Lecture 5
Preliminary Version
Contents
• Bernoulli Equation – Terminology
• Bernoulli Equation - Examples
• Dimensional Analysis
• Dimensional Homogeneity
• Drag on a Sphere / Stokes Law
• Self Similarity
• Dimensionless Drag / Drag Coefficient
• Dynamic Similarity
• Model Testing
• Drag on bodies in Flows with Free
Surfaces
• Froude Number
• Skin-Friction Drag / Wave Drag
What Did We Do In Last Lecture?
Derivation of Bernoulli Equation
p


2
v 2   g z  const.
• Constant along streamline
• All terms have units of pressure
• p is the physically existing pressure
• 2nd term is kinetic energy per unit volume
associated with mean motion of flow
• 3rd term hydrostatic pressure
What Did We Do In Last Lecture?
Keep in mind analogy …
Epot + Ekin = const.
Finished previous lecture
with …
BERNOULLI EQUATION
p
v2

 z  const .  H
 g 2g
Alternative Form
p

2
v 2   g z  const .
(7)
• Equations apply along a single stream line.
• Constant for each streamline can be
different.
Continued...
•
Eq. (7) from previous lecture / slide …
p
•

2
v 2   g z  const.
(7)
If Eq. (7) integrated along streamline
between any two points indicated by suffixes
1 and 2 one gets ...
p1 

2
v12   g z1  p2 

2
v22   g z2
This is form in which we will usually apply
Bernoulli Equation most of the time.
Continued...
•
Note: For compressible fluid integration of Euler’s
equation ...
1 dp
dv
dz
v  g 0
 ds
ds
ds
can only be partially completed to give ...
dp v 2
 g  2 g  z  0
Now need relation between density and pressure to
continue.
Terminology for Bernoulli Equation
p
(1)


2
v2
(2)

gz

(3)
const .
(4)
(STATIC) PRESSURE
(1) =
The physically existing pressure in fluid. Whenever
reference is made to “pressure” without further
qualification then interpret as this one.
DYNAMIC PRESSURE
(2) =
Not a physically existing pressure. If z=0, such that Elev.
Press. Vanishes, then Dyn. Press. is the difference
between Tot. Press. And Stat. Press.
ELEVATION PRESSURE
(3) =
Can often be made to vanish, i.e. when streamline
horizontal such that we can chose z=0. If streamline not
horizontal, but if change in height is small such that
Elev. Press. is small compared to other terms one usually
neglects it.
TOTAL PRESSURE
(4) =
This has physical significance that it is the pressure at
which fluid comes to rest. Assume z=0, V=0 then one
gets Stat. Press=Tot. Press.
Terminology for Bernoulli Equation

p
•

2
v2

gz

const .
Note that the STATIC PRESSURE is NOT the
pressure in a static fluid (that is called the hydrostatic
pressure).
•
Neither is the STATIC PRESSURE the pressure
where the fluid is at rest (that is called the stagnation
pressure).
Terminology for Bernoulli Equation
•
For ...
p1 

2
v12
  g z1  p2 
2
v22   g z2
v2  0 and z1  z2
With:
p1 
•


2
v12  p2
Hence sum of static pressure at station 1 plus
dynamic pressure at station 1 is equal to static
pressure at station 2 where fluid is at rest
(stagnates). Thus we call ...
p1 

2
v12  Stagnation Pressure
(or Pitot Pressure)
Final Terminology
•
When Bernoulli written in form ...
p
g
(1)

v2
2g
(2)

z

const.
(3)
(4)
… such that all terms in units of ‘LENGTH’
then one calls ...
(1) =
STATIC HEAD
(2) =
DYNAMIC HEAD
(3) =
ELEVATION HEAD
(4) =
TOTAL HEAD
… and finally ...
(1) + (2) =
STAGNATION HEAD
And recall that ...
Pressure measured relative to atmospheric
pressure is called ...
GAUGE PRESSURE
or
GAGE PRESSURE
… as distinct from
ABSOLUTE PRESSURE
Example 1:
Pitot-Static Tube / Prandtl Tube
Continued …
(1)
( 2)
p1
p2
v1  ?
v2  0
Question: What is value of v1?
Solution:
•
Bernoulli for points (1) and (2)
p1 
•

2
v12   g z1  p2 

2
v22   g z2
Neglect hydrostatic pressure; horizontal streamline …
p1 

2
v12
  g z1  p2 
p1 

2
v12  p2 

2

2
v22   g z2
v22
Continued …
•
(1)
( 2)
p1
p2
v1  ?
v2  0
Point (2) is stagnation point where velocity is zero …
p1 

