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DIMENSIONAL ANALYSIS
AND
HYDRAULIC MODELS
1
Textbook: Understanding Hydraulics by Less Hamil ISBN 10 0 – 333-77906 -1
This chapter explores the difference between units and dimensions. It then shows how the
analysis of dimensions can be used to derive the equations that govern hydraulic phenomena.
In some cases it is possible to obtain dimensionless groupings of variables, such as the Reynolds
and Froude numbers, that have a particular hydraulic significance. Since such groupings are
dimensionless, they do not change with the size or scale of the hydraulic system concerned.
This leads to the concept of hydraulic models, where scaled down versions of a system are used
to predict the performance of a real thing. Examples include the analysis of the head discharge
characteristics of unusually shaped weirs and the determination of the equations and
performance characteristics of pumps and turbines. Thus dimensional analysis is a powerful and
useful tool that can be used to investigate and obtain solutions to real problems. The questions
include: (Textbook: Understanding Hydraulics by Less Hamil) ISBN 10 0 – 333-77906 -1
1) What is the difference between units and dimensions?
2) What is the difference between fundamental dimensions and secondary dimensions?
3) What is dimensional homogeneity and why it is important?
4) How can a hydraulic model be used to predict the performance of the real thing?
5) What is meant by hydraulic similarity?
6) What are scale effects?
7) Why do we use sometimes distorted models?
8) How do you go about undertaking a hydraulic model investigation?
Units and Dimensions
1) What is the difference between units and dimensions?
A meter is a unit. It is a unit of length. Length is a fundamental dimension, which could also be
expressed in other units such as mm, inches, feet yard, km or miles. So a dimension can be
expressed in many different units.
There are only three (3) fundamental dimensions; Mass, Length and Time. Everything else can
be expressed in M, L, and T.
2) What is the difference between fundamental dimensions and secondary dimensions?
Fundamental dimensions is a dimension that cannot be broken down into component parts, it
stand on its own identity like Mass, Length and Time; while Secondary dimensions are a
combination of two or more fundamental dimensions to form another dimension.
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UNITS AND DIMENSIONS : (Fluid Mechanics, Volume 1 by J.F.DOUGLAS & R.D. MATTHEWS,3rd Edition)
The systems of units which still remain of importance in various parts of the world are the foot-poundsecond system (fps), the centimeter –gram-second system(cgs) and the meter-kilogram-second
system(MKS), but at present the SI units are the preferred system
Table 1: page 3 (Fluid Mechanics, Volume 1 by J.F.DOUGLAS & R.D. MATTHEWS,3rd Edition)
Quantity
fps
Absolute
cgs
Technical Absolute
MKS
Technical Technical
Length
ft
ft
cm
cm
m
Time
sec
sec
sec
sec
sec
Mass
lb-mass
slug
g-mass
981gm
9.81kg
Force or weight
Poundal
lb-force
dyne
g-force
kg-force
metric tonne=103 kg = 2205lb
1 slug=32.2 lb-mass
1g-force=981 dynes
1lb-force =32.2 poundals
The fps and the cgs system: the systems appear in two forms(The absolute system and technical
system).In the absolute system, the unit of mass is a fundamental unit and the unit of force is derived
using Newton’s second law of motion, whereas in the technical system, the unit of force is the
fundamental unit and the unit of mass is derived using
Newton’s second law.
In the MKS absolute units so far as mechanics is concerned, correspond with SI units and it seems
possible that MKS Technical units may continue in use for sometime alongside SI units.
Force = mass (acceleration)
In solving problems, it is essential to keep to one system of units only. if the data are in the different
systems, they should be converted immediately to the system selected.
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The System International of Units
Fundamental Units:
Length: meter (m)
Mass: kilogram (kg)
Electric current: ampere ((A)
Absolute temperature: Kelvin (k)
Luminous intensity: candela ( cd)
All other units are derived from these fundamental units.
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Time: second(s or sec)
Table II, (page 4, Fluid Mechanics, Volume 1 by J.F.DOUGLAS & R.D. MATTHEWS, 3rd Edition)
Quantity
Geometrical
Angle
Length
Area
Volume
First moment of area
Second moment of
area
Strain
Kinematic
Time
Velocity, Linear
Acceleration, Linear
Velocity, Angular
Acceleration, Angular
Volume rate of
Discharge
Dynamic
Mass
Force
Weight
Defining Equation
Dimensions
Unit
Symbol
Arc/Radius(a ratio)
(including all linear
Measurement)
Length x Length
Area x Length
Area x Length
Area x Length2
M0L0T0
L
radian
meter
rad
m
L2
L3
L3
L4
square meter
cubic meter
meter cubed
meter to fourth power
m2
m3
m3
M4
Extension/Length
L0
a ratio
Distance/time
Linear velocity/Time
Angle/time
Angular Velocity/Time
Volume/Time
T
LT-1
LT-2
T-1
T-2
L3T-1
second
meter per second
meter per second squared
radians per second
radians per second squared
cubic meters per second
s
ms-1
ms-2
rad s-1
rad s-2
m3s-1
Force/Acceleration
Mass x acceleration
M
MLT-2
kilogram
newton= kilogram
meter/second2
newton= kilogram
meter/second2
Kilogram per cubic meter
newtons per cubic meter
a ratio
newtons per square
meter=pascal
newtons per square
meter=pascal
newtons per square
meter=pascal
newtons seconds
kilogram-meter squared
kg
N=kgms-
kilogram meter per seconds
newton meter=joule
Joule/second=watt
kilogram per meter second=10
poise
meter squared per second
kilogram per second squared
Ns
Nm=j
JS-1=W
Kgm-1s-1
Force
-2
MLT
Mass Density
Specific Weight
Specific gravity
Pressure(intensity)
Mass/Volume
Weight/Volume
Density/Density of water
Force/Area
ML-3
ML-2T-2
M0L0T0
ML-1T-2
Stress
Force/Area
ML-1T-2
Elastic Modulus
Stress/Strain
ML-1T-2
Impulse
Mass Moment of
inertia
Momentum, Linear
Work,Energy
Power
Viscosity,Dynamic
Force x Time
Mass x Length 2
MLT-1
ML2
Mass x Linear Velocity
Force x Distance
Work/Time
Shear stress/Velocity
gradient
Dynamic viscosity/Density
Energy /Area
MLT-1
ML2T-2
ML2T-3
ML-1T-1
Viscosity,kinematic
Surface Tension
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L2T-1
MT-2
2
N=kgms2
Kg m-3
Nm-3
Nm-2=Pa
Nm-2=Pa
Nm-2=Pa
Ns
Kg m2
m2s-1
Kgs-2
Principle of Dimensional Homogeneity
Dimensional homogeneity means that the dimensions of each additive term on both
sides of equations must be equal. The principle of homogeneity of dimensions can be
used to:
