Dimensional Analysis And Similitude
Similitude is a concept applicable to the testing of engineering models.
A model is said to have similitude with the real application(Prototype) if the two share,
Geometric similarity:
The model is the same shape as the prototype(scaled)
L L
  p   m
h h
Kinematic similarity : Fluid flow of both the model and prototype
must undergo similar time rates of change motions.
 v1   v1 
  p   m
(fluid streamlines are similar)
 u1   u1 
 v2   v2 
  p   m
 u2   u2 
Dynamic similarity : Ratios of all forces acting on corresponding fluid particles and
boundary surfaces in the two systems are constant. (ie: when forces at corresponding points of the
model and the prototype are similar)
v1
u1
h
Fbx
Fax
v2
u2
 Fa x   Fa x 

 

 Fa  p  Fa  m
y 
y 


Fby
Fay
Prototype
L
v1
h
u1
 Fbx   Fbx 

 

 Fb  p  Fb  m
y 
y 


Fbx
Fax
Fay
Fby
v2
u2
Model
L
If the two systems are dynamically similar , its implies that the two systems are kinematically and geometrically similar
Dr. Neel Gunasekera
Dimensional Analysis And Similitude
Similitude is a concept applicable to the testing of engineering models in
order to predict the actual systems by extrapolating the results of the
experiment .
Factors that is required to be consider
•No of variables ( parameters ) that should be selected for testing ?
•How do we select the most effective minimum no of parameters from the
selected list ?
• Can be used the concept of non-dimensional parameters
•Ex: In flow through pipes,
• If we parameterized the experimental results based on the Reynolds
Number then we can incorporate effects of density, diameter, velocity and
the kinematic viscosity in a single non-dimensional variable (only a number).
Dr. Neel Gunasekera
Reynolds Law of Similarity
Re 
dv

