Similitude and Dimensional Analysis -Hydraulics engineering

Hydraulics Engineering
Similitude and
Dimensional Analysis
Dr. M. Mubashir Qureshi

Similitude and Model Analysis
 Similitude is a concept used in testing of Engineering
Models.
 Usually, it is impossible to obtain a pure theoretical
solution of hydraulic phenomenon.
 Therefore experimental investigations are often
performed on small scale models, called model
analysis.
 A few examples, where models may be used are
spillway, energy dissipation structures, river channels,
ships in towing basins, hydraulic turbines, centrifugal
pumps etc. and to study such phenomenon as the
action of waves and tides on beaches, soil erosion, and
transportation of sediment etc.

Model Analysis
 Model: is a small scale replica of the actual structure.
 Prototype: the actual structure or machine.
 Note: It is not necessary that the models should be
smaller than the prototype, they may be larger than
prototype.
Prototype Model
Lp3
Lp1
Lp2
Fp1
Fp3
Fp2
Lm3
Lm1
Lm2
Fm1
Fm3
Fm2

Model Analysis
 Model Analysis is actually an experimental method of
finding solutions of complex flow problems.
 The followings are the advantages of the model analysis
 The performance of the hydraulic structure can be predicted in
advance from its model.
 Using dimensional analysis, a relationship between the variables
influencing a flow problem is obtained which help in conducting
tests.
 The merits of alternative design can be predicted with the help
of model analysis to adopt most economical, and safe design.
 Note: Test performed on models can be utilized for
obtaining, in advance, useful information about the
performance of the prototype only if a complete
similarity exits between the model and the prototype.

Ubrug Spillway, Jatiluhur Dam, Indonesia

Similitude-Type of Similarities
 Similitude: is defined as similarity between the model
and prototype in every respect, which mean model and
prototype have similar properties or model and
prototype are completely similar.
 Three types of similarities must exist between model
and prototype.
 Geometric Similarity
 Kinematic Similarity
 Dynamic Similarity

 Geometric Similarity: is the similarity of shape. It is said to exist
between model and prototype if ratio of all the corresponding
linear dimensions in the model and prototype are equal. E.g.
p p p
r
m m m
L B D
L
L B D
= = =
◼ Where: Lp, Bp and Dp are Length, Breadth, and diameter of
prototype and Lm, Bm, Dm are Length, Breadth, and diameter of
model.
◼ Lr= Scale ratio
◼ Note: Models are generally prepared with same scale ratios in every
direction. Such a model is called true model. However, sometimes
it is not possible to do so and different convenient scales are used
in different directions. Such a models is call distorted model

 Kinematic Similarity: is the similarity of motion. It is said to exist
between model and prototype if ratio of velocities and acceleration
at the corresponding points in the model and prototype are equal.
E.g.
1 2 1 2
1 2 1 2
;p p p p
r r
m m m m
V V a a
V a
V V a a
= = = =
◼ Where: Vp1& Vp2 and ap1 & ap2 are velocity and accelerations at
point 1 & 2 in prototype and Vm1& Vm2 and am1 & am2 are velocity
and accelerations at point 1 & 2 in model.
◼ Vr and ar are the velocity ratio and acceleration ratio
◼ Note: Since velocity and acceleration are vector quantities, hence
not only the ratio of magnitude of velocity and acceleration at the
corresponding points in model and prototype should be same; but
the direction of velocity and acceleration at the corresponding
points in model and prototype should also be parallel.

 Dynamic Similarity: is the similarity of forces. It is said to exist
between model and prototype if ratio of forces at the
corresponding points in the model and prototype are equal. E.g.
( )
( )
( )
( )
( )
( )
gi vp p p
r
i v gm m m
FF F
F
F F F
= = =
◼ Where: (Fi)p, (Fv)p and (Fg)p are inertia, viscous and gravitational
forces in prototype and (Fi)m, (Fv)m and (Fg)m are inertia, viscous
and gravitational forces in model.
◼ Fr is the Force ratio
◼ Note: The direction of forces at the corresponding points in model
and prototype should also be parallel.

Types of forces encountered in fluid Phenomenon
 Inertia Force, Fi: It is equal to product of mass and acceleration in
the flowing fluid.
 Viscous Force, Fv: It is equal to the product of shear stress due to
viscosity and surface area of flow.
 Gravity Force, Fg: It is equal to product of mass and acceleration
due to gravity.
 Pressure Force, Fp: it is equal to product of pressure intensity and
cross-sectional area of flowing fluid.
 Surface Tension Force, Fs: It is equal to product of surface
tension and length of surface of flowing fluid.
 Elastic Force, Fe: It is equal to product of elastic stress and area
of flowing fluid.

Dimensionless Numbers
 These are numbers which are obtained by dividing the
inertia force by viscous force or gravity force or
pressure force or surface tension force or elastic force.
 As this is ratio of once force to other, it will be a
dimensionless number. These are also called non-
dimensional parameters.
 The following are most important dimensionless
numbers.
 Reynold’s Number
 Froude’s Number
 Euler’s Number
 Weber’s Number
 Mach’s Number

 Reynold’s Number, Re: It is the ratio of inertia force to the viscous force
of flowing fluid.
. .
Re
. .
. . .
. . .
Velocity Volume
Mass Velocity
Fi Time Time
Fv Shear Stress Area Shear Stress Area
QV AV V AV V VL VL
du VA A A
dy L

   
   
= = =
= = = = =
2
. .
. .
. .
. .
Velocity Volume
Mass Velocity
Fi Time TimeFe
Fg Mass Gavitational Acceleraion Mass Gavitational Acceleraion
QV AV V V V
Volume g AL g gL gL

 
 
= = =
= = = =
◼ Froude’s Number, Re: It is the ratio of inertia force to the gravity force
of flowing fluid.

 Eulers’s Number, Re: It is the ratio of inertia force to the pressure force
of flowing fluid.
2
. .
Pr . Pr .
. .
. . / /
u
Velocity Volume
Mass Velocity
Fi Time TimeE
Fp essure Area essure Area
QV AV V V V
P A P A P P

 
 
= = =
= = = =
2 2
. .
. .
. .
. . .
Velocity Volume
Mass Velocity
Fi Time TimeWe
Fg Surface Tensionper Length Surface Tensionper Length
QV AV V L V V
L L L
L

  
   

= = =
= = = =
◼ Weber’s Number, Re: It is the ratio of inertia force to the surface
tension force of flowing fluid.

 Mach’s Number, Re: It is the ratio of inertia force to the elastic force of
flowing fluid.
2 2
2
. .
. .
. .
. . /
: /
Velocity Volume
Mass Velocity
Fi Time TimeM
Fe Elastic Stress Area Elastic Stress Area
QV AV V L V V V
K A K A KL CK
Where C K

  


= = =
= = = = =
=

Model Laws or similarity Laws
 We have already read that for dynamic similarity ratio of
corresponding forces acting on prototype and model should be equal.
i.e
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
g pv s e Ip p p p p p
v s e Ig pm m m mm m
F FF F F F
F F F FF F
= = = = =
( ) ( )
( )
( )
( )
( )
Thus dynamic similarity require that
v g p s e I
v g p s e Ip p
Iv g p s e mm
F F F F F F
F F F F F F
FF F F F F
+ + + + =
+ + + +
=
+ + + +
◼ Force of inertial comes in play when sum of all other forces is not
equal to zero which mean
◼ In case all the forces are equally important, the above two
equations cannot be satisfied for model analysis

Model Laws or similarity Laws
 However, for practical problems it is seen that one
force is most significant compared to other and is
called predominant force or most significant force.
 Thus for practical problem only the most significant
force is considered for dynamic similarity. Hence,
models are designed on the basis of ratio of force,
which is dominating in the phenomenon.
 Finally the laws on which models are designed for
dynamic similarity are called models laws or laws of
similarity. The followings are these laws
 Reynold’s Model Law
 Froude’s Model Law
 Euler’s Model Law
 Weber’s Model Law
 mach’s Model Law

Reynold’s Model Law
 It is based on Reynold’s number and states that
Reynold’s number for model must be equal to the
Reynolds number for prototype.
 Reynolds Model Law is used in problems where viscous
forces are dominant. These problems include:
 Pipe Flow
 Resistance experienced by submarines, airplanes, fully immersed
bodies etc.
( ) ( )Re Re
1
: , ,
m mP P
P m
P m
P P r r
rP
m m
m
P P P
r r r
m m m
V LV L
or
V L V L
V L
V L
where V L
V L
 





= =
= =
 
 
 
= = =

 The Various Ratios for Reynolds’s Law are obtained as
r
r
r
P P P r
m m m r
P P
r
m m
2
r
r
sin /
Velocity Ratio: V =
L
T L /V L
Time Ratio: Tr=
T L /V V
V / Vr
Acceleration Ratio: a =
V / Tr
Discharge Ratio: Q
Force Ratio: F =
P m
mP P
m P m
P
m
P P
r r
m m
VL VL
ce and
LV
V L
a T
a T
A V
L V
A V
m
  
 
 

   
= =   
   
= =
= =
= =
= =
2 2 2
2 2 2 3
r r rPower Ratio: P =F .V =
r r r r r r r r r r r r
r r r r r r r
a Q V L V V L V
L V V L V
  
 
= = =
=

 Q. A pipe of diameter 1.5 m is required to transport an oil of
specific gravity 0.90 and viscosity 3x10-2 poise at the rate of
3000litre/sec. Tests were conducted on a 15 cm diameter pipe using
water at 20oC. Find the velocity and rate of flow in the model.
p p p p pm m m
m m
2
2
p 2
For pipe flow,
According to Reynolds' Model Law
V D DV D
D
900 1.5 1 10
3.0
1000 0.15 3 10
3.0
Since V
/ 4(1.5)
1.697 /
3.0 5.091 /
5.
m m
m p p p
m
p
p
p
m p
m m m
V
V
V
V
Q
A
m s
V V m s
and Q V A
  
   

−
−
=  =
 
= =
 
= =
=
 = =
= = 2
3
091 / 4(0.15)
0.0899 /m s

=
◼ Solution:
◼ Prototype Data:
◼ Diameter, Dp= 1.5m
◼ Viscosity of fluid, μp= 3x10-2 poise
◼ Discharge, Qp =3000litre/sec
◼ Sp. Gr., Sp=0.9
◼ Density of oil=ρp=0.9x1000
=900kg/m3
◼ Model Data:
◼ Diameter, Dm=15cm =0.15 m
◼ Viscosity of water, μm =1x10-2 poise
◼ Density of water, ρm=1000kg/m3n
◼ Velocity of flow Vm=?
◼ Discharge Qm=?

Froude’s Model Law
 It is based on Froude’s number and states that Froude’s
number for model must be equal to the Froude’s number
for prototype.
 Froude’s Model Law is used in problems where gravity
forces is only dominant to control flow in addition to
inertia force. These problems include:
 Free surface flows such as flow over spillways, weirs, sluices,
channels etc.
 Flow of jet from orifice or nozzle
 Waves on surface of fluid
 Motion of fluids with different viscosities over one another
( ) ( )e e
/ 1; : ,
m mP P
P m
P P m m P m
P P P
r r r r
m mP
m
m
V VV V
F F or or
g L g L L L
V V L
V L where V L
V LL
V
L
= = =
= = = =
 
 
 

 The Various Ratios for Reynolds’s Law are obtained as
r
P P P r
m m m
P P
r
m m
2 2 5/2
r
sin
Velocity Ratio: V
T L /V L
Time Ratio: Tr=
T L /V
V / Vr
Acceleration Ratio: a = 1
V / Tr
Discharge Ratio: Q
Force Ratio: Fr=
mP
P m
pP
r
m m
r
r
rP
m r
P P
r r r r r
m m
r r
VV
ce
L L
LV
L
V L
L
L
La T
a T L
A V
L V L L L
A V
m a
=
= = =
= = =
= = = =
= = = =
=
( )
2 2 2 2 3
3
2 2 2 3 2 7/2
Power Ratio: Pr=Fr.Vr=
r r r r r r r r r r r r r r r
r r r r r r r r r r r r
Q V L V V L V L L L
L V V L V L L L
    