1. To check whether the equation has been correctly formed;
2. To establish the form of an equation relating a number of variables;
3. To assist in the analysis of experimental results.
Also, if an equation truly expresses a proper relationship between variables in a physical
process, it will be dimensionally homogenous; i.e., each of its additive terms will have the
same dimensions.
NOTE:
The principle of Dimensional Homogeneity should be well understood by the
students before learning the different methods of Dimensional analysis. The
different methods are Rayleigh’s method, Indicial method, Pi Buckingham Theorem,
Matrix Method.
Example (Dimensional Homogeneity)
ILLUSTRATIVE PROBLEM
A useful theoretical equation for computing the relation between pressure, velocity, and altitude
in an steady flow of nearly inviscid, nearly incompressible fluid, with negligible heat transfer and
shaft work is the Bernoulli’s relation named after Daniel Bernoulli. The equation is shown below:
p0 = p + 1/2ρV2 + ρgZ
Where:
Po=stagnation pressure
P=pressure in moving fluid
V=velocity
Ρ=density
Z=altitude
G=gravitational acceleration
1. Show that the above equation satisfies the principle of dimensional homogeneity, which
states that all additive terms in a physical equation, must have the same dimensions.
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Solution:
a) p0 = p + 1/2ρV2 + ρgZ
b) ( ML-1T -2) =( ML-1T -2) + (ML-3) (LT-1)2 + (ML-3) (LT -2) (L)
c) (ML-1T -2) =( ML-1T -2) Left hand side=Right Hand Side, hence the equation is dimensionally
homogenous.
2.) Show in the above equation that consistent units results without the use of additional
conversion factors in S.I. units.
a) (N/m2) =(N/m2) + (Kg/m3)( m2/s2) +(Kg/m3)( m/s2)(m)
b) (N/m2) =(N/m2)
3) Show in the above equation that consistent units results without additional conversion
factors in B.G. system of units.
a) (lb/ft2) = (lb/ft2) + (Slugs/ft3) (ft2/s2) + (Slugs/ft3) +( ft/s2)(ft)
b) (lb/ft2) = (lb/ft2)
H) TUTORIAL PROBLEMS
1. Show that the following equations satisfy the principle of Dimensional
Homogeneity?
a) τo = hf (ρgA)/PL
b) Q = CAo √2gh
c)
V= R2/3 S1/2/n
d) The equation for the discharge (Q) over a sharp crested rectangular weir Is:
Q = 0.667 CD L (2g)1/2 H1/2
where : L = Length of weir
g = acceleration due to gravity
H= Head over the weir
CD = Coefficient of discharge (dimensionless)
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Chapter 2
Dimensional Analysis and Hydraulic Similitude
I. Dimensional Analysis
A.HISTORY OF DIMENSIONAL ANALYSIS (Fluid Mechanics by Frank M. White, Fourth Edition, Page 280).
Historically, the first person to write extensively about units and dimensional reasoning in physical
relations was Euler in 1765. Euler’s ideas were far ahead of his time, as were those of Joseph Fourier,
whose 1822 book, Analytical Theory of Heat, outlined what is now called the principle of Dimensional
Homogeneity and even developed some similarity rules for heat flow. There were no further significant
advances until Lord Rayleigh’s book in 1877, Theory of sound, which proposed a” method of
Dimensions” and gave several examples of Dimensional analysis. The final breakthrough which
established the method as we know it today is generally credited to E. Buckingham in 1914, whose
paper outlined what is now called the Buckingham Pi Theorem for describing Dimensionless parameters.
However, it is now known that a Frenchman, A. Vaschy, in 1892, and a Russian, D Riabouchinsky, in 1911
had independently published papers reporting results equivalent to Pi Theorem. Following Buckingham’s
paper, P.W. Bridgman published a classic book in 1922, outlining the general Theory of Dimensional
Analysis. The subject continues to be controversial because, there is so much art and subtlety in using
Dimensional Analysis. Thus, since Bridgman, there have been at least 24 books published on the
subject.. There will be probably more, but seeing the whole list might make some fledgling authors think
twice. Nor Dimensional analysis limited to neither Fluid Mechanics nor engineering. Specialized books
have been written on the application of Dimensional Analysis to Metrology, Astrphysics, economics,
building scale models, chemical processing pilot plants, social sciences, biomedical sciences, pharmacy,
fractal geometry, and even the growth of plants.
B. Definitions:
1. Dimensional Analysis
is a method of Dimension. It is a mathematical
technique used in research work for design and for conducting model tests. It deals
with the dimension of the physical quantities involved in the phenomenon. All
physical quantities are measured by comparison which is made with respect to an
arbitrary fixed value. Length L, Mass M and time T are three fixed Dimensions which
are of importance to Fluid Mechanics. If in any problem in Fluid Mechanics, heat is
involved, then temperature is also taken as fixed dimension. These fixed Dimensions
are called fundamental dimension or fundamental quantity.
2. Dimensional Analysis
is a method of reducing the number and complexity of
experimental variables which affect a given physical phenomenon , by using a sort
of compacting technique.(Fluid Mechanics by Frank M. White, Fourth Edition, page
278)
3. Dimensional Analysis
is the analysis of the basic relationships of the various
physical quantities involved in the motion and in the dynamic action of the fluid.
