Rayleigh's method- Theory and Examples Buckingham Pi Theorem- Theory and Examples Model and Similitude Forces on Fluid Dimensionless Numbers Model laws Distorted models

1. MAIN TOPICS
Rayleigh's method- Theory and Examples
Buckingham Pi Theorem- Theory and Examples
Model and Similitude
Forces on Fluid
Dimensionless Numbers
Model laws
Distorted models
Chapter :
Dimensional Analysis, Similitude, and Model laws
Fluid Mechanics

2. Mass, Length ,Time are
fundamental Quantities.
Derived quantities
possess more than one
fundamental dimensions

3. Rayleigh's method of dimensional analysis is a conceptual tool used in an
engineering. This form of dimensional analysis expresses a functional
relationship of some variables in the form of an exponential equation.
Question : RAYLEIGH’S METHOD Steps
Step 1:
Establish relationship between dependent and independent variable.
Step2:
Setup an equation between dependent and independent variable. The
dependent variable is expressed as a product of all the independent variables
raised to unknown integer exponents. (a,b,c,d….. are unknown integer
exponents)

4. Step 3:
Form a tabular column representing the variables in the equation, their units
and dimensions.
Step 4:
Use the dimension in step 3 to obtain the unknown value a,b,c,d,…. in step 2
by using the principles of dimensional homogeneity of the variables.
Step 5:
Substitute the unknown values of a,b,c,d… in the equation which was formed
in step 2.

5. Example.1:
Find the expression for the drag force on smooth sphere of
diameter D, moving with a uniform velocity V in a fluid of density
ρ and dynamic viscosity µ. Apply RAYLEIGH’S METHOD
Solution:
so we have a smooth sphere of diameter D and this sphere is
thrown into a fluid with a velocity V. The fluid has a density ρ and
viscosity μ. as the sphere moves through the fluid there will be a
generation of drag force F “force which is against the motion of
the sphere”. so we understand that the drag force F depends on
D,V,ρ and μ.

6. Step 1: Relation between dependent and independent variable.

7. Step 2: Set up of the equation

8. Step 3: Set up Tabular column

9. Step 4: Use dimension from step 3 to solve the unknown variables
a,b,c,d

10. Step 5: Substituting the value of a, b, c, d in step 2

11. Example.2 : Find the expression for time period of a simple pendulum. Time
period T depends on length of pendulum L and g. Apply RAYLEIGH’S METHOD
The time period of a simple pendulum:
It is defined as the time taken by the pendulum to finish one full oscillation and
is denoted by “T”

12. Example. 3 : Find the expression for power P developed by pump which
depends on Head H, Discharge Q and specific weight ω of fluid.
Apply RAYLEIGH’S METHOD

14. Question : Buckingham’s Pi Theorem
The dimensions in the previous examples are analyzed using Rayleigh's
Method.
 Alternatively, the relationship between the variables can be obtained
through a method called Buckingham's π theorem.
Buckingham ' s Pi theorem states that: If there are n variables in a problem
and these variables contain m primary dimensions (for example M, L, T) the
equation relating all the variables will have (n-m) dimensionless groups.
Buckingham referred to these groups as π groups.
The final equation obtained is in the form of : π1 = f(π2, π3 ,….. πn-m )

15. Steps of Buckingham’s Pi Theorem
Step 1: List all the variables that are involved in the problem.
Step 2: Express each of the variables in terms of basic dimensions.
Step 3: Determine the required number of pi terms.
Step 4: Select a number of repeating variables, where the number required is
equal to the number of reference dimensions.
Step 5: Form a pi term by multiplying one of the non repeating variables by
the product of the repeating variables, each raised to an exponent that will
make the combination dimensionless.
Step 6: Repeat Step 5 for each of the remaining non repeating variables.
Step 7: Check all the resulting pi terms to make sure they are dimensionless.
Step 8: Express the final form as a relationship among the pi terms, and think
about what it means.

16. 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;
1. • As far as possible, dependent variable should not be selected as repeating
variable
2. • 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
3. • The repeating variables selected should form a dimensionless group
4. • The repeating variables together must contain all three fundamental dimension
i.e., MLT
5. • 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, , (ii) l, v, or (iii) d, v, μ.

17. Example 4:

19. Step 4:

20. Step 5:

21. Step 6:

22. Step 7:

23. Step 8:

24. Example 5:
A thin rectangular plate having a width w and a height h is located so that it
is normal to a moving stream of fluid.
Assume that the drag, D, that the fluid exerts on the plate is a function of w
and h, the fluid viscosity, μ ,and ρ, respectively, and the velocity, V, of the
fluid approaching the plate. Determine a suitable set of pi terms to study
this problem experimentally.

30. 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 , and bulk modulus of air k. Express the functional relationship
between the variables and the resisting force.
Example 6:

34. Question : 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.

35. MODEL ANALYSIS

36. Model Analysis is actually an experimental method of finding solutions
of complex flow problems
The followings are the advantages of the model analysis
• Using dimensional analysis, a relationship between the variables influencing a
flow problem is obtained which help in conducting tests
• The performance of the hydraulic structure can be predicted in advance from its
model
• 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

37. 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

38. 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.
Note: Models are generally prepared with same scale ratios in every
direction. Such models are called true models. However, sometimes it is not
possible to do so and different convenient scales are used in different
directions. Thus, such models are called distorted model

39. 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.

40. 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.

41. For the fluid flow problems, the forces acting on the fluid
mass may be any one, or a combination of the several of
the following forces:
Inertia force,
Viscous force,
Gravity force,
Pressure force,
Surface tension force,
Elastic force,
Question : Forces acting on the fluid

42. Inertia Force (Fi): Fi=ma
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 the fluid flow problems.
Viscous Force (Fv): Fv = τ A = (μ · ∂u/∂y)A
It is equal to the product of shear stress 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.
Gravity Force (Fg): Fg = mg
It is equal to the product of mass and acceleration due to gravity of the
flowing fluid.
It is present in case of open surface flow.

43. Pressure Force (Fp): Fp = pA
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.
Surface Tension Force (Fs): Fs = σL
It is equal to the product of surface tension and length of surface or
flowing fluid.
Elastic Force (Fe): Fe = KL2
It is equal to the product of elastic stress and area of the flowing fluid.
- For a flowing fluid, all the above forces may not always be present. And
also the forces, which are present in a fluid flow problem, are not of equal
magnitude. There are always one or two forces, which dominate the other
forces. These dominating forces govern the flow of fluid.

44. Question : DIMENSIONLESS NUMBERS
Dimensionless numbers are the 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 Number

45. μ = Dynamic Viscosity
ν = Kinematic Viscosity
Reynold’s Number, Re:
It is the ratio of inertia force to the viscous force of flowing fluid.

46. Froude’s Number, Fe:
It is the ratio of inertia force to the gravity force of flowing fluid.

47. Eulers’s Number, Eu:
It is the ratio of inertia force to the pressure force of flowing fluid.

48. Weber’s Number, We:
It is the ratio of inertia force to the surface tension force of flowing fluid.

49. Mach’s Number, M:
It is the ratio of inertia force to the elastic force of flowing fluid.

50. Question : MODEL LAWS OR SIMILARITY LAWS
We have already learned that for dynamic similarity, ratio of corresponding
forces acting on prototype and model should be equal i.e.

51. However, for practical problