnv ??????
or dnvvdsuu )()(
2121 ?????
ds???
as
so
dnvvdsuuds )()( 2121 ?????
As the top and bottom of the dashed line approach
the vortex sheet,,become the
velocity components tangential to the vortex sheet
immediately above and below the sheet,
0?dn 21,uu
dsuuds )( 21 ???
or )( 21 uu ???
The local jump in the tangential velocity across the
vortex sheet is equal to the local sheet strength,
Philosophy of airfoil theory for inviscid,
incompressible flows,
Step 1,Replace the airfoil surface with a vortex
sheet of strength )(s?
Step 2,Find a suitable distribution of such that
the wall boundary condition can be satisfied,That is,
the combination of the free stream flow and the
vortex sheet will make the vortex sheet(the surface
of the airfoil) a streamline of the flow,
)(s?
Step 3,Calculate the circulation around the airfoil,
and then get the lift by Kutta-Joukowski theorem
??? ds? ??? ?? VL ?
Note 1,There are no general analytical solution
for an airfoil with arbitrary shape and thickness,This
should be solved numerically with suitable digital
computers,Vortex panel method (Sec,4.9)
)(s?
Note 2,Physical significance of the vortex sheet
which has been used to replace the surface of the
airfoil surface,Boundary layer is a highly viscous
region,the vorticity inside the boundary layer is finite,
Step 4,Approximation for a thin airfoil,shift the
vortex sheet from the airfoil surface to the camber
line of the airfoil,The upper and lower part of the
vortex sheet are coincide together,
This time,Find a suitable distribution of such
that the wall boundary condition can be satisfied,
That is,the combination of the free stream flow and
the vortex sheet will make the vortex sheet(camber
line of the airfoil) a streamline of the flow,
)(s?
Note 3,After the thin airfoil approximation,it is
possible to give a closed-form analytical solution of, )(s?
4.5 The Kutta Condition
For potential flows,different choice of
gives different lifting flow around circular
cylinder,And it is the same to the situation
of airfoils,
?
Two different flows around a same airfoil at
the same attack angle
The nature knows how to pick a right solution,We
need an additional condition that fixes for a given
airfoil at a given attack angle,
?
Experimental results for the development of the flow
field around an airfoil which is set into motion from an
initial state of rest,
(a)
(b)
(c)
Experimental results demonstrate that the flow is
smoothly leaving the top and bottom surface of the
airfoil at the trailing edge,
If the flow is smoothly leaving the top and bottom
surface of the airfoil at the trailing edge,then the
circulation is the value the nature adopts,
2?
The condition,that the flow is smoothly leaving the
top and bottom surface of the airfoil at the trailing
edge,which is a physical observation,is called as,
Kutta Condition
Kutta condition used in theoretical analysis
2
2
2
1 2
1
2
1 VpVp
aa ?? ???
The pressure at both the top and bottom surface
immediately adjacent to point a (trailing edge),
21 VV ?
1,For a given airfoil at a given angle of attack,
the value of circulation around the airfoil
is such that the flow leaves the trailing edge
smoothly
2,If the trailing-edge angle is finite,then the
trailing edge is a stagnation point,
3,If the trailing edge is cusped,then the
velocities leaving the top and bottom
surfaces at the trailing edge are finite and
equal in magnitude and direction,
?
At the trailing edge (TE),we have,
21)()( VVaTE ??? ??
0)()( 21 ???? VVaTE ??
For finite-angle trailing edge,
For cusped trailing edge,
0)()( 21 ???? VVaTE ??
0)( ?TE?For trailing edge,
4.6 Kelvin’s circulation theorem and the
starting vortex
Examination of Kutta condition in a detailed
way,
0??DtD
Kelvin’s Circulation
Theorem
Explanation for the generation of the
circulation around an airfoil with Kelvin’s
theorem,
From kelvin’s theorem,
0??DtD
? 021 ????
243 ?????
? 34 ????
The circulation around the airfoil is equal and opposite
to the circulation around the starting vortex,
4.7 Classical thin airfoil theory,
the symmetrical airfoil
Where are we in the road map of this chapter?
Under the assumption of thin airfoil,the
vortex is distributed along the mean camber
line,
what will be the condition for the variation
of? )(s?
The mean camber line should be a streamline of the
combined flow and Kutta condition is satisfied at the
trailing edge,i.e.,0)( ?TE?
further approximation of the placement of
the vortex sheet
what will be the condition for the variation
of? )(x?
