Lecture 17
Final Version
Contents
• Lift on an airfoil
• Dimensional Analysis
• Dimensional Homogeneity
• Drag on a Sphere / Stokes Law
• Self Similarity
• Dimensionless Drag / Drag
Coefficient
1
Recall : Cylinder with Circulation in a Uniform Flow
• Without performing calculation, can see that a uniform flow around a fix cylinder
gives no net lift or drag on cylinder since pressure distribution on surface is
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 :
Line vortex Arbitrary
at origin constant

U
And set :
C  K ln R

R2 
r
  U  sin   r    K ln
r 
R

(1) 
Velocity
Components
Doublet
at origin
(1)

ur 
2

1 
R
 U  cos 1  2 
r 
r 

uq =
¶y
¶r
æ
R2 ö
K
ç
÷
= - U¥ sin q ç1+ 2 ÷
+
÷
çè
÷
r ø
r
2
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
3
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



On cylinder
surface
Upstream
undisturbe d
flow
U2S = ur2 + u2q
• Substituting for surface flow speed :
uR = 0 ,
uq = - 2U¥ sin q +
with
K
R
… difference in pressures between surface and undisturbed free stream
pS  p
1
  U 2
2
2



4
K
K
2

sin   
1  4 sin  
RU 

 R U 
In particular for non-rotating cylinder
where K=0:
pS  p 
Def.: Pressure
1
 U 2 1  4 sin 2 
2
Coefficient
Cp 

pS  p
 1  4 sin 2 
1
 U 2
2

 (1)

(2)


Only top half of cyl. shown.
4
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
5
if any of these cases may not be realizable in reality or not!.
Continued...
Equation (1) …
pS  p
1
  U 2
2
2



4
K
K
2

sin   
1  4 sin  
R
U
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 6
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?
------------------------End of Recall--------------------
7
Circulation and Lift for Aerofoil Applications
• If a thin symmetric aerofoil is placed at zero incidence in an inviscid,
irrotational and imcompressible uniform flow, the flow pattern shown
in Fig. (1) below ensues. There is no circulation and the aerofoil does
not generate lift. (This case is analogous to the cylinder with   0)
Fig. (1)
• In case of cylinder, can generate vorticity by spinning cylinder. For
airfoil section this can be achieved by setting it at incidence or by
using a non-symmetric shape (which shape to get lift? … and to get
negative lift?). Placed at incidence, flow past a thin symmetric
aerofoil is shown in Fig. (2).
Fig. (2)
Clearly airfoil experiences upward force - compare flow speeds on
upper and lower surfaces. We have seen that this type of flow speed
differential can be modelled by using line vortices which yield
circulation and hence lift. (In Fig. (2) line vortices would have
negative K to give clockwise velocity contribution.)
8
Continued...
• If we wish to calculate lift per unit span on aerofoil section using
Kutta-Joukowski Lift Theorem ...
L
   U 
b
 : Density Air
U  : Flow Speed
 : Circulation
… then need to know value of circulation for a given aerofoil at a
specific flow speed and for particular angle of incidence. Key to
finding unique value of circulation lies in modelling flow at trailing
edge of aerofoil. Consider Fig. (3a-c) below ...
Fig. (3)
Evidently the correct value
  Kutta has been used in Fig. (3c).
The KUTTA CONDITION ...
… states that flow from upper and lower surfaces must leave trailing
edge with same speed. The Kutta condition thus determines correct
value of circulation when performing a calculation of flow around a
lifting aerofoil.
9
Continued...
• Increasing either
 : Angle of Incidence
increases
or U  : Flow Speed
Kutta
and hence the lift.
Is there a limit to how large one can make angle of incidence and
hence Kutta ?
