UNIT-3Dimensional analysis and model studies

Dimensional analysis & model
studies

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.

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

 
  
 




• 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
LV
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

   

  
   

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
  


  
  
 
  
    
  
 
 
 
 
   
  
   
   

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 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
that 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.

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.
 
 
 
 
 
 
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.

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 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.

• 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.

• 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 LV V V
K A K A KL C
K
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 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

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 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
 






 
 
 
 
 
  

• 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
  
 
  


• 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 20o
C. 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/m3
n
 Velocity of flow Vm=?
 Discharge Qm=?

• 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

• 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

  





   
  
   
   

 


   
  
   
   
  

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
  
   
 
 
 

• 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
    
   
   
  

• 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=?

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 =?

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.
• 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

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
  

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

  

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

   

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
 Technique called Buckingham Pi Theorem
 Arranges parameters into lesser number of
dimensionless groups of variables
 Based on Mass-Length-Time System (MLT)
 Let X1, X2, X3, … , Xn be n dimensional variables
 We can write a dimensionally homogeneous
equation relating these variables as…
  0
,...,
,
,
, 4
3
2
1 
n
X
X
X
X
X
f

 
 
k
n
k
n










,...,
0
,...,
,
2
1
2
1


 Technique called Buckingham Pi Theorem
 We can rearrange this equation into the following
where f is another function and each is an
independent dimensionless product of some of
the X’s


 Steps in Buckingham Pi Theorem:
 Let us focus on an example as we work through
the steps: Drag force (FD) on submerged sphere
as it moves through a stationary, viscous fluid
 STEP 1: Identify all variables and count the
number of variables (n)
n = 5   0
,
,
,
, 


V
D
F
f D

 STEP 2: List the dimensions of each variable in the
MLT system and find the number of fundamental
dimensions (m)
m = 3 (M, L, T)
LT
M
L
M
T
L
V
L
D
T
ML
FD






 3
2

 STEP 3: Find the reduction number, k
k = Usually equal to m (cannot
exceed m,
rarely less than m)
k = try to find m dimensional
variables that cannot be formed
into a dimensionless
group

 STEP 3:
m = 3 (M, L, T)
So k = m = 3!
LT
M
T
L
L
L
M
DV 
 3


 STEP 4: Determine n-k (This is the number of
dimensionless groups needed!)
n-k = 5-3 = 2
 Step 5: Select k variables to be primary
(repeating) variables that contain all m (M, L, T)
dimensions
V
D


 Step 5: Select k variables to be primary
(repeating) variables that contain all m (M, L, T)
dimensions
 Step 6: Equate exponents of each dimension on
both sides of Pi term (M0
L0
T0
):
D
c
b
a
c
b
a
F
V
D
V
D
2
2
2
2
1
1
1
1







  0
0
0
1
1
1
3
1
1
1
1
1
T
L
M
LT
M
T
L
L
L
M
V
D
c
b
a
c
b
a






















 


 Step 6: Working with the 1st
Pi Term:
 
1
1
1
1
1
1
1
1
1
1
1
1
1
0
0
0
1
1
1
3
1
1
1
1
1
Re
1
,
1
,
1
0
1
:
0
1
3
:
0
1
:















































DV
V
D
c
b
a
c
T
c
b
a
L
a
M
T
L
M
LT
M
T
L
L
L
M
V
D
c
b
a
c
b
a







 Step 6: Working with 2nd
Pi Term:
 
2
2
2
2
1
2
1
1
1
1
1
1
1
1
0
0
0
2
1
1
1
3
2
2
2
2
2
2
,
1
,
2
0
2
:
0
1
3
:
0
1
:
V
D
F
F
V
D
b
a
c
c
T
c
b
a
L
a
M
T
L
M
T
ML
T
L
L
L
M
F
V
D
D
D
c
b
a
D
c
b
a

















































 Step 7: Rearrange the Pi groups as desired:
   
 
 
Re
Re
2
2
2
2
1
1
2
2
1






V
D
F
V
D
F
D
D










UNIT-3Dimensional analysis and model studies

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