The mean camber line should be a streamline of the
combined flow and Kutta condition is satisfied at the
trailing edge,i.e.,0)( ?c?
condition expressed by velocity components
0)(,???? swV n
:,nV?
:)(sw?
Component of the free stream velocity
normal to the camber line,
Component of the velocity induced by
vortex sheet normal to the camber line,
? ?? ?dxdzVV n ??? ??? 1,t a ns i n ?
For small angle of attack and small camber,
? ? 1t a n,1 ???? dxdz?
so that
)(,dxdzVV n ?? ?? ?
:)(sw?
:)( xw
Component of the velocity induced by
vortex sheet normal to the camber line,
Component of the velocity induced by
vortex sheet normal to the chord line,
For thin airfoil,the approximation bellow is consistent
with the thin airfoil theory
)()( xwsw ??
incremental normal velocity induced by the
vortex sheet placed on the camber line
3c o s2 ??
?
r
dsdV
in ??
2
0
c o s?
xxr ??
1c o s?
dxds ?
1
32
0 c o s
c o sc o s
2
1
?
???
? ? ?
??
TE
LE
in xx
dx
V
Where,if the shape of the camber line is given,
are functions of x, 321,,???
For small cambered airfoil,are small values,
that means 321,,???
1co sco sco s 321 ??? ???
? ???
TE
LE
in xx
dx
V
02
1 ?
?
1co sco sco s 321 ??? ???
? ???
TE
LE
in xx
dx
V
02
1 ?
?
Simplification introduced above is equivalent to
satisfying the boundary conditions on the x axis
instead of the mean camber line,
Conclusion,
Velocity dw at ponit x induced by the elemental
vortex segment at point ξ
)(2
)(
??
???
?
??
x
ddw
Velocity w at ponit x induced by all elemental vortex
segments at along the chord line is obtained by
integrating dw from ξ=0 to ξ=c
? ???
c
x
dxw
0 )(2
)()(
??
???
The mean camber line should be a streamline of the
combined flow and Kutta condition is satisfied at the
trailing edge,i.e.,
0)( ?c?
0)(,???? swV n
0)(,???? swV n
? )(,dxdzVV n ?? ?? ? )()( xwsw ?
?
? ?
?
??
c
x
dxw
0 )(2
)()(
??
???
?
0
)(2
)(
0
?
?
??
?
?
?
?
? ? ?
?
c
x
d
dx
dzV
??
????
?
?
?
?
?
? ??
? ?? dx
dzV
x
dc ?
?
???
? 0 )(
)(
2
1
?
?
?
?
?
?
??
?
??
dx
dz
V
x
dc
?
?
???
? 0 )(
)(
2
1
Fundamental equation of thin airfoil theory
It is simply a statement that the camber line is
a streamline of the flow,
Note,referring to the textbook(page 270),
Analysis for a symmetric airfoil
Step 1,Fundamental equation for a symmetric airfoil
?
?
???
? ?
?
??
V
x
dc
0 )(
)(
2
1 ?
0?
dx
dz
Step 2,Transform ξ into θ
)c o s1(
2
?? ?? c )c o s1(2 0???
cx
Note,ξ is a dummy variable,x is a fixed
point
?
)c o s1(
2
?? ?? c
?
??? dcd s i n
2
?
Step 3,Rewritten the fundamental equation of thin
airfoil theory in the arguments of
0,??
?
?
???
? ?
?
??
V
x
dc
0 )(
)(
2
1
? )c o s1(
2 ?? ??
c ??? dcd s i n
2? )c o s1(2 0???
cx
?
??
????
?
?
???? V
d
0 0c o sc o s
s i n)(
2
1
Note,pay attention to the limits of integral
Step 4,An rigorous solution of the equation above
for can be obtained from complex variable
analysis,but it is beyond the scope of this textbook,)(??
The solution is,
?
????
s in
)c o s1(2)( ??
?V
After the solution been given,what should
we do now? Is it a true for our problem? How
to prove it? What is the principle to be based
to prove?
Step 5,Verification for the solution
There are two conditions which the solution
must satisfy,they are wall condition and
Kutta condition,
Wall condition………
?
??
????
?
?
???? V
d
0 0c o sc o s
s i n)(
2
1
Substituting the solution into the wall
condition given above
?? ???? ?
??
??
??
?
?
??
????
? 0 00 0 c o sc o s
)c o s1(
c o sc o s
s i n)(
2
1 dVd
Now we have to prove that
?
??
??
?
? ?
?
? ?
?
??
