potential flows

Fluid Dynamics ME 5313 / AE 5313

Potential Flows (revised 10/24/2013)

Instructor: Dr. Albert Y. Tong Department of Mechanical and Aerospace Engineering

The University of Texas at Arlington

Potential Flows 2

Velocity Potential

For irrotational flow, 0uζ = ∇× =

But

Thus will satisfy u i jx yφ φφ ∂ ∂

= ∇ = +∂ ∂

uxφ∂

=∂

vyφ∂

=∂

( ) 0φ∇× ∇ ≡

and

Potential Flows 3

Velocity Potential

2 2

0

zv ux y

x y y x

ζ

φ φ

∂ ∂≡ −∂ ∂

∂ ∂= − =∂ ∂ ∂ ∂

φ

Irrotational

: velocity potential

∴

Potential Flows 4

Some Properties of

2φ

φ

1φA

B

( )B B

ABA A

u dl udx vdyΓ = ⋅ = +∫ ∫

Potential Flows 5

Some Properties of φ

B

B AA

dφ φ φ= = −∫

uxφ∂

=∂

vyφ∂

=∂

B

ABA

dx dyx yφ φ ∂ ∂

Γ = + ∂ ∂ ∫

Existence of => Irrotational Flow φ

and

Potential Flows 6

Some Properties of

For incompressibility, we have 0u∇⋅ =

0u vx y∂ ∂

+ =∂ ∂

0x x y y

φ φ ∂ ∂ ∂ ∂ + = ∂ ∂ ∂ ∂ 2 2

22 2 0

x yφ φ φ∂ ∂+ = ∇ =

∂ ∂

must satisfy the Laplace Equation

is a harmonic function

φφ

φ

Potential Flows 7

Stream Function

Definition: vxψ∂

≡ −∂

uyψ∂

≡∂

Examine the continuity equation,

0u vux y∂ ∂

∇ ⋅ = + =∂ ∂

(for incompressible flow)

0 ?x y y x

ψ ψ ∂ ∂ ∂ ∂ + − = ∂ ∂ ∂ ∂

and

ψ

Potential Flows 8

Stream Function 2 2

0x y y x x y y x

ψ ψ ψ ψ ∂ ∂ ∂ ∂ ∂ ∂ + − = − ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ =

The continuity equation is automatically satisfied.

ψ

ψ => Incompressible flow Existence of

Potential Flows 9

Stream Function

So what about irrotationality?

0?uζ = ∇× =

zv ux y

ζ ∂ ∂= −∂ ∂

x x y yψ ψ ∂ ∂ ∂ ∂ = − − ∂ ∂ ∂ ∂

2 22

2 2x yψ ψ ψ

∂ ∂= − + = −∇ ∂ ∂

ψ

Potential Flows 10

Stream Function 2

zζ ψ=> = −∇0zζ =

2 0ψ∴ ∇ =

For irrotational flow,

is a harmonic function ψ

ψ

Potential Flows 11

Summary of and

φ ψ 2 2 0φ ψ∇ = ∇ =i)

ii)

iii)

iv)

v)

are both harmonic,

For incompressible flows,

For irrotational flows,

Existence of

Existence of

and

φ =>

ψ =>Irrotational

Incompressible

must be harmonic

must be harmonic

φ

φ ψ

ψ

Potential Flows 12

Property of Stream Function

Aψ

BψB

A

Flow across AB ( )B

A

udy vdx= −∫

Potential Flows 13


B

A

dy dxy xψ ψ ∂ ∂ = − − ∂ ∂

∫

B

A

dx dyx yψ ψ ∂ ∂

= + ∂ ∂ ∫

B

B AA

dψ ψ ψ= = −∫Streamlines and equipotential lines are orthogonal

Potential Flows 14


Cφ =

0d dx dyx yφ φφ ∂ ∂

= = +∂ ∂

dy udx vφ

= −

Equipotential Lines:

dy udx v

= −

φ

u v

Streamlines and equipotential lines are orthogonal.

