Lecture 16
Final Version
Contents
• Combinations of Solutions: Solid
Bodies in a Potential Flow
(Rankine Oval etc.)
• Cylinder in Uniform Flow
• Cylinder with Circulation in a
Uniform Flow
• Pressure Distribution Around the
Cylinder
• Kutta-Joukowski Lift Theorem
• Circulation and Lift for Aerofoil
Applications
Design Project
•
•
•
Instructions now on www as PDF file. (Instructions
should also appear as hardcopies via your pigeon hole
Deadline for submission extended until Friday Week
12 (= Fri. 19 Jan. 2007)
Submission sheet will appear on www soon-ish (and
also via pigeon holes)
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.
Continued...
How Does Flow Speed Vary Along Surface Streamline?
• Recall velocity components in cartesian coord. from Eqs. (3)/(4) :
u  U 
m
cos
r
v
m
sin 
r
U 2  v  u 2  v2
• Flow speed is:
2
2
2
2
U U   m cos    m sin    U 2  2U  m cos  m2 cos 2   m2 sin 2 
r
r
r
r

 r

2

2U  m
m2
U 
cos  2 sin 2   cos2 
r
r
2


2m
m2 
 U 1 
cos  2 2 
r
U
r U 


2




U 2
2U  m
m2

cos  2
r
r
NOW RECALL: a 
 a2 2 a

U  U 1  2 
cos 
r
 r

2
2

(6)
• To find flow speed on body surface, VS , evaluate Eq. (6) subject to
r
a    
sin 
(5) for surface streamline. This gives ...
m
U
Continued...
• Note: Since we know surface flow speed, we can evaluate static
pressure at any point on surface. Using Bernoulli equation along central
streamline that divides into surface streamline and, as usual, ignoring
gravitational term...
1
1
p   U 2  pS   VS2
2 




 2 

Upstream
undisturbe d
flow


On surface
1
1
2
 U
 pS  p   VS2
2
2
1
2

V
2
S
p

p
V
pS  p
S

S


1
 2
2
1
1
1
2
2
2
U
 U
 U
 U
2
2
2
pS  p
VS2
1
2
1
2
U
 U
2
Hence we get the non-dimensional pressure coefficient :
ps  p
VS2
Cp 
 1
2
1
2
U
 U
2
Continued...
Uniform Flow and Sink
(instead of Source)
Stream Function
Source
(previous case)
Sink
 y
 x
 y
 x
Cart. Coord.
  x, y   U  y  m tan 1  
  x, y   U  y  m tan 1  
Pol. Coord.
 r ,   U  r sin   m
 r ,   U  r sin   m
• Only difference: plus sign(s) changed into minus sign(s)!
Hence, for sink expect a very similar analysis as above for source.


Source
Sink
S.P.
• Note: In real world (inviscid) flow pattern for sink would not be observed! Flow
would initially follow body contour but (due to viscosity) detach at separation
points indicated by S.P. in sketch for sink. The phenomenon of SEPARATION will
be covered later. At this stage learn that...
Potential flows do NOT model ALL features of a real flow!!!
This has lead to potential flow often being termed IDEAL FLOW.
•Obvious question now is what happens for ...
Uniform Flow + Source + Sink
We consider ‘symmetric’ case where:
Strength
Source
Sink
m
Location
 c, 0
m
c, 0
We cannot have both at origin now! What would happen if both at
origin?
• By considering the two sketches on previous slide we can anticipate
shape of surface streamline and resulting body...
… an oval.
• Using superposition, can readily write stream function for this flow:
  x, y   U  y  m tan 1 
y 
y 
1 
  m tan 
 (1)
xc
xc
  