(Hydraulics by Wisler, King and Woodburn, Fifth edition, page318).
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4. Dimensional Analysis
is a mathematical method which is of considerable
value in problems which occur in Fluid Mechanics. All physical quantities can be
expressed in terms of certain primary quantities which in mechanics are Length
(Fluid Mechanics, Volume 2, J.F.Douglas and R.D. Matthews, 3rd Edition.)
5. Dimensional Analysis
is a powerful tool for deriving dimensional relationship
of a hydraulic physical phenomenon.(Hydraulics in Civil and Environmental
Engineering by Andrew Chadwick and John Morfett, Third Edition, page 342).
6. Dimensional Analysis
also forms the basis for the design and operation of
physical scale models which are used to predict the behavior of their full sized
counterparts called the “prototypes” (Civil Engineering Hydraulics by Nalluri and
Featherstone, 4th Edition).
C. Purpose of Dimensional Analysis
1. To reduce the variables involved in a physical Hydraulic phenomenon and group
them in dimensionless forms for theoretical reference in the creation of a model of a
full scale hydraulic structure.
D. Benefits of Dimensional Analysis
1. Enormous savings in time and money;
2.
It helps in thinking and planning for an experiments and theory. It suggest
dimensionless ways of writing equations before we waste money on computer time to
find solutions It suggest variables which can be discarded ; sometimes dimensional
analysis will immediately reject variables, and at other times ,it group them off to the
side, where a few simple tests will show them unimportant.
3. Dimensional analysis provides scaling laws which can convert data from a cheap
small model to design information for an expensive, large prototype.
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Pi –Buckingham Theorem of Dimensional Analysis
There are several methods of reducing a number of dimensional variables into smaller
number of dimensionless groups. One of the method is the one proposed by
Buckingham in 1914 and is now called the Buckingham pi Theorem. The name pi comes
from the mathematical notation Π, meaning a product of variables. The dimensionless
groups found from the theorem are power products denoted by Π1, Π2, Π3, etc. The
method allows the pi’s to be found in sequential order without resorting to free
exponents.
A. Procedure of Buckingham Pi Theorem(Frank M. White)
1. List and count the number of variables involved in the problem. If any important variables
are missing, Dimensional analysis will not be accurate or fail.
2. List the dimensions of each variable according to MLT.
3. Select the variables that will be used as repeating variables. These repeating variables are
always repeated in each Π terms. The choice of repeating variables is governed by the
following considerations:
a) As far as possible, the dependent variable should not be selected as
repeating variables ;
b) The repeating variables should be chosen in a such a way that one
variable contains geometric property, second variable contains flow
property and third variable contains fluid property;
-Variables with Geometric Property:
Length (L), Diameter (D), Height (h), etc.
-Variables with flow property:
Velocity (V), Acceleration (a), etc.
-Variables with fluid property:
Viscosity (μ), Density (ρ), etc.
c) No two repeating variables should have the same
dimensions.
NOTE: In most fluid mechanics problems, the choice of repeating variables may be :
i) d,V, ρ ; ii) l,V, ρ; iii) l,V,μ; or
iv) d, V,μ.
4. Generally, if heat is not included in the physical phenomenon, the three (3)
fundamental dimensions are only M, L and T.Subtract the total number of known
variables to three (3) and this will determine the number of Pi terms.
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5. Write the Pi terms in equation of the form f= (П1, П2, П3, etc)
6. Solve each Pi terms by using the principle of Dimensional Homogeneity.
7. Substitute the Pi terms in Equation mentioned in No.5.
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B.
ILLUSTRATIVE PROBLEM (Hydraulics by Chadwick and Morfett, page 342).
As part of the development programme, scale model test have been carried out on a new
hydraulic machine. The experimental Team has presented the following data. Thrust Force
F, the flow velocity V, dynamic viscosity μ, and density ρ, of the fluid. A typical size of the
system, L, is also given.
Two questions must be posed namely:
a) How to analyze or plot the data in the most informative way and
b) How to relate the performance of the model to that of the working prototype.
Given variables:
F, V, μ, ρ, L
SOLUTION: By Pi-Buckingham Theorem.
1.
2.
3.
4.
5.
6.
7.
Number of Fundamental Dimensions is Three (M, L, and T).
Number of known Variables is five.
Number of Pi – Terms = 5 – 3, hence there will be Two Pi-terms.
In function form, the equation is f(ρ, V, L,F, μ) =0
The repeating variables are ρ, V, L.
The final equation after evaluating each П terms should be in the form f(П1, П 2).
As each П is dimensionless, equate the equation as follows:
a) f (π, π 2) = MoLoTo
π1= ρa , Vb, Lc, F
(ML-3)a (LT -1)b Lc (MLT -2) = MoLoTo
8. Evaluate the value of the exponent a, b and c.
Power of M ; a + 1 = 0,hence a = -1
Power of T ; -b – 2 = 0
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b=-2
Power of L; -3a + b +c =0
c = -2
SO;
π1 = F/ ρ V2 L2
ALSO
:
π 2= ρa , Vb, Lc , μ = MoLoTo
After similar evaluation: π2 = μ / ρVL
9.
Finally,
f (π1, π 2) = f (F/ ρV2L2, μ / ρVL)
There are number of important points to be made about the final
result of Buckingham Theorem.
1. Two groups have emerged from the analysis,
F/ ρV2L2 and μ/ρVL.Both groups are dimensionless. For conciseness,
dimensionless groups are referred to as П groups.
2. Dimensionless groups are independent of units and of scale. П1 and П 2,
are therefore both applicable to the model and the prototype.
3.
Both П groups represent ratios of forces:
Force = mass x acceleration
Mass of a body= ρ Volume= ρ L3,
and acceleration = Velocity /time= LT -2
hence, F= (ρ L3) ( LT -2 ) = (ρL4 T -2 ) , therefore , F= ρV 2L2
4. So П1 is the ratio of thrust force to the inertia force.
5.