   
= = = =
= = =

 Q. In the model test of a spillway the discharge and velocity of flow
over the model were 2 m3/s and 1.5 m/s respectively. Calculate the
velocity and discharge over the prototype which is 36 times the
model size.
( ) ( )
( )
2.5 2.5p
m
2.5 3
For Discharge
Q
36
Q
36 2 15552 /sec
r
p
L
Q m
= =
=  =
p
m
For Dynamic Similarity,
Froude Model Law is used
V
36 6
V
6 1.5 9 /sec
r
p
L
V m
= = =
=  =
◼ Solution: Given that
◼ For Model
◼ Discharge over model, Qm=2 m3/sec
◼ Velocity over model, Vm = 1.5 m/sec
◼ Linear Scale ratio, Lr =36
◼ For Prototype
◼ Discharge over prototype, Qp =?
◼ Velocity over prototype Vp=?

Numerical Problem:
 Q. The characteristics of the spillway are to be studied by means of a geometrically
similar model constructed to a scale of 1:10.
 (i) If 28.3 cumecs, is the maximum rate of flow in prototype, what will be the
corresponding flow in model?
 (i) If 2.4m/sec, 50mm and 3.5 Nm are values of velocity at a point on the spillway,
height of hydraulic jump and energy dissipated per second in model, what will be the
corresponding velocity height of hydraulic jump and energy dissipation per second in
prototype?◼ Solution: Given that
For Model
◼ Discharge over model, Qm=?
◼ Velocity over model, Vm = 2.4 m/sec
◼ Height of hydraulic jump, Hm =50 mm
◼ Energy dissipation per second, Em =3.5 Nm
◼ Linear Scale ratio, Lr =10
◼ For Prototype
◼ Discharge over model, Qp=28.3 m3/sec
◼ Velocity over model, Vp =?
◼ Height of hydraulic jump, Hp =?
◼ Energy dissipation per second, Ep =?

p 2.5 2.5
m
2.5 3
p
m
For Discharge:
Q
10
Q
28.3/10 0.0895 /sec
For Velocity:
V
10
V
2.4 10 7.589 /sec
r
m
r
p
L
Q m
L
V m
= =
= =
= =
=  =
p
m
p 3.5 3.5
m
3.5
For Hydraulic Jump:
H
10
H
50 10 500
For Energy Dissipation:
E
10
E
3.5 10 11067.9 /sec
r
p
r
p
L
H mm
L
E Nm
= =
=  =
= =
=  =

Classification of Models
 Undistorted or True Models: are those which are
geometrically similar to prototype or in other words if the scale
ratio for linear dimensions of the model and its prototype is same,
the models is called undistorted model. The behavior of prototype
can be easily predicted from the results of undistorted or true
model.
 Distorted Models: A model is said to be distorted if it is not
geometrically similar to its prototype. For distorted models
different scale ratios for linear dimension are used.
 For example, if for the river, both horizontal and vertical scale
ratio are taken to be same, then depth of water in the model of
river will be very very small which may not be measured
accurately.
◼ The followings are the advantages of distorted models
◼ The vertical dimension of the model can be accurately measured
◼ The cost of the model can be reduced
◼ Turbulent flow in the model can be maintained
◼ Though there are some advantage of distorted models, however the results of such models
cannot be directly transferred to prototype.

