DIMENSIONAL
ANALYSIS
AND SIMILARITY
DIMENSIONS AND UNITS



Dimensions: Measure of a physical quantity, e.g:
length, time, mass
Units: Assignment of a number to a dimension, e.g
: mm, m, kg, g
7 primary dimensions:

Mass
M
(kg)

Length
L
(m)

Time
t
(sec)

Temperature
T
(K)

Current
I
(A)

Amount of Light C
(cd)

Amount of matter N
(mol)



In Fluid Mechanics – only 3 considered (M, L & T)
All non-primary dimensions can be formed by a
combination of the primary dimensions
E.g:
{Velocity} = {Length/Time} = {L/T} = {LT-1}
{Force} = {Mass Length/Time} = {ML/T2} = {MLT-2}
DIMENSIONAL ANALYSIS
Dimensional analysis is a method for reducing the
number and complexity of experimental variables
that affect a given physical phenomenon, by using a
sort of compacting technique.
Primary purposes of dimensional analysis
To generate nondimensional parameters that help in
the
design
of
experiments
(physical
and/or
numerical) and in reporting of results.
To
obtain
scaling
laws
so
that
prototype
performance
can
be
predicted
from
model
performance.
To predict trends in the relationship between
parameters.
The benefits of dimensional analysis
 Saving in time and money
 Helps
planning
for
coming
experiment
or
theory/simulation (before we spend money on
computer analysis/simulation)
 Provides scaling laws that can convert data from
a cheap, small model to design information for an
expensive, large prototype.
 E.g : The force, F on a particular body shape
immersed in a fluid stream is depend on the body
length, L, fluid velocity, V, fluid density, 
and fluid viscosity, µ or F = f(L, V, , µ ).
 Because of the geometry and flow condition are so
complicated, the theory fail to yield the solution
for F.
 Therefore, the function of f(L, V, , µ ) must be
find experimentally (or numerically). If we want to
predict the effect of L to the F, we have to run the
experiment for 10 length L. For each L we need 10
values of V, 10 value of  and 10 value µ. Or in
other words we have to run about 10 000 experiments
which involved a very high cost. If each experiment
is cost about RM…, totally we have to spend about
RM….
 When using the dimensional analysis, we could reduce
the variable involved becomes, F/(V2L2) = g (VL/µ).
The experiment need to run only for 10 value of Re
not for each single variable of L, V,  or µ.
The  Buckingham Theorem
There are some method can be used in predicting the
nondimensional
parameters/group
Rayleigh,
Buckingham, Step–by-step (Ipsen).
The most popular method – the Theorem
Buckingham (Edgar Buckingham 1867-1940)
of

Based on two theorem of  Buckingham :
i. If a problem involving m variables, it can be
reduced to a relationship among independent
dimensionless products, where n is the minimum
number of reference dimensions required to
describe the variables.
ii.Each of  group is consist of n repeating
variables and one non-repeating variable.
7 steps in doing  Buckingham analysis
1. List and count m variables involved. If any
important variables are missing, dimensional
analysis will fail.
2. List the dimensions of each variable and count n
where n = number of primary dimension involved
(according to MLT OR FLT system)
3. Determine the  group exist where  = m – n
4. Select n repeating variables (based on the
guideline in selecting the repeating variables)
5. Form the group of ’s with combining the
repeating variables with one of the non-repeating
variable.
6. Check all the resulting  terms to make sure they
are dimensionless.
7. Express the final form as a relationship among
the  terms as : 1 = Φ(2, 3,…, m-n)
Guidelines for choosing Repeating Variables