Potential Flows 15


Cψ =

0d dx dyx yψ ψψ ∂ ∂

= = +∂ ∂

Streamlines:

dy vdx uψ

=

v− u

Potential Flows 16


dy dy u vdx dx v uφ ψ

⋅ = −

1dy dydx dxφ ψ

⋅ = −

∴ Equipotential Lines are orthogonal to Streamlines

Potential Flows 17

Complex Potential

Reminiscent of Cauchy-Riemann relations

u vx y y xφ ψ φ ψ∂ ∂ ∂ ∂

= = = = −∂ ∂ ∂ ∂

If we have a complex function, F(z), and if F(z) is analytic,

( ) ( , ) ( , )F z x y i x yφ ψ= +i.e.

By definitions:

Potential Flows 18

Complex Potential

The real part is a valid velocity potential for a particular flow field and the imaginary part is also automatically a valid stream function for the same flow field. F(z) is called Complex Potential

Potential Flows 19

Complex Velocity

( )F z iφ ψ= +

( ) dFW zdz

= (complex velocity)

Since F(z) is analytic, is direction independent dFdz

( )dF F Fdz x iy

∂ ∂= =∂ ∂

Potential Flows 20

Complex Velocity

u iv= −

F ix x x

φ ψ∂ ∂ ∂= +

∂ ∂ ∂

Proof: 1

( )F F Fiiy i y y∂ ∂ ∂

= = −∂ ∂ ∂

i iy yφ ψ ∂ ∂

= − + ∂ ∂

( )i v iu= − +

u iv= −

Potential Flows 21

Complex Velocity

( )dF F F u ivdz x iy

∂ ∂∴ = = = −

∂ ∂

In polar coordinates: ( ) irW u iu e θ

θ−= −

Complex conjugate of W, W u iv= +( )( )WW u iv u iv= − +

2 2u v= +

Potential Flows 22

Uniform Flow

( )F z Az=Consider

i) If A is real

( )dF z A u ivdz

= = −

0u Av

=> == x

iy

Horizontal uniform flow (back)

Potential Flows 23

Uniform Flow

ii) If A is pure imaginary, A iC=

( )dF z iC u ivdz

= = −

0uv C

=> == −

iy

x

Vertical uniform flow (back)

Potential Flows 24

Uniform Flow

iii) If A is complex, 1 2A C iC= +

1 2( )dF z C iC u iv

dz= + = −

1

2

u Cv C

=> == −

iy

x α

Uniform flow at an angle to the x-axis

α

Potential Flows 25

Uniform Flow iA Ue α−=

(cos sin )U iα α= −

u iv= −

cosu U α=

sinv U α=

α( ) iF z Ue zα−=

( )F z Uz=

( )F z iUz= −

0α =

2πα =

at

at

(case i)

(case ii)

Potential Flows 26

Source and Sink

Consider: ( ) lnF z C z=

( ) (ln )F z C r iθ∴ = +

where C is a real constant

lnC rφ = Cψ θ=

iz re θ=ln lnz r iθ= +

ψ

φand

Potential Flows 27

Source and Sink

( )

( )

( ) ( )i iri

r

dFW zdzCW zzC CW z e u iu e

re rCur

θ θθθ

− −

=

=

= = = −

=> = 0uθ =and

Potential Flows 28

Source and Sink

For C > 0

r 2

2 2

rm r uCm r Cr

π

π π

= ⋅

= =

( ) ln ln2mF z C z zπ

= =

Potential Flows 29

Source and Sink

For C < 0

r 2 2Cm r Cr

π π= =

( ) ln ln2mF z C z zπ

= =

(Suction sink)