Classification of Models
 Scale Ratios for Distorted Models
( )
( )
( )
r
r
P
P
Let: L = Scale ratio for horizontal direction
L =Scale ratio for vertical direction
2
Scale Ratio for Velocity: Vr=V /
2
Scale Ratio for area of flow: Ar=A /
P P
H
m m
P
V
m
P
m r V
m
P P
m
m m
L B
L B
h
h
gh
V L
gh
B h
A
B h
=
=
= =
= = ( ) ( )
( ) ( ) ( ) ( ) ( )
3/2
PScale Ratio for discharge: Qr=Q /
V
r rH V
P P
m r r r r rH V V H
m m
L L
A V
Q L L L L L
A V
= = =

Distorted model
 Q. The discharge through a weir is 1.5 m3/s. Find the discharge
through the model of weir if the horizontal dimensions of the
model=1/50 the horizontal dimension of prototype and vertical
dimension of model =1/10 the vertical dimension of prototype.
( )
( )
( ) ( )
3
p
r
r
3/2
P
3/2
Solution:
Discharge of River= Q =1.5m /s
Scale ratio for horizontal direction= L =50
Scale ratio for vertical direction= L =10
Since Scale Ratio for discharge: Qr=Q /
/ 50 10
V
P
H
m
P
V
m
m r rH
p m
L
L
h
h
Q L L
Q Q
=
=
=
 = 
3
1581.14
1.5/1581.14 0.000948 /mQ m s
=
 = =

Distorted model
 Q. A river model is to be constructed to a vertical scale of 1:50 and a
horizontal of 1:200. At the design flood discharge of 450m3/sec, the
average width and depth of flow are 60m and 4.2m respectively. Determine
the corresponding discharge in model and check the Reynolds’ Number of
the model flow.
𝐷𝑖𝑠𝑐ℎ arg 𝑒 𝑜𝑓 𝑅𝑖𝑣𝑒𝑟 = 𝑄 𝑝 = 450𝑚3
/𝑠
𝑊𝑖𝑑𝑡ℎ = 𝐵𝑝 = 60𝑚 𝑎𝑛𝑑 𝐷𝑒𝑝𝑡ℎ = 𝑦𝑝 = 4.2 𝑚
Horizontal scale ratio= Lr 𝐻 =
𝐵 𝑃
𝐵 𝑚
= 200
Vertical scale ratio= Lr 𝑉 =
𝑦 𝑃
𝑦 𝑚
=50
Since Scale Ratio for discharge: Qr=QP/𝑄 𝑚
= 𝐿 𝑟 𝐻 𝐿 𝑟 𝑉
3/2
∴ 𝑄 𝑝/𝑄 𝑚 = 200 × 503/2
= 70710.7
⇒ 𝑄 𝑚 = 450/70710.7 = 6.365 × 10−3
𝑚3
/𝑠

Distorted model
( )
( )
m
VL
Reynolds Number, Re =
4
/ 60/ 200 0.3
/ 4.2/50 0.084
0.3 0.084 0.0252
2 0.3 2 0.084 0.468
0.0252
0.05385
0.468
Kinematic Viscosity of w
m
m m
m p r H
m p r V
m m m
m m m
m
m
L R
Width B B L m
Depth y y L m
A B y m
P B y m
A
R
P

 
 
 
=
= = = =
= = = =
= =  =
= + = +  =
= = =
6 2
6
ater = =1 10 /sec
4 4 0.253 0.05385
Re 54492.31
1 10
>2000
Flow is in turbulent range
m
m
VR


−
−

    
= = =   
   


Dimensional Analysis
 Introduction: Dimensional analysis is a mathematical
technique making use of study of dimensions.
 This mathematical technique is used in research work
for design and for conducting model tests.
 It deals with the dimensions of physical quantities
involved in the phenomenon. All physical quantities are
measured by comparison, which is made with respect
to an arbitrary fixed value.
 In dimensional analysis one first predicts the physical
parameters that will influence the flow, and then by,
grouping these parameters in dimensionless
combinations a better understanding of the flow
phenomenon is made possible.
 It is particularly helpful in experimental work because
it provides a guide to those things that significantly
influence the phenomena; thus it indicates the
direction in which the experimental work should go.

Types of Dimensions
 There are two types of dimensions
 Fundamental Dimensions or Fundamental Quantities
 Secondary Dimensions or Derived Quantities
 Fundamental Dimensions or Fundamental
Quantities: These are basic quantities. For Example;
 Time, T
 Distance, L
 Mass, M

Types of Dimensions
 Secondary Dimensions or Derived Quantities
 The are those quantities which possess more than one
fundamental dimension.
 For example;
 Velocity is denoted by distance per unit time L/T
 Acceleration is denoted by distance per unit time square L/T2
 Density is denoted by mass per unit volume M/L3
 Since velocity, density and acceleration involve more
than one fundamental quantities so these are called
derived quantities.

