Lecture 6
Final Version
Contents
• Dimensional
Analysis to Evaluate Drag
on Bodies Moving in Homogeneous Fluid
at Low Reynolds Number
• Dimensional Analysis to Evaluate Drag
on Bodies Moving in Homogeneous Fluid
at High Reynolds Number
• Dynamic Similarity and Model Testing
(Start)
What Did We Do In Last Lecture?
Bernoulli Equation, Nomenclature
Examples Bernoulli Equation
Example 1: Pitot-Static / Prandtl Tube
Example 2: Water Flowing in Open Channel
Example 3: Venturi Tube
Example 4: Beaver Burrow
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, ρ
Fluid Viscosity, µ
Write
D = F (d ,V , ρ , µ )
(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 ...
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 ρ
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 , ρ , µ )
(1) - repeated
reduces to ...
D = F (d ,V , µ )
(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 , µ )
(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 ⋅ µ1
Units:
kg m
s2
m3 kg kg m 4
=
Units: m ⋅ 3 ⋅
s sm
s4
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 , µ )
N=
Units:
Dimensions:
kg m
s2
M L T −2
M : Mass
m
L
m
s
L T −1
L : Length
(2) - repeated
kg
ms
M L−1 T −1 (3a-d)
T : Time
WANT !!!
α, β , γ
such that
α
β
D ∝ V ⋅ d ⋅µγ
gives
M L T −2
M L T −2
Find by subst. Eq.(3a-d) into Eq.(4) ...
(4)
Continued...
D ∝ V α ⋅ d β ⋅µγ
ML T
−2
=
(L T )
−1 α
(L )
β
(ML
(4) - repeated
−1
T
)
−1 γ
Collect corresponding terms ...
M L T −2
= Mγ
Lα + β −γ
T −α −γ
(5)
By comparing exponents ...
• … of M on left and right hand side of Eq.(5)
1=γ
(6a)
• … of L on left and right hand side of Eq.(5)
1=α + β −γ
(6b)
• … of T on left and right hand side of Eq.(5)
− 2 = −α − γ
(6c)
• Substitut Eq.(6a) into Eq.(6c) ...
− 2 = −α −1
α =1
• Substitut Eq.(6a) and Eq.(6d) into Eq.(6b) ...
1 = 1+ β −1
β =1
(6d)
Continued...
• In summary we get...
D ∝ V α ⋅ d β ⋅µγ
(4) - repeated
… where ...
α =1
β =1
γ =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 µ
3π for sphere. Must be obtained
from experiments or theory
is the ...
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 α ⋅ d β ⋅µγ
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
(2a )2
gives 4 a
2
(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 µ is given
by ...
D = 3π V ⋅ d 2 ⋅ µ
(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
N=
kg m
s2
• 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 ...
ρ
p∞ + V∞2 = ps
2
ps − p∞ =
ρ
2
V∞2
(1)
• 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 =π⎜ ⎟
⎝2⎠
• Since
2
d : Diameter
PRESSURE= DRAG/AREA
⎛d ⎞
Drag ≈ ( ps − p∞ ) π ⎜ ⎟
⎝2⎠
With Eq. (1)
• Divide
(5)
Drag ≈
ρ
2
V∞2
d
π ⎛⎜ ⎟⎞
⎝2⎠
2
2
through by right hand side and DEFINE the
DRAG COEFFICIENT CD
CD =
Drag
ρ
2
V∞2
d
π ⎛⎜ ⎞⎟
⎝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
ρ
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.
Dimensional Analysis to Evaluate Drag on
Bodies Moving in Homogeneous Fluid at
High Reynolds Number
• In previous section derived Stokes law for drag acting on sphere
at Re<<1.
• For very small Re: Inertia Forces << Viscous Forces. This
enabled neglecting fluid density in dimensional analysis.
• Now carry out dimensional analysis for Re>>1, I.e. for case when
we can no longer neglect density of fluid
• This time represent drag in non-dimensional form, i.e. by drag
coefficient CD ...
CD
Dimensionless
= F (V , d , µ , ρ )
(1)
These quantities must be
‘mixed together’ by function
F in a dimenensionless group.
• Assume dimensionless group of form ...
Π ∝ V α d β µγ ρδ
• Arbitrarily
set ...
… and find ...
(2)
α =1
β,γ ,δ
Will discuss consequences of setting α = 1at end.
Need to do this because only have 3 dimensions (M,L,T) in
problem. This can only produce (see for low Re number) 3
simultaneous equations; but have 4 unknowns α , β , γ , δ
Continued...
• As with low Re case, look at ...
Quantity
Symbol
Units
Dimensions
Velocity
V
ms
Diameter
d
m
L1 T −1
L
Viscosity
µ
kg m s
M 1 L−1 T −1
Density
ρ
kg m 3
M 1 L−3
CD
Dimensionless
M 0 L0 T 0
Drag
• On previous slide assumed ...
