Aerodynamics Chapter 4 r

8/19/2019 Aerodynamics Chapter 4 r

1/63

Copyright by Dr. Zheyan Jin

Zheyan JinSchool of Aerospace Engineering and Applied MechanicsTongji

University

Shanghai, China, 200092

AerodynamicsAerodynamics


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Chapter 4 Incompressible Flow Over AirfoilsChapter 4 Incompressible Flow Over Airfoils

4.1 Introduction

The aerodynamic consideration of

wings could be split into two parts:

(1). The study of the section of

a wing- an airfoi l.

(2). The modification of such

airfoi l properties to account

for the complete, finite wing.

Ludwig Prandtl and Göttingen (1912-1918):


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4.2 Airfoil Nomenclature

Mean camber line: is the locus of the points midway between upper and

lower surfaces of an airfoil as measured perpendicular to the mean camber

line.

Leading and trailing edges: the most forward and rearward points of the

man camber line.

Chord line: the straight l ine connecting the leading and trailing edges.


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Thickness: is the height of profile measured normal to the mean camber line.

Leading-edge radius: is the radius of a circ le, tangent to the upper and

lower surface, with i ts center located on a tangent to the mean camber line

drawn through the leading edge of this l ine.

Camber: is the maximum distance between the mean camber line and the

chord measured normal to the chord.


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Four-digit series: (for example, NACA 2412)

1. The first digit is the maximum camber in hundredths of chord.

2. The second digit is the location of maximum camber along the chord from the

leading edge in tenths of chord.

3. The last two digits give the maximum thickness in hundredths of chord.

Five-digit series: (for example, NACA 23012)

1. The first digit when multiplied by 3/2 gives the design lift coefficient in tenths.

2. The next two digit when divided by 2 give the location of maximum camber along

the chord from the leading edge in hundredths of chord.

3. The final two digits give the maximum thickness in hundredths of chord.

NACA airfoils (National Advisory Committee for Aeronautics):


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6-series: (for example, NACA 65-218)

1. The first digit simply identifies the series.

2. The second gives the location of minimum pressure in tenths of chord from the

leading edge.

3. The third digit is the design lift coefficient in tenths.

4. The last two digits give the maximum thickness in hundredths of chord.

Many of the large aircraft companies today design their own special purpose

airfoil; for example the Boeing 727,737,747, 757, 767, and 777 all have specially

designed Boeing airfoils.

Such capability is made possible by modern airfoil design computer programs

utilizing either panel techniques or direct numerical finite-difference solutionsof the governing partial differential equations for the flow field.

NACA airfoils:


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4.3 Airfoil Characteristics

At low-to-moderate angles of attack,

cl varies linearly withα.

The slope of this straight line is

denoted by a0 and is called lift slope.

The value ofα when lift equals zero

is called the zero-lift angle of attack.

Lift coefficient:

d

dcl0a

lc

0L

max,lc

Stall due to flow

separation


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Drag coefficient:


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At low to moderate angles of attack Cl-α curve is linear. The flow moves

slowly over the airfoil and is attached over most of the surface. At highangles of attack, the flow trends to separate from the top surface.

●

Cl,max occurs prior to stall

●

Cl,max is dependent on Re=ρvc/µ●

Cm,c/4 is independent of Re except for largeα

●

Cd is dependent on Re

●

The linear portion of the Cl-α curve is independent of Re and can be

predicted using analytical methods.

Aerodynamic center. There is one point on the airfoil about which the

moment is independent of angle of attack.


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4.4 Vortex Filament

Consider 2-D/point vortices of same strength duplicated in every planeparallel to z-x plane along the y-axis from to .

The flow is 2-D and is irrotational everywhere except the y-axis.

Y-axis is the straight vortex filament and may be defined as a line.

x

z

y

●

Definition: A vortex filament is a straight or

curved line in a fluid which coincides with the

axis of rotation of successive fluid elements.

Г


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4.4 Vortex Filament

●

Helmholt’s vortex theorems:

1. The strength of a vortex filament is constant along its length.

Proof: A vortex filament induces a velocity field that

is irrotational at every point excluding the filament.

Enclose a vortex filament with a sheath from which

a slit has been removed. The vorticity at every point

on the surface=0. Evaluate the circulation for the

sheath.