2
v12  p2 
p1 
•

2

2
v22
v12  p2
Solve for velocity …
v1 
2

 p2  p1 
Continued …
(1)
( 2)
p1
p2
v1  ?
v2  0
Stagnation
Pressure
•
Free-Stream
Static Pressure.
Solve for velocity …
v1 
2

 p2  p1 
[Pressure here
Equal to p1 at
point (1) since
it is assumed
that probe does
not affect flow.]
Continued …
Pitot-Static Tube / Prandtl Tube
• This type of probe/methodology is used in my Vortex /
•
Wind tunnel Lab L5 to measure free stream velocity via
pressures!
In lab I will assume that you are familiar with this!
• In lab will work with arrangement similar to that
shown below. The difference being that aeroplane
is replaced by a circular cylinder, and probe not in
wake of cylinder but in region where flow is not
affected by presence of cylinder.
Pitot-Static /
Prandtl Tube
Example 2:
•
Water flowing in open channel at depth of 2m and
velocity 3m/s. It then flows downs contracting chute
into another channel where depth is 1m and velocity
10m/s. Assuming frictionless flow, determine difference
in elevation of the channels.
(1)
2m 3m/s
y?
1m
( 2)
10m/s
Solution:
•
Bernoulli for points (1) and (2)
p1 
•

2
v12
  g z1  p2 

2
v22   g z2 (A)
Since (1) and (2) on free surface:
p1  p2  patm.
Eq. (A) becomes...

2
v12
  g z1 

2
v22   g z2
(B)
Continued ...
(1)
2m 3m/s
y?
1m
•
2
v12
  g z1 

2
v22   g z2
•
From sketch we see... z1  y  2 m
•
Introduce this into (B) ...

2
v12
  g  y  2 

2
(B)
v22   g z2 (C)
Solve (C) for y ...
y
•
10m/s
Last line from previous page repeated...

•
( 2)


1 2 2
v2  v1   z2  2 m 
2g
(D)
With given values ...
v1  3 m/s , v2  10 m/s ,
z1  1 m
Continued...
•
Eq. (D) yields...
y
2
2






10
m/s

3
m/s
 1 m  2 m 
2
2  9.81 m/s
1
 3.64 m
Example 3: VENTURI TUBE
Example 3: VENTURI TUBE
•
Venturi tube is device where flow rate in a pipe line is
measured by narrowing a part of tube. In narrowed
part flow velocity increases. By measuring resultant
decreasing pressure flow rate in pipe line can be
determined.
 m (with    )
m
w
(3)
D2 with Area, A2
(2)
w
Water
z2
z1
(1)
D1 with Area, A1
z0
Solution:
•
Bernoulli for points (1) and (2)
p1 
•
w
2
V12
  w g z1  p2 
w
2
V22   w g z2 (1)
Mass conservation yields...
A1 V1  A2 V2
A1
V2 
V1
A2
(2)
Continued ...
•
Substituting Eq. (2) into Eq. (1) gives...
w
 w  A1
2

 V1    w g z2
p1 
v12   w g z1  p2 
2
2  A2 
•
Rearrange ...
  A 2 
V12   1   1  p1  p2   w g  z1  z2 
  A2 

2


w
V1 
p1  p2   w g  z1  z2 

2

 w   A1 
   1

2   A2 


This solution is maximum velocity V1 which would
exist for frictionless flow. In reality there is
viscosity present and one introduces a discharge
coefficient cd such that ...
V1 actual  cd V1 ideal
•
To apply above equation still need to evaluate pressure
difference ...
Continued ...
•
How to get pressure difference ….
h
m
h
l
Water,  w
z2  z1
(2)
(1)
•
Use pressure-depth ideas developed earlier. Summing
up all pressure contributions on path from (1) to (2) ...
p1   w g  z2  z1    w gl   w gh   m gh
  m gh   m gh   w gl  p2
p1  p2   w g  z2  z1    w gh   m gh
p1  p2   w g  z2  z1     w   m gh
Example 4:
•
Beavers with engineering degrees discovered use of
Bernoulli principle to ventilate burrows. If mount is
built at one entrance wind velocity increases, creating a
pressure differential that induces air flow through
burrow. Calculate the pressure difference between
entrance and exit for the friendly beaver...
V  22 km/h
Streamline
( 2)
V  18 km/h
(1)
p2  p1
air  1.225 kg/m3
Solution:
•
Bernoulli for points (1) and (2)
p1 
•

2
v12
  g z1  p2 

2
v22   g z2
(A)
Assume negligible elevation change, hence: z1  z2
p1 
•

2
Eq. (A) becomes...
v12
 p2 

2
v22
Solve (B) for pressure difference ...
(B)
Continued...
•
Last equation from previous slide repeated...
p1 
 2

v1  p2  v22
2
2
p  p1  p2 
•
(C)
With given values...
v1  18 km/h  5 m/s
•

v22  v12 
2

v2  22 km/h  6.11 m/s
Eq. (C) yields ...