In general, if the source or sink is at z0

0( ) ln( )2mF z z zπ

= − 0z

Potential Flows 30

Vortex

For C pure imaginary φ

ψ

C iD= where D is real

( ) lnF z D iD rθ= − +

Dφ θ=> = − and lnD rψ =

𝐹𝐹(𝑧𝑧) = 𝐶𝐶𝐶𝐶𝐶𝐶𝑧𝑧 = 𝑖𝑖𝑖𝑖(𝐶𝐶𝐶𝐶𝑙𝑙 + 𝑖𝑖𝑖𝑖)

Potential Flows 31

Vortex

( )

( )

( ) ( )

ii

ir

dFW zdzC C iDW z ez re r

W z u iu e

θθ

θθ

−

−

=

= = =

= −

0ru=> = and Durθ = −

Potential Flows 32

Vortex

0 00 0

D uD u

θ

θ

> ⇒ < ⇒

< ⇒ > ⇒

if clockwise

anti-clockwise 2

0

u dl u rdπ

θ θΓ = ⋅ =∫ ∫

2 2D r Dr

π πΓ = − ⋅ ⋅ = −

2D

πΓ

=> = −

Potential Flows 33

Vortex

( ) lnF z iD z∴ =

( ) ln2

F z i zπΓ

= −

00

Γ > ⇒Γ < ⇒

anti-clockwise

clockwise

θ

For a vortex located at z0 𝐹𝐹(𝑧𝑧) = −𝑖𝑖𝛤𝛤2𝜋𝜋 ln(𝑧𝑧 − 𝑧𝑧0)

Potential Flows 34

Superposition of Two Sources

-a a

m m

y

x

( ) ln( ) ln( )2 2m mF z z a z aπ π

= + + −

Potential Flows 35


( ) ln( )( )2mF z z a z aπ

= + −

2 2( ) ln( )2mF z z aπ

= −

( ) 1 1( )2

dF z mW z u ivdz z a z aπ

= = + = − − +

-a a

m m

y

x

Potential Flows 36


Along the y-axis (i.e. x=0)

1 1( )2mW z

iy a iy aπ

= + − +

2 2

2( )2m i yW z u iv

a yπ −

= = − +

0u=> =2 2

22m yv

a yπ

= + and

Potential Flows 37


2 2 2 22

2m yva y a yπ

= ⋅ ⋅+ +

Note:

2m

rπ2sinθ

2 2a y+

a mθ

y

The y axis is a streamline

Potential Flows 38

Two-Dimensional Dipole

-a a

m -m

y

x

source sink


= + − −

Potential Flows 39


-a a

m -m

y

x

source sink Limiting process:

0a →m →∞ma πµ=

where is the strength of the dipole (doublet) µ

Potential Flows 40


ln( ) ln 1 az a zz

+ = +

ln ln 1 azz

= + +

ln( ) ln . . .az a z H O Tz

∴ + = + +

Note: 2 3

ln(1 ) ...2! 3!α αα α+ = + + +

...!

n

nα

ln(1 ) . . .H O Tα α+ = +Similarly,

ln( ) ln . . .az a z H O Tz

− = − +

Potential Flows 41



= + − −

As a result,

( ) ln . . . ln . . .2m a aF z z H O T z H O T

z zπ = + + − − +

0,lim ( )

a ma

maF zz zπµ

µπ→ =

= =

Potential Flows 42


Flow pattern:

2 2

( )( )( )

x iyF zz x iy x yµ µ µ −

= = =+ +

2 2

xx yµφ∴ =+ 2 2

yx yµψ −

=+

and

Potential Flows 43


i) Streamlines: 0ψ ψ= = constant

0 2 2

yx yµψ ψ −

= =+

2 2

0

0yx y µψ

+ + =

2 22

0 02 2x y µ µ

ψ ψ

+ + =

Circles with centers

0

0,2µψ

−

Radius 02

µψ

Potential Flows 44


i) Equipotential Lines: 0φ φ= = constant

0 2 2

xx yµφ φ= =+

2 2

0

0xx y µφ

+ − =

2 22

0 02 2x yµ µ

φ φ

− + =

Circles with centers

0

,02µφ

Radius 02

µφ

Potential Flows 45

Two-Dimensional Dipole y

x

Streamlines

Equipotential lines

Potential Flows 46

Flow in a Sector

Consider: ( ) nF z Uz= 1n >

(Note: 1n = i.e. ( )F z Uz= is a uniform rectilinear flow)

