058:0160
Jianming Yang
Fall 2012
Chapter 8
1
Chapter 8: Potential Flow
1 Introduction
For high Re external flow about
streamlined bodies viscous effects
are confined to boundary layer
and wake region. For regions
where the BL is thin i.e. favorable
pressure gradient regions,
Viscous/Inviscid interaction is
weak and traditional BL theory
can be used. For regions where
BL is thick and/or the flow is separated, i.e. adverse pressure gradient regions more
advanced boundary layer theory must be used including viscous/Inviscid interactions.
For internal flows at high Re viscous effects are always important except near the
entrance. Recall that vorticity is generated in regions with large shear. Therefore, outside
the B.L. and wake and if there is no upstream vorticity then 𝜔 = 0 is a good
approximation.
Note that for compressible flow this is not the case in regions of large entropy gradient.
Also, we are neglecting noninertial effects and other mechanisms of vorticity generation.
058:0160
Jianming Yang
Chapter 8
2
Fall 2012
Potential Flow Theory:
1) Determine 𝜙 from solution to Laplace equation
∇2 𝜙 = 0
B.C.:
at 𝑆𝐵 : 𝐔 ∙ 𝐧 = 0 →
𝜕𝜙
𝜕𝑛
=0
at 𝑆∞ : 𝐔 = ∇𝜙
2) Determine 𝐔 from 𝐔 = ∇𝜙 and 𝑝(𝑥) from Bernoulli equation
Therefore, primarily for external flow application we now consider inviscid flow theory
(𝜇 = 0) and incompressible flow with 𝜌 = 𝑐𝑜𝑛𝑠𝑡.
Euler Equation
∇∙𝐔=0
𝜌
𝐷𝐮
𝐷𝑡
= 𝜌𝐠 − ∇𝑝 →
𝐔 ∙ ∇𝐔 = ∇
𝜌
𝜕𝐮
𝜕𝑡
𝑈2
2
𝜌
𝜕𝐮
𝜕𝑡
+ 𝜌𝐔 ∙ ∇𝐔 = −∇(𝑝 + 𝜌𝑔𝑧)
− 𝐔 × (∇ × 𝐔) = ∇
+ ∇ (𝑝 + 𝜌
𝑈2
2
𝑈2
2
−𝐔×𝛚
+ 𝜌𝑔𝑧) = 𝜌𝐔 × 𝛚
If 𝛚 = 0, i.e., ∇ × 𝐔 = 0, then 𝐔 = ∇𝜙
058:0160
Jianming Yang
Fall 2012
𝜌
𝜕(∇𝜙)
𝜕𝜙
𝜕𝑡
1
𝜕𝑡
𝜕𝜙
∇ (𝜌
𝜌
Chapter 8
3
𝜕𝑡
+ ∇ (𝑝 + 𝜌∇𝜙 ∙ ∇𝜙 + 𝜌𝑔𝑧) = 0
2
1
) + ∇ (𝑝 + 𝜌∇𝜙 ∙ ∇𝜙 + 𝜌𝑔𝑧) = 0
2
1
+ 𝑝 + 𝜌∇𝜙 ∙ ∇𝜙 + 𝜌𝑔𝑧 = 𝐵(𝑡)
2
Bernoulli’s Equation for unsteady incompressible flow
Continuity equation shows that GDE for 𝜙 is the Laplace equation which is a 2nd order
linear PDE ie superposition principle is valid. (Linear combination of solution is also a
solution)
∇ ∙ 𝐔 = ∇ ∙ ∇𝜙 = ∇2 𝜙 = 0
𝜙 = 𝜙1 + 𝜙2
∇2 𝜙1 = 0, ∇2 𝜙2 = 0 → ∇2 𝜙1 + ∇2 𝜙2 = 0 → ∇2 (𝜙1 + 𝜙2 ) = 0 → ∇2 𝜙 = 0
Techniques for solving Laplace equation:
1) superposition of elementary solution (simple geometries)
2) surface singularity method (integral equation)
3) FD or FE
4) electrical or mechanical analogs
5) Conformal mapping ( for 2D flow)
6) Analytical for simple geometries (separation of variable etc)
058:0160
Jianming Yang
Fall 2012
2 Elementary Plane-Flow Solutions
Recall that for 2D we can define a stream function such that:
u  y
v   x
 z  vx  u y 