Circulation Ad V sd V S

C


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4.4 Vortex Filament

●



CC

0sdVor0sdV-0V

，

Sheath is irrotational. Thus

ad cb

dc ba

0sdVsdVsdVsdV

or


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4.4 Vortex Filament

●



c

d

d

c

b

a

d

c

b

a

sdVsdVsdV

0sdVsdV

However,

ac

d b

0sdVsdV

as it constitutes the integral across the slit.

Thus,


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4.4 Vortex Filament

●



2. A vortex filament cannot end in a fluid; it must extend to the

boundaries of the fluid or form a closed path.

3. In the absence of rotational external flow, a fluid that is irrotational

remains irrotational.

4. In the absence of rotational external force, if the circulationaround a path enclosing a definite group of particles is initially zero,

it will remain zero.

5. In the absence of rotational external force, the circulation around

a path that encloses a tagged group of elements is invariant.


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4.5 Vortex Sheet

An infinite number of straight vortex filamentplaced side by side from a vortex sheet. Each

vortex filament has an infinitesimal strength

γ(s):

γ(s) is the strength of vortex

sheet per unit length along s.

r 2v

for 2-D (point vortex)


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4.5 Vortex Sheet

A small portion of the vortex sheet of strength

γds induces an infinitesimally small velocity dV

at a field point P(r, θ).

Thus

r

ds

2mentvortexfilav

so

r

dsdv

2

P


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4.5 Vortex Sheet

CirculationГ around a point vortexis equal to the strength of the vortex.

Similarly, the circulation around the

vortex sheet is the sum of the

strengths of the elemental vortices.

Therefore, the circulationГ for a

finite length from point ‘a’ to point ‘b’

on the vortex sheet is given by:

Across a vortex sheet, there is a discontinuous change in the tangential

component of velocity and the normal component of velocity is preserved.

b

)(a

dss


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4.5 Vortex Sheet

AsΔn→0, we get

n()(

][

2121

2112

Box

）wwsuus

sunwsunwld v

ld v

)()(

21

21

uusuus

Γ =(u1-u2) states that the local jump in tangential velocity across

the vortex sheet is equal to the local sheet strength.


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4.6 Kutta Conditions:

1. For a given airfoil at a given angle of attack, the value of Г

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 inmagnitude and direction.

0)( lu V V TE

2V

021 V V

1V

a

At point a:

Finite angleCusp

2V

021 V V

1V a

At point a:


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4.7 Bound Vortex and Starting Vortex:

The question might arise: Does a real airfoil flying in a real fluid

give rise to a circulation about itself?

The answer is yes.

When a wing section with a sharp T.E is put into motion, the fluid

has a tendency to go around the sharp T.E from the lower to the

upper surface. As the airfoil moves along vortices are shed from

the T.E which from a vortex sheet.


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●

Helmholt’s theorem:

IfГ=0 originally in a flow it remains zero.

●

Kelvin’s theorem:

Circulation around a closed curve

formed by a set of continuous fluid

elements remains constant as the fluid

elements move through the flow,

DГ/Dt=0.


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Both the theorems are satisfied by the

starting vortex and bound vortex

system.

In the beginning,Г1 =0 when the flow

is started within the contour C1.

When the flow over the airfoil is

developed,Г2 within C2 is still zero

includes the starting vortexГ3 and the

bound vortexГ4 which are equal and

opposite.


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4.8 Fundamental Equation of Thin Airfoil Theory

Thin airfoil theory is based on the assumption that under certain conditionsan airfoil section may be replaced by its mean camber line (mcl).

Experimental observation:

If airfoil sections of the same mcl but different thickness functions are tested

experimentally at the same angle of attack, it is found that the lift Lˈ

and thepoint application of the lift for the different airfoil sections are practically the

same provided that

(1) maximum airfoil thickness (t/c) is small;

(2) Camber distribution (z/c)max = m is small;

(3) Angle of attack (α) is small.


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This observation permitted the formulation of thin airfoil theory because it

allowed the airfoil to be replaced by the mcl.

V

α

mcl

V

α

mcl

Thin airfoi l theory

The problem now is to find, theoretically the flow of an ideal fluid around

this infinitely thin sheet (mcl) flying through the air at the velocity V ͚ at an

angle of attackα.



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Any solution must satisfy:

(1) Equation of continuity

(2) Irrotational condition

(3) Outer b.c.- Flow at infinity must be undisturbed(4) Inner b.c. – mcl must be a streamline

(5) In addition, since the thin airfoil is being supported in level flight there

must be a lift Lˈ

acting on the airfoil.