Uniform
flow
Source at (  c , 0 )
Sink at ( c , 0 )
• Second and third terms can be combined using:
  

tan 1    tan 1    tan 1 
1





To give a more concise form for stream function


2c y

2
2
2
 x  y c 
  x, y   U  y  m tan 1 
Continued...
• From either of the two forms of S.F. on previous slide, one can determine velocity
components
u



xc
xc
 U  m 


2
2
y
 x  c 2  y 2 
 x  c  y
v



1
1
 my 

2
2
2
2
x




x

c

y
x

c

y


(2)
(3)
• Now find stagnation points, where u=v=0. From Eq. (3) one sees that
when y=0 then v=0.
• Substitute y=0 into Eq. (2) and then find value of x which gives that
u=0.
• After some manipulation the solutions for x are:

2m 
x   c 1 

 cU  
1
2
 L
• Hence, stagnation points at:
 L, 0
and
L, 0
•Now determine value of S.F. for surface streamline from Eq (1).
 y 
1  y 
  m tan 

 xc
 xc
  x, y   U  y  m tan 1 
It can be seen that this is trivial and that
 0
S
(1) - repeated
Continued...
• Rankine Oval then looks like
...
• We already determined value
of L. Can find points of
maximum
velocity
and
minimum
pressure
at
shoulders +/-h, of oval using
similar methods. All these
parameters are a function of
the...
… basic dimensionless parameter
d
m
U c
In summary one obtains
L 
2m 

 1 
c  U  c 
1
2
umax
2 m U  c 
 1
U
1  h2 c2
•As one increases dimensionless
parameter d from zero to large values,
oval shape increases in size and
thickness from flat plate of length 2c to
huge, nearly circular ‘cylinder’. Here
think of increase when
c  const. and U   const
•All Rankine ovals, except very thin
ones, have large adverse pressure
gradient on leeward surface. Thus,
boundary-layer will separate in rear,
broad wake flow develops, inviscid
pattern unrealistic in that region.
 ha 
h

 cot 
c
 2m U c 

Overall Strategy for Plotting Streamlines from
Stream Function was...
Write down stream function for flow by appropriately
combining individual solutions for source, sink and line
vortex as a sum:
  x, y   1 x, y    2  x, y    3  x, y  
Note: Huge choice as far as selction of parameters is concerned!
Souce strength, vortex direction of rotation, strength … ...

Calculate expressions for vel. components u, v from
u


, v
y
x

Determine coord.of stagnation point(s) via u=0 , v =0.

Determine value of stream function passing through
(stagnation) point by substituting coordinates of
(stagnation) point(s) into the stream function.

Set stream function equal to the value you have
determined for point in question.

Determine values of x, y (or r,  ) that satisfy this
expression and plot to obtain streamline.

Choose new point x,y
From previous it should be obvious how one can find stream function for a
cylinder (circle) in a uniform flow ...
Cylinder in a Uniform Flow
•Turn Rankine oval into circle by allowing ...
m

U c
c0
• Achieved by moving source and sink closer to origin ...
Limit (c=0) would ultimately ‘cancel’ pair!
• Ensure their influence remains by allowing m to increase in size.
Necessary limit is...
c0
• Recall that Rankine
oval had S.F. ...
• Now let c  0
with
c m  const.


2c y

2
2
2
 x  y c 
  x, y   U  y  m tan 1 
then the argument of tan-1 goes to zero...
• Noting that tan 1    for small  gives
  x, y   U  y  m
2c y
x2  y2  c2
  2mc
• Define DOUBLET STRENGTH:

Stream Function for Cylinder Flow
  x, y   U  y 
y

x2  y 2



Uniform
flow
DOUBLET
at origin
(1)
Note: Merging of source and sink as above produces structure known as
DOUBLET.
Continued...
• More convenient to work in polar coordinates! S.F. can be written ...
 r ,   U  r sin  
 sin 
r


 U  sin   r 
U



r 
• From Eq. (1) can get velocity components in usual way...