V
dV
0 0c o sc o s
)c o s1(
Fortunately,there is a standard integral given
as
0
0
0 0 s i n
s i n
c o sc o s
c o s
?
??
??
??? ndn
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
??
?
??
??
??
?
?
?
??
??
?
?
0
0
0
0
0
0
c o sc o s
c o s
c o sc o s
c o sc o s
)c o s1(
ddV
dV
??
?
?
?
? ??? VV )0(
So,the solution satisfies the fundamental
equation of thin airfoil theory
Kutta condition… … …
?
????
s in
)c o s1(2)( ??
?V
0
02)(
?? V???
the value is undetermined
With L’Hospital’s rule(罗毕塔法则 ）
0
c o s
s i n2)( ???
? ?
???? V
So,the solution satisfies Kutta condition
0)( ???
Step 6,Calculation of the characteristics for thin
airfoils
???
c
d
0
)( ???
)c o s1(2 ?? ?? c ??? dcd s i n2? )c o s1(2 0??? cx
?
???
?
????
0
s in)(
2
dc
?
????
s in
)c o s1(2)( ??
?V
?? ? ???? cVdcV ?????
?
0
)c o s1(
?
Lift per unit span,(Kutta-joukovski)
2???? ???? VcVL ????
The lift coefficient
Sq
Lc
l
?
?
?
)1(cS ?
??
?
???
2
)1(5.0 2
2
??
??
??
cV
Vc
c l
Lift slope =
?
?
2?
d
dc l
Moment coefficient about the leading edge
Incremental lift contributed by the elemental
vortex segment dξ is, ?? ?? dVdL ?
This incremental lift creates a moment about
the leading edge )( dLdM ???
The total moment about the leading edge(per
unit span) due to the entire vortex sheet is
?? ???????
cc
dVdLM
00
)()( ??????
2
2 ??cqM
????
(Problem 4.4)
Moment coefficient
22,
????????
?? cq
M
Scq
Mc LELE
lem
as
2
2 ll corc ?? ????
so
4,
l
lem
cc ??
The moment coefficient about the quarter-
chord point is
4,4,
l
lemcm
ccc ??
04,?cmc
?
4,
l
lem
cc ??
The center of pressure is at the quarter-chord
point for a symmetric airfoil
As the moment coefficient about the quarter-
chord point of a symmetric airfoil is
independent of the attack angle,that is always
equal to zero,then,the quarter-chord point of
a symmetric airfoil is called as aerodynamic
center,
Summary,referring back to the textbook
Experiment results for a symmetric airfoil
1,??2?
lc
2,Lift slope ?2?
3,The center of pressure and the aerodynamic
center are both located at the quarter-chord
point,
Theoretical results for a symmetric airfoil
4.8 The cambered airfoil
Thin airfoil theory for a cambered airfoil is a
generalization of the method for a symmetric
airfoil,That means,the result for a
symmetrical airfoil is a special case of the
cambered airfoil,To keep the mean camber
line of a cambered airfoil be a streamline of
the flow,the condition is
?
?
?
?
?
? ??
? ?? dx
dzV
x
dc ?
?
???
? 0 )(
)(
2
1
as the value of the camber is not zero,it
makes the analysis more difficult than in the
case of symmetric airfoil,
After we use the same transform
)c o s1(
2
?? ?? c )c o s1(2 0???
cx
We obtain
?
?
?
?
?
? ??
? ?? dx
dz
V
d
?
??
????
?
?
0 0c o sc o s
s i n)(
2
1
find a solution for from the equation
above,at the same time,the solution of
must satisfy the Kutta condition
)(??
)(??
0)( ???
Step 1,Introduction of a rigorous solution
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
1
0 s in
s in
)c o s1(
2)(
n
n nAAV ?
?
?
??
At the very first,let us make a closed comparison
between the styles of the solution of symmetric
airfoil and the solution of cambered airfoil,
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
1
0 s in
s in
)c o s1(
2)(
n
n nAAV ?
?
?
??
?
????
s in
)c o s1(2)( ??
?V
Symmetric airfoil
Cambered airfoil
The form of the first term in the solution of a
cambered airfoil is nearly the same to the
solution of the symmetric airfoil,This term
can be looked as the skeleton for the
solutions for both symmetrical or cambered
airfoil,
The Fourier sine series can be looked as a
fine tuning of the solutions,so that the
camber can be taken in to account,
Step 2,To find the solution of is equal to find
the specific values of all the coefficients,
)(??
0A nA
Substituting the solution into the fundamental
equation,
?