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And П2 is the ratio of inertia force to the viscous force.
ILLUSTRATIVE PROBLEM 2:
Textbook: Understanding Hydraulics by Less Hamil ISBN 10 0 – 333-77906 -1- (page362)
The theoretical discharge (Q) over a sharp crested rectangular weir is:
Q = 2/3 b (2g)1/2 (H)3/2
where b = is the width of the weir crest ,g is gravity, and H is the head above the crest . It is
expected that ρ the density and μ the dynamic viscosity of water should be included in the
analysis since they are important variables. Using Pi-Buckingham Theorem, investigate whether
or not ρ and μ do influence the theoretical discharge (Q) over weir.
SOLUTION:
1) Define the dimensions of each variable in the phenomenon:
No.
1
2
3
4
5
6
Quantity
Discharge
Width
Gravity
Head
Density
Dynamic Viscosity
Variables
Q
b
g
H
ρ
μ
Unit
m3/s
m
m/s2
m
Kg/m3
Kg/ms
Dimension
L3 T-1
L
LT-2
L
ML-3
ML-1T-1
2) The quantities can be written in the form of a functional relationship
f ( q,b,g,H,ρ,μ ) = 0
3)
4)
5)
6)
7)
Number of Fundamental Dimensions is Three (M, L, and T).
Number of known Variables is six.
Number of Pi – Terms = 6 – 3, hence there will be 3 Pi-terms.
The repeating variables are g,H and ρ.
The final equation after evaluating each П terms should be in the form:
f(П1, П 2, П 3 ) =0.
8) As each П is dimensionless, equate the equation as follows:
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f (П1, П 2, П3) = MoLoTo
9) π1 = ga , Hb, ρc ,Q
π2 = ga , Hb, ρc ,b
π3 = ga , Hb, ρc ,μ
For π 1 ; a = -1/2 . b = -5/2 and c = 0
π1 =.
Q
.
For π 1 ; a = -1/2 . b = -3/2 and c = - 1
π1 =.
g1/2 H5/2
μ
.
1/2
3/2
ρg H
For π2 : a = 0 ,b = -1 and c =0
Π2 =. b .
H
10) Therefore
f(π1, π2, π 3 ) =0 becomes f( .
Q
g1/2 H5/2
.,. b ., .
H
μ
. } =0
1/2
3/2
ρg H
11) Any of the terms can be rearranged , combined or inverted , so:
.
Q
g1/2 H5/2
. = f’{ . b . , .
μ
.}
1/2
3/2
H
ρg H
12) Or
Q = f’{. b g1/2 H5/2 . , . ρ g1/2 H3/2 . }
H
μ
Q = f’{ b g1/2 H3/2 , . ρ g1/2 H3/2 . }
μ
The first term in the brackets is recognizable as the weir discharge equation, while the second
indicates that ρ and μ do affect the discharge.
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Reason: Since they do not appear in the weir equation: Q = 2/3 b (2g)1/2 (H)3/2 , their effect
has to be included in the Coefficient of discharge, that is why the new weir equation becomes
Q = CD2/3 b (2g)1/2 (H)3/2 .
Also, V = (gH)1/2 , then the second term becomes ρHV/μ , which is the Reynolds Number(Re)
.Thus:
Q = f’{ b g1/2 H3/2 , Re }
C. TUTORIAL PROBLEMS (Pi – Buckingham Theorem)
Problem 1
Hydraulic machine
The quantities which are usually considered in a Dimensional analysis of
machines are:
Hydraulic
a. The power (P) and rotational speed (N) of the machine;
b. The pressure head (H) generated by the machine;
c. The corresponding discharge (Q);
d. The typical machine size (D) and the roughness (ks);
e. The fluid characteristics (ρ) and (μ).
Derive dimensionless groups for the Hydraulic machine the relation of power with the
other physical quantities of the Hydraulic machine.
Note: represent H= gH. And Use ρ, N and D as repeating variables.(See Hydraulics in
Civil and Environmental Engineering by Chadwick and morfett,3rd Edition)
Answer:
(π1= P/ρN3D5)
(π3= gH/N2D2)
(π2= Q/ND3)
(π4 = ρND2/ μ)
(π5= ks /D)
Problem 2
Find an expression for the Drag Force on a smooth sphere of diameter D,
moving with a uniform velocity V in a fluid of density ρ and dynamic viscosity
μ.Let repeating variables be ρ,V,D.
Problem 3
Obtain an expression for the pressure gradient (∆p) in a circular pipeline, of
Diameter D, length L, effective roughness, k, conveying an incompressible fluid of
density, ρ, dynamic viscosity, μ, at a mean velocity, V, as a function of non
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Dimensional groups. Use π- Buckingham Theorem. Repeating Variables are ρ, V
and D. See Fluid Mechanics by Nalluri and Featherstone on Dimensional
Analysis).
Problem 4
The capillary rise h of a liquid in a tube varies with the tube diameter d, gravity g, fluid density
ρ, surface Tension σ and the contact angle Ө.
Find dimensionless statement of this relation. (See Fluid Mechanics by Frank M. White.)
Problem 5(Handbook, Civil Engineering Calculations by Tyler Hicks, PAGE 6.27).
The velocity of raindrop in still air is known or assumed to be a function of gravitational
acceleration, drop diameter of the raindrop, dynamic viscosity of the air, and the density of
both the water and the air. Develop a dimensionless parameters associated with these
phenomenon.
Quantity
V= Velocity of the raindrops
g=gravitational acceleration
D=diameter of drops
μa= air viscosity
ρw =water density
ρa =air density
Units
LT-1
LT-2
L
ML-1T-1
ML-3
ML-3
Repeating Variables: g,D, μa
Problem 6
The theoretical discharge (Q) over a sharp crested rectangular weir is:
Q = 2/3 b (2g)1/2 (H)3/2
where b = is the width of the weir crest ,g is gravity, and H is the head above the crest . It is
expected that ρ the density and μ the dynamic viscosity of water should be included in the
analysis since they are important variables. Using Pi-Buckingham Theorem, investigate whether
or not ρ and μ do influence the theoretical discharge (Q) over weir. Use g, H and μ as the
repeating variables.