( )( ) cos sinnF z Ur n i nθ θ= +

cosnUr nφ θ=> =

sinnUr nψ θ=θ

Potential Flows 47

Flow in a Sector

cosnUr nφ θ= sinnUr nψ θ=and

0θ = and nπθ =At 0ψ⇒ = so it is a streamline

2nπ

2nπ

0ψ =

0ψ =

Potential Flows 48

Flow in a Sector

2nπ

2nπ

0ψ =

0ψ =

1 cos( )nru nUr n

rφ θ−∂

= =∂

0 :θ = 0 ;uθ =1 0n

ru nUr −= >

:nπθ = 0 ;uθ = 1 0n

ru nUr −= − <

:2nπθ = 1 ;nu nUrθ

−= − 0ru =

𝑢𝑢𝑖𝑖 =1𝑙𝑙𝜕𝜕𝜕𝜕𝜕𝜕𝑖𝑖 = −𝐶𝐶𝑛𝑛𝑙𝑙𝐶𝐶−1sin(𝐶𝐶𝑖𝑖)

Potential Flows 49

Uniform Flow about a Circular Cylinder

ab− 1b

−1a

m m m− m−| | 1z =

, 1a b >1 1( ) ln( ) ln ln( ) ln

2mF z z b z z a z

b aπ = + + + − − − −

1( )( ) ln

12 ( )

z b zm bF z

z a za

π

+ + = − −

Potential Flows 50


Along the unit circle, iz e θ=

( )

( )

( )

( )

1 1

( ) ln1 12

i i i i

i i i i

e b e e a em b aF z

e a e e a ea a

θ θ θ θ

θ θ θ θπ

− −

− −

+ + − − = − − − −

ab− 1b

−1a

m m m− m−| | 1z =

, 1a b >

Potential Flows 51


Recall: 2 2zz real x y= = +

( ) ( )1 11 1( ) ln

2

i i i i i ie b e e e ae em b aF z

real quantity

θ θ θ θ θ θ

π

− − − + + − − =

1 1

( ) ln2

i i i ie b e e a em b aF z

real quantity

θ θ θ θ

π

− − + + + − − + =

Potential Flows 52


Note: cos sinie iθ θ θ= +cos sinie iθ θ θ− = −

1 12cos 2cos( ) ln

2

b am b aF z

real quantity

θ θ

π

+ + − − + ⇒ =

Potential Flows 53


( )F z real iφ ψ= = +

0ψ⇒ =

1z = (unit circle) is a streamline

( )( ) ln2mF zπ

= real quantity

Therefore,

Potential Flows 54


Now, let constant a b= →∞ma=and

1( ) ln 1 ln 1 ln 12m z zF z

a az aπ = + + + − −

1ln 1 ln( 1)az

− − − −

𝐹𝐹(𝑧𝑧) =𝑚𝑚2𝜋𝜋 𝐶𝐶𝐶𝐶 �

𝑎𝑎 �1 + 𝑧𝑧𝑎𝑎� 𝑧𝑧 �1 + 1

𝑎𝑎𝑧𝑧�

−𝑎𝑎 �1 − 𝑧𝑧𝑎𝑎� 𝑧𝑧 �1 − 1

𝑎𝑎𝑧𝑧��

Potential Flows 55


1 1( ) . . .2m z zF z i H O T

a az a azπ

π = + + + + +

2 2lim ( )2a

m zF za azπ→∞

= +

.m consta=

1m za zπ

= +

constant=U

Potential Flows 56


1( )F z U zz

∴ = +

Far away: 1z >>

( )F z Uz→ An uniform horizontal flow (from left to right)