• For a flat plate with incidence  the lift experienced is
L
  U   cU   
b
where c is length of plate. Non-dimensional lift coefficient given by
L
CL 
 2 
1
 U 2 b c
2
• This result (as with calculations for aerofoils) is achieved by using a
line of vortices - a vortex sheet - ‘within’ aerofoil section to generate
circulation, rather than a single line vortex as used for cylinders in our
earlier considerations.
10
Continued...
• Qualitative comparison of pressure coefficient for NACA 0012 airfoil at incidence
with one of the earlier graphs for pressure coefficient of a rotating cylinder.
(Comparison included here to highlight where corresponding points / regions are
and to practice how to read such graphs.)
NACA
airfoil
Upper wing surface
Lower wing surface
Upper cyl. half
Rotating
cylinder
Lower cyl. half
• Note: In order to get negative pressure coefficient on top half of
cylinder and (i.e. upward lift) need to reverse sense of rotation of line
vortex used in example for flow around rotating cylinder.
11
Dimensional Analysis and Model Testing
•Introduction to Dimensional Analysis
Consider drag D of sphere ….
On what quantities does it depend?
Diameter, d
Flow Speed, V
Fluid Density, 
Fluid Viscosity, 
Write
D  F d ,V ,  ,  
(1)
• Note: Eq.(1) reads … Drag, D, is a Function of ...
What does the above mean in terms of the
measurements we have to carry out to collect
data for all possible spheres in all types of fluids?
12
Continued ...
WE NEED ...
1 page for Drag as
function of 2 variables
(e.g. velocity and
diameter)
d increases
from curve
to curve
1 page for
each value
of 
1 book for Drag as
function of 3 variables
(e.g. velocity, diameter,
density)
Shelf of books for Drag as a
function of 4 variables
(velocity, diameter, density,
viscosity)
If we want 10 data points per curve, at £10 each
experiment, this will cost...
10 10 10 10  £10  £100,000
THERE MUST BE
A BETTER WAY !?!?
13
Continued...
DRAG
GOAL IS TO COMPRESS SHELF
OF BOOKS INTO ONE SINGLE
GRAPH...
4 Independent
Experimental Parameters
How Could We Possibly
Achieve This?
14
•Dimensional Analysis for Re<<1
(Note: Later we will relax this restriction and look at larger Re.)
What does imposed restriction mean?
Re  1
Rate of Change of Momentum 
Inertia Force 
 i.e.
  1
Viscous Force
 Viscous Force 
Inertia Forces  Viscous Forces
Viscous Forces are the dominant forces!
• Inertia Forces are associated with density of Fluid
• Consequently, if Inertia Forces << Visc. Forces then,
to a good approximation DRAG DOES NOT
DEPEND ON DENSITY of FLUID!
Thus, Eq.(1)...
D  F d ,V ,  ,  
(1) - repeated
reduces to ...
D  F d ,V ,  
(2)
15
Continued...
So, we are restricted to flow with..
L
A
M
I
N
A
R
LOW RE NUMBER
Restrictions exclude ...
T
U
R
B
U
L
E
N
T
HIGH RE NUMBER
16
Continued...
• The expression Eq.(2) ...
D  F d ,V ,  
(2) - repeated
… represents a VERY general statement!!!
• CRUCIAL NEXT STEP:
Ensure that function F has such a form that
one ends up with same dimensions
on both sides of equal sign.
• Hence, we may NOT choose a function that produces a
non-sense statement where units are for instance ...
D  d 2  V 3  1
Units:
kg m
s2
m3 kg kg m 4
 4
Units: m  3 
s sm
s
2
• QUESTION:
How Do I Have To Choose Exponents
Such That Units AreThe Same on
Both Sides Of Equation?
17
Continued...
Answer question by determining conditions for
exponents under which one gets same units on both
sides of equation ...
D  F d ,V ,   (2) - repeated
N
Units:
Dimensions:
kg m
s2
M L T 2
M : Mass
m
L
m
s
L T 1
L : Length
kg
ms
M L1 T 1 (3a-c)
T : Time
WANT !!!
,  , 
such that


