DIMENSIONAL ANALYSIS
In many experimental set-ups we may not know
how the "system" under investigation will behave.
For a given problem it may be impossible to test
for every conceivable condition.
Therefore, a limited number of selected tests
must be performed and from these tests, for
example using a model of a structure, the real fullscale structure has to be designed.
As a guide to what "selected tests" should be
carried out we must make some educated
guesses as to which parameters are important.
This only comes from experience.
However, one useful method of finding out which
parameters are important and how they react with
each other is by the use of DIMENSIONAL
ANALYSIS.
All equations that describe a system must have
their UNITS balanced. For example
F = m . a
The dimensional units of force (F) must balance
the dimensional units of mass (m) and
acceleration (a).
In other words DIMENSIONAL HOMOGENEITY
exists on either side of the equation. Hence,
F = [ M L T-2 ]
m = [M]
a = [ L T-2 ]
where [ ] denotes "units of" and the fundamental
units are;
M = mass
L = length
T = time
Typical quantities and their units
Mass
m
M
Length
L
L
Time
t
T
Force
F
M L T-2
Velocity
U
L T-1
Pressure
P
M L-1 T-2
Acceleration
a
L T-2
Gravity
g
L T-2
Density
ρ
M L-3
Dynamic viscosity
μ
M L-1 T-1
Kinematic viscosity
v
L2 T-1
Discharge
Q
L3 T-1
Specific weight
σ
M L-2 T-2
The Buckingham π theorem
In many applications simple expressions
involving just one quantity are not relevant and so
we must then consider forming NONDIMENSIONAL GROUPS and examine how these
groups relate to one another.
One method of forming these groups is known as
the BUCKINGHAM π THEOREM.
This theorem states that:If there is a relationship between n parameters
(e.g. velocity, pressure, length, etc) we may write
this as
g ( a1, a2, a3, ......... , an ) = 0
where g is the function of the various parameters.
If m is the number of dimensions involved in this
expression then (n - m) "independent
dimensionless ratios" or "π parameters" may be
expressed.
Hence;
G ( π1, π2, π3, ......... , πn-m ) = 0
or
π1 = G1 ( π2, π3, ......... , πn-m )
We cannot predict G or G1 and the relationship
between the independent π parameters must be
found experimentally.
How to solve a problem
(1) List the parameters involved (if too many are
chosen they will be discarded during the
analysis)
(2) Evaluate the dimensionless π parameters
(3) Determine experimentally the relationship
between the π parameters.
Example
To find how the force F acting on a smooth
sphere of diameter D depends on the velocity U of
the sphere and the density ρ and viscosity μ of
the fluid through which it is moving.
Hence, we wish to know how;
F = f ( U, D, ρ, μ )
We may rewrite this as:
g ( F, U, D, ρ, μ ) = 0
so that we have 5 parameters (n = 5).
Now we select the primary dimensions for these
parameters which are: M, L and T. Hence, there
are 3 primary parameters (m = 3).
Note that:
D = [L]
F = [ M L T-2 ]
ρ = [ M L-3 ]
U = [ L T-1 ]
μ = [ M L-1 T-1 ]
The number of π groups will then be (n - m)
groups. Hence, 2 dimensionless groups will result
so that
G ( π1, π2 ) = 0
We can find these two groups by choosing what
are termed the REPEATING VARIABLES. These
must not include the dependent variable (F in this
case) but together they must contain all of the
primary variables (that is; M, L and T). Hence, we
may choose;
U, D and ρ
Then we use these repeating variables with each
of the remaining variables, in turn, to establish
the non-dimensional π parameters.
Hence;
π1 = Ua Db ρc F
= [ L ] a [ L ] b [ M ] c [ M L ] = M0 L0 T0
T
L3
T2
Equating exponents of M, L and T gives;
M: c + 1 = 0
L: a + b - 3c + 1 = 0
T: - a - 2 = 0
Hence;
 c = -1
 b = -2
 a = -2
π1 = U-2 D-2 ρ-1 F =__F___
ρ U2 D2
Similarly,
π2 = Ud De ρf μ
= [ L ] d [ L]
T
e
[ M ] f [ M ] = M0 L0 T0
L3
LT
Again, equating exponents gives;
M: f + 1 = 0
L: d + e - 3f - 1 = 0
T: - d - 1 = 0
Hence;
π2 = U-1 D-1 ρ-1 μ =
 f = -1
 e = -1
 d = -1
μ__
UDρ
As a result, the function would be in the form;
G(
F
, μ ) = 0
ρ U2 D2 U D ρ
or
F
= G1 (
μ )
ρ U2 D2
UDρ
The nature of the above relationship may then be
determined experimentally.