Never pick the dependent variable.
Never
pick
parameters
that
are
already
dimensionless.
Never pick two parameters with the same
dimensions or with dimensions that differ by
only an exponent.
Chosen repeating parameters must represent all
the primary dimensions.
Pick simple parameters over complex parameters.
Chosen
repeating
parameters
must
not
by
themselves be able to form a dimensionless
group.
Example :
Consider flow of an incompressible fluid of
density,  and viscosity, µ through a long,
horizontal section of round pipe of diameter, D.
The velocity profile is uniform with V is the
average velocity across the pipe cross section.
Because of frictional forces between the fluid and
the pipe wall, there exists a shear stress, tw on
the inside pipe wall. The shear stress is also
constant down the pipe in the fully developed
region. We assume some constant average roughness, 
height along the inside wall of the pipe.
Develop a nondimensional relationship between shear
stress, τ and the other parameters in the problem.
Step 1 - List and count m variables involved
There
are
six
parameters
(dimensional
variables,
nondimensional variables, and dimensional constants) in
this problem; Therefore m = 6.
They are listed in functional form or, τ = f(, µ, D, V,
)
Step 2 – Determine the number of primary dimensions,
n
List the dimensions of each parameter and writing
each dimension with exponents since this helps with
later algebra.
Variables
Symbol
Dimension
Shear Stress
τ
[M1L-1T-2]
Density

[M1L-3]
Velocity
V
[L1T-1]
Diameter
D
[L1]
Viscosity
µ
[M1L-1T-1]
Surface Roughness

[L1]
Then identify the number of n, where n is a number
of primary dimensions involved. In this problem the
n = 3 (M, L and T).
Step 3 - Determine the  group exist
where  = m – n
In this problem, the number of ’s predicted is 3
or  = 6 – 3 = 3
(1, 2, 3)
Step 4 - Select n repeating variables.
We need to choose the repeating parameters base on n.
So, in this problem we need to choose 3 repeating
parameters since n=3.
Since this is often the hardest (or at least the most
mysterious) part of the method of repeating variables,
there are several guidelines about choosing repeating
parameters that can be followed.
From the guidelines, we choose , D, V
variables/parameters for this problem.
as repeating
Guidelines for choosing Repeating Variables
• Never pick the dependent
•
•
•
•
τ = f (ρ, µ, D, V, ε )
Variables
Symbol
Dimension
Shear
Stress
τ
[M1L-1T-2]
Density

[M1L-3]
Velocity
V
[L1T-1]
Diameter
D
[L1]
Viscosity
µ
[M1L-1T-1]
Surface
Roughness

[L1]
Chosen r/v
•
variable.
Never pick parameters that
are already dimensionless.
Never pick two parameters
with the same dimensions or
with dimensions that differ
by only an exponent.
Chosen repeating parameters
must represent all the
primary dimensions.
Pick simple parameters over
complex parameters whenever
possible.
Chosen repeating parameters
must not by themselves be
able to form a dimensionless
group.
Variables
Symbol
Dimension
Shear
Stress
τ
[M1L-1T-2]
Density

[M1L-3]
Velocity
V
[L1T-1]
Diameter
D
[L1]
Viscosity
µ
[M1L-1T-1]
Equate the exponents of each Surface
primary dimension
Roughness
independently to solve for a,
b and c.

[L1]
Chosen r/v
where a, b and c are constant exponents that need to be
determined.
M :
L :
T :
0 = 1 + a
0 = - 1 - 3a + b + c
0 = -2-c
a = -1, b = 0, c = -2
Therefore,
𝜋1 = 𝜏 𝜌
𝜏
𝜋1 = 2
𝜌𝑉
−1
𝐷
0
𝑉
−2
Variables
Symbol
Dimension
Shear
Stress
τ
[M1L-1T-2]
Density

[M1L-3]
Velocity
V
[L1T-1]
Diameter
D
[L1]
Viscosity
µ
[M1L-1T-1]
Surface
Roughness

[L1]
Chosen r/v
Solve for a, b and c.
M :
L :
T :
0 = 1 + a
0 = - 1 - 3a + b + c
0 = - 1 - c
a = -1, b = -1, c = -1
Therefore,
𝜋2 = 𝜇 𝜌
−1
𝐷
−1
𝑉
𝜇
𝜌𝐷𝑉
𝜋2 = = =𝑅𝑒
𝜌𝐷𝑉
𝜇
−1
Variables
Symbol
Dimension
Shear
Stress
τ
[M1L-1T-2]
Density

[M1L-3]
Velocity
V
[L1T-1]
Diameter
D
[L1]
Viscosity
µ
[M1L-1T-1]
Surface
Roughness