If the Reynolds number (Re) of two systems are equal, we can
assume that the flow behavior in both systems will be similar.
This means that the ratio of inertial forces to viscous forces is the
same in both cases, leading to similar flow patterns, whether it's
laminar or turbulent flow.
So, if the Reynolds numbers match, the fluid dynamics (such as
velocity distribution, pressure drop, and flow patterns) in the
two systems will behave similarly, even if the size or scale of the
systems differ.
This principle allows engineers and scientists to use small-scale
models (like in wind tunnels or water channels) to predict the
behavior of larger systems accurately, provided the Reynolds
number is matched.
Dr. Neel Gunasekera
Model Testing
Resistance R to the motion of a completely submerge body is given by ,
𝑅 = 𝜌𝑉 2 𝐿2 𝜙
𝑉𝐿
𝜈2
Where, 𝜈: 𝐾𝑖𝑛𝑒𝑚𝑎𝑡𝑖𝑐 𝑉𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦, 𝑉: 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑡𝑕𝑒 𝑓𝑙𝑜𝑤
𝐿: 𝐿𝑒𝑛𝑔𝑡𝑕 𝑜𝑓 𝑡𝑕𝑒 𝑏𝑜𝑑𝑦 𝑎𝑛𝑑 𝜌: 𝑑𝑒𝑛𝑠𝑖𝑡𝑦
If the resistance of a 1:8 scale air-ship model when tested in water at 12 m/s
is 220N, what will be the resistance in air of the air ship at the corresponding
speed?
𝜋1 =
𝑅
𝜌𝑉 2 𝐿2
𝑉𝐿
𝜋2 = 𝜈 = 𝜌DV/
The model study satisfy the Reynolds law of similarity,
Given data 𝜈-air / 𝜈-water =13 and 𝜌-water/ 𝜌-air = 810
𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑖𝑛𝑔 𝜋2,
𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑖𝑛𝑔 𝜋1,
𝑉𝑝 =𝑉𝑚
𝜋1 =
𝐿𝑝
𝜈𝑝
𝐿𝑚
𝜈𝑚
𝑅
𝜌𝑉 2 𝐿2
=12x1/8x13=19.5m/s
Resistance = 45.9N
Dr. Neel Gunasekera
Non Dimensional Parameters
Non-dimensional parameters are used to compare different systems and scale
models to real-world situations. They eliminate units of measurement (like
meters or seconds) and focus on ratios, so you can compare systems of different
sizes or types more easily.
For example, the Reynolds number is a non-dimensional parameter that helps
predict whether the flow will be smooth or turbulent, regardless of the size of the
system. If two systems have the same Reynolds number, their flow behavior
should be similar, even if one is a small model and the other is a real-life
situation.
Basically, these parameters help us understand and predict how things will
behave without worrying about specific units or sizes.
Dr. Neel Gunasekera
Non Dimensional Parameters
Dr. Neel Gunasekera
Non Dimensional Parameters
Dr. Neel Gunasekera
Non Dimensional Parameters
Dr. Neel Gunasekera
Non Dimensional Parameters
Dr. Neel Gunasekera
Dimensional Analysis
Why do we need dimensional analysis ?
Experiments, especially at full scale, can be expensive and difficult to carry out.
The data collected from these experiments can be complex and challenging to
interpret.
Dimensional analysis provides a strategy for choosing relevant data and how it should
be presented.
Allows us to create functional relation ship between dependent and independent
variables using non dimensional parameters in a systematic way.
Dimensional analysis provide a scientific basis for modelling and scaling.
It provides a way to plan and carry out experiments, and enables one to scale up
results from model to prototype.
Dr. Neel Gunasekera
Dimensional Analysis
What is dimensional analysis ?
Concepts:
•Dimensions and Units
•Fundamental Dimensional Quantities and Derived Quantities
•Dimensional Reasoning and Dimensional Homogeneity
•Dimensionless Quantities
Dr. Neel Gunasekera
Dimensional Analysis
Concepts:
•Dimensions and Units
A dimension is a measure of physical quantity (with out numerical values) and Unit is
a way to assign a number to that dimension.
Length is a dimension that is measured in units such as m ,mm, feet, miles etc.
It is necessary to know the magnitude of each dimension (property) and the system of
units (SI, US), to complete the description of the physical situation.
The system of units chosen does not affect the real size of the object (only the numerical
value of its measurement) – width of the object (w) = 304.8 mm =12 inch = 1 feet
• Fundamental Dimensional Quantities and Derived Quantities
The fundamental dimensional quantities are L , T , M , θ and A to represent mass,
length, time, temperature and charge respectively.
Dimensions of all the other quantities can be derived in terms of these primitive
dimensions [M] [ L] [ T] [θ] [A]
Dr. Neel Gunasekera
Dimensional Analysis
Concepts:
Dimensions of Derived Quantities
Dr. Neel Gunasekera
Dimensional Analysis
Concepts:
A moving object could be described in terms of its mass, length, area or volume,
velocity and acceleration. The temperature ,density and viscosity of the medium
through which it moves – would also be of importance.
The measures of these properties used to describe the physical state of the object or the
system are known as its dimensions.
The numerical values has been used to quantify the dimensions.
Property
Dimension
Unit
Value
Radius
r
[L]
[m]
0.03 m
Mass
m
[M]
[Kg]
0.25 Kg
Volume
v
[L3]
[m3]
Surface Area
A
[L2]
[m2]
Velocity
v
[LT-1]
[ms-1]
Acceleration
a
[LT-2]
[ms-2]
Density of Medium
ρ
[ML-3]
[Kgm-3]
900 Kgm-3
Viscosity of Medium
γ
[L2T-1]
[m2s-1]
0.0025 m2s-1
Surface Tension
η
[MT-2]
[Kgs-2]
0.023Kgs-2
Acceleration of Gravity
g
[LT-2]
[ms-2]
9.8ms-2
Bouncy force
B
[MLT-2]
[kgms-2]
Temperature of Medium
T
[θ]
[K]
2ms-1
350 K
Dr. Neel Gunasekera
Dimensional Analysis
Concepts:
Dimensional Reasoning and Dimensional Homogeneity
Dimensional reasoning :
For an equation describing a physical situation to be true,
The two sides must be equal both numerically and dimensionally.
ie: The all terms are of the same kind and have the same dimensions
This Property is called the Dimension Homogeneity
EX: Bernoulli's Equation
2
2
p1 V1
p V