( x )  ( y )   2  0
x
y
i.e.  2  0
Also recall that  and  are
orthogonal.
u   y  x
v   x   y
d   x dx   y dy  udx  vdy
d   x dx   y dy  vdx  udy
i.e.
dy
u
1
 
dx  const
v dy
dx  const
Chapter 8
4
058:0160
Jianming Yang
Chapter 8
5
Fall 2012
Uniform stream
u  U    y   x  const
v  0   x   y
  U x
  U y
Note:  2   2  0 is satisfied.
i.e.
𝐔 = ∇𝜙 = 𝑈∞ 𝐢
Say a uniform stream is at an angle  to the x-axis:
𝑢 = 𝑈∞ cos 𝛼 =
𝜕𝛹
𝜕𝑦
=
𝜕𝜙
𝜕𝑥
𝑣 = 𝑈∞ sin 𝛼 = −
𝜕𝛹
𝜕𝑥
=
𝜕𝜙
𝜕𝑦
After integration, we obtain the following expressions for the stream function and
velocity potential:
𝛹 = 𝑈∞ (𝑦 cos 𝛼 − 𝑥 sin 𝛼 )
𝛷 = 𝑈∞ (𝑥 cos 𝛼 + 𝑦 sin 𝛼 )
058:0160
Jianming Yang
Chapter 8
6
Fall 2012
2D Source or Sink:
Imagine that fluid comes out radially at origin with uniform rate in all directions.
(singularity at origin where velocity is infinite)
Consider a circle of radius 𝑟 enclosing this source. Let 𝑣𝑟 be the radial component of
velocity associated with this source (or sink). Then, form conservation of mass, for a
cylinder of radius 𝑟, and unit width perpendicular to the paper,
𝑄 = ∫𝐴 𝐔 ∙ 𝑑𝐀 = (2𝜋𝑟)(𝑏)𝑣𝑟
or
𝑣𝑟 =
 vr 
𝑄
2𝜋𝑏𝑟
m
, v  0
r
Where: 𝑚 =
𝑄
2𝜋𝑏
is the convenient constant with unit velocity × length
(𝑚 > 0 for source and 𝑚 < 0 for sink). Note that 𝐔 is singular at (0,0) since 𝑣𝑟 → ∞
In a polar coordinate system, for 2-D flows we will use:
𝐔 = ∇𝜙 =
𝜕𝜙
𝜕𝑟
+
1 𝜕𝜙
𝑟 𝜕𝜃
And:
∇ ∙ 𝐔 = 0 or
1 𝜕
𝑟 𝜕𝑟
(𝑟𝑣𝑟 ) +
1 𝜕
𝑟 𝜕𝜃
(𝑣𝜃 ) = 0
058:0160
Jianming Yang
Chapter 8
7
Fall 2012
i.e.:
 1 

r r 
1 

v  Tangential velocity 

r 
r
Such that ∇ ∙ 𝐔 = 0 by definition.
Therefore,
m  1 
vr  

r r r 
1 

v  0 

r 
r
i.e.
  m ln r  m ln x 2  y 2
v r  Radial velocity 
  m  m tan 1
y
x
058:0160
Jianming Yang
Chapter 8
8
Fall 2012
Doublets:
The doublet is defined as:
ψ  m(1   2 )  1   2  

m
tan θ  tan θ

1
2
tan (θ  θ )  tan (  ) 
1
2
m
1  tan θ tan θ
1
2
sink
source
sink
source
r sin 
r sin 
; tan  2 
r cos   a
r cos   a
r sin 
r sin 


2ar sin 
tan( )  r cos   a r cos   a  2
r sin 
m 1  r sin 
r  a2
r cos   a r cos   a
tan 1 
For small value of a

a 0

  lim  m tan 1 (
2ar sin  
2amy
y
)




r 2  a 2 
r2
x2  y 2
  2am ( Doublet Strength)

 cos 
r
058:0160
Jianming Yang
Fall 2012
Chapter 8
9
By rearranging:
y
 2
 2
2
ψ 2

x

(
y

)