(1)
Continued...
ur 
u 
1 
 U  cos
r 
2

R
1 

2 

r 



  U  sin 
r
(2)
Where
we
used...
2

R
1 

2 

r 

R 
(3)

U
• CHECK that this flow really does represent a cylinder in uniform flow.
Stagnation points:
Eq. (3) :
 0
 


u  0
u  0
• Substitute these angles into Eq.(2) and set ur  0
to get ...
For   0 :

R2 
0  U  cos 0 1  2 
r 

R2
 1 2
r
 r R
For    :

R2 
0  U  cos  1  2 
r 

R2
 1 2
r
 r R
R, 0
Hence,... Stagnation points:
R,  
and
• Surface S.L. VALUE by substituting one stag. point into Eq. (1)...

 R, 0  U  sin 0  R 

 
  0
U  R 
• Now get equation for Surface S.L. by equating Eq. (1) to zero ...
r2 

U

r 

U
or
As required surface S.L. is circle with radius R 
rR
 U
PLOT

Continued...
Uniform Flow +
Doublet
= Flow over a Cylinder
• Velocity Components on cylinder surface are obtained, by setting
r=R, from...

R2 
Eq. (2) ur  U  cos 1  2 
r 


R2 
Eq. (3) u   U  sin  1  2 
r 



ur  0
u   2 U  sin 
Might have expected to find that radial flow component is zero on
surface - flow cannot pass through (solid) cylinder wall!
• Note also that maximum flow speed occurs at
3
2
2
where it is  2 U  and  2 U 


and  
respectively.
Means in both cases (top and bottom half of cyl.) flow is from left to right! On
top negative value as velocity points in clockwise (negative angle) direction. On
bottom in anti-clockwise (positive angle) direction.
• Finally, note symmetry of flow about both the x- and y- axes. What
does this tell you about the pressure distribution on the cylinder
surface… remember the Bernoulli Equation!
Continued...
• Since we can now mathematically describe …
...we can, in principle, also describe flow through an arbitrary array of
cylinders as, for instance, the flow shown in the photo below. We
simply need to put several doublets in our uniform flow.
Cylinder with Circulation in a Uniform Flow
• Without performing calculation, can see in preceding flow no net lift or drag on
cylinder since pressure distribution on surface symmetric about x- and y-axis..
• In order to generate lift need to break symmetry. Achieved by introducing line
vortex of strength, K, at origin which introduces circulation
.
G = 2p K
• Note that this does not violate the flow around cylinder: line vortex produces a u
component of velocity only. Hence, we are still adhering to condition that flow
cannot pass through cylinder boundary.
• Working from S.F. for cylinder in uniform flow additional inclusion of line vortex
gives:
 r ,   U  r sin  
 sin 
 K ln r  C
r
 




Uniform
flow
Use result that radius
R 
of resulting cylinder is :
(1) 
Line vortex Arbitrary
at origin constant
Doublet
at origin

And set :
U
C  K ln R

 
  K ln r  K ln R
 r ,   U  sin   r 
U r

 

 1
  K ln r  ln R 
 U  sin  r 
 U r 


R2 
r
  U  sin   r    K ln
r 
R

Velocity
Components

ur 
uq =
2

1 
R
 U  cos 1  2 
r 
r 

-
¶y
¶r
(1)
æ
R2 ö K
÷
= - U¥ sin q çç1+ 2 ÷
+
÷
çè
÷
r ø
r
Continued...
• So, on surface (r=R), velocity components are:
ur  0
u   2U  sin  
• Surface Stagnation points also need:

K
R
u  0
sin  
K
2 RU 
Note: By setting vortex strength zero (K=0), recover flow over cylinder in
uniform flow with stagnation points at   0,
• Plotting,… Choose value for K,… Now first get value of S.F. for r=R,... then
set S.F. equal to that value,… then compile table r vs. angle… This gives
particular streamline through stagnation points.
Then choose any other point in flow field not on stagnation streamline,…
determine value of S.F. for this point,… set S.F. equal to that value,… then
compile table r vs. angle… This gives streamline through the chosen particular
points… Then choose another point in flow field… etc (compare flow chart
from beginning of lecture). For various values of K the following, flow fields
emerge...
K 0
K 1
K 2
K 3
Continued...
• Can now also describe flow through an arbitrary array of cylinders when
each of them is rotating! (Note: In photo below cylinders are not
rotating)
Pressure Distribution Around the Cylinder
• To evaluate press. on cyl. surface use Bernoulli Eq. along S.L. that
originates far upstream where flow is undisturbed. Ignoring grav.
forces:
1
1
2  p  U 2
p   U 
S
S
2
2