Summary of the procedure
(1) List all possible parameters involved in the
problem (e.g U, D, ρ, F, g, μ etc)
(2) Select the set of primary dimensions (e.g. M,
L and T). Note that sometimes we can use
force F as a primary dimension. There are
other dimensions used in some physical
systems such as the temperature θ used for
heat transfer and the charge q used in
electrical systems.
(3) Write the functional relationship.
For example; g ( U, D, ρ, F, g ) = 0
(4) Select the repeating variables. Note;
(a)
Don’t use the dependent quantity
(e.g. Force).
(b)
The variables should contain all m
dimensions of the problem (M, L and T
must be in the repeating variables).
(c)
Choose one variable from the geometry
of the system (e.g. D), one variable from
the fluid properties (e.g. ρ) and one
variable from the flow (e.g. U).
(5) Write the π parameters in terms of exponents,
e.g.
π1
= Ua Db ρc F
and substitute for the dimensions
π1
= [ L ] a [ L ] b [ M ] c [ M L ] = M0 L0 T0
T
L3
T2
(6) Equate the exponents of M, L and T and solve
the simultaneous equations
(7) Substitute back the values of these exponents
into the π parameter
(8) Repeat for all π parameters
(9) Establish the functional relationship in terms
of π parameters.
G ( π1, π2, ..... , πn-m ) = 0
or solve for one π parameter
π1 = G1 ( π2, π3, ...... , πn-m )
Example
As a fluid flows through a circular pipe (diameter
D, length l) there is a pressure loss ΔP due to
friction between the fluid and the rough pipe wall
(the characteristic height of the pipe wall
roughness is e). What is the relationship between
ΔP and the relevant geometrical, fluid and flow
parameters?
Answer
Assume that
ΔP = f ( ρ, U, D, l, μ, e )
which may be rewritten as
g ( ΔP, ρ, U, D, l, μ, e ) = 0
Here, we have 7 parameters (n = 7) and if we
select the 3 primary dimensions of M, L and T (m
= 3) we end up with (n - m) π parameters, that is 4
π parameters.
Choosing ρ, U and D as the repeating parameters
we have the following results for π1, π2, π3 and π4;
π1 parameter
π1 = ρa Ub Dc ΔP
= [ M ] a [ L ] b [ L ] c [ M ] = M0 L0 T0
L3
T
L T2
M: a + 1 = 0
L: - 3a + b + c - 1 = 0
T: - b - 2 = 0
 a = -1
 c = 0
 b = -2
π1 = ρ-1 U-2 D0 ΔP = ΔP
ρ U2
π2 parameter
Hence,
π2 = ρd Ue Df μ
= [ M ] d [ L ] e [ L ] f [ M ] = M0 L0 T0
L3
T
LT
M: d + 1 = 0
L: - 3d + e + f - 1 = 0
T: - e - 1 = 0
Hence,
π2 =
μ__
ρUD
 d = -1
 f = -1
 e = -1
π3 parameter
π3 = ρg Uh Di l
= [ M ] g [ L ] h [ L ] i [ L ] = M0 L0 T0
L3
T
M: g = 0
L: - 3g + h + i + 1 = 0
T: - h = 0
Hence,
π3 =
 g = 0
 i = -1
 h = 0
l
D
π4 parameter
π4 = ρj Uk Dl e
= [ M ] j [ L ] k [ L] l L = M0 L0 T0
L3
T
M: j = 0
L: - 3j + k + l + 1 = 0
T: - k = 0
Hence,
π4 = e
D
 j = 0
 l = -1
 k= 0
From these π
relationship is
parameters
the
functional
G ( π1, π2, π3, π4 ) = 0
or
ΔP = G1 ( μ
,
ρ U2
ρUD
l ,
D
e )
D
Non-dimensional groups
Several non-dimensional groups of variables
regularly occur in fluid mechanics. In many cases
the variable group has the name of the engineer
or scientist associated with it.
For example, the group
Re = ρ U l
μ
is known as the REYNOLDS NUMBER, after
Osborne Reynolds (1880s) studied the change
from laminar to turbulent flow in a pipe. The same
number is also relevant to many other types of
flow.
Here, U is the flow velocity and l is a
characteristic length.
Note that Re is the ratio of
in the flow.
Inertia forces
Viscous forces
In aerodynamics PRESSURE COEFFICIENTS are
used. These are non-dimensional representations
of pressure of the form
Cp =
ΔP_
½ ρ U2
where ΔP is taken to be the local pressure minus
a reference pressure value (usually the static
pressure).
Another non-dimensional group, related to free
surface flows (i.e. flow at an air-water interface), is
known as the FROUDE NUMBER (after William
and Robert Froude).