(
)
2
x y
2
2

and center at (0, +/-R) i.e. circles
2
are tangent to the origin with center on y axis. The flow direction is from the source to
the sink.
It means that streamlines are circles with radius R 
058:0160
Jianming Yang
Chapter 8
10
Fall 2012
2D vortex:
Suppose that value of the 𝛹 and 𝜙 for the source are reversal.
𝑣𝑟 = 0
𝑣𝜃 =
1 𝜕𝜙
𝑟 𝜕𝜃
=−
𝜕𝛹
𝜕𝑟
=
𝐾
𝑟
Purely circulatory flow with 𝑣𝜃 → 0 like 1/𝑟. And
𝜙 = 𝐾𝜃
𝛹 = −𝐾 ln 𝑟
2D vortex is irrotational everywhere except at the origin where 𝐔 and 𝐔 × ∇ are infinity.
Circulation is defined by 𝛤 = ∮𝐶 𝐮 ∙ 𝑑𝐬. For potential flow 𝛤 = 0 in general.
However, this is not true for the point vortex due to the singular point at vortex core
where 𝐔 and 𝐔 × ∇ are infinity.
If singularity exists: Free vortex 𝑣𝜃 = 𝐾⁄𝑟
2𝜋
2𝜋 𝐾
𝛤 = ∮0 𝑣𝜃 𝐞̂𝜃 ∙ 𝑟𝑑𝜃𝐞̂𝜃 = ∮0
𝑟
𝑟𝑑𝜃 = 2𝜋𝐾
𝐾=
𝛤
2𝜋
Note: for point vortex, flow still irrotational everywhere except at origin itself where 𝐔 →
∞, i.e., for a path not including (0,0) 𝛤 = 0
The point vortex singularity is important in aerodynamics, since, extra source and sink
can be used to represent airfoils and wings.
058:0160
Jianming Yang
Chapter 8
11
Fall 2012
An infinite row of vortices:
1
1
2𝜋𝑦
𝛹 = −𝐾 ∑∞
𝑖=1 ln 𝑟𝑖 = − 2 𝐾 ln [2 (cosh 𝑎 − cos
Where 𝑟𝑖 is radius from origin of ith vortex.
2𝜋𝑥
𝑎
)]
For |𝑦| ≫ 𝑎 the flow approaches uniform flow with
𝑢=
𝜕𝛹
𝜕𝑦
=±
𝜋𝐾
𝑎
(+: below x axis; −: above x axis)
Note: this flow is due to infinite row of vortices and
there isn’t any pure uniform flow
Vortex sheet:
From afar (i.e. |𝑦| ≫ 𝑎) looks that thin sheet occurs which is a velocity discontinuity.
Define the strength of vortex sheet 𝛾 =
2𝜋𝐾
𝑎
2𝜋𝐾
𝑑𝛤 = 𝑢𝑙 𝑑𝑥 − 𝑢𝑢 𝑑𝑥 = (𝑢𝑙 − 𝑢𝑢 )𝑑𝑥 =
𝑑𝑥
𝑎
𝑑𝛤
i.e. 𝛾 = = Circulation per unit span
𝑑𝑥
Note: There is no flow normal to the sheet so that
vortex sheet can be used to simulate a body surface. This is the basis of airfoil theory
where we let 𝛾 = 𝛾(𝑥 ) to represent body geometry.
058:0160
Jianming Yang
Chapter 8
12
Fall 2012
3 Potential Flow Solutions for Simple Geometries
Circular cylinder (without rotation):