Methodology of Dimensional Analysis
 The Basic principle is Dimensional Homogeneity, which
means the dimensions of each terms in an equation on
both sides are equal.
 So such an equation, in which dimensions of each term
on both sides of equation are same, is known as
Dimensionally Homogeneous equation. Such equations
are independent of system of units. For example;
 Lets consider the equation V=(2gH)1/2
 Dimensions of LHS=V=L/T=LT-1
 Dimensions of RHS=(2gH)1/2=(L/T2xL)1/2=LT-1
 Dimensions of LHS= Dimensions of RHS
 So the equation V=(2gH)1/2 is dimensionally
homogeneous equation.

Methods of Dimensional Analysis
 If the number of variables involved in a physical phenomenon are
known, then the relation among the variables can be determined
by the following two methods;
 Rayleigh’s Method
 Buckingham’s π-Theorem
 Rayleigh’s Method:
 It is used for determining expression for a variable (dependent)
which depends upon maximum three to four variables
(Independent) only.
 If the number of independent variables are more than 4 then it is
very difficult to obtain expression for dependent variable.
 Let X is a dependent variable which depends upon X1, X2, and X3 as
independent variables. Then according to Rayleigh’s Method
X=f(X1, X2, X3) which can be written as
X=K X1
a, X2
b, X3
c
Where K is a constant and a, b, c are arbitrary powers which are
obtained by comparing the powers of fundamental dimensions.

Rayleigh’s Method
 Q. The resisting force R of a supersonic plane during flight can be
considered as dependent upon the length of the aircraft l, velocity V, air
viscosity μ, air density ρ, and bulk modulus of air k. Express the functional
relationship between the variables and the resisting force.
-2 1 1 1 3 1 2
( , , , , ) , , , , (1)
Where: A = Non dimensional constant
Substituting the powers on both sides of the equation
( ) ( ) ( ) ( )
Equating the powers of MLT on both
a b c d e
a b c d e
R f l V K R Al V K
MLT AL LT ML T ML ML T
   
− − − − − −
=  =
=
sides
Power of M 1
Power of L 1 - -3 -
Power of T 2 - - -2
c d e
a b c d e
b c e
 = + +
 = +
 − =
◼ Solution:

Rayleigh’s Method
Since the unkown(5) are more than number of equations(3). So expressing
a, b & c in terms of d & e
1- -
2- -2
1- 3 1-(2- -2 ) 3(1- - )
1-2 2 3-3 -3 2-
Substituting the values
d c e
b c e
a b c d e c e c c e e
c e c c e e c
=
=
= + + + = + + +
= + + + + + =
2 2 2 1 2 2 2
2 2
2
2 2
2
in (1), we get
( )( )c c e c c e e c c c c e e e
c e
R Al V K Al V l V V K
K
R A l V
Vl V
K
R A l V
Vl V
     


 

 
 
− − − − − − − − − −
= =
    
=     
     
   
=    
   

Buckingham’s π-Theorem:
 Buckingham’s π-Theorem: Since Rayleigh’s Method becomes
laborious if variables are more than fundamental dimensions (MLT),
so the difficulty is overcome by Buckingham’s π-Theorem which
states that
 “If there are n variables (Independent and Dependent) in a
physical phenomenon and if these variables contain m fundamental
dimensions then the variables are arranged into (n-m)
dimensionless terms which are called π-terms.”
 Let X1, X2, X3,…,X4, Xn are the variables involved in a physical
problem. Let X1 be the dependent variable and X2, X3, X4,…,Xn
are the independent variables on which X1 depends.
Mathematically it can be written as
X1=f(X2 ,X3 ,X4 ,Xn) which can be rewritten as
f1(X1,X2 X3 X4 Xn)=0
 Above equation is dimensionally homogenous. It contain n
variables and if there are m fundamental dimensions then it can be
written in terms of dimensions groups called π-terms which are
equal to (n-m)
 Hence f1(π1 π2 π3,… πn-m)=0