Problem 7 Textbook: Understanding Hydraulics by Less Hamil ISBN 10 0 – 333-77906 -1- (page363)
17
In a turbulent flow, the headloss when the liquid flows through a smooth pipe is assumed to
depen upon the quantities below. Determine the form of the dimensionless equation using PiBuckingham Theorem. Use V,D, ρ as repeating variables
hf =Head loss
V= Velocity
g=gravitational acceleration
D=diameter
μ = dynamic viscosity
ρw =water density
LT-1
LT-1
LT-2
L
ML-1T-1
ML-3
I. Hydraulic Similitude
A. INTRODUCTIONS (Fluid Mechanics by Frank M. White)
So far we have learned about dimensional homogeneity and the Pi Theorem Method, using
power products, for converting homogenous physical relations to dimensionless form. This is
straightforward mathematically, but there are certain engineering difficulties which need to be
discussed.
First, we have more or less taken for granted that the variables which affect the process can be
listed and analyze. Actually selection of the important variables requires considerable judgment
and experience. The engineer must decide, e.g., whether viscosity can be neglected. Are there
significant temperature effects? Is surface tension important? What about wall roughness? Each
Pi group which is retained increases the expense and effort required. Judgment in selecting
variables will come through practice and experience.
Once the variables are selected and Dimensional analysis is performed, the experimenter seeks
to achieve similarity between the model tested and the prototype to be designed. With
sufficient testing, the model data will reveal the desired dimensionless function between
variables.
Each evaluated Pi terms can now be used to determine the complete similarity between the
model and the prototype, either graphically or analytically. But instead of complete similarity,
the engineering literature speaks of particular type of similarity, technically known as
SIMILITUDE.
B. Definitions
1. Similitude is defined as the similarity between the model and the prototype in
every respect which means that the model and the prototype are completely
similar. Similarity between model and prototype may take three (3) different
forms: geometric similarity; kinematic similarity and dynamic similarity.
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2. Similitude is the principles and theory of predicting the behavior of full scale
hydraulic structures or prototype by conducting an experiment on a scale model
of the prototype.
3. Similitude is the principle used in predicting the performance of a full scale
hydraulic structures (such as Dams, spillways, etc.) or full scale hydraulic machines
(such as turbines and pumps).The prediction is done by actually constructing a
model of the structures or machines and tests are performed on them to obtain
the desired information before manufacturing a full scale structures.
The model is the small replica of the actual structure or
machine .The actual structure or machine is called prototype.
HYDRAULIC SIMILARITY :
( Textbook: Understanding Hydraulics by Less Hamil ISBN 10 0 – 333-77906 -1- (page365)
The Relationship between model and prototype performance is determined by the laws of hydraulic
similarity. Since there are many laws and model cannot comply with all of them, simultaneously, the
model will not reflect totally the performance of the prototype. Some error will be incurred, which is
referred to as SCALE EFFECT. But by careful design of the model , by using a reasonably large scale,
and by judicious interpretation of the results ( experience helps), scale effects can be minimized. To
design a successful hydraulic model and use it effectively to predict the performance of the prototype
requires knowledge of similarity. There are three types of hydraulic similarity that must be considered.
There are three types of hydraulic similarity that must be considered.
C. The three types of similarity in Hydraulic Similitude
1. Geometric Similarity implies similarity of form.
a) A model is geometrically similar to the prototype, if the ratios of all
homologous lengths, areas and volumes,
(Homologous means the point of comparison between the model and prototype
have the same relative location) between the model and the prototype are equal.
(Hydraulics by Wisler, King and Woodburn, 3rd Edition).
b) A model or prototype is geometrically similar, if and only if, all body dimensions
in all three coordinates have the same linear scale ratio. ( Fluid Mechanics by
Frank M. White)
The quantities involved in geometric similarity are length, area and volume.
i)Ratio of homologous length between model and prototype:
Lm/Lp = Lr
where:
Lr is the scale ratio between model and prototype
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ii) Ratio of homologous area between model and
prototype.
Am/Ap=Lm2/Lp2=Lr2
iii) Ratio of homologous volume
Vm/Vp=Lm3/Lp3=Lr3
2. Kinematic Similarity implies similarity of motion.
a) Kinematic similarity of model to prototype is attained if the paths of
homologous moving particles are geometrically similar and if the ratios of
velocities of the various homologous particles are equal. (Hydraulics by Wisler,
King and Woodburn, 3rd Edition).
Kinematic similarity introduces the concept of time as well as length. The ratio of
the time required for homologous particle to travel homologous distances in
model and prototype is
Tm/Tp = Tr
For acceleration, am/ap = ar
b) Kinematic similarity requires that the model and prototype have the same length
scale ratio and the same time scale ratio. The result is that the velocity scale ratio
will the same at all homologous points of the model and prototype ( Frank M.
White) .
3. Dynamic Similarity
Dynamic similarity means the similarity of forces between the model and the
prototype. Dynamic similarity is said to exist between the model and the
prototype, if the ratios of the homologous forces between the two are equal.
Also the directions of the corresponding forces must be the same.
Dynamic similarity exists when the model and the prototype have the same length
scale ratio, time scale ratio, and force scale ratio.
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The requirements therefore for Dynamic similarity must include geometric and
kinematics similarities, in addition to force scale ratio similarity. Without the two,
there can be no dynamic similarity.
D. Types of forces acting in moving fluid for purposes of Dynamic
Similarity.
In Fluid flow problems, the forces acting on a fluid mass may be anyone, or a
combination of the several of the following forces.