In the previous we derived the following equation for the doublet:
𝛹=−
𝜆𝑦
𝑥 2 +𝑦 2
=−
𝜆 sin 𝜃
𝑟
When this doublet is superposed with a uniform flow
parallel to the x- axis, we get:
𝛹 = 𝑈∞ 𝑟 sin 𝜃 −
𝜆 sin 𝜃
𝑟
= 𝑈∞ (1 −
𝜆 1
𝑈∞ 𝑟 2
) 𝑟 sin 𝜃
Where: 𝜆 is the doublet strength which is determined from the kinematic body boundary
condition that the body surface must be a stream surface. Recall that for inviscid flow it is
no longer possible to satisfy the no slip condition as a result of the neglect of viscous
terms in PDEs.
The inviscid flow boundary condition is: 𝑟 = 𝑅, 𝐔 ∙ 𝐧 = 0, i.e. 𝑣𝑟 (𝑟 = 𝑅) = 0.
𝑣𝑟 = 𝐔 ∙ 𝐧 =
1 𝜕𝛹
𝑟 𝜕𝜃
= 𝑈∞ (1 −
𝜆 1
𝑈∞
𝑟2
) cos 𝜃 = 0
→ 𝜆 = 𝑈∞ 𝑅2
If we replace the constant 𝜆⁄𝑈∞ by a new constant 𝑅2 , the above equation becomes:
𝑅2
1 𝜕𝛹
𝑅2
𝛹 = 𝑈∞ (1 − 2 ) 𝑟 sin 𝜃 and 𝑣𝑟 =
= 𝑈∞ (1 − 2 ) cos 𝜃
𝑟
𝑟 𝜕𝜃
𝑟
This radial velocity is zero on all points on the circle 𝑟 = 𝑅. That is, there can be no
velocity normal to the circle 𝑟 = 𝑅. Thus this circle itself is a streamline.
058:0160
Jianming Yang
Chapter 8
13
Fall 2012
We can also compute the tangential component of velocity for flow over the circular
cylinder. From equation,
𝑣𝜃 = −
1 𝜕𝛹
𝑟 𝜕𝑟
= −𝑈∞ (1 +
𝑅2
𝑟2
) sin 𝜃
On the surface of the cylinder 𝑟 = 𝑅, the tangential and
radial components of velocity are:
𝑣𝜃 = −2𝑈∞ sin 𝜃
𝑣𝑟 = 0
How about pressure 𝑝? look at the Bernoulli's equation:
𝑝
𝜌
1
𝑝∞
2
𝜌
+ (𝑣𝑟2 + 𝑣𝜃2 ) =
1
2
+ 𝑈∞
2
We can get the following non-dimensional form:
𝐶𝑝 (𝑟, 𝜃) =
𝑝−𝑝∞
1
2
𝜌𝑈∞
2
=1−
𝑣𝑟2 +𝑣𝜃2
2
𝑈∞
For the circular cylinder being studied here, at the surface, the only velocity component
that is non-zero is the tangential component of velocity. Using 𝑣𝜃 = −2𝑈∞ sin 𝜃, we get
at the cylinder surface the following expression for the pressure coefficient:
𝐶𝑝 = 1 − 4 sin2 𝜃
Where 𝜃 is the angle measured from the rear stagnation point (at the intersection of the
back end of the cylinder with the x- axis).
058:0160
Jianming Yang
Chapter 8
14
Fall 2012
From pressure coefficient we can calculate the fluid force on the cylinder:
1
2
𝐅 = − ∫𝐴 (𝑝 − 𝑝∞ )𝐧𝑑𝑠 = − 𝜌𝑈∞
∫𝐴 𝐶𝑝 (𝑅, 𝜃)𝐧𝑑𝑠
2
𝑑𝑠 = (𝑅𝑑𝜃)𝑏 (𝑏 =span length)
2𝜋
1
2
𝐅 = − 𝜌𝑈∞
𝑏𝑅 ∫0 (1 − 4 sin2 𝜃)(cos 𝜃 𝐢 + sin 𝜃 𝐣)𝑑𝜃
2
𝐶𝐿 = 1
2
Lift
2 2𝑏𝑅
𝜌𝑈∞
𝐶𝐷 = 1
2
Drag
2 2𝑏𝑅
𝜌𝑈∞
𝐅∙𝐣
=1
2 2𝑏𝑅
𝜌𝑈∞
2
=1
𝐅∙𝐢
2 2𝑏𝑅
𝜌𝑈∞
2
1
2𝜋
1
2𝜋
= − ∫0 (1 − 4 sin2 𝜃) sin 𝜃 𝑑𝜃 = 0
2
= − ∫0 (1 − 4 sin2 𝜃) cos 𝜃 𝑑𝜃 = 0 (dÁlembert paradox)
2
𝐶𝐿 = 𝐶𝐷 = 0 is due to symmetry of 𝐶𝑝 . The viscous effects were confined to thin
boundary layer and it has negligible effect on the inviscid flow for high Re flow about
streamlined bodies. A circular cylinder is not a streamlined body (like as airfoil) but
rather a bluff body. Thus, as we shall determine now the potential flow solution is not
realistic for the back portion of the circular cylinder flows.
For general cases:
𝐶𝐷 = 𝐶𝐷 form drag + 𝐶𝐷 skin friction
𝐶𝐷 form drag :
drag due to viscous modification of pressure distribution
𝐶𝐷 skin friction :
Drag due to viscous shear stress
For stramlined bodies, 𝐶𝐷 skin friction > 𝐶𝐷 form drag .
For bluff bodies, 𝐶𝐷 form drag > 𝐶𝐷 skin friction
058:0160
Jianming Yang
Chapter 8
15
Fall 2012
Circular cylinder with circulation:
The stream function associated with the flow over a circular cylinder, with a point vortex
of strength 𝛤 placed at the cylinder center is:
𝛹 = 𝑟𝑈∞ sin 𝜃 −
𝑅2
𝑟
𝑈∞ sin 𝜃 −
Γ
2𝜋
ln 𝑟
The radial and tangential velocity is given by:
𝑣𝑟 =
1 𝜕𝛹
𝑟 𝜕𝜃
= 𝑈∞ (1 −
𝑅2
𝑟
2 ) cos 𝜃
𝑣𝜃 = −
1 𝜕𝛹
𝑟 𝜕𝑟
= −𝑈∞ (1 +
𝑅2
Γ
𝑟
2𝜋𝑟
2 ) sin 𝜃 +
On the surface of the cylinder (𝑟 = 𝑅):
𝑣𝑟 =
1 𝜕𝛹
𝑟 𝜕𝜃
= 𝑈∞ (1 −
𝑅2
𝑅
2 ) cos 𝜃 = 0
𝑣𝜃 = −
1 𝜕𝛹
𝑟 𝜕𝑟
= −2𝑈∞ sin 𝜃 +
Γ
2𝜋𝑅
2𝜋
𝛤 = ∮𝐶 𝐮 ∙ 𝑑𝐬 = ∮0 𝑣𝜃 𝑅𝑑𝜃, i.e., vortex strength is circulation
Next, consider the flow pattern as a function of 𝛤. To start lets calculate the stagnation
points on the cylinder i.e.:
𝑣𝜃 = −2𝑈∞ sin 𝜃 +
Note: 𝐾 =
𝛤
2𝜋
,𝛽=
Γ
2𝜋𝑅
=0
→
sin 𝜃 =
𝐾
𝑈∞ 𝑅
So, the location of stagnation point is function of 𝛤.
Γ
4𝜋𝑈∞ 𝑅
=
𝐾
2𝑈∞ 𝑅
=
𝛽
2
058:0160
Jianming Yang
𝛽=
Chapter 8
16
Fall 2012
𝐾
𝑈∞ 𝑅
=
𝛤
2𝜋𝑈∞ 𝑅
0 (sin 𝜃 = 0)
1(sin 𝜃 = 0.5)
2 (sin 𝜃 = 1)
>2(sin 𝜃 > 1)
𝜃𝑠 (stagnation point)
0,180
30,150
90
Is not on the circle but where 𝑣𝜃 = 𝑣𝑟 = 0
For flow patterns above (except a), we should expect to have lift force in – 𝑦 direction.
058:0160
Jianming Yang
Chapter 8
17
Fall 2012
Summary of stream and potential function of elementary 2-D flows:
In Cartesian coordinates:
𝑢=
𝜕𝛹
𝜕𝑦
=
𝜕𝜙
𝜕𝑥
𝑣=−
𝜕𝛹
𝜕𝑥
=
𝜕𝜙
𝜕𝑦
In polar coordinates:
𝑣𝑟 =
𝜕𝜙
𝜕𝑟
=
1 𝜕𝛹
𝑟 𝜕𝜃
𝑣𝜃 =
1 𝜕𝜙
𝑟 𝜕𝜃
=−
𝜕𝛹
𝜕𝑟
Flow
Uniform Flow
Source (𝑚 > 0) or Sink (𝑚 < 0)
Doublet
𝜙
𝑈∞ 𝑥
𝑚 ln 𝑟
λ cos 𝜃
𝑟
Vortex
90 Corner flow
𝐾𝜃
1
2
2)
(
𝐴
𝑥
−
𝑦
2
Solid-Body rotation
Doesn’t exist
𝛹
𝑈∞ 𝑦
𝑚𝜃
λ sin 𝜃
−
𝑟
−𝐾 ln 𝑟
𝐴𝑥𝑦
𝜔𝑟 2
1
2
These elementary solutions can be combined in such a way that the resulting solution can
be interpreted to have physical significance; that is, represent the potential flow solution
for various geometries. Also, methods for arbitrary geometries combine uniform stream
with distribution of the elementary solution on the body surface.
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Some combinations of elementary solutions give body geometries of practical importance
Body name
Elemental combination
Flow Patterns
Rankine Half Body
Uniform stream+source
Rankine Oval
Uniform stream+source+sink
Kelvin Oval
Uniform stream+vortex pair
Circular Cylinder without Uniform stream+doublet
circulation
Circular Cylinder with
circulation
Uniform stream+doublet+vortex
Keep in mind that this is the potential flow solution and may not well represent the real
flow especially in region of adverse pressure gradient.
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The Kutta – Joukowski lift theorem:
Since we know the tangential component of velocity at any point on the cylinder (and the
radial component of velocity is zero), we can find the pressure field over the surface of
the cylinder from Bernoulli’s equation:
𝑝
1
𝑝
1 2
+ (𝑣𝑟2 + 𝑣𝜃2 ) = ∞ + 𝑈∞
𝜌
Therefore: 𝑝 = 𝑝∞ +
=
1
2
(𝑝∞ + 𝜌𝑈∞
2
1
2
𝜌[𝑈∞
2
−
𝜌Γ2
8𝜋2 𝑅
1
−
2
(𝑣𝑟2
𝜌
+ 𝑣𝜃2 )]
= 𝑝∞ +
2
2
2 ) − 2𝜌𝑈∞ sin 𝜃 +
𝜌Γ2
2
1
2
𝜌 [𝑈∞
2
𝜌𝑈∞ Γ
𝜋𝑅
