This document discusses dimensional analysis and model analysis. It begins by introducing dimensional analysis as a technique that uses the dimensions of physical quantities to understand phenomena. It then describes the two types of dimensions: fundamental dimensions like length, time, and mass, and secondary dimensions that are combinations of fundamental ones, like velocity. It presents the methodology of dimensional analysis, which requires equations to be dimensionally homogeneous. Two common methods are described: Rayleigh's method and Buckingham's pi-theorem. Rayleigh's method can be used for up to 4 variables while Buckingham's pi-theorem groups variables into dimensionless pi-terms. Model analysis is introduced as an experimental method using scale models, and the importance of similitude or

1. DIMENSIONAL ANALYSIS AND
SIMILITUDE & MODEL
ANALYSIS

2. 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.

3. 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

4. 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.

5. 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.

6. 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.

7. 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:

8. 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
     


 

 
 
         
 
 
   
  
   
   
 
 
 
  
  
  
  
 

9. 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)

10. BUCKINGHAM’S Π-THEOREM:
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))

11. 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 shouldn’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.

12. BUCKINGHAM’S Π-THEOREM:
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

 
  
 




13. BUCKINGHAM’S Π-THEOREM:
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
( ) ( )
Equating the powers of MLT on both sides, we get
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 -1
b b
l V
lV

   

  
   

14. BUCKINGHAM’S Π-THEOREM:
3 1 3 3 3 1 2
3
3 3
3 3 3 3
3 3
( ) ( )
Equating the powers of MLT on both sides, we get
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
  


  
  
 
  
    
  
 
 
 
 
   
  
   
   

15. 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 ships
in towing basins, air planes in wind tunnel, hydraulic
turbines, centrifugal pumps, spillways of dams, river
channels etc and to study such phenomenon as the

16. 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 that the prototype, they may be larger than
prototype.
Prototype Model
Lp3
Lp1
Lp2
Fp1
Fp3
Fp2
Lm3
Lm1
Lm2
Fm1
Fm3
Fm2

17. 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

18. 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

19. SIMILITUDE-TYPE OF SIMILARITIES
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

20. SIMILITUDE-TYPE OF SIMILARITIES
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.

21. SIMILITUDE-TYPE OF SIMILARITIES
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.
 
 
 
 
 
 
g
i v
p p p
r
i v g
m m m
F
F 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.

22. 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.

23. 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

24. DIMENSIONLESS NUMBERS
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 V
A A A
dy L

   
  
 
  
    
2
. .
. .
. .
. .
Velocity Volume
Mass Velocity
Fi Time Time
Fe
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.

25. DIMENSIONLESS NUMBERS
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 Time
E
Fp essure Area essure Area
QV AV V V V
P A P A P P

 
 
  
   
2 2
. .
. .
. .
. . .
Velocity Volume
Mass Velocity
Fi Time Time
We
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.

26. DIMENSIONLESS NUMBERS
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 Time
M
Fe Elastic Stress Area Elastic Stress Area
QV AV V L V V V
K A K A KL C
K
Where C K

  


  
    


27. LECTURE # 13

28. 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 p
v s e I
p p p p p p
v s e I
g p
m m m m
m m
F F
F F F F
F F F F
F F
    
   
 
 
 
 
Thus dynamic similarity require that
v g p s e I
v g p s e I
p p
I
v g p s e m
m
F F F F F F
F F F F F F
F
F 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

29. 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

30. 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 m
P P
P m
P m
P P r r
r
P
m m
m
P P P
r r r
m m m
V L
V L
or
V L V L
V L
V L
where V L
V L
 






 
 
 
 
 
  

31. REYNOLD’S MODEL LAW
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
m
P P
m P m
P
m
P P
r r
m m
VL VL
ce and
L
V
V L
a T
a T
A V
L V
A V
m
  
 
 

   
 
   
   
 
 
 
 
2 2 2
2 2 2 3
r r r
Power 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
  
 
  


32. REYNOLD’S MODEL LAW
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 p
m m m
m m
2
2
p 2
For pipe flow,
According to Reynolds' Model Law
V D D
V 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=?

33. REYNOLD’S MODEL LAW
Q. A ship 300m long moves in sea water, whose density is 1030 kg/m3.
A 1:100 model of this ship is to be tested in a wind tunnel. The velocity
of air in the wind tunnel around the model is 30m/s and the resistance of
the model is 60N. Determine the velocity of ship in sea water and also
the resistance of ship in sea water. The density of air is given as
1.24kg/m3. Take the kinematic viscosity of sea water and air as 0.012
stokes and 0.018 stokes respectively.
 Solution:
 For Prototype
 Length, Lp= 300m
 Fluid = sea water
 Density of sea water, ρp= 1030 kg/m3
 Kinematic Viscosity, νp=0.018 stokes
=0.018x10-4 m2/s
 Let Velocity of ship, Vp
 Resistance, Fp
 For Model
 Scale ratio = Lp/Lm=100
 Length, Lm= Lp/100 = 3m
 Fluid = air
 Density of air, ρm= 1.24 kg/m3
 Kinematic Viscosity, νm=0.012 stokes
=0.012x10-4 m2/s
 Velocity of ship, Vm=30 m/s
 Resistance, Fm = 60 N

34. REYNOLD’S MODEL LAW
For dynamic similarity between model and prototype, the
Reynolds number for both of them should be equal.
 
 
4
4
2 2
2 2 2 2
2 2
0.012 10 3
30 0.2 /
0.018 10 300
Resistance= Mass Acceleration= L V
L V 1030 300 0.2
369.17
1.24 3 30
L V
369.17 60 22150.2
p m
p m
p m m p
p p
m
m
p
L
VL VL
V V
L
Vp m s
Since
F
Thus
F
F N

  





   
  
   
   

 


   
  
   
   
  

35. 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 m
P P
P m
P P m m P m
P P P
r r r r
m m
P
m
m
V V
V V
F F or or
g L g L L L
V V L
V L where V L
V L
L
V
L
  
   
 
 
 

36. FROUDE’S MODEL LAW
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=
m
P
P m
p
P
r
m m
r
r
r
P
m r
P P
r r r r r
m m
r r
V
V
ce
L L
L
V
L
V L
L
L
L
a 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
    
   
   
  

37. FROUDE’S MODEL LAW
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 2.5
p
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=?

38. 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 =?

39. FROUDE’S MODEL LAW
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
 
  
 
  

40. 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.
Undistorted 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.

41. 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
P
Scale Ratio for discharge: Qr=Q /
V
r r
H V
P P
m r r r r r
H V V H
m m
L L
A V
Q L L L L L
A V
  

42. 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 r
H
p m
L
L
h
h
Q L L
Q Q



  
3
1581.14
1.5/1581.14 0.000948 /
m
Q m s

  

43. 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.
 
 
   
3
r
r
3/2
r P
3/2
arg 450 /
60 4.2
Horizontal scale ratio= L =200
Vertical scale ratio= L =50
Since Scale Ratio for discharge: Q =Q /
/ 200 50 7
V
p
p p
P
H
m
P
V
m
m r r
H
p m
Disch e of River Q m s
Width B m and Depth y m
B
B
y
y
Q L L
Q Q
 
   



   
3 3
0710.7
450/1581.14 6.365 10 /
m
Q m s

   

44. 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





 
   
  
   

   