?
?
?
?
? ??
? ?? dx
dz
V
d
?
??
????
?
?
0 0c o sc o s
s i n)(
2
1
???
?
???
?
??? ?
?
?
?
1
0 s ins in
)c o s1(2)(
n
n nAAV ??
???
dx
dz
dnAdA
n
n
??
?
?
?
?
? ??
?
?
?
??
???
???
??
?
??
1
0
0
0
0
0
c o sc o s
s ins in1
c o sc o s
)c o s1(1
0
0
0 0 s i n
s i n
c o sc o s
c o s
?
??
??
??? ndn
?
??
0
0 0
0
c o sc o s
)c o s1(1
A
dA
?
?
?? ?
??
??
?
from
We can have
and as
0
0 0
c o s
c o sc o s
s i ns i n
??
??
????
n
dn
??
??
then
dx
dz
dnAdA
n
n
??
?
?
?
?
? ??
?
?
?
??
???
???
??
?
??
1
0
0
0
0
0
c o sc o s
s ins in1
c o sc o s
)c o s1(1
Reduced to
dx
dz
nAA
n
n ??? ?
?
?
??
1
00 c o s
or
?
?
?
???
1
00 c o s)(
n
n nAAdx
dz
??
It is the transformed version of the
fundamental equation of thin airfoil theory
Note,the equation
?
?
?
?
?
? ??
? ?? dx
dzV
x
dc ?
?
???
? 0 )(
)(
2
1
is evaluated at a given point x along the chord
hence the equation
?
?
?
???
1
00 c o s)(
n
n nAAdx
dz
??
is also evaluated at a given point x along the
chord,here,and correspond to the
same point x along the chord,
dxdz 0?
is a function of, And dxdz
0? )c o s1(2 0???
cx
)( 0?fdxdz ?
Step 3,Investigation of Fourier cosine series
expansion
The general form of the Fourier cosine series
expansion representing a function of f(θ)
over an interval 0≤ θ ≤π is given by
?
?
?
??
1
0 c o s)(
n
n nBBf ??
If an integration is taken on the both side over
the interval 0≤ θ ≤π,then
?????
???
dnBdBdf
n
n? ???
?
?
??
1
00
0
0
c o s)(
?
0)( 0
0
??? Bdf ???
?
?
??
?
?
?? 00 )(1 dfB
If we multiplies cos(θ) on the both sides,and
an integration is taken on the both side over
the interval 0≤ θ ≤π,then
?????
???
??
?
dnBdB
df
n
n? ??
?
?
?
??
1
00
0
0
c o sc o sc o s
c o s)(
?
??????
??
dBdf ?? ??
0
1
0
c o sc o s0c o s)(
)(0c o sc o s
0
nmifdnm ??? ???
?
)(
2
c o sc o s
0
nmifdnm ??? ????
?
??????
??
dBdf ?? ??
0
1
0
c o sc o s0c o s)(
?
???
?
?
?? 01 c o s)(2 dfB
In the same way,we can prove
???
?
?
?? 0 c o s)(2 dnfB n
Conclusion,if a function of f(θ) over an
interval 0≤ θ ≤π is given by
?
?
?
??
1
0 c o s)(
n
n nBBf ??
The coefficients and should be
0B nB
??
?
?
?? 00 )(1 dfB
???
?
?
?? 0 c o s)(2 dnfB n
Step 4,Solution of and
0A nA
as
?
?
?
????
1
000 c o s)()(
n
n nAAfdx
dz
???
With the use of the results for the investigation
of Fourier series expansion,we can get a direct
expression of and
0A nA
0
0
0
1 ?
?
?
?
??? ddxdzA
or
0
0
0
1 ?
?
?
?
??? ddxdzA
and
0
0
0c o s
2 ??
?
?
?? dndxdzA n
What shall we keep in mind? and what we have
to think about at this moment?
Please back to our textbook at page 276,
after we get the solution of,then we
are ready to obtain expressions for the
aerodynamic coefficients for a cambered
airfoil
)(??
Step 1,Calculation of the total circulation of the
vortex sheet
?? ???
?
???????
00
s in)(
2
)( dcd
c
???
?
???
?
??? ?
?
?
?
1
0 s ins in
)c o s1(2)(
n
n nAAV ??
???
?
?
?
?
?
?
?
?
???? ? ? ?
?
?
?
? ?
?????
0
1
0
0 s ins in)c o s1( dnAdAcV
n
n
?
?
?
?
?
?
??
?
?