D  V  d 
(4)
gives
M L T 2
M L T 2
Find by subst. Eq.(3a-c) into Eq.(4) ...
18
Continued...
D  V   d  
ML T
2

L T 
1 
L 

ML
(4) - repeated
1
T

1 
Collect corresponding terms ...
M L T 2
 M
L   
T  
(5)
By comparing exponents ...
• … of M on left and right hand side of Eq.(5)
1
(6a)
• … of L on left and right hand side of Eq.(5)
1     
(6b)
• … of T on left and right hand side of Eq.(5)
 2    
• Substitut Eq.(6a) into Eq.(6c) ...
 2    1
 1
(6c)
(6d)
• Substitut Eq.(6a) and Eq.(6d) into Eq.(6b) ...
1  1   1
 1
19
Continued...
• In summary we get...
D  V   d  
(4) - repeated
… where ...
 1
 1
 1
•This is the ONLY possible solution for the three
simultaneous linear equations Eqs.(6a-c)!
•It is the ONLY possible solution that ensures ...
DIMENSIONAL HOMOGENEITY
This solution ...
D  const  V d 
3
is the ...
for sphere. Must be obtained
from experiments or theory
STOKES’ LAW
Recall, that it is only valid for low Re!
20
Continued...
STOKES’ LAW
Recall, that it is only valid for low Re!
D  const  V d 
• While
we assumed Re<<1 experiments show that Stokes
law is, in fact, valid for Re<2.
• For flow regime where Stokes law is valid drag is
proportional to velocity. Hence, doubling velocity results in
double drag. We will later see that this is not the case for
higher Reynolds numbers.
• The constant in Stokes law can, in principle be obtained
from one single experiment.
• Think about all this an ‘let it sink in’… We have
determined formula for drag forces acting on sphere
without knowing anything about the physics of the flow.
The only thing we did was ensuring dimensional
homogeneity! Of course the whole strategy can only yield
correct results if we have identified all parameters involved
in problem.
21
A note on ….
Self Similarity
• Recall that we used a function of type...
D  V   d   
for the dimensional analysis. This is called a power-law
relation.
• A common
view is that scaling or power-law relations are
nothing more than the simplest approximations to the
available experimental data, having no special advantages
over other approximations. ...
… IT IS NOT SO !
Scaling laws give evidence of a very deep property of the
phenomena under consideration their ...
… SELF SIMILARITY
Such phenomena reproduce themselves, so to speak, in time
and space.
From: G.I. Barenblatt, Scaling, self-similarity, and intermediate
asymptotics. Cambridge University Press, 1996
22
Continued...
‘Reproducing themselves’ means ...
…
that
wake
behind an inclined
flat plate looks the
same as flow ...
… the wake of a
grounded tankship.
Power laws are ‘Magnifying Glasses’…
Example:
(I)
Going from a 2
to
gives
2a 2
4 a2
(II)
Apart from scale factor 4 Eq.(II)
is the same as Eq.(I)
23
Continued...
Some Background Info:
Classic example illustrating how powerful dimensional
analysis can be ...
Explosion of Atomic Bomb
r t 
Ground
Ground
100 m
• By measuring radius r as a function of time, t, G.I. Taylor was
able to deduce energy released when bomb explodes by means
of dimensional analysis alone from analyzing freely available
cine films of explosions.
• The figure was considered ‘Top Secret’ back in the 1940s
• Taylor’s result caused ‘much embarrassment in US government
circles.
G. I. Taylor
24
Continued...
TIP:
• Use requirement for dimensional homogeneity as a quick
check for correctness of unfamiliar equations!
Example
• Someone claims that drag force, D, acting on sphere with diameter
d moving with velocity V through a fluid of viscosity  is given
by ...
D  3 V  d 2  
(Formula is wrong!)
Use dimensional arguments to show that this formula cannot be
right!
Solution
• Left-hand side of equation is:
D is a force, hence, dimensions are
[N]=
ékg . m/s2 ù= M L
êë
úû T2
([..] tells that you take the dimension of the unit)
• Right-hand side of equation is:
Velocity  Diameter 2  Viscosity
L
T
´
2
(L)
´
M
LT
=
M L2
T2
Different dimensions on both sides of the
equation. Hence, formula cannot be right!
25
Continued...
Briefly recall where we were coming from and where
we are heading for….
DRAG
GOAL WAS…
TO COMPRESS SHELF OF BOOKS WITH DRAG
DATA INTO ONE SINGLE GRAPH...
4 Independent
Experimental Parameters
We are not there yet but
we are getting closer...
26
Dimensionless Drag / Drag Coefficient
p  p
Stagnation
Point
Flow
ps , V  0
V  V
• Apply Bernoulli along streamline to stagnation point ...
 2
 2
(1)
p  V  ps
2
ps  p 
2
V
• Pressure
in wake must be approximately equal to
pressure in free stream
pw  p
(2)
• If one neglects viscosity then drag arises only because of
different pressures on ‘front’ and back of sphere.
With Eq. (2)
p  ps  pw
(3)
p  ps  p
(4)
27
Continued...
• Pressure
forces act on area approximately equal to
CROSS-SECTIONAL AREA
d 
A  
2
• Since
2
d : Diameter
PRESSURE= DRAG/AREA
d 
Drag   ps  p    
2
With Eq. (1)
• Divide
(5)
Drag 

2
V2
d 
 
2
2
2
(6)
through by right hand side and DEFINE the
DRAG COEFFICIENT CD
CD 
Drag

2
V2
d
  
2
2
• CD is a non-dimensional number
• CD is
a non-dimensional representation of
the drag force
28
Continued...
Notes:
• Main
assumption was to neglected viscosity. This means we are
dealing with Reynolds numbers for which Stokes’ law is NOT
applicable. Hence, we have considered high Reynolds numbers, i.e.
Re>>1. Only for these an extended wake exists.
• As right-hand and left-hand side approximately equal in Eq.(6) the
drag coefficient must be of order 1 under the assumptions (Re>>1)
we made.
On previous page defined drag coefficient for a sphere. This
definition can be extended to include bodies of arbitrary shape...
General Definition of Drag (and Lift) Coefficient
Drag
CD
Coefficient of

Lift
CL
Drag or Lift

2
V2 Area
Notes:
• Carefully check exact definitions of quantities such as CD, CL or Re
Before using data found in literature! Definitions may vary!
• Usually one uses projected area (cross-sectional), i.e. area one sees
when looking towards body from upstream for CD. But for CL for
airfoils one uses one uses the planform area.
29