(6) Since Lˈ=ρV ͚Г (Kutta-Joukowski Theorem), any theoretical

analysis must introduce a circulationГ around the airfoil section of

sufficient magnitude to satisfy the Kutta condition that the flow leave

the TE smoothly.



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Summary:

Therefore in thin airfoil theory the mcl is replaced by a vortex sheet

of varying strengthγ(s) such that the above conditions aresatisfied and our aim is to determine this ‘γ’ distribution.

Thin airfoil theory stated as a problem says for a vortex sheet

placed on the mcl in a uniform flow of V ͚ determineγ(s) such that

the mcl is a streamline subject to the conditionγ(TE)=0.



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Principle: Mean camber line is a streamline of the flow.

Velocity induced by a 2-D vortex is whereГ is the

strength of the 2-D vortex. Similarly the velocity induced by the vortex

sheet of infinitesimal length ds is given by

ê2

)(

Vd P r

dss

ê2

êvVr



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To force the mean camber line to be a streamline, the sum of all

velocity components normal to the mcl must be equal to zero.Consider the flow induced by an elemental vortex sheet ds at point

P on the vortex sheet.

ê2

)(Vd Pr dss

r

dssd w PP

2

cos)(cosvd '

Thus, the velocity normal to the mcl is：

whereβ is the angle made by dvp to the normal at P, and r is the

distance from the center of ds to the point P.



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The induced velocity due to the vortex sheet representing the entire

mcl is given by.

)sin(V n, V

TE

LE

' cos)(

2

1

r

dsswP

Now determine the component

of the free stream velocity

normal to the mcl.

whereα is the angle of attack and ε is the angle made by the tangentat point P to the x-axis.



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The slope of the tangent line at point P is given by:

))tan(sin

)tan

1

,

1

dx

dzV V

dx

dz

n

（

（

0))(tan(sincos)(

2

1 1TE

LE

dxdz

V dsr

sr

tan)tan( dx

dz

or

In order that the mcl is a streamline. 0)( ,' nP V sw or



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After changing these variables and making the small angle approximationfor sin and tan, and upon rearrangement we get:

)()(

2

1 c

00 dx

dzV dx

x x

x

within thin airfoil theory

approximation s→x,

ds→dx,cosβ =1 and

r →(x0-x), where x

varies from 0 to c, and

x0 refers to the point P.



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In order to facilitate analytic solution, we do a variable transformation

such that:

)cos1(2

)cos1(2

00

c

x

c x

The following analysis is an exact solution to the flat plate or an

approximate solution to the symmetric airfoil. The mean camber linebecomes the chord and hence:

4.9 Flat Plate at an Angle of Attack

V dx x x x

dx

dz

c

00

)(

2

1

0


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Here we simply state a rigorous solution forγ(θ) as:

V d 00 )cos1()cos1(

2

c

sin2

c)(

2

1

sincos12)(

V

d V d

00

00 )cos(cos

)cos1()cos(cos

sin)(21

θ =0 at LE andθ =π at TE andθ increases in CW,

dx=(sinθdθ)c/2


We can verify this solution by substitution as follows:

V d

00 )cos(cos

sin)(

2

1


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Thus, it satisfies the equation:

0

0

00 sin

sin

)cos(cos

cos

nd

n

We now use the following result to evaluate the above integral.


V

V d d

V d

V )0(

)cos(cos

cos

)cos(cos

1

)cos(cos

)cos1(

00

00

00

V d

00 )cos(cos

sin)(

2

1


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Whenθ=π,0112)( V

In addition, the solution forγ also satisfies the Kutta condition.


0cos

sin

2)(

V

By using L’Hospital’s rule, we get

Thus it satisfies the Kutta condition.


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Where s is along the mcl.

By using thin airfoil approximation:

TE

LE dssV V )(L

'

TE

LE

c

dx xV dx xV L0

' )()(

d cV dx xV c

00' sin)(

2

1)(L

Using the transformation x= (1-cosθ) c/2

4.10 2-D lift coefficient for a thin/symmetrical airfoil


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Substituting the solution:

sin

)cos1(2)(

V

2,2'

d

dC and

S q

LorC ll


0

' )cos1(22

1d V cV L

0

2'

)cos1( d cV L

2'

V c L

)1(22

2'

c

V

L

S q L 2'

2d

dC lshows that lift curve is linearly proportional to the angle of attack.