1.225 kg/m 3
6.11 m/s2  5 m/s2
p  p1  p2 
2
kg m
2
N
s
 7.55
 7.55
 7.55
 7.55 Pa
2
2
2
ms
m
m
kg

Example 5: Homework/ Background Reading
•
A hurricane is a tropical storm formed over ocean by
low atmospheric pressures. As hurricane approaches
land, inordinate ocean swells (very high tides)
accompany hurricane. A Class-5 hurricane features
winds in excess of 240 km/h, although wind velocity at
center “eye” is very low.
In sketch below atmospheric pressure 310 km from eye
is 101,591 Pa (558.8 mm Hg at point 1, generally
normal for ocean) and winds are calm. Hurricane
atmospheric pressure at eye at point (3) is 74,500 Pa
(762 mm Hg).
(I) Estimate ocean swell at eye of the hurricane at (3)
(II)Estimate ocean swell at (2) where wind velocity is
310 km/h.
Eye
Hurricane
Calm
Ocean
(1) Level
(2)
h2
(3)
h1
Ocean
Continued ...
Assumptions
•
Steady, incompressible, irrotational (so that Bernoulli
applicable)
(Note: This is certainly a very questionable
assumption for a highly turbulent flow)
•
The effect of water drifted into air is negligible
Properties
Density air
 A  1.225 kg/m
Density sea water
3
 SW  1023 kg/m 3
Density Mercury
 M  13,560 kg/m 3
Continued, Cengel p. 199 ...
Solution, Part (I):
•
Reduced atmospheric pressure over the water causes to
rise. Thus decreased pressure at (3) relative to (1)
causes ocean water to rise at (3). Hence...
p   ghHg   ghSW
 Hg ghHg   SW ghSW
hSW
hSW 
13,560 kg/m 3
1023 kg/m
3
 Hg

hHg
 SW
(A)
762 mm Hg  558.8 mm Hg 
 2.693 m
•
This result is equivalent to the storm surge at the eye of
the hurricane since the velocity there is negligible and
there are no dynamic effects.
Continued...
Solution Part (II):
•
Here we have to take the high wind speed into account
and need Bernoulli. We write down Bernoulli equation
for points A and B...
p A v A2
pB vB2

 zA 

 zB
g 2 g
g 2 g
•
Since zA=zB and vB=0 we have...
p A v A2
p

 B
g 2 g
g
2
 240,000 m/s 


pB  p A v 2A  3600
  226 .5 m


g
2g
2  9.81 ms -2
•
Density value to be used is air density in
hurricane...
 Hur 
PAir
Patm, Air
 atm, Air 
74,500 Pa
1.225 kgm 3
101,591Pa
 0.898 kgm 3
Continued...
•
Using relation (A) from Part (I) we get ...
 Air
hdynamic 
h
 SW Air
hSW 
0.898 kg/m 3
1023 kg/m
3
 226.5 m
 0.2 m
•
Therefore pressure at (2) is 0.2m seawater column
lower than pressure at point (3) due to high wind
velocities, causing ocean to rise an additional 0.2 m.
Then total storm surge at point (2) becomes...
h2  h1  h dynamic  2.693 m  0.2 m  2.893 m
DISCUSSION
Problem involves highly turbulent flow and intense
breakdown of streamlines. Thus, applicability of
Bernoulli equation in Part (II) is questionable.
Furthermore, flow in eye of storm is not irrotational,
and Bernoulli constant changes across streamlines.
Bernoulli analysis can be thought of as limiting, ideal
case, and shows that rise of seawater due to highvelocity winds cannot be more than 2.893m.
Dimensional Analaysis 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?
Continued ...
WE NEED ...
d increases
from curve
to curve
1 page for Drag as
function of 2 variables
(e.g. velocity and
diameter)
1 page for
each value
of 
1 book for Drag as
function of 2 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 !?!?
Continued...
DRAG
GOAL IS TO COMPRESS SHELF
OF BOOKS INTO ONE SINGLE
GRAPH...
4 Independent
Experimental Parameters
How Could We Possibly
Achieve This?
•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!
• Intertia 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)
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
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?
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 
gives
M L T 2
M L T 2
Find by subst. Eq.(3a-c) into Eq.(4) ...
(4)
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
• Substitut Eq.(6a) and Eq.(6d) into Eq.(6b) ...
1  1   1
 1
(6c)
(6d)
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!
Continued...
• 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.
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
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)
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
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, units are
kg m
N 2
s
• Right-hand side of equation is:
Velocity  Diameter 2  Viscosity
m

s
m 
2

kg

ms
kg m 2
s2
Different dimensions on both sides of the
equation. Hence, formula cannot be right!
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...
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)
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
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
(6)
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.