Π ∝ V1
dβ
M 0 L0 T 0
(L T )
1
−1 1
µγ
ρδ
( L ) β (M 1 L−1 T −1 )γ (M 1 L−3 )δ
• Multiply out and collect terms ...
M 0 L0 T 0 ∝ M γ +δ L1+ β −γ −3δ T −1−γ
• Compare exponents ...
• M on l.h.s. and r.h.s : 0 = γ + δ
• L on l.h.s. and r.h.s : 0 = 1 + β − γ − 3δ
• T on l.h.s. and r.h.s : 0 = − 1 − γ
• Solve system (a) -(c) ...
δ = 1 (d)
• (c) in (a) ...
• (c) and (d) in (b)
β =1
and we
assumed
(a)
(b)
γ = − 1 (c)
α =1
Continued...
• In summary we get ...
Π ∝ V d µ
1
1
−1
ρ
1
=
Vdρ
where we used ν = µ ρ
• But what if we had chosen α
µ
=
Vd
ν
= Re
The
dimensionless
drag is a function
of the Reynolds
number!!!
to be different from 1 ?
Would have ended up with ...
Π ∝ Reκ
Thus, CD could involve terms containing any power of Re
• In any case one can conclude that ...
C D = F (Re )
For uniform homogeneous flow past a
particular, but arbitrary, shape of body the nondimensional forces (i.e. drag coefficient) depend
O-N-L-Y on Re
Continued...
Hence, our shelf of books, from beginning of
section, can be replaced by ONE single curve
displaying ...
C D vs. Re
CD
Re =
CD =
ρ
2
Vd
ν
Drag Force
V 2 × Cross. Sect. Area
Drag Force = C D ×
ρ
2
V 2 × Cross. Sect. Area
Continued...
NOTE:
• At low Re, i.e. at Re<<1, one has Stokes Law ...
D = 3π V d µ
D ∝ V
So, doubling velocity results in double drag!
• At higher Reynolds
numbers, in range between about
10<Re<500000, graph on previous page shows that...
1≈
CD ≈ 1
ρ
2
D≈
ρ
2
D
V 2 × Cross. Sect. Area
V 2 × Cross. Sect. Area
D∝ V2
So, doubling velocity results in quadrupling
the drag!
Look back to section where we defined drag coefficient. There
argued that its value must be order 1. Hence, it should not come
as a surprise that graph, which is obtained from experiments,
shows this.
Continued...
Finally also note that ...
•…
decreasing drag coefficient, with increasing Reynolds
number does (generally) NOT imply that drag forces on
body decrease!!!
• To see this assume that speed of an object increases from V1
to V2=2V1 …
• At low Re (Stokes Law) this results in an increase of the
drag force from F1 to F2=2F1 …
THEN,... Re1 =
L V1
Re 2 =
ν
L V2
ν
=
L (2V1 )
ν
= 2 Re1
WHILE,...
CD (Re1 ) =
ρ
2
C D (Re 2 )
=
F1
V12 × A
ρ
2
F2
V22 × A
=
(2 F1 )
ρ
(2V1 ) 2 × A
2
2
F1
= ×
4 ρV2 ×A
1
2
1
= × C D (Re1 )
2
Hence, drag force has doubled but drag
coefficient has halved!
Nevertheless, there is a region where drag force really
decreases as drag coefficient decreases. For cylinders/spheres
this happens in Re range between about 200,000 and 500,000
where drag coefficient suddenly decreases sharply. Will discuss
this later.
Continued...
And now really, really finally note that ...
• The
Dimensional Analysis we performed is a simplified
version of what is known as the ...
Buckingham Π - Theorem
or just
Π - Theorem
Look at details concerning this as homework!
You will find it in most introductory books on
fluid dynamics.
Dynamic Similarity and Model Testing
What are the required conditions for
model experiments?
• Want to test scale model of a prototype in wind tunnel.
• Question: Why would any of the measured model data be of any
relevance to the prototype? After all they are of different size!
• So, what conditions need to be satisfied such that wind tunnel data
tell us anything about the prototype?
One Needs:
•Geometric Similarity
Model has the same shape as prototype
•Kinematic Similarity
Additionally it is required that ...
Fluid Velocity For Model
at corresponding points = Const.
Fluid Velocity For Prototype
•Dynamic Similarity
Additionally to Geo. Sim. and Kin. Sim. it is required
that: Non.-dimensional forces are the same for model and
prototype, e.g. ...
(C D )m = (C D ) p
Know that for body in homogeneous fluid drag coefficient is
function of Re number o-n-l-y. Hence, Dyn. Sim. only possible if
(Re )m = (Re ) p
Dyn, Sim. Hard
to achieve!