Potential Flows 57


Near Field:

( ) UF zz

→ (a dipole)

1( )F z U zz

= +

1i iU re er

θ θ− = +

Potential Flows 58


( ) ( )1( ) cos sin cos sinF z U r i ir

θ θ θ θ = + + −

2 2

1 1cos 1 sin 1U r irr r

θ θ = + + −

2

1cos 1Urr

φ θ ∴ = +

2

1sin 1Urr

ψ θ = −

Potential Flows 59


2

1cos cosru U Ur rφ θ θ∂ = = + − ∂

2

1cos 1ru Ur

θ = −

2

1sin 1u Urθ θ = − +

𝑢𝑢𝑖𝑖 =1𝑙𝑙𝜕𝜕𝜕𝜕𝜕𝜕𝑖𝑖

=1𝑙𝑙𝑛𝑛𝑙𝑙(−𝑠𝑠𝑖𝑖𝐶𝐶𝑖𝑖) �1 +

1𝑙𝑙2�

Potential Flows 60


ab− 2Rb

−2R

a

m m m− m−| |z R=

Remarks:

i) Extension to radius = R

2

( ) RF z U zz

= +

Potential Flows 61


ii) Combining a dipole and a uniform flow

Consider:

( )F z Uzzµ

= +

i iUre er

θ θµ −= +

( ) ( )cos sin cos sinUr i irµθ θ θ θ= + + −

Potential Flows 62


( ) cos sinF z Ur i Urr rµ µθ θ = + + −

φ ψ

For constant on ψ = r R= , we have

sinURRµψ θ = − =

constant

Potential Flows 63


URRµ

⇒ =

2URµ =

For ψ = constant, 0URRµ

− =

2 2

( ) UR RF z Uz U zz z

⇒ = + = +

Potential Flows 64

Hydrodynamic Forces on a Cylinder in a steady 2-D Uniform Flow

(i) : force in the x-direction xF

0PU∞

θ

y

x

Potential Flows 65


cosxF p Rdπ

πθ θ

−= − ⋅∫

02 cosxF p R d

πθ θ= − ⋅∫

θ

P

cosP θ

sinP θ

RdθR

sinyF p R dπ

πθ θ

−= − ⋅∫

Potential Flows 66


Bernoulli Equation:

2 2 201 1( )2 2

pp u v U constρ ρ ∞+ + = + =

2 20

22

112

p p u vUUρ ∞

∞

− += −

dFW u ivdz

= = −

2u

Potential Flows 67


2 2W u iv WW u v= + ⇒ = +2

( )dF d RW U zdz dz z∞

= = +

2

2(1 )RW Uz∞= −

22

2(1 )iRW U er

θ−∞= −

2| (1 )i

z RW U e θ

∞== −2| (1 )i

z RW U e θ−

∞=

⇒ = − and

Potential Flows 68


2 2 2 2 2(1 )(1 )i iWW U e e u vθ θ∞

−= − − = +

2 2 2 2 2(1 1)i iu v U e eθ θ∞

−+ = − − +

2 22 2 22 (1 )

2

i ie eu v Uθ θ

∞

−++ = −

2 2 22 (1 cos 2 )u v U θ∞

+ = −

Potential Flows 69


2 2

2 2(1 cos 2 )u vU

θ∞

+⇒ = −

2 22

2 4sinu vU

θ∞

+=

20

21 4sin1

2

p p

Uθ

ρ∞

−= −

Potential Flows 70


0

212

p p

Uρ∞

−

2π π θ

A

B,D

C A C

B

D

Potential Flows 71


2 20

1 (1 4sin )2

U Pρ θ∞ − +

2 20

0

12 (1 4sin ) cos2xF R U P d

π

θ ρ θ θ∞ = − − + ∫

0

2 cosxF p R dπ

θ θ= − ⋅∫

Potential Flows 72


0 000

cos sin | 0P d Pπ π

θ θ θ= =∫Similarly,

2

0

1 cos 02

U dπ

ρ θ θ∞ =∫We have left:

2 2

0

12 4sin ( )cos2

R U dπ

θ ρ θ θ∞− −∫

2π

π

cosθ

θ

Potential Flows 73


Examine: 2

0sin cos d

πθ θ θ∫

3

0

sin | 03

πθ= = 0xF∴ =

Similarly, 0yF =

0xF = is because of the neglect of viscosity.