[L1]
Chosen r/v
Solve for a, b and c.
M :
L :
T :
0 = a
0 = 1 - 3a + b + c
0 = - c
a = 0, b = -1, c = 0
Therefore,
𝜋3 = 𝜀 𝜌
𝜀
𝜋3 =
𝐷
0
𝐷
−1
𝑉
0
Variables
Symbol
Dimension
Shear
Stress
τ
[M1L-1T-2]
Density

[M1L-3]
Velocity
V
[L1T-1]
Diameter
D
[L1]
Viscosity
µ
[M1L-1T-1]
Surface
Roughness

[L1]
Chosen r/v
Solve for a, b and c.
Step 6
Check all the resulting 𝜋 terms to make sure they
are dimensionless.
CHECK!!!
𝜏
𝑀
𝐿3 𝑇 2
𝜋1 =
= 2 𝑥
𝑥 2
2
𝜌𝑉
𝐿𝑇
𝑀 𝐿
=1
𝜌𝑉𝐷
𝑀𝐿−3 𝐿 𝐿𝑇 −1
𝜋2 =
=
=1
−1
−1
𝜇
𝑀𝐿 𝑇
𝜀 𝐿
𝜋3 = =
=1
𝐷 𝐿
Step 7
Express the final form as a relationship among the
𝜋 terms as :
𝜋1 = Φ 𝜋2 , 𝜋3 , … . , 𝜋𝑚−𝑛
𝜏
𝜌𝑉𝐷 𝜀
=Φ
,
2
𝜌𝑉
𝜇 𝐷
𝜏=
𝜌𝑉 2 Φ
𝜀
𝑅𝑒,
𝐷
Remember…
The method of repeating variables properly predicts
the functional relationship between
dimensionless groups.
However, the method of repeating variables cannot
predict the exact mathematical form of the
equation.
Example 2
A thin rectangular plate having a width w and a
height h is located so that it is normal to a
moving stream of fluid. Assume the drag, that the
fluid exerts on the plate is a function of w and h,
the fluid viscosity and density, and respectively,
and the velocity V of the fluid approaching the
plate. Determine a suitable set of pi ( 𝜋) terms to
study this problem experimentally.
Tutorial
It is desired to determine the wave height when
wind blows across a lake. The wave height, H, is
assumed to be a function of the wind speed, V,the
water density, the air density, the water depth, d,
the distance from the shore, and the acceleration
of gravity, g, as shown in Fig. P7.7. Use d, V, and
as repeating variables to determine a suitable set
of pi (𝜋) terms that could be used to describe this
problem.
SIMILARITY
Normally, in practical, most of fluid flow problem
are too complex, also deals with big scale
system/objects (e.g air planes, cars, ships, wind
turbines etc.)
The equations are either unknown or too difficult
to solve.
Experimentation is the only method of obtaining
reliable information.
In most experiments, geometrically-scaled models
are used (to save time and money).
However, experimental conditions and results must
be properly scaled so that results are meaningful
for the full-scale prototype.
3 types of similarity :
i. Geometric Similarity,
ii.Kinematic Similarity and
iii.Dynamic Similarity
Complete Similarity is achieved only if all 3
conditions are met. This is not always possible,
e.g., river hydraulics models.
Geometric Similarity
-
the model must have the same shape as the prototype.
each length dimension must be scaled by the same
factor.
E.g : For above one-tenth-scale model of prototype wing
the model length are 1/10 as large , but the angle of
attack is remain the same : 10 not 1.
Model nose is 1/10 as large
Model surface roughness is 1/10 as large
If the prototype is constructed with protruding
fasteners, the model also should have homologous
protruding fasterners 1/10 as large.
Kinematic Similarity
Velocity at any point in the model must be
proportional (velocity scale ratio must be the same
for both scale model and prototype).
Dynamic Similarity
All forces in the model flow scale by a constant
factor to corresponding forces in the prototype
flow.
Example :
The aerodynamic drag of a new sports car is
to be predicted at a speed of 50.0 mi/h at
an air temperature of 25°C. Automotive
engineers build a one fifth scale model of
the car to test in a wind tunnel. It is
winter and the wind tunnel is located in an
unheated building; the temperature of the
wind tunnel air is only about 5°C.
Determine how fast the engineers should run
the wind tunnel in order to achieve
similarity between the model and the
prototype.