1. Inertia Force,(Fi)
It is equal to the product of mass and acceleration of the flowing
fluid and acts in the direction opposite to the direction of
acceleration. It always exists in fluid flow problem. In formula;
F=ma
2. Viscous force, (Fv)
It is equal to the product of shear stress (τ) due to viscosity and surface
area of the flow. It is present in Fluid flow problems, where viscosity is
having an important role to play. In formula:
Fv = μ du/dy (A)
Viscosity (μ) defined:
Viscosity is that property of fluid which determines the amount of its
resistance to shearing stress. A perfect would have no viscosity. There is no
perfect fluid, but gasses show less variation in viscosity than liquids. Water is
one of the least viscous of all liquids, whereas glycerin, heavy oil and molasses
are liquids having comparatively high viscosities.
3. Gravity Force (Fg)
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It is equal to the product of mass and acceleration due to the gravity of
the flowing fluid. It is present in case of open surface flow.
Fg= mg
4. Pressure Force (Fp)
It is equal to the product of pressure intensity and cross sectional area of
the flowing fluid .It is present in case of pipe flow. In formula
Fp= pA
5. Surface Tension Force( Fs)
It is equal to the product of surface tension per unit length and length of
surface of the flowing liquid.In formula:
Fs= surface tension per unit length (Length)
Fs= σL
6.
Elastic Force Fe
It is equal to the area of elastic stress and the area of the flowing liquid. In
formula:
F=stress (Area)
E. Relation of Different kinds of forces acting in flowing liquid to
Dynamic similarity between the model and the prototype.
For dynamic similarity between the model and the prototype, the ratio of the
homologous forces between the model and the prototype should be equal.
It means that for dynamic similarity between the model and the prototype, the
dimensionless numbers should be the same between the two.
But since it is difficult to satisfy the condition that all dimensionless numbers
(Reynolds’s No., Froude No., Weber No., Euler’s No. and Mach No.), it is sufficient
that MODELS are designed on the basis of the ratio of the
FORCES DOMINATING THE FLUID FLOW.
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F. The Model Laws in Dynamic Similarity
1.Reynold’s Model Law
It is defined as the ratio of inertia force to the viscous force of the flowing fluid.
Derive the Reynold’sNumber (Re= ρVL/μ)
Inertia force:
a) Fi= ma:
mass=density (volume)
and
acceleration= velocity/time
b) Fi= ρ (Volume) (Velocity/time)
c) Fi= ρ (Volume / time) (Velocity)
d) Fi= ρ (Discharge) (Velocity);
but Q=AV
e) Fi= ρ(AV)(V)
f) Fi = ρ A (V) 2
Viscous Force:
a)Fv= Shear stress x Area ; but τ= μdu/dy; where du/dy V/L
b) Fv= μ (V/L)(A)
Definition of Reynolds’s Number is Re= Fi/Fv
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Substitute evaluated values, hence:
Re= ρ (AV) 2 / μ (V/L) (A)
Re= ρ (V) L / μ.
Notation;
Ρ= density
V= Velocity of flowing liquid
L = Length of section
Μ= Dynamic viscosity
2. Froude’s Model Law
The Froude Number is defined as the square root of the ratio of the
inertia force to the gravity force of the flowing liquid.
Fe = √Fi/Fg
but Fi = ρA(V) 2 and
a. Gravity Force (Fg) = mass x gravity
b.Fg= (density x volume) x gravity
c.Fg = ρ x Volume x g
d.Fg = ρ x L3 x g
e.Fg = ρ x L2 x L x g
f.Fg = ρ x A x L x g :
substitute evaluated values:
Fe =√ ρ (AV) 2 / ρ x A x L x g
Fe = √ (V) 2 / L g
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Fe = V / √ L g
where :
V= Velocity
L= Length
G= gravitational pull of the earth
3.Euler’s Model Law
Derive the Euler’s Number Eu = V √ρ / √p
Euler’s Number is defined as the ratio of inertia force to the
pressure force of the flowing fluid.
a.
Eu = √Fi/Fp
b. Fi = ρA(V) 2
and
Fp = pA
c. Substitute to definition of Euler’s Number.
Eu = √ ρA(V) 2 / √ pA
d. Eu = √ ρA(V) 2 / √ pA
e. Eu = √ ρ (V) 2 / √ p
, finally
f. Eu = V √ρ / √p
4. Weber’s Model Law
Derive the Weber’s Number We = V / √σ/ρL
Weber’s Number is defined as the ratio of inertia force to the surface
Tension force of the flowing fluid. In formula We = √Fi/Fs
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a. Fi = ρA(V) 2
and
b. Fs = Surface Tension per unit length x Length
Fs = σL
c.Substitute inertia force and surface tension force to definition of Weber Number;
We = √ ρA(V) 2 / √ σL , finally
We = V / √σ/ρL
4.Mach Model Law
Derive the Mach’s Number M = V √ρ / √fs
Mach’s Number is defined as the ratio of inertia force to the elastic
force of the flowing fluid. In formula M = √Fi/elastic force
a. Fi = ρA(V) 2
and
b. Elastic Force = elastic stress x A
C.Substitute inertia force and elastic force to definition of Mach’s Number;
We = √ ρA(V) 2 / √ fs(A) , finally
M = V √ρ / √fs
Note:
Velocity of sound in fluid= √elastic stress/ density of fluid, mathematically,
C=√fs/ρ, hence
M= V/C
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G. Application of the different Model Laws:
1. Reynold’s Model Laws; (Re= ρVL/μ)
Reynold’s Model Law is the theoretical principle in which models are based on Reynold’s
Number. Models based on Reynold’s Number include;
i. pipe flow;
ii. Resistance experienced by submarines, airplanes, fully immersed
bodies, etc.
For Dynamic similarity between models and prototype, the
Reynold’s Number between the model and the prototype must be
equal
Tutorial Problem 1 (Reynold’s Model Law)
A pipe of diameter 1.50 meters is required to transport an oil of specific gravity of 0.90,
viscosity 3x10 -2 poise at the rate of 300 liters per second.
Test were conducted on a 15 cm diameter pipe using water at 200 C.
Find the velocity and rate of flow in the model, using viscosity of water equal to 0.01 poise at
20 0 C.
