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Lecture 15
Final Version
Contents
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Solutions to Laplace Equation
Line Source/Sink
Line Vortex
The Circulation
Combinations of Solutions: Solid
Bodies in a Potential Flow
(Rankine Oval etc.)
What Did We Do In Last Lecture?
Flow Around a Corner of Arbitrary Angle
zxiy
F  z   R x, y   i I  x, y 
One can interpret real part, R(x,y), of complex
function, F(z), as velocity potential and imaginary
part, I(x,y), as stream function of a two-dimensional
flow.
Derivative, F’=u-iv, corresponds to complex
conjugate of velocity, u+iv.
dF
 F '  u  x, y   i v  x, y 
dz
SOLUTIONS TO LAPLACE EQUATION
• We learnt that in inviscid, irrotational and incompressible
flow
Laplace equation determines flow field. For
instance, in 2D all that is required is to solve:
 2
x 2

 2
y 2
(1)
0
subject to boundary conditions (usually no flow through a
solid boundary) obtained when a body (e.g. an aerofoil) is
placed in a flow which is uniform infinitely far upstream..
• To satisfy boundary conditions, one often constructs a
composite solution. Note that if 1 and 2 are solutions
to Eq. (1) then function 1   2  is also a solution since:
2
x
2
1  2  

2
y
2
1  2 
 21
x
2

 21
y
2

 22
x
2

 22
y
2


0
0
0
(2)
Continued …
• In general then, if i where i  1, 2, 3 are each solutions
to Laplace equation, so must be
total  A1 1  A2 2  A3 3  
where Ai i  1, 2, 3 are constants.
• This process of constructing a solution is known as
LINEAR SUPERPOSITION.
Functions i , are usually well-known solutions to Laplace
equation and are ‘building bricks’ of a composite solution.
We construct a solution according to Eq. (2), finding
necessary constants Ai such that boundary conditions are
satisfied.
Some useful solutions to the Laplace Equation
as examples
• We now identify some useful solutions to Laplace
equation. Remember that for a potential function to be
a solution to Laplace equation velocity must satisfy
conditions of zero vorticity and imcompressibility (the
mass-continuity equation).
• If fluid is incompressible and because we are dealing
with two-dimensional flows, then there must also exist a
stream function. Below we will identify both the velocity
potential,   x, y  , and the stream function,   x, y  , for
particular flows.
Example: Uniform Flow (Revision, see earlier examples!)
Paralell to the x-axis….
Need to project free stream
velocity onto radial and
circumferential
unit
vector!
Minus sign mathematically
negative
sense,
i.e.
clockwise
U   U  sin 
U r  U  cos
In cartesians ... V  x, y   U  i  0 j
In polar coordinates ...
V r ,   U  cos ir  U  sin  i
• Have already seen in previous example that this flow has zero
vorticity and that it satisfies the mass-continuity equation.
• We found that velocity potential and stream function are...
Velocity Potential
In cartesians ...
Stream Function
  x, y   U  x  C1   x, y   U  y  C2
In polar coordinates ...
  x, y   U  r cos  C1
  x, y   U  r sin   C2
Example: Line Source/Sink
• Suppose z-axis were a sort of thin-pipe manifold through which issued
fluid at a total rate Q uniformly along its length b. Looking at xy plane,
we would see a cylindrical outflow or line source as sketched below. If
flow is perfectly uniform then flow speed is entirely in radial direction,
i.e. there is no circumferential velocity.
Questions and Solutions:
(Q1) Determine radial flow velocity ur as function of the radial
coordinate r.
(A1) Apply mass-continuity to a cylindrical control volume of radius r
which has source as its centre line. Surface area of control
volume through which fluid has to pass is...
ACV  2 r b
Since, ...
Area x Flow Velocity = Volume Flow Rate Through Area
one gets...
ur 
m
Q
2 b
Q
Q

ACV 2 r b

m
r
(1)
Will need this
repeatedly. Keep
this result in mind!
: Source Strength; Volumetric flow rate (Q)
per unit length and divided by
2
Continued...
(Q2) Is this an irrotational flow? Think about
deformation of fluid elements as they move
away from the origin ...
(A2) Need to verify that vorticity is zero. In polars,
vorticity is ...
 