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Calculation of Moment Coefficient:

c

LE dL x M 0

'' )(

c

LE d V x M 0

')(

c

LE xdx xV M 0

' )(


c

LE dx xV x M 0

'

))((

0

' sin2

)cos1(2sin

)cos1(2d

ccV V M LE

)2

()2

(2

)cos1(2

22

2

0

22

2'

cqcV d

cV M LE


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2c

2

''

,

cq

M

Scq

M LE LE LE m

2c l

4c ,

l LE m

C


'

4/

'

4/

'

cc LE L M M

4/ccc 4/,, lcm LE m

C i ht b D Zh Ji


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4/,c cm


4/, cc l LE m

is equal to zero for all values ofα.

c/4 is the aerodynamic center.

Aerodynamic center is that point on an airfoil where moments

are independent of angle of attack.

0c 4/m, c

C i ht b D Zh Ji


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(A)

4.11 Thin Airfoil Theory for Cambered Airfoil

d c

dx

c x

sin2

)cos1(2

where dz/dx is the slope of mcl at x0.

For symmetrical airfoil, mcl is a straight line and hence dz/dx=0 everywhere.

On the other hand, for a cambered airfoil dz/dx varies from point to point.

)(

)(

2

1 c

00 dx

dz

V dx x x

x

As before, we do a variable transformation given by:

C i ht b D Zh Ji


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(B)


Equation (A) becomes

Such a solution forγ(θ) will make the camber line astreamline of the flow.

However, as before a rigorous solution of Equation (B) for γ(θ)

is beyond the scope of this course:

)(coscos

sin)(

2

1 c

00 dx

dzV d

1

0 sin)sin

cos1(2)(

n

n n A AV

C i ht b D Zh Ji


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01 0 00 0

0

)cos(cos

sinsin1

)cos(cos

)cos1(1

xn

n

dx

dzd

n Ad

A

The first integral can be evaluated from the standard from given in

equation

Substitute this solution in equation (B)

0

0

00 sin

sin

)cos(cos

cos

nd

n

Copyright by Dr Zheyan Jin


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00 0

cos)cos(cos

sinsin

nd n

The first term becomes:

The remaining integrals can be obtained from another standard form,

which is given below:

0

00

00

0

0

00

0

)coscos

cos(1)coscos

(1

)coscos

cos1(

1

A

d Ad A

d A



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1

00 cos)(n

n n A Adx

dz

The second term becomes:

Therefore Equation (B) becomes:

1

0

10

0

coscoscos

sinsin1

n

n

n

n n Ad n A

Upon rearrangement the slope at a point P on the mcl is given by:

01

00 cos xn

ndx

dzn A A



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nn B A

d dz

dz B A

000 )(

1)(

Where,

10 cos)(n

n n B B f

,,2,1

cos)(2

)(

1

0

00

n

d n f B

d f B

n

From Fourier series:



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cc

d c

dx x00

sin)(2

)(

]sinsin)sin([

sinsin22

)cos1(22

sin)(2

sinsin

)cos1(2)(

100

0

01

0 0

0

1

0

n

n

n

n

c

n

n

d n A AcV

d n AV c

d AV c

d c

n A AV

From thin airfoil theory:

Evaluation of Circulation :



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]2

[2c

]2

[

]2

[

)1(0

)1(2

sinsin

10

''

10

2'

10

10

A Acq

L

sq

L

A AcV V L

A AcV

n

n

d n A

l

n

n

Using:

Cl is normalized by theα as seen by the chord connecting the LE and

TE of the mcl. c is the chord connecting the LE and TE of the mcl.



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d

dx

dz

d dx

dzd dx

dz

A A

)1(cos)(1

2c

cos)(2

))(1

(2c

2c

0

l

00l

10l

d

dx

dz

d

d

L

L Ll

l

00

00

)1(cos1

)(2)(c

c

Note that as in the case of symmetric airfoil, the theoretical lift slope

for a cambered airfoil is 2π. It is a general result from thin airfoil

theory that dcl/dα=2π for any shape airfoil.

Also,

zero lift angle of attack



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c LE

c

dx x xcV scq

M

dx x xV

02

'

LEm,

0

'

LE

)(2

c

)(M

01

0

20,

1

0

sinsin)cos1()cos1(c

]sinsin

cos)1([2)(

sin2

)cos1(2

n

n LE m

n

n

d n Ad A

n A AV

d cdx

c x

As before we do a variable transformation from x toθ.