It is known as the d’ Alembert’s paradox

Potential Flows 74

Flow about a Circular Cylinder with Circulation

2

( ) ( ) ln2

R iF z U z z Cz π

Γ= + + +

Γ

RU

Potential Flows 75


( ) (Re Re ) (ln )2

i i iF z U R i Cθ θ θπ

− Γ= + + + +

2 cos ln2 2

iUR R Cθθπ πΓ Γ

= − + +

0 ln 02i R CψπΓ

= ⇒ + =

ln2iC RπΓ

⇒ = −

@ r R=

Potential Flows 76


2

( ) ln2

R i zF z U zz Rπ

Γ ∴ = + +

2

2

1( ) 12

R iW z Uz zπ

Γ= − +

2

221

2i iR iU e e

r rθ θ

π− − Γ

= − +

Potential Flows 77


( ) irW u iu e θ

θ−= −

2

2 2i i iR ie U e e

r rθ θ θ

π− − Γ

= − +

2 2

2 21 cos 1 sin2

i R Re U i Ur r r

θ θ θπ

− Γ

= − + + +

Potential Flows 78


2

2(1 )cosrRu Ur

θ⇒ = −

2

2(1 )sin2

Ru Ur rθ θ

πΓ

= − + −

and

Potential Flows 79


On r = R:

0ru =and

(1 1)sin2

u URθ θ

πΓ

= − + −

2 sin2

UR

θπΓ

= − −

Potential Flows 80


At the stagnation point, 0ru uθ= =

0 2 sin2

u URθ θ

πΓ

= = − −

sin4 UR

θπΓ

= −

Potential Flows 81


(iii) 4 URπΓ >

No Solution (Stagnation point detached from the cylinder)

(ii) 4 URπΓ ≤

2 solutions: 1sin4 UR

θπ

− Γ = −

0,Γ =(i) 0,θ π=

Potential Flows 82

Blasius’ Theorem (I)

: arbitrary closed contour which encloses the object. 0

2

2x yC

i dFF iF dzdz

ρ − = ∫

0C

iC0Cx

y

Potential Flows 83

Blasius’ Theorem (I) Application: Consider the case of Uniform flow over a

cylinder with circulation

2

( ) ln2

R i zF z U zz Rπ

Γ = + +

2

2( ) 12

dF R iW z Udz z zπ

Γ= = − +

Potential Flows 84


2 4 2 22

2 4 2 2 2

21 14

R R i U RUz z z z zπ π

Γ Γ= − + − + −

2 2 4 2 2 2

22 4 2 2 3

24

R U R U i U i URUz z z z zπ π π

Γ Γ Γ= − + − + −

2 22 22 2

2 2( ) 1 2 12 2

R i i RW z U Uz z z zπ π

Γ Γ = − + + −

Potential Flows 85


0

2

2x yC

iF iF W dzρ− = ∫

1(2 ) ( )2i i b residuesρ π= ∑

Singularities:

1i UbπΓ

⇒ =

0z =

Potential Flows 86


0xF∴ =

yF Uρ= Γ

0x yF F⇒ = = when for the non circulating case.