 h1  2  2  h2
g 2 g
g 2 g
[ ML1T 2 ]
[ LT 1 ]2
LHS 

 [ L]
[ ML3 ][ LT 2 ] [ LT 2 ]
LHS  [ L]  [ L]  [ L]
Dimensional Homogeneity can be used to make a rapid check of any physical
phenomenon by an algebraic analysis of dimensional equation (before substitution
of the numerical data).
Dr. Neel Gunasekera
Dimensional Analysis
Concepts:
Dimensionless Quantities
Can be defined as a quantity with the dimensions of “1” , no M, L, T.
[ex: Ratio or Percent]
Eg: Rd = d / l  [L]/[L] =1 [Strain] = [ Extension] / [Initial Length] = L 1-1 = 1
•Dimensional quantities can be made dimensionless by normalizing them with respect
to another dimensional quantity of the same dimensionality.
[ ex: Reynolds No : ρDV/μ, Froude No : V/(gy)1/2 ]
•Dimensionless variables and equations are independent of units ( scalable )
•Relative importance of term can be easily estimated
•Scale is automatically built into the dimensionless expressions
•Reduces the complexity of the problem -Reduces many problems to a single,
normalized problem.
Dr. Neel Gunasekera
Dimensional Analysis
Methods of Dimensional Analysis
1. Indicial Method
2. Buckingham Pi
( Step by Step method or method of repeating variables)
3. Hunsker and Rightmire
4. Matrix Method
Dimensional Analysis
Method of repeating variables based on Buckingham π Theorem
If there are n variables in a problem and these variables contain m
primary dimensions (e.g. M, L, T), the equation relating the variables
will contain n-m dimensionless groups”. Buckingham referred to
these dimensionless groups as π1, π2, …, πn-m, and the final equation
obtained is:
π1 = φ (π2, π3,…, πn-m) Or φ(π1 ,π2, π3,…, πn-m)=0.
Dr. Neel Gunasekera
Dimensional Analysis
Work flow of Method of repeating variables
Step 1: List all the variables that are involved in the problem
Step 2: Express each of the variables in terms of reference dimensions-MLT
Step 3: Determine the required number of π terms
Step 4: Select a number of repeating variables, where the number required is
equal to the number of reference dimensions.
Step 5: Form π terms: multiply non-repeating variables by repeating variables
to get dimensionless numbers.
Step 6: Repeat Step 5 for each of the remaining non-repeating variables.
Step 7: Check all the resulting π terms to make sure they are dimensionless.
Step 8: Express the final form as a relationship among the π terms, and think
about what it means.
Dr. Neel Gunasekera
Dimensional Analysis
EX1 : The thrust F of a screw propeller is known to depend upon the diameter d.
speed of advance v, fluid density ρ, revolutions per second N, and the coefficient
of viscosity μ of the fluid. Find an expression for F in terms of these quantities.
Dr. Neel Gunasekera
Dimensional Analysis
EX1 : The thrust F of a screw propeller is known to depend upon the diameter d.
speed of advance V, fluid density ρ, revolutions per second N, and the coefficient