𝜌𝑈 Γ
2
And
𝐴 = (𝑝∞ + 𝜌𝑈∞
− 2 2) 𝐵 = ∞
2
8𝜋 𝑅
𝜋𝑅
Calculation of Lift:
2𝜋
𝐿 = −𝑅𝑏 ∫0 (𝐴 + 𝐵 sin 𝜃 + 𝐶sin2 𝜃) sin 𝜃 𝑑𝜃
− (−2𝑈∞ sin 𝜃 +
Γ
2𝜋𝑅
2
) ]
sin 𝜃 = 𝐴 + 𝐵 sin 𝜃 + 𝐶sin2 𝜃
2
𝐶 = −2𝜌𝑈∞
2𝜋
= −𝑅𝑏 ∫0 (𝐴 sin 𝜃 + 𝐵 sin2 𝜃 + 𝐶sin3 𝜃)𝑑𝜃
=
2𝜋
−𝑅𝑏 ∫0 𝐵sin2 𝜃𝑑𝜃
𝜃
1
2𝜋
2
4
0
= −𝐵𝑅𝑏 ( − sin 2𝜃)|
= −𝐵𝑅𝑏𝜋 = −𝜌𝑈∞ Γ𝑏
This is an important result. It says that clockwise vortices (negative numerical values of
Γ) will produce positive lift that is proportional to Γ and the free stream speed with
direction 90 degrees from the stream direction rotating opposite to the circulation. Kutta
and Joukowski generalized this result to lifting flow over airfoils. Equation 𝐿 = −𝜌𝑈∞ Γ𝑏
is known as the Kutta-Joukowski theorem.
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2𝜋
Drag: 𝐿 = −𝑅𝑏 ∫0 (𝐴 + 𝐵 sin 𝜃 + 𝐶sin2 𝜃) cos 𝜃 𝑑𝜃
2𝜋
= −𝑅𝑏 ∫0 (𝐴 cos 𝜃 + 𝐵 sin 𝜃 cos 𝜃 + 𝐶sin2 𝜃 cos 𝜃)𝑑𝜃
2𝜋
= −𝑅𝑏(𝐴 sin 𝜃 + 12𝐵 cos2 𝜃 + 13𝐶sin3 𝜃)| = 0
0
This contrast between potential flow theory and realistic drag is the dÁlembert Paradox,
which was solved by Prandtl (1904) with his boundary layer theory, i.e. viscous effects
are always important very close to the body where the no slip boundary condition must
be satisfied and large shear stress exists which contributes the drag.
Lift for rotating Cylinder:
We know that 𝐿 = −𝜌𝑈∞ Γ𝑏, so the lift coefficient is
𝐶𝐿 = 1
2
𝐿
2 2𝑏𝑅
𝜌𝑈∞
=−
𝛤
𝑈∞ 𝑅
Experiments have been performed that simulate the previous
flow by rotating a circular cylinder in a uniform stream. In this
case 𝑣𝜃 = 𝑅𝜔 which is due to no slip boundary condition.
- Lift is quite high but not as large as theory (due to viscous
effect ie flow separation)
- Note drag force is also fairy high
Velocity ratio 𝑎𝜔⁄𝑈∞
Flettner (1924) used rotating cylinder to produce forward motion.
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4 Method of Images
The method of image is used to model “slip” wall effects by constructing appropriate
image singularity distributions.
2D Plane Boundaries:
m
m
2
2
Q  ln x 2   y  a   ln x 2   y  1
2
2