 Properties of π-terms:
 Each π-term is dimensionless and is independent of system of
units.
 Division or multiplication by a constant does not change the
character of the π-terms.
 Each π-term contains m+1 variables, where m is the number of
fundamental dimensions and also called repeating variable.
 Let in the above case X2, X3, X4 are repeating variables and if
fundamental dimensions m=3 then each π-term is written as
Π1=X2
a1. X3
b1. X4
a1 .X1
Π2=X2
a2. X3
b2. X4
a2 .X5
.
Πn-m=X2
a(n-m). X3
b(n-m). X4
a(n-m) .Xn
Each equation is solved by principle of dimensionless homogeneity and values
of a1, b1 & c1 etc are obtained. Final result is in the form of
Π1=(Π2, Π3, Π4 ,…, Π(n-m))
Π2=(Π1, Π3, Π4 ,…, Π(n-m))

Methods of Selecting Repeating Variables
 The number of repeating variables are equal to number
of fundamental dimensions of the problem. The choice
of repeating variables is governed by following
considerations;
 As far as possible, dependent variable should’t be selected as
repeating variable
 The repeating variables should be chosen in such a way that one
variable contains geometric property, other contains flow property
and third contains fluid property.
 The repeating variables selected should form a dimensionless group
 The repeating variables together must have the same number of
fundamental dimension.
 No two repeating variables should have the same dimensions.
 Note: In most of fluid mechanics problems, the choice of repeating
variables may be (i) d,v ρ, (ii) l,v,ρ or (iii) d, v, μ.

 Q. The resisting force R of a supersonic plane during flight can be
considered as dependent upon the length of the aircraft l, velocity
V, air viscosity μ, air density ρ, and bulk modulus of air k. Express
the functional relationship between the variables and the resisting
force.
1 2 3
( , , , , ) ( , , , , , ) 0
Total number of variables, n= 6
No. of fundamental dimension, m=3
No. of dimensionless -terms, n-m=3
Thus: ( , , ) 0
No. Repeating variables =m=3
Repeating variables = ,
R f l V K f R l V K
f
l
   

  
=  =
=
1 1 1
1
2 2 2
2
3 3 3
3
,
π-terms are written as
a b c
a b c
a b c
V
Thus
l V R
l V
l V K

 
  
 
=
=
=

 Now each Pi-term is solved by the principle of dimensional
homogeneity
1 1 1 3 1 2
1
1 1
1 1 1 1
1 1
( ) ( )
Equating the powers of MLT on both sides, we get
Power of M: 0=c +1 c =-1
Power of L: 0=a +b -3c +1 2
Power of T: 0=-b -2 b =-2
o o o a b c
term M L T L LT ML MLT
a
 − − −
−  =

 = −

 -2 -2 -2
1 1 2 2
2 1 2 3 2 1 1
2
2 2
2 2 2 2
( ) ( )
Power of M: 0 1 -1
Power of L: 0 -3 -1 1
Pow
o o o a b c
R
l V R
L V
term M L T L LT ML ML T
c c
a b c a
  

 − − − −
=  =
−  =
= +  =
= +  = −
2 2
-1 -1 -1
2 2
er of T: 0 - -1 -1b b
l V
lV

   

=  =
 =  =

3 1 3 3 3 1 2
3
3 3
3 3 3 3
3 3
( ) ( )
Power of M: 0 1 -1
Power of L: 0 -3 -1 0
Power of T: 0 - - 2 -2
o o o a b c
term M L T L LT ML ML T
c c
a b c a
b b
 − − − −
−  =
= +  =
= +  = −
=  =
 0 -2 -1
3 2 2
1 2 3 2 2 2
2 2
2 2 2 2
( ) , , 0
, ,
K
l V K
V
Hence
R K
f f or
l V lV V
R K K
R l V
l V lV V lV V
  


  
  
 
  
    
=  =
 
= = 
 
   
=  =   
   

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