0Γ =

( )( )x yi UF iF i Uρπ ρπΓ

∴ − = − = − Γ

Potential Flows 87

Blasius’ Theorem (I) - Proof

iC

0Cx

yP

Pdy

dx−Pdy

Pdx−dy

dxv

u

( )dm udy vdx ρ= −

Potential Flows 88


Force Balance:

0 0

( )xC C

F Pdy udy vdx uρ− − = −∫ ∫ Eq. (1)

In the x-direction:

iC

0Cx

yP

Pdy

dx−Pdy

Pdx−dy

dxv

u

( )dm udy vdx ρ= −

0C

m u ndsρ= ⋅∫

back

Potential Flows 89


0n ds⋅ =

x yn n i n j= +

0x yn dx n dy⇒ + =

yx nndy dx

λ= − =

( )n dyi dxjλ= −

𝑑𝑑𝑠𝑠 = 𝑑𝑑𝑑𝑑𝑖𝑖 + 𝑑𝑑𝑑𝑑𝑑𝑑

Potential Flows 90


1 2 2dy dx dsλ− = + =

dyi dxjnds−

∴ =

m u ndsρ= ⋅∫

( )( ) dyi dxjui vj ds

dsρ −

= + ⋅ ⋅∫( )udy vdxρ= −∫

Potential Flows 91


Bernoulli Equation: 12

dPu u G Bρ

⋅ + − =∫

0 0

( )yC C

F Pdx udy vdx vρ− + = −∫ ∫ Eq (2)

In the y-direction:

Neglect body force Pρ

back

Potential Flows 92


1 '2

Pu u Bρ

⋅ + =

'12

P u u Bρ= − ⋅ +

12

dPu u G Bρ

⋅ + − =∫

Potential Flows 93


Equation (1) becomes:

0 0

2 21 ( ) ( )2x

C C

F u v dy udy vdx uρ ρ= + − −∫ ∫

0

2 21 ( )2x

C

F uvdx u v dyρ = − − ∫Similarly, Equation (2) becomes:

0

2 21 ( )2y

C

F uvdy u v dxρ = − + − ∫

Potential Flows 94


Consider: 0

2

2 C

i W dzρ∫ ( )W u iv= −

0

2( ) ( )2 C

i u iv dx idyρ= − +∫

0

2 2( ) 2 ( )2 C

i u v i uv dx idyρ = − − + ∫

Potential Flows 95


0

2 2 2 2( ) 2 ( ) 22 C

i u v dx uvdy i u v dy uvdxρ = − + + − −

∫

0

2 2 2 21 1( ) ( )2 2C

uvdx u v dy i uvdy u v dxρ = − − + − − ∫

x yF iF= −

0

2

2 C

i W dzρ= ∫

Potential Flows 96

Blasius’ Theorem (II)

0

2Re2 C

M zW dzρ = −

∫

iC

0Cx

y

M

Potential Flows 97

Conformal Transformation

Joukowski Transformation

Potential Flows 98


z x iy= + iζ ξ η= +2zζ =( )f zζ =

iy

x

iη

ξ

z plane planeζ

Potential Flows 99


For 2( )f z z= we have

( )2 2 2( ) 2i x iy x y ixyζ ξ η= + = + = − +2 2 2x y xyξ η∴ = − =

Potential Flows 100


(i) Harmonic function remains harmonic ?

2 2 2 2

2 2 2 20 0x yφ φ φ φ

ξ η⇒

∂ ∂ ∂ ∂+ = + =

∂ ∂ ∂ ∂i.e.

Issues for consideration:

?

Potential Flows 101


(a) f is analytic

(b) 0dfdz

≠

(ii) Complex velocities preserved or changed?

( )( ) dF zW zdz

=

( ) ( )dF d dWd dz dzζ ζ ζζζ

= ⋅ = ⋅

Therefore, the velocity is not preserved.

Potential Flows 102


(iii) Are the source and vortex strengths preserved?

( )C

W z dz im= Γ +∫. . . ( ) ( )

C

L H S u iv dx idy= − ⋅ +∫

( ) ( )C

udx vdy i udy vdx= + + −∫. . .im R H S= Γ + =

u dl⋅ = Γ

m

Potential Flows 103


( ) ( )z zC C

dW z dz W dzdzζζ= ⋅∫ ∫

z zimΓ +( )

C

W dς

ζ ζ= ∫

imζ ζ= Γ +

z zm mζ ζ∴Γ = Γ =and

The source and vortex strengths are preserved.

