potential flows
DESCRIPTION
potential flows fluid mechanicsTRANSCRIPT
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
Property of Stream Function
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
Property of Stream Function
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
Property of Stream Function
Cψ =
0d dx dyx yψ ψψ ∂ ∂
= = +∂ ∂
Streamlines:
dy vdx uψ
=
v− u
Potential Flows 16
Property of Stream Function
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
Superposition of Two Sources
( ) 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
Superposition of Two Sources
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
Superposition of Two Sources
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
( ) ln( ) ln( )2 2m mF z z a z aπ π
= + − −
Potential Flows 39
Two-Dimensional Dipole
-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
Two-Dimensional Dipole
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
Two-Dimensional Dipole
( ) ln( ) ln( )2 2m mF z z a z aπ π
= + − −
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
Two-Dimensional Dipole
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
Two-Dimensional Dipole
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
Two-Dimensional Dipole
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
Uniform Flow about a Circular Cylinder
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
Uniform Flow about a Circular Cylinder
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
Uniform Flow about a Circular Cylinder
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
Uniform Flow about a Circular Cylinder
( )F z real iφ ψ= = +
0ψ⇒ =
1z = (unit circle) is a streamline
( )( ) ln2mF zπ
= real quantity
Therefore,
Potential Flows 54
Uniform Flow about a Circular Cylinder
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
Uniform Flow about a Circular Cylinder
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
Uniform Flow about a Circular Cylinder
1( )F z U zz
∴ = +
Far away: 1z >>
( )F z Uz→ An uniform horizontal flow (from left to right)
Potential Flows 57
Uniform Flow about a Circular Cylinder
Near Field:
( ) UF zz
→ (a dipole)
1( )F z U zz
= +
1i iU re er
θ θ− = +
Potential Flows 58
Uniform Flow about a Circular Cylinder
( ) ( )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
Uniform Flow about a Circular Cylinder
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
Uniform Flow about a Circular Cylinder
ab− 2Rb
−2R
a
m m m− m−| |z R=
Remarks:
i) Extension to radius = R
2
( ) RF z U zz
= +
Potential Flows 61
Uniform Flow about a Circular Cylinder
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
Uniform Flow about a Circular Cylinder
( ) 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
Uniform Flow about a Circular Cylinder
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
Hydrodynamic Forces on a Cylinder in a steady 2-D Uniform Flow
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
Hydrodynamic Forces on a Cylinder in a steady 2-D Uniform Flow
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
Hydrodynamic Forces on a Cylinder in a steady 2-D Uniform Flow
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
Hydrodynamic Forces on a Cylinder in a steady 2-D Uniform Flow
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
Hydrodynamic Forces on a Cylinder in a steady 2-D Uniform Flow
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
Hydrodynamic Forces on a Cylinder in a steady 2-D Uniform Flow
0
212
p p
Uρ∞
−
2π π θ
A
B,D
C A C
B
D
Potential Flows 71
Hydrodynamic Forces on a Cylinder in a steady 2-D Uniform Flow
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
Hydrodynamic Forces on a Cylinder in a steady 2-D Uniform Flow
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
Hydrodynamic Forces on a Cylinder in a steady 2-D Uniform Flow
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
Flow about a Circular Cylinder with Circulation
( ) (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
Flow about a Circular Cylinder with Circulation
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
Flow about a Circular Cylinder with Circulation
( ) 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
Flow about a Circular Cylinder with Circulation
2
2(1 )cosrRu Ur
θ⇒ = −
2
2(1 )sin2
Ru Ur rθ θ
πΓ
= − + −
and
Potential Flows 79
Flow about a Circular Cylinder with Circulation
On r = R:
0ru =and
(1 1)sin2
u URθ θ
πΓ
= − + −
2 sin2
UR
θπΓ
= − −
Potential Flows 80
Flow about a Circular Cylinder with Circulation
At the stagnation point, 0ru uθ= =
0 2 sin2
u URθ θ
πΓ
= = − −
sin4 UR
θπΓ
= −
Potential Flows 81
Flow about a Circular Cylinder with Circulation
(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
Blasius’ Theorem (I)
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
Blasius’ Theorem (I)
0
2
2x yC
iF iF W dzρ− = ∫
1(2 ) ( )2i i b residuesρ π= ∑
Singularities:
1i UbπΓ
⇒ =
0z =
Potential Flows 86
Blasius’ Theorem (I)
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
Blasius’ Theorem (I) - Proof
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
Blasius’ Theorem (I) - Proof
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
Blasius’ Theorem (I) - Proof
1 2 2dy dx dsλ− = + =
dyi dxjnds−
∴ =
m u ndsρ= ⋅∫
( )( ) dyi dxjui vj ds
dsρ −
= + ⋅ ⋅∫( )udy vdxρ= −∫
Potential Flows 91
Blasius’ Theorem (I) - Proof
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
Blasius’ Theorem (I) - Proof
1 '2
Pu u Bρ
⋅ + =
'12
P u u Bρ= − ⋅ +
12
dPu u G Bρ
⋅ + − =∫
Potential Flows 93
Blasius’ Theorem (I) - Proof
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
Blasius’ Theorem (I) - Proof
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
Blasius’ Theorem (I) - Proof
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
Conformal Transformation
z x iy= + iζ ξ η= +2zζ =( )f zζ =
iy
x
iη
ξ
z plane planeζ
Potential Flows 99
Conformal Transformation
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
Conformal Transformation
(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
Conformal Transformation
(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
Conformal Transformation
(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
Conformal Transformation
( ) ( )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
Joukowski Transformation
( )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
Joukowski Transformation
(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
Joukowski Transformation
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
Joukowski Transformation
Special case:
2 cosx c θ= 0 2θ π≤ ≤
0y =
𝑙𝑙 = 𝑐𝑐
Potential Flows 109
Joukowski Transformation
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
Joukowski Transformation
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
Joukowski Transformation
2cz ζζ
= +
( ) ( ) dW z Wdzζζ=
3. As (an identity mapping) ζ →∞ z →∞and
( ) ( )W z W ζ= when 1ddzζ→
Potential Flows 112
Joukowski Transformation
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
Joukowski Transformation
Consider, iη
ξ
a'ξ'iη
α
Potential Flows 121
Joukowski Transformation
2' '
'( ) aF Uζ ζζ
= +
' ie αζ ζ=
2
( ) ( )i iaF U e eα αζ ζζ
−= +
Potential Flows 122
Joukowski Transformation
22
22
( )2 2
2 2
i iz z aF z U c e ez z c
α α−
⇒ = + − + + −