Given :
For Prototype;
1. Diameter of prototype(Dp)=1.50 m
2. Viscosity of oil (μp)
=3x10 -2 poise
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3. Q for prototype (Qp)
=300 liters/sec
=3.0 m3/s
=0.90
4.Specific gravity for oil
Relative Density of oil = Specific gravity x density of water= 0.90 x 1000 kg /m3
For Model;
1. Diameter of model (Dm)
2. Viscosity of water (μm)
3. Density of water (ρm)
=15 cm = 0.15m
=0.01 poise
= 1000kg/m3
SOLUTION:
Principle of solution:
For pipe flow, the dynamic similarity will be obtained if the Reynold’s number in
the model and the prototype are equal.
(Re) model= (Re) prototype;
By definition,
Re= ρ (V) L / μ
a. (ρV L / μ )model= ( ρV L / μ) prototype,
For pipes, Linear Dimension (L) = Diameter (D)
b. Vm / VP = ρm/ ρp x Dm/Dp
Substitute given values
c. Vm / VP = ( 900/1000) x ( 1.5/0.15) x (1x10 -2/3x10 -2)
d. Vm / VP = 3.0
or
e. Q = AV;
Vm = 3.0 (VP)
for prototype,
hence
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Vp= Q/A ,
Vp = 3.0
/ (П/4) (D)2
Vp = 1.697 m/s
f. Vm = 3.0 (VP)
hence
Vm = 3.0(1.697) = 5.091 m/s
g. Finally, Q = AV,
Q= (П/4) (D)2 x 5.091 ,
Q= 0.0899 m3/s = 89.9 l/s
TUTORIAL PROBLEMS (Reynold’s Model Law)
1. The ratio of lengths of a sub-marine and its model is 40:1. The speed of
sub-marine (prototype) is 12 m/s. The model is to be tested in a wind tunnel.
Find the speed of air in wind tunnel. Also determine the ratio of the
drag (resistance) between the model and its prototype. Take the value of
kinematic viscosities for seawater and air as 0.012 stokes and 0.016 stokes
respectively. The density for seawater and air is given as 1030 kg/m3 and
1.24 kg/m3 respectively.
2. A ship 300 meters 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 30 m/s and the resistance of the model is 60 N.
Determine the velocity of the ship in sea water and also the resistance of the ship
in sea water. The density of air is given is given as 1.24 kg/m3. Take the kinematic
viscosity of sea water and air as 0.012 stokes and 0.018 stokes, respectively. 1
stokes = 1 cm2/s.
3. A 10 km pipeline, 750 mm in diameter, is to be used to convey oil, Dynamic
viscosity (μ) is (0.008 kg/m s) and density (ρ) is 850 kg/m3. The design discharge is
450 l/s. The pipeline will incorporate booster pumping stations at suitable intervals.
As part of the design procedure, a model study is to be carried out. A model scale 1:
50 have been selected, and air is to be used as the model fluid. The air has a density
of 1.20 kg/m3 and air viscosity of 1.80 x10 -5kg/m s.
At what mean air velocity will the model be correctly simulating the flow of oil?
If the head loss in the model is 10 meters for the full pipe length, what will be
the head loss in the prototype?
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4.Water is flowing through a pipe of diameter 30 centimeters at a velocity of 4.00
m/s. Find the velocity of oil flowing in another pipeline of diameter 10
centimeters, if the condition of dynamic similarity should be satisfied between
the two pipes. The viscosity of water and oil is given as 0.01 poise and 0.025 poise.
The specific gravity of oil is 0.80.
5a. The hydraulic head loss in a simple pipeline is assumed to depend on the
following quantities:
Density (ρ), dynamic viscosity (μ), diameter (D), Length (L) and roughness ks of
the pipe, and the typical velocity (V) of flow. Assume the repeating variables
to be ρ, V, D.
Develop the appropriate dimensionless groups to describe the flow.
5b. A 10 km pipeline, 750 mm in diameter, is to be used to convey oil, Dynamic
viscosity (μ) is (0.008 kg/m s) and density (ρ) is 850 kg/m3. The design discharge is
450 l/s. The pipeline will incorporate booster pumping stations at suitable intervals.
As part of the design procedure, a model study is to be carried out. A model scale 1:
50 have been selected, and air is to be used as the model fluid. The air has a density
of 1.20 kg/m3 and air viscosity of 1.80 x10 -5kg/m s.
i)
ii)
At what mean air velocity will the model be correctly
simulating the flow of oil?
If the head loss in the model is 10 meters for the full pipe
length, what will be the head loss in the prototype?
Answer:
5a. П1= hf/D
П3= ks/D
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П2= μ/ρDV
П4 =L/D
Question 2b
8 marks
The ratio of lengths of a sub-marine and its model is 40:1. The speed of sub-marine
(prototype) is 12 m/s. The model is to be tested in a wind tunnel. Find the speed of
air in wind tunnel. Also determine the ratio of the drag (resistance) between the
model and its prototype. Take the value of kinematic viscosities for seawater and
air as 0.012 stokes and 0.016 stokes respectively. The density for seawater and air is
given as 1030 kg/m3 and 1.24 kg/m3 respectively.
Given :
For Prototype;
1. Speed (VP)
=10 m/s
2. Kinematic viscosity (νp) =0.012 x10-4 m2/s: Note 1 stoke = cm2/s
3. Density ( ρP)
=1000 kg/m3
For Model;
1. Speed (VP) =____ m/s
2. Kinematic viscosity (νp) =0.016 x10-4 m2/s: Note 1 stoke = cm2/s
3. Density ( ρP)
=1.24 kg/m3
SOLUTION:
Part(a) : Principle of solution:
For dynamic similarity , Reynold’s number in the model and the
prototype are equal
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(Re) model= (Re) prototype;
By definition, Re= (V) L / ν
(V) L / ν)model= (V) L / ν) prototype,
Vm = νm/ νp x LP/Lm ( VP)
Substitute given values
Vm =400 m/s
Part (b) Ratio of drag force:
Inertia force:
a) Fi= ma:
mass=density (volume)
b) Fi= ρ (Volume) (Velocity/time)
c) Fi= ρ (Volume / time) (Velocity)
d) Fi= ρ (Discharge) (Velocity);
e) Fi= ρ(AV)(V)
f) Fi = ρ A (V) 2
g) Fi = ρ L2 V2
and Fr = Fi = ( ρ L2 V2)P = 467.22
Fm (ρ L2 V2)m
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but Q=AV
and
acceleration= velocity/time
II. FROUDE MODEL LAW: Fe = V / √ L g
Froude Model Law is the law in which the models are based on Froude Number which
means for dynamic similarity between the model and the prototype, the Froude Number
for them should be equal.
