Joukowski Mapping

AOE 5104Advanced Aero- and Hydrodynamics

Dr. William Devenport andLeifur Thor Leifsson

2

Why Circular Cylinders?

Mapping

3

The Problem of the AirfoilTerminology

Chord cα

V∞

Lift l per unit span

cVlCl 2

21

∞

=ρ

Γ−= ∞Vl ρ

Lift coefficient

Kutta Joukowski Thm.

Invariance of Circulation under Mapping

z-plane ζ-plane

ζζ

ddzzWW )()(~ =

Γ Γ=Γ~

Loop

Loop

qi

dW

dzdzdW

dzzWiq

loop

loop

loop

~~

)(~

)(~

)(

+Γ=

=

=

=+Γ

∫

∫

∫

ζζ

ζζ

4

The Problem of the AirfoilConsider the ideal flow past a given airfoil at a fixed angle of attack

These flows differ only by…

To choose the realistic flow solution we employ what is know as the ‘Kutta’ condition, that the flow leave smoothly from the trailing edge. The Kutta condition is an empirical observation that results from the tendency of the viscous boundary layer to separate at a salient edge.

Wedge trailing edge –stagnation point at t.e. u

u

Cusp trailing edge –Velocity (and pressure) same on both sides

5

Symbol Conventions

x

iη

ξ

ζ=ζ(z)iyz-plane

dzdFzW

zF

=)(

)(

ζ-plane

ζζζ

ζζ

ddzW

dFdW

zFF

==

=~

)(~))(()(~

Initial Flow Mapped Flow Mapping

Critical at

0=dzdζ

iη

ξ

z= z(ζ)ζ-plane

ζζ

ζ

dFdW

F~

)(~)(~

= Critical at

0=ζd

dz

x

iyz-plane

dzdW

dzdFzW

zFzFζ

ζ~)(

))((~)(

==

=

Our Mappings to this Point

The JoukowskiMapping

6

Joukowski Mapping ζζ /2Cz +=0>= realC

ζ-plane z-plane

-C C -2C 2C

Effects on Space

Critical Points?

⎟⎠⎞

⎜⎝⎛

+−

=⎟⎟⎠

⎞⎜⎜⎝

⎛+−

CzCz

CC

22

2

ζζ

Behavior at ∞?

7

Consider a Series of Circles Cutting the Right-Hand Critical Point

1. ζ1=0

2. Re{ζ1}=0, Im{ζ1}>0

3. Re{ζ1}<0, Im{ζ1}=0

4. Re{ζ1}<0, Im{ζ1}>0

ζ-plane

Cζ1

a

‘a’ adjusted so circle always cuts right-hand critical point

8

1

2

)(~ζζ

ζζα

α

−+= ∞−

∞

ii eaVeVF

ζζ /2Cz +=

1. ζ1=0

ζ-plane z-plane

Circle coincident with mapping circle

The Flat Plate

9

ζζ /2Cz +=

2. Re{ζ1}=0, Im{ζ1}>0 Circle centered on imaginary axis

The Circular ArcIm{ζ1} controls camber

ζ-plane z-plane

ζ-plane z-plane

10

ζζ /2Cz +=

3. Re{ζ1}<0, Im{ζ1}=0 Circle centered on negative real axis

The Symmetric AirfoilRe{ζ1} controls thickness

ζ-plane z-plane

ζ-plane z-plane

11

ζζ /2Cz +=

4. Re{ζ1}<0, Im{ζ1}>0 Circle centered in 2nd quadrant

The Cambered AirfoilRe{ζ1} controls thickness.

Im{ζ1} controls camber

ζ-plane z-plane

ζ-plane z-plane

12

Mapping an Airfoil Flowζ-plane

C

ζ1

a

δ

m β>0

β

δ

ζζ

i

i

aeCme

−=−=

1

1

z-plane

dzdW

dzdFzW

zFzFζ

ζ~)(

))((~)(

==

=

2C

αV∞

)(2)()(~

)(log2

)(~

12

1

2

11

2

ζζπζζζ

ζζπζζ

ζζ

αα

αα

−Γ

−−

−=

−Γ

−−

+=

∞−∞

∞−∞

ieaVeVW

ieaVeVFi

i

e

ii

z=ζ+C2 / ζ

Γ ?

13

Results for Liftζ-plane

C

z-plane

2Cα

V∞

z=ζ+C2/ ζ

c

2

2

1ζζC

ddz

−=ζ1

aβ

)sin(8)sin(4

221

2

βαπρ

βαρπρ

+==

+=Γ−=

∞

∞∞

ca

cVlC

aVVl

l

4/4

cCaCc

Ca

≈≈≥≥

Whereand a and c increase slowly with camber and thickness

and

1. The lift on an airfoil varies as the sine of the angle of attack or, for small angles, linearly with angle of attack

2. The primary (and almost exclusive) influence of camber in controlling the zero lift angle of attack -β

3. The lift curve slope at zero angle of attack is 2π for a flat plate, and increases weakly with increasing thickness and camber

for a thin airfoil

14

Obtaining the Pressure Distributionζ-plane

C

z-plane

2Cα

V∞

z=ζ+C2/ ζ

c

2

2

1ζζC

ddz

−=

)(2)()(~

12

1

2

ζζπζζζ

αα

−Γ

−−

−= ∞−∞

ieaVeVWi

i

dzdWzW ζζ )(~)( =

ζ1

a

1. Choose a set of points on the circle2. For these points determine

Velocity on the circleDerivative of mapping Airfoil coordinates

3. Evaluate Cp on the airfoil using Bernoulli

4. Plot Cp vs x, i.e Cp|airfoil vs Re{zairfoil}

θζζ icirc ae+= 1|

)|(~circW ζ

2

2

|1

circcirc

Cddz

ζζ−=

circcircairfoil

Cz|

||2

ζζ +=

22

222 /)|(1/)(1| ∞∞ −=−= VdzdWVzWC

circcircairfoilp

ζζ

)sin(4 βαπ +−=Γ ∞aVwith

β