1 r u  1 ur

r r
r 
(2)
Substituting in values for radial and azimuthal
velocity gives, ...
 
m
1 r  0 1  r
 

0
r r
r 


 
0
0
Continued...
(Q3) Is mass-continuity satisfied?
(A3) Evidently it will be satisfied from our derivation of
radial and circumferential velocity. However, to
show this formally, substitute these into masscontinuity equation. In polars mass-continuity is ...
1  r ur  1 u ?

0
r r
r 
(3)
Note symmetry / similarity of Eq.
(2) and Eq. (3) in co- ordinates!
Substituting in values for radial and azimuthal
velocity gives, ...
 m
 r  
1
r  1 0
  

0
r r
r 




0
0
CONCLUDE: Because irrotational and mass continuity
satisfied Laplace equation must be satisfied ...
… everywhere in plane EXCEPT at origin, location of
source, where radial flow speed, is infinite- Eq. (1)
previous slide.
Note: A sink is opposite of source - fluid is sucked into
‘hole’. Same formulation as for source is used except that
source strength becomes -m
Continued...
(Q4) Evaluate velocity potential and stream function!
(A4) From their definitions, using polars ...
ur 
Stream Function :
ur 
Velocity Potential :
• From (4a)
From (4b)
1 
r 
m 1 


r r 
 0

r

r
and u  
and u 

m 

r r
From (5b)

0
1 
r 
(4a,b)
(5a,b)
…now integrate 
  m  f1r 
…now integrate 
  c1  g1 
  m
Compatibility required, as usual, 
and neglecting constant ...
• From (5a)

r
(6a)
…now integrate    m ln r  f 2  
1 
…now integrate 
r 
Compatibility required, as usual,
and neglecting constant ...

  c2  g2 r 
  m ln r
(6b)
Continued...
Velocity Potential
  m ln r
Stream Function
  m
  m ln r3
  m ln r2
 m

4
 m
 0

8
  m ln r1
= ur
Again, note:
• Stream function constant along streamlines (here in radial direction).
• Lines of constant potential are perpendicular to streamlines
• For what should be obvious reasons, polar coordinates have been
used to determine the velocity potential and the stream function.
Can convert both into cartesian forms ...
Continued...
Conversion of stream function and velocity potential into
cartesian forms.
• Suppose ‘red dot’ moves on circle around origin of xy coordinate
system...
• One has ...

r x y
2
2
2
1
and
 y
 x
  tan 1 
• Added velocity components are...
u  ur cos    u sin  
v  ur sin    u cos 
For our example circumferential velocity does not exis, hence...
v  ur sin  
u  u cos  
r
with...
one gets ...
And so ...
cos   
u  ur
x
r
x m x mx

 2
r r r
r
u 
mx
x 2  y 2 
sin   
v  ur
v 
y
r
y m y my

 2
r r r
r
my
x 2  y 2 
(7a,b)
Continued...
• Velocity
potential and stream function can either be found using
integration of velocity components in Eq. (7a,b) from previous slide or
by using solutions obtained earlier (Eq. (6a,b)) and performing a direct
change of variables in polar-coordinate forms (remember that these
are scalar quantities).
Velocity Potential, Eq. (6b)
Stream Function Eq. (6a)
  m ln r
  m



Change of Variables
  m ln  x 2  y 2  2
1

y
  m tan 1  
 x
• Can use these forms to find velocity potential and stream function due
to a source/sink which is not located at origin. If source/sink was
located at (x0 , y0 ) then same analysis can be carried out using
translated axes and following results apply:




  m ln   x  x0 2   y  y0 2


1
2



  y  y0  



x

x

0 
  m tan 1
Example : Line Vortex
 2
Laplace Equation for velocity potential
x
2

 2
y
2
0
• For irrotational, incompressible flow from previous example we can
show that stream function also satisfies the Laplace equation, i.e. that...
 2  2
 2 0
2
x
y
u
• Recall...

y
and v  
Hence, Eq. (1) equivalent to...