Upstream
undisturbe d
flow

• Re-arranging...
pS  p 
On cylinder
surface
 U2 
1
 U 2 1  S2 
 U 
2


• Substituting for flow speed ( uR  0 , u   2U  sin 

pS  p 


1
u2  u2
 U 2 1  R 2 
2
U


gives...
  1  U2 1  u2 




2
1
 U 2 1   2U  sin  K / R 2 U 2
2

K
)
R
U 2 

… difference in pressures between surface and undisturbed free stream
pS  p
1
  U 2
2
2

 K 
4K
2

sin   
1  4 sin  
R
U
R
U



 
In particular for non-rotating cylinder
where K=0:
pS  p 
1
 U 2 1  4 sin 2 
2
Def.: Pressure
Coefficient
Cp 

pS  p
 1  4 sin 2 
1
 U 2
2



(1)
(2)


Only top half of cyl. shown.
Continued...
Qualitative behaviour of
2

 K 
pS  p
4K
2

Cp 
 1  4 sin  
sin   
1
R
U
R
U
2





 U

2



for various values of K RU  .
(  0 : Rear of cyl. ,    2  1.57 : Top of cyl.,
    3.14 : Front of cyl. ,   3 2  4.71: Bott. of cyl. )
• Best way of interpreting above graphs is to think of flow velocity and radius being constant
while vortex strength is increasing from one plot to next.
• When plotting graphs I did not explicitly specify velocity or radius! I simply used different
numeric values for K RU 
in order to illustrate behaviour of graph. I have not considered
if any of these cases may not be realizable in reality or not!.
Continued...
Equation (1) …
pS  p
1
  U 2
2
2

 K 
4K
2

sin   
1  4 sin  
RU 
R
U







… can be used to calculate net lift and drag acting on cylinder!
Sketch (A)
Sketch (B)
• In Sketch (b) ...
L  p  sin    pS  p  sin 
D  p  cos   pS  p  cos
• Hence, integration around cylinder surface yields total L and D ...
L
2
p
0
S
 p  sin  b R d
D
2
p
0
S
 p  cos b R d
where b is width (into paper) of cylinder. Substituting for pressure using
Eq. (1), and integrating (most terms drop out), leads to following results:
1
L    U 2
2
 4K 

  b R
R
U


  2 U K  b
Or, lift per unit width:
D0

L
   U  2  K     U  
b
Thus, drag zero…
a remarkable result!


Kutta  Joukowski
Lift Theorem
d' Alembert' s
Paradox
Continued...
• Net lift is indicated in sketch below. ... Note that if a line vortex is used which
rotates in mathematically positive sense (anti-clockwise) then resulting lift is
negative, i.e. downwards.
U
L
   U 
b
L
• Final notes: How is lift generated? ... From sketch above and from
pressure profiles plotted earlier it is evident how this is physically
achieved… Breaking of the flow symmetry in x-axis means that flow
round lower part of cylinder is faster than round top - this means that
pressure is lower round bottom and so a net downward force results.
Notice that symmetry in y-axis is retained … symmetry of pressure
on left-hand and right-hand faces is retained and so there is no net
drag force. Keep in mind that our analysis was for an ideal fluid (i.e.
there is no viscosity). In a real flow would fore-aft symmetry be
retained?
• Lastly, since lift is proportional to circulation, we wish to make
circulation large to generate a large lifting force. In applications of
above flow this is achieved by spinning cylinder to produce large
vorticity… but is there a limit to how much circulation we should
produce?