Thus，

Determination of moment coefficient:



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0

2

0

2

2sin

2

cos

d

d

),,3(0

)2(4

)1(0

sincossin

),,2(0

)1(2

sinsin

0

0

n

n

n

d n

n

n

d n

Using the following definite integrals:



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422c

422

c

sincossinsinsincosc

210,

2100,

10 00

2

00

0,

A A A

A A A A

d n And n And Ad A

LE m

LE m

n

LE m

)(44

c

2

22

2c

2c

21210

,

10

A A A A A

A A

l LE m

l



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214

,

21

4,

,

''

4

'

4c

)(44

c

4

cc

4

A A

A Ac

c L M M

cm

llcm

LE m

c LE

)(

c4

1)(

c44

cc

2121

,

A Ac

c A Acc

x

c x

ll

cp

lcp LE m

The location of center of pressure:



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)(c4

121 A A

c

c xl

cp

As the angle of attack changes, the center of pressure also changes.

Indeed, as the lift approaches zero, xcp moves toward infinity; that is, it

leaves the airfoil.

For this reason, the center of pressure is not always a convenient

point at which to draw the force system on an airfoil.

Rather, the force-and-moment system on an airfoil is more

conveniently considered at the aerodynamic center .



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py g y y



TE

LE

TE

LE ul

ul

dssV L

ds p p L

ds p pdL

)(

cos)(

)1(cos)(

'

'

'

TE

LE

TE

LE)()( dssV ds p p ul

Equating (A) and (B) and setting cosɳ

≈ 1, we get

Relationship between pressure on mcl and :

(A)

(B)

or

)()( sV p pul

(1)



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py g y y



))((2

)(2

1)(

2

1 2222

luluul

uull

uuuu p p

wu pwu p

2

lu uuV

Using Bernoulli’s equation:

(2)

(3)

From vortex sheet theory:

)(u sulu

From (1), (2) and (3)



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py g y y



V s

V V sc

q

uuuu

c

q

p pc

q p p

q p pc

u pl p

luul

u pl p

ulu pl p

ulu pl p

)(22)(c

))((2

1

c

c

c

2,,

,,

,,

,,

i.e., within thin airfoil approximation, the average of top and bottom

surface velocities at any point on the mcl is equal to the freestreamvelocity.

Pressure coefficient difference between lower surface and upper surface:



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py g y y


4.12 Summary

b

adss y x )(

2

1),(

b

adss)(

A vortex sheet can be used to synthesize the inviscid, incompressible

flow over an airfoil. If the distance along the sheet is given by s and

the strength of the sheet per unit length is γ(s), then the velocity

potential induced at point (x,y) by a vortex sheet that extends from

point a to point b is

The circulation associated with this vortex sheet is

21 uu

Across the vortex sheet, there is a tangential velocity discontinuity, where

Vortex Sheet:



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4.12 Summary

0)TE(

The Kutta condition is an observation that for a airfoil of given shape

at a given angle of attack, nature adopts that particular value of

circulation around the airfoil which results in the flow leaving smoothly

at the trailing edge.

If the trailing-edge angle is finite, then the trailing edge is a stagnation

point.

If the trailing edge is cusped, then the velocities leaving the top and

bottom surfaces at the trailing edge are finite and equal in magnitudeand direction.

Kutta Condition:

In either case:



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4.12 Summary

)()(

2

1 c

0 dx

dzV

x

d

Thin airfoil theory is predicated on the replacement of the airfoil by the

mean camber line.

A vortex sheet is placed along the chord line, and its strength adjusted

such that, in conjunction with the uniform freestream, the camber linebecomes a streamline of the flow while at the same time satisfying the

Kutta condition.

The strength of such a vortex sheet is obtained from the fundamental

equation of thin airfoil theory:

Thin airfoil theory:



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4.12 Summary

Symmetrical airfoil

1. cl=2πα.

2. Lift slope = dcl

/dα

=2π

.3. The center of pressure and the aerodynamic center are

both at the quarter-chord point.

4. cm,c/4=cm,ac=0.

Results of thin airfoil theory:



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4.12 Summary

Cambered airfoil

1.

2. Lift slope = dcl /dα=2π.

3. The aerodynamic center is at the quarter-chord pint.

4. The center of pressure varies with the lift coefficient.

Results of thin airfoil theory:

0

00 )1(cos1

2 d dx

dzcl



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Aerodynamics Chapter 4 r