(1)

x
v u

0
x y
Which is the statement that vorticity is zero!
• This means that stream function for source/sink flow from previous
example satisfies Laplace Eq.!!! (Recall, we started our previous example
from velocity potential satisfying Laplace Eq.)
• Hence, this stream function could represent velocity potential.
• Therefore define new flow where roles of velocity potential and stream
function have been swapped around, i.e ...
Velocity Potential
  K
Stream Function
   K ln r
where K is a constant (vortex strength). Velocity components are now
easily found...
Radial Velocity Component
ur 

0
r
Circumferential Velocity Component
u 
1  K

r  r
Continued...
ur  0
u 
K
r
• This is a flow exclusively in tangential direction, with flow speed inversely
proportional to (radial) distance from vortex centre. The larger the value of K the
higher velocity at any particular radius r - this is why K is called vortex strength.
Flow is akin to that seen around a plug hole… What would be a better
representation of a flow near/down a plug hole?
• Note that flow is not a rigid body motion with constant angular velocity and
speed directly proportional to radial distance from axis of rotation!
K
V.P. :    K

2
K
3
8
K
S.F.:    K ln r

4
K
 0
A
B
u 
K
r1
u 
K
r2
Along line AB one has 1/r velocity
‘decay’. Speed downwards on interval
B0 and upwards on 0A. Velocity
infinite at origin; of course this cannot
happen in reality. In reality viscosity
removes the singularity at origin.

8
Continued...
AGAIN… Note from a physical standpoint that flow is
IRROTATIONAL everywhere … except at origin where
circumferential velocity is infinite and so is vorticity.
Important here…. … to realize
what irrotational means!
… Put a piece of paper on our irrotational, potential vortex
and watch it... What you observe as it goes around is...
The EXCLAMATION MARK … … does NOT change its orientation
in space as it moves around centre of the vortex! … This is because the
flow is irrotational; its vorticiy is zero...!
THE CIRCULATION
• For irrotational line vortex ‘exclamation mark did not rotate’ - in the
sense of changing its orientation in space. Had it changed its orientation
then flow would have been rotational. It is THIS rotation that vorticity
deals with.
• Nevertheless, exclamation mark moved around centre of vortex. So
there was some sort of rotary motion. This aspect of flow is connected
with the CIRCULTATION.
• For a region of flow...
CIRCULATION IS DEFINED as the
clockwise integral around the
closed curve (that encloses the region) of the
flow velocity component along the curve.

 V cos  ds   V  d s
C
C
• Hmhhh,… wouldn’t I expect this integral to be zero?
• No it is not zero! Convince yourself by looking at the example ...
u ,  x
v,  y
v, y
u ,  x
   v    y    u    x    v    y    u    x 
 v y  u x  v y  ux  2 u x  2 v y  0
(1)
So, … integral (i.e. circulation) gives a finite value! Reason is that
plus/minus signs always appear in such combinations that one always
adds up (positive) terms!
• Note, the larger u and v (i.e. the faster I move around curve) the larger
circulation.
• Similar sums as above in Eq.(1) could be written down for line vortex...
• Circulation for line vortex will be constant no matter around which one
of the circular streamlines I integrate. Velocity decrease as 1/r but
circumference of circle increases proportional to r.
Continued...
• Two slides ago where we defined circulation we had…

V cos ds  V  d s
C
C
• So, then...
 u   dx 
   
  v  d s   v    dy  
C
C
 w   dz 
   


• For velocity potential one has:
• Hence, ...
u


x
 u dx  v dy  w dz 
, v

y

, w

z

u dx  v dy  w dz 
dx 
dy 
dz  d
x
y
z



Total Differential, 1st year Maths
• So, then, ...
   
C
(A)
C
End
d   Start
 End  Start
Circulation only depends on values of velocity potential
at start and end points of the path along
which we integrate!
!
OK,… Then let’s calculate the circulation for our
line vortex and for our source!
(i) For LINE VORTEX the velocity potential was found to be ...
  K
As we move along curve C surrounding the line vortex the angle
changes as...
 :  0  2
Hence,...
End
   Start  K 2  K 0
 2 K
(ii) For LINE SOURCE the velocity potential was found to be ...
  m ln r
When we move along curve C surrounding line source then radius...
rStart  rEnd
Hence,...
End
   Start
 m ln rEnd  m ln rStart  0
Continued...
NOTES:
• Value of circulation does not depend on the shape of closed curve, C. To find
results above, we could chose a circular or non-circular path round the vortex
(i) and source (ii).
• In general, value of circulation denotes net algebraic strength of all vortex
filaments contained within the closed curve. This can be seen by re-writing
the line integral into a surface integral using Stokes theorem...
 