Similar results can be obtained for
dipoles and vortices
Multiple Boundaries:
The method can be extended for multiple boundaries by
using successive images.
airfoil in ground effect with image airfoil of opposite circulation
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5 Complex variable and conformal mapping
This method provides a very powerful method for solving 2-D flow problems. Although
the method can be extended for arbitrary geometries, other techniques are equally useful.
Thus, the greatest application is for getting simple flow geometries for which it provides
closed form analytic solution which provides basic solutions and can be used to validate
numerical methods.
Function of a complex variable
Conformal mapping relies entirely on complex mathematics. Therefore, a brief review is
undertaken at this point.
A complex number 𝑧 is a sum of a real and imaginary
part; 𝑧 = real + 𝑖 imaginary
𝑖 = √−1
The term 𝑖, refers to the complex number
𝑖 = √−1,
𝑖 2 = −1, 𝑖 3 = −𝑖, 𝑖 4 = 1
Complex numbers can be presented in a graphical
format. If the real portion of a complex number is
taken as the abscissa, and the imaginary portion as the
ordinate, a two-dimensional plane is formed.
𝑧 = real + 𝑖 imaginary = 𝑥 + 𝑖 𝑦
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- A complex number can be written in polar form using Euler's equation;
𝑧 = 𝑥 + 𝑖 𝑦 = 𝑟𝑒 𝑖𝜃 = 𝑟(cos 𝜃 + 𝑖 sin 𝜃)
Where:
𝑟2 = 𝑥2 + 𝑦2
- Complex multiplication:
𝑧1 𝑧2 = (𝑥1 + 𝑖 𝑦1 )(𝑥2 + 𝑖 𝑦2 ) = (𝑥1 𝑥2 − 𝑦1 𝑦2 ) + 𝑖 (𝑥1 𝑦2 + 𝑦1 𝑥2 ) = 𝑟1 𝑒 𝑖𝜃1 ∙ 𝑟2 𝑒 𝑖𝜃2
= 𝑟1 𝑟2 𝑒 𝑖(𝜃1+𝜃2)
- Conjugate: 𝑧 = 𝑥 + 𝑖 𝑦 𝑧̅ = 𝑥 − 𝑖 𝑦 𝑧𝑧̅ = 𝑥 2 + 𝑦 2
- Complex function: 𝑤(𝑧) = 𝑓(𝑧) = (𝑥, 𝑦) + 𝑖 (𝑥, 𝑦)
The complex function 𝑤(𝑧) is analytic and has a unique derivative 𝑑𝑤/𝑑𝑧 independent
of the direction of differentiation within the complex z-plane. This condition leads to the
Cauchy-Riemann conditions:
𝜕𝜙
𝜕𝑥
=
𝜕𝛹
𝜕𝑦
and
𝜕𝜙
𝜕𝑦
=−
𝜕𝛹
𝜕𝑥
Recall that the velocity potential and stream function were shown to satisfy this
relationship as a result of their othogonality. Thus, complex function 𝑤 =  + 𝑖
represents 2-D flows.
𝜕2 𝜙
𝜕𝑥 2
=
𝜕2 𝛹 𝜕2 𝜙
,
𝜕𝑦𝜕𝑥 𝜕𝑦 2
=−
𝜕2 𝛹
𝜕𝑥𝜕𝑦
, i.e.
𝜕2 𝜙
𝜕𝑥 2
+
𝜕2 𝜙
𝜕𝑦 2
= 0 and similarly for 𝛹. Therefore if analytic
and regular also harmonic, i.e., satisfy Laplace equation.
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Application to potential flow
𝑤(𝑧) =  + 𝑖 Complex potential
where : velocity potential, : stream function
𝑑𝑤
𝑑𝑧
=
𝜕𝜙
𝜕𝑥
+𝑖
𝜕𝛹
𝜕𝑥
= 𝑢 − 𝑖 𝑣 = (𝑢𝑟 − 𝑖 𝑢𝜃 )𝑒 −𝑖𝜃 Complex velocity
Chapter 8
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Implication:
-Angles are preserved between the intersections of any two lines in the physical
domain and in the mapped domain.
-The mapping is one-to-one, so that to each point in the physical domain, there is one
and only one corresponding point in the mapped domain.
For these reasons, such transformations are called conformal.
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Conformal mapping
The real power of the use of complex variables for flow analysis is through the
application of conformal mapping: techniques whereby a complicated geometry in the
physical 𝑧-domain is mapped onto a simple geometry in the 𝜁-plane (circular cylinder)
for which the flow-field solution is known. The flow-field solution in the 𝑧-plane is
obtained by relating the 𝜁 -plane solution to the 𝑧 -plane through the conformal
transformation 𝜁 = 𝑓(𝑧) (or inverse mapping 𝑧 = 𝑔(𝜁)).
Transformation of a circle of radius a in the z-plane into an ellipse in the z-plane by
means of the Zhukhovsky transformation.
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The streamlines and equipotential lines of the ζ-plane become the streamlines of
equipotential lines of the z-plane.
Laplace equation in the z-plane transforms into Laplace equation is the ζ-plane.
Velocities in two planes are proportional.
Two independent theorems concerning conformal transformations are:
(1)
Closed curves map to closed curves
(2)
Rieman mapping theorem: an arbitrary closed profile can be mapped onto the
unit circle.
Many transformations have been investigated and are compiled in handbooks. Some
elementary transformations:
a) linear: 𝑤 =
𝑎𝑧+𝑏
𝑐𝑧+𝑑
, 𝑎𝑑 − 𝑏𝑐 ≠ 0
b) corner flow: 𝑤 = 𝐴𝑧 𝑛
c) Joukowsky: 𝑤 = 𝜁 + 𝑐 2 ⁄𝜁
d) exponential: 𝑤 = 𝑒 𝑛
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