Potential Flows 104


( )zζ ζ=

2cz ζζ

= +

iη

ξ

planeζ −

iy

x

2caa

−

2caa

+

z plane−r a=r b=

r c=

a b c> >

where c is a real

Potential Flows 105


(i) cζ > Consider a circle with r = a iae a cθζ = >

2cz ζζ

= +2

i icae ea

θ θ−= +2 2

( ) cos ( )sinc ca i aa a

θ θ= + + −

Potential Flows 106


2

( ) coscx aa

θ∴ = +

2

( )sincy aa

θ= −

2 2

( ) cos ( )sinc cz a i aa a

θ θ= + + −

x iy= +

Potential Flows 107

Joukowski Transformation 2 2

2 22 2

2 2sin cos 1

( ) ( )

x yc ca aa a

θ θ+ = = ++ −

2caa

+Major axis = along x

2caa

−Minor axis = along y

Potential Flows 108


Special case:

2 cosx c θ= 0 2θ π≤ ≤

0y =

𝑙𝑙 = 𝑐𝑐

Potential Flows 109


2cz ζζ

= +

give the same point on the z - plane ζ2cζ

and

2cζζ

=Set

Then we have,

2 2 2

2

c c cz zc

ζζ ζ ζ

= + ⇒ = +

Potential Flows 110


Remarks on Joukowski’s Transformation

1. It is double-valued 2

, cζζ⇒

2. map to the entire z-plane. cζ ≥

same z point.

map to the entire z-plane. cζ ≤

Potential Flows 111


2cz ζζ

= +

( ) ( ) dW z Wdzζζ=

3. As (an identity mapping) ζ →∞ z →∞and

( ) ( )W z W ζ= when 1ddzζ→

Potential Flows 112


4. Singular Point: 0ζ =

5. Two critical points: when cζ = ±

U∞U∞

0dzdζ

=

Potential Flows 113

Flow Around Ellipses

iy

x

2caa

−

2caa

+

U

iη

ξ

aU

2cz ζζ

= +

planeζ − z plane−

Note: a c>

Potential Flows 114

Flow Around Ellipses 2

( ) ( )aF Uζ ζζ

= +2cz ζζ

= +

2 2 0z cζ ζ⇒ − + =

2 2( )2 2z z cζ = ± −

ζ →∞ z ζ→;

Potential Flows 115

Flow Around Ellipses

22 2

2 2

( ) ( )2 2

( )2 2

z z aF z U cz z c

∴ = + − + + −

This is the complex potential for uniform flow U over an ellipse with major axis, , and minor axis, .

2caa

+2ca

a−

Potential Flows 116

Modified Joukowski Transformation

Major axis is on the imaginary axis

Consider: 2cz ζζ

= −

Similar to the previous results: 2

( ) coscx aa

θ= −2

( )sincy aa

θ= +Corresponds to aζ =

Potential Flows 117

Modified Joukowski Transformation A vertically oriented ellipse iy

x

2caa

+

2caa

−

iη

ξ

a

Potential Flows 118

Modified Joukowski Transformation

2 2( )2 2z z cζ = + +

22 2

2 2

( ) ( )2 2

( )2 2

z z aF z U cz z c

= + + + + +

iy

x

2caa

+

2caa

−

iη

ξ

a

Potential Flows 119

Joukowski Transformation In general:

iy

xα

Potential Flows 120


Consider, iη

ξ

a'ξ'iη

α

Potential Flows 121


2' '

'( ) aF Uζ ζζ

= +

' ie αζ ζ=

2

( ) ( )i iaF U e eα αζ ζζ

−= +

Potential Flows 122


22

22

( )2 2

2 2

i iz z aF z U c e ez z c

α α−

⇒ = + − + + −

potential flows

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