V  d s   d A
C
S
• Circulation is an important concept in fluid mechanics: it is circulation
which (together with onset of flow speed) develops lift (Magnus Force) on
a spinning cylinder, spinning cricket ball and an aerofoil section.
• In potential flow, line vortex is used to model circulation that occurs in real
life.
• Use of word ‘Circulation’ to label integral used for its definition is slightly
misleading.
It does not necessarily mean that fluid elements are moving around in circles
within this flow field!
Rather, when circulation exist it simply means that line integral is finite.
For example, if airfoil below is generating lift the circulation taken around a
closed curve enclosing the airfoil will be finite, although the fluid elements
are by no means executing circles around the airfoil (as clearly seen from
streamlines sketched in figure).
Simplified version of ‘Wing Example’
on shown on left...
COMBINATIONS OF SOLUTIONS: SOLID
BODIES IN A POTENTIAL FLOW
• Recall: Can use PRINCIPLE OF SUPERPOSITION for velocity potential.
• In addition, have shown that for incompressible, irrotational flow, stream function
also satisfies Laplace Eq. So can similarly construct flow solutions by combining
S.F. associated with uniform flow, source/sink flow and line-vortex flow.
• In fact, we will almost exclusively use stream function here because we are
interested in pattern of streamlines; once we find stream function, we can use fact
that it is constant along streamlines to plot out streamlines.
Uniform Flow and Source: THE RANKINE BODY
What happens if we combine...


?
   uniform flow   source
Cartesian Coordinates:
y
  x, y   U  y  m tan 1 
(1)
Polar Coordinates:
 r ,   U r sin   m
(2)
 x

• (1)/(2) represent complete descriptions of flow field. But what does it
look like?...
•To graph lines of constant  , first look for STAGNATION POINTS.
There ... V  0
Thus, both velocity components must be zero… Differentiate to get
expressions for velocity components ...
Continued...
y
• Cartesian Coordinates:   x, y   U  y  m tan 1 
 x

m

x
r cos
U

cos
u
 U  m 2

U

m



2
2
r
x  y 
y
r
m

y
r sin 
sin 


m

m

v
2
2
2
r
x  y 
x
r
• For v=0

ur 

m
U
 m U  , 0 
u  
and
u  0 need
Since m, r positive choose...
 

 U  sin 
r
  0,  , 2 , 3 , ...
to get a solution for
m
m
 ur  U  cos    U  
r
r

x
 r ,   U  r sin   m
1 
m
 U  cos 
r 
r
For
u 0
m
u  U 
x
STAGNATION POINT at
• Polar Coordinates:
(4)
y0
Eq. (4) requires ...
...with this Eq. (3) gives...
(3)
STAGNATION POINT at
ur 0

r
ur  0
m
U
m U  ,  
Continued...
• In both cases same location for Stagnation Point ...
Cartesian Coordinates
 m U  , 0 
Polar Coordinates
m U  ,  
We repeated ourselves to demonstrate that either coordinate system can be used. In
general choose the one that makes the analysis easiest.
• Now use for distance between origin and stagnation point:
a
m
U
• Find S.L. that arrives at stagnation point and divides there. Using ...
  const.
along this S.L., use a known point - the stagnation point - to evaluate constant.
With Eq (2) from above ...
m
 r ,   U  r sin   m   s  U 
sin   m  m
U
where suffix s denotes ‘along particular streamline through stagnation point’.
• Streamline found by equating Eq (2) to the constant and rearranging...
 m  U r sin   m
m     a    
r



U sin 
sin 


PLOT
(5)

Continued...
Plotting stagnation streamline: r 

 2
3 2
 0
with
r
a    
sin 
r
a 2
a 2
r sin    a
etc. ...
• Stagnation streamline defines shape of (imaginary) solid half-body
which may be fitted inside streamline boundary; remember flow does
not cross streamline … or solid boundary. Call this special S.L.
SURFACE STREAMLINE.
• Body shape named after Scottish engineer W.J.M. Rankine (18201872).
• Now only have stagnation/surface. To get other S.L.’s, choose point,
determine constant for S.L. through point and then sketch particular
S.L. through this point by compiling a table as above.