Froude model law is applicable when the gravity force is the only predominant force, which
control the flow in addition to the force of inertia.
Froude Model Law is applied in the following fluid flow problems:
1.
Free Surface Flows such as flow over spillways, weirs, sluices, channels etc.
2.
Where waves are likely to be formed on surfaces of fluid
3.
Flow of jet from orifice or nozzle
4.
Where fluids of different densities flow over one another.
DIFFERENT SCALE RATIOS BASED FROM FROUDE MODEL LAW:
For dynamic similarity between the model and the prototype,
a. Scale ratio for length (Lr)
(Fe) model = (Fe) prototype, but,
Fe = V / √ L g
(V / √ L g) model = (V / √ L g)prototype
If the test on the model are performed on the same place, where the prototype is to
operate, then the value of gravity will be the same, and
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(V / √ L) model = (V / √ L)prototype, simplify the equation;
Vm/Vp= √Lm/Lp, vice versa, Vp / Vm= √Lp / Lp
Let Lr = scale ratio then √ Lr = VP / Vm = √Lp / Lp
b. Scale ratio for time (Tr) = Tp / Tm
Time = Length/Velocity
Tr = Tp / Tm
Tr = (L/V)prototype / (L/V) model
Tr = (Lp/Lm) x ( Vm / Vp)
Tr = Lr x 1/√ Lr, finally for scale ratio for time
Tr = √ L r
c. Scale ratio for acceleration, ar = ap/am
acceleration = velocity/time
ar = √ Lr x 1/ √ Lr
d. Scale ratio for discharge Qr =Qp/Qm
Q r = ( Lr)2.5
e. Scale ratio for force.
Force = mass x acceleration
Fr = Lr3
f. Scale ratio for power
Power = work per unit of time =F x L/ T
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Pr = Lr 3.5
ILLUSTRATIVE PROBLEM (Froude Model Law)
A spillway is to be built to a geometrically similar scale of 1/50 across a flume of 60
cm width. The prototype is 15.00meters high and maximum head on it is expected to
be 1.50 meters.
i. What height of model and what head on the model should be used?
Ii.What flow per meter length of the prototype should be used?
Given values:
For Prototype
(Hp) Height of Prototype = 15.00 meters
(Hp*)Head on Prototype = 1.50 meters
For model
(Bm)Width of model = 60 cm = 0.60 meters
(Qm)Flow over model= 12 l/s =0.012 m3/s
(pm)Pressure in model = - 20 cm=0 .20 m
(Lr) Scale ratio for length = 1:50
SOLUTION
I. Scale ratio for length
a. Lr= Hp/Hm for comparison between the prototype and the model
35
b.
50 = 15 / Hm
; Hm =15/50,
and
Hm = 0. 30 m
c. Head on model
Hm* = 1.5/50 = 0.03m
d. Width of Prototype
Bp = Lr x Bm = 50 x0.60 = 30 m
2. Scale ratio for discharge Qr =Qp/Qm
Q r = ( Lr)2.5 = Qp/Qm
Qp/Qm= (50) 2.5
Qp= (50) 2.5 (Qm)
Qp = 17,677 (12)
= 212,132.04 l/s
Discharge per meter length of prototype = Qp / length of prototype
= 212,132.04/ 30
= 7,071.07 l/s.
TUTORIAL PROBLEMS: Froude Model Law
1. In 1 in 40 model of a spillway, the velocity and discharge are 2.00m/s
and 2.50m3/s .Find the corresponding velocity and discharge in the
prototype.
36
2A ship model of scale 1/50 is towed through sea water at a speed of 1.00
m/s. A force of 2.0 N is required to tow the model. Determine the speed of
the ship and the propulsive force on the ship, if prototype is subjected to
wave resistance only.( This means that the Froude Number for model and
prototype should be the same.)
III. Euler’s model law (Eu = V √ρ / √p)
Euler’s model law is the law in which the models are designed on Euler’s Number. This
means that for dynamic similarity between the model and the prototype, the Euler’s
Number for model and Prototype should be equal.
Euler’s model law is applicable when the pressure forces are alone predominant in
addition to the inertia force.
Euler’s model law is applied to fluid problems, where flow is taking place in a closed pipe in
which case turbulence is fully edveloped so that the viscous force are negligible and gravity
force and surface tension is absent. This law is also used where cavitations takes place.
IV. Weber model law, (We = V / √σ /ρL)
Weber law is the law in which models are based on Weber’s Number. As defined, Weber’s
Number is the ratio of square root of inertia force to surface tension force. Hence where
surface tension effects dominate the flow of fluid in addition to inertia force, the dynamic
similarity between the model and the prototype is obtained by equating the Weber Number
of the model to the Weber number of the prototype. Hence according to this law, (We)
model = (We) prototype
Weber Model Law is applicable in;
i.
ii.
iii.
iv.
Capillary rise in narrow passages
Capillary movement of water in soil
Capillary waves in channels
Flow over weirs for small heads.
V. Mach model law, (M = V / √fs /ρ)
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Mach model law is the law in which the models are designed based on Mach number.
Mach number is the ratio of square root of inertia force to elastic force. Hence where
the forces due to elastic compression dominate the flow of fluid in addition to inertia
force, the dynamic similarity is obtained by equating the Weber number between the
model and the prototype.
38