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 ??