10
12
s ins in
)c o s1(
0
0
nfo r
nfo r
dn
d
?
???
???
?
?
??
?
??
? ???
? 10 2 AAcV
??
Step 2,After the total circulation being evaluated,
the lift per unit span is
??
?
??
? ?????
???? 10
2
2
AAcVVL ????
Step 3,The lift coefficient is
? ?102 2
)1(5.0
AA
cV
Lc
l ??
?
?
??
?
?
000
1 ?
??
???? d
dx
dzA
00 0c o s
2 ??
?
??? dn
dx
dzA
n
?
?
?
?
?
?
??? ? 0
0
0 )1( c o s
1
2 ??
?
??
?
d
dx
dz
c l
?
?
2??
d
dcs l o p eL i f t l
?? 2?ddc l
From thin airfoil theory for any shape airfoil
Step 4,Definition of zero lift angle
)( 0??? Lll
d
dcc ??
?
)(2 0??? Llc ???
?
?
?
?
?
?
??? ? 0
0
0 )1( c o s
1
2 ??
?
??
?
d
dx
dz
c l
)(2 0??? Llc ???
Comparison between two expressions of the lift
coefficient
We have
0
0
00 )1( c o s
1 ??
?
?
?
? ???? ddxdzL
For symmetric airfoil
00 ??L?
?
?
?
?
?
? ????
22
2
10,
AAAc
lem
?
Step 5,Moment coefficient
? ?102 AAc l ?? ??
??
?
??
? ???? )(
44 21,
AAcc llem ?
For symmetric airfoil
4,llem cc ??
Moment coefficient about the quarter-chord point
)(
4 124,
AAc cm ?? ?
It is independent to the attack angle,it depends on
the shape of the camber line of the airfoil,
Thus,the quarter-chord point is the theoretical
location of the aerodynamic center for a cambered
airfoil
Step 6,Location of the pressure center
l
lemLE
cp c
cc
L
M
x,??
?
?
??
??
?
??
? ???? )(
44 21,
AAcc llem ??
?
?
?
?
?
?
??? )(1
4 21
AA
c
c
x
l
cp
?
4.9 Lifting flows over arbitrary bodies,the
vortex panel numerical method
Advantages for thin airfoil theory,and its
limit for applications,
Method suited to calculated the
aerodynamic characteristics of bodies of
arbitrary shape,thickness and orientation,
the vortex panel method
Comparison between the vortex panel
method and source panel method,lifting
body and nonlifting body,
Philosophy of the vortex panel method
Covering the body surface with a vortex sheet,
Find the strength distribution to make the surface a
streamline of the flow,
Approximate the vortex sheet by a series of straight
panels,
Let the vortex strength per unit length be
constant over a given panel,
)(s?
If the number of vortex panels is n,there will be n
unknowns to be solved,that is
Solve the n unknowns,such that the body surface
becomes a streamline of the flow and that the Kutta
condition is satisfied,
n????,,,,321 ??
Algorithm of the vortex panel method
Velocity potential introduced at P due to the jth panel
j
j
jpjj ds???? ???? 2
1
j
j
jpjj ds???? ???? 2
1
where is a constant over the jth panel
j?
j
j
pj xx
yy
?
?
? ? 1t a n?
and
Velocity potential introduced at P due to all the panels
? ??
??
????
n
j
j
j
pj
j
n
j
j dsP
11
2
)( ?
?
?
??
Put P at the control point of the ith panel
Put P at the control point of the ith panel,that means
point P is located at,then ),(
ii yx
ji
ji
ij xx
yy
?
?
? ? 1t a n?
and
? ?
?
??
n
j
j
j
ij
j
ii dsyx
1
2
),( ?
?
?
?
At the control points,the normal component of the
velocity is zero,that is
0,??? nn VV
iin VnVV ?c o s,??? ???
??
? ?
?
?
?
??
n
j j
j
i
ijj
n dsnV
1
2
?
?
?
0
2
c o s
1
?
?
?
? ? ?
?
?
n
j
j
j
i
ijj
i ds
n
V
?
?
?
?
j
j i
ij
ji dsn? ?
?
?
?
,J 0
2
c o s
1
,?? ?
?
?
n
j
ji
j
iV J
?
?
?
0
2
c o s
1
,?? ?
?
?
n
j
ji
j
iV J
?
?
?
It represents n equations with n unknowns
Kutta condition should be satisfied at the trailing edge,
0)( ?TE?
?
01 ?? ?ii ?? ? 1??? ii ??