Fr =
U__
(g l )
It may also be quoted as
Fr2 =
Froude number is the ratio of
in the flow.
U2
gl
Inertia forces
Gravity force
Model studies
The main usefulness of non-dimensional numbers
is in model studies.
In many design situations we cannot predict how
a structure is going to repsond to the fluid flow
over or around it.
e.g. Turbines, Spillways
Flows around bridges, buildings, etc
It is not practical to test full-scale models of these
structures and, therefore, we must rely upon
small-scale models.
If we ensure that the non-dimensional number for
the full-scale case is the same as that for the
small-scale model then we may relate the data
from the model tests to the real structure.
How do we do this?
We must ensure that the model and the test
conditions are such that the forces, moments and
dynamic loads are scaled correctly relative to the
full-scale structure.
What are the criteria to ensure similarity between
the model and full-scale?
(1) Geometric similarity - model and prototype (i.e.
full-scale) to be of the same shape and all linear
dimensions scaled accordingly.
(2) Kinematic similarity - the flows around the
model and prototype should be related in both
magnitude and direction by a scale factor. The
flow patterns must be the same for both model
and prototype.
(3) Dynamic similarity - this exists when the
forces involved in both the model and prototype
are related in magnitude by a constant scale
factor at all points in the flow (i.e. viscous forces,
pressure forces and surface tension forces must
all be considered).
When dynamic similarity exists the data from
model studies may be used with confidence to
predict forces on the prototype.
To do this we use the non-dimensional numbers
(Re, Fr, Cp, etc).
Thus, if a model and a prototype are geometrically
similar and the flows are kinematically similar,
then we can consider Reynolds number for
example;
Remodel = Reprototype
ρUl
μ
model
= ρUl
μ
prototype
There is now no reason why we can't use a
different media to test the model in from that
experienced by the prototype (for example, water
rather than air), provided the Reynolds number
similarity is maintained.
Similarly, if we consider the Froude number;
Frmodel = Frprototype
U
(g l)
=
model
U_
(g l)
prototype
For the case of the sphere moving through a fluid
that we looked at earlier, we found;
π1 =
F
ρ U2 D2
and
π2 =
μ__
UDρ
Since π2 is a non-dimensional number it may also
be expressed as the Reynolds number;
UDρ
μ
Hence;
F
= G1 ( U D ρ )
ρ U2 D2
μ
If we were to test a model sphere to predict the
force on a large prototype sphere then
(a)
Geometric and kinematic similarity must
be arranged
(b)
Dynamic similarity must be ensured, i.e.
ρUD
μ
model
= ρUD
μ
prototype
If this is true then
F
ρ U2 D2
=
model
F__
ρ U2 D2
prototype
and so the forces measured on the model may be
used to predict the forces on the prototype.
Example
The drag of a large spherical transducer is to be
predicted, based on wind tunnel tests using a
model sphere. The model is 0.2m in diameter and
the prototype is 0.4m.
If the prototype is going to be towed at 2 m/s in
water, determine the velocity at which the wind
tunnel model should be tested.
If the drag of the model is measured as 10N
estimate the drag of the prototype.
Note that;
ρair = 1.2 kg/m3
μair = 18.15x10-6 kg/ms
ρwater = 1000 kg/m3
μwater = 1.3x10-3 kg/ms
Solution
For a sphere we have already shown that
F
= f
ρ U2 D2
(UDρ)
μ
Now, for dynamic similarity
(Re)m = (Re)p
ρm Um Dm = ρp Up Dp
μm
μp
i.e.
Hence,
Um
.
= 2 .
0.4 . 103 . 18.15 x 10-6
0.2
1.2
1.3 x 10-3
= 46.54 m/s
Dp
Dm
. ρp . μm
ρm
μp
= Up
In addition, for dynamic similarity we have
ρm
Fm
=
Fp
_
Um2 Dm2
ρp Up2 Dp2
Hence,
ρp . Up2 . Dp2
ρm
Um2
Dm2
Fp = Fm .
= 10 .
103 . 22
. 0.42
1.2
46. 542
0.22
= 61.56 N
Note:
(1) The forces on the model and prototype are
not the same but the non-dimensional
numbers are.
(2) The model tests may be performed in a
different environment as long as the flow
conditions are dynamically similar (i.e. the
Reynolds numbers are matched).
(3) Similarity may be applied to all branches of
fluid mechanics, such as
(a) Immersed objects (e.g. wind loading on a
building or bridge immersed in the earth’s
atmospheric boundary layer).
(b) Surface flows (e.g. spillways or ship
design).
(c) Machines (e.g. Pelton wheel, fans and
pumps).
(d) Rivers and channels.