of viscosity μ of the fluid. Find an expression for F in terms of these quantities.
Step 1: List all the variables that are involved in the problem
Can’t express one in terms of another (ex1: diameter, Area) , *(ex2:(μ, ρ, υ) - υ= μ/ ρ]
 (q1 , q2 , q3 ,...,qn )  0
q1   (q2 , q3 ,...,qn )
Dependent
Variable
F   (d ,V ,  , N ,  )
n=6
Independent Variables
Step 2: Express each of the variables in terms of reference dimensions -MLT
Variable
[F]
[d]
[v]
[ρ]
Dimensions
[MLT-2]
[L]
[LT-1] [ML-3]
[N]
[μ]
[T-1]
[ML-1T-1]
m=3 ,for M, L, T
Step 3: Determine the required number of π terms
π terms :r = n-m
Variables-n=6, Ref. Dim m=3
r=3 (π terms )
 1   ( 2 ,  3 ) or  ( 1 ,  2 ,  3 )  0
Dr. Neel Gunasekera
Dimensional Analysis
Variable
[F]
[d]
[V]
[ρ]
[N]
[μ]
Dimensions
[ML-1T-1]
[L]
[LT-1]
[ML-3]
[T-1]
[ML-1T-1]
Step 4: Select r repeating variables, where the number required is equal to the number
of reference dimensions.
F   (d ,V ,  , N ,  )
A. Do not include dependent variable
B. All reference dimensions (MLT) must be represented in the repeating variables
C. Each must be dimensionally independent
1. Cannot be combined to form one
2. Cannot from dimensionless numbers
d[L]
d[L]
F [MLT-2]
d[L]
V[LT-1]
V[LT-1]
V[LT-1]
V[LT-1]
ρ  [ML-3]
μ[ML-1T-1]
N  [T-1]
N  [T-1]
N=V/d  [T-1]
B
A
C-1
Dr. Neel Gunasekera
Step 5: Form π terms:
Dimensional Analysis
F   (d , v,  , N ,  )
multiply non-repeating variables by repeating variables to get dimensionless numbers.
•Can raise repeating variable by any power
[d]a [v]b [ρ]c [F] = [1]
•Non repeating variable must be raised to power of 1
Repeating Variable Non-Repeating Variable
d[L]
F [MLT-2]
v[LT-1]
μ[ML-1T-1]
ρ  [ML-3]
N  [T-1]
repeating variables
non-repeating variables
Equation for --π1
[d]a [v]b [ρ]c [F] = [1]
[L] a [LT-1] b [ML-3] c [MLT-2] = M0 L0 T0
M c+1=0 
c=-1
T -b-2=0 
b=-2
L a+b-3c-1=0  a=-2
F
1  2 2
v d
Dr. Neel Gunasekera
Dimensional Analysis
Step 6: Repeat Step 5 for each of the remaining non-repeating variables.
Repeating Variable Non-Repeating Variable
d[L]
F [MLT-2]
v[LT-1]
μ[ML-1T-1]
ρ  [ML-3]
N  [T-1]
Equation for --π2
[d]p [v]q [ρ]r [N] = [1]
[L] p [LT-1]q [ML-3] r[T-1] = M0 L0 T0
p=1 q=-1 r = 0
Nd
2 
v
Equation for --π3
[d]x [v]y [ρ]z [μ] = [1]
[L] x [LT-1]y [ML-3] z [ML-1T-1] = M0 L0 T0
x = -1 , y= -1, z= -1
3 

dv
Dr. Neel Gunasekera
Dimensional Analysis
Step 7: Check all the resulting π terms to make sure they are dimensionless.
F
[ MLT 2 ]
1  2 2 
 [1]
3
1 2
2
v d
[ ML ][ LT ] [ L]
Nd [T 1 ][ L]
2 

 [1]
1
v
[ LT ]

[ ML1T 1 ]
3 

 [1]
3
1
dv [ ML ][ L][ LT ]
Dr. Neel Gunasekera
Dr. Neel Gunasekera
Dimensional Analysis
Step 8: Express the final form as a relationship among the π terms, and think about
what it means.
F   (d , v,  , N ,  )
 1   ( 2 ,  3 )
 Nd  
F

  
,
2 2
v d
 v dv 
 Nd  

F  v d  
,
 v dv 
2
2
Dr. Neel Gunasekera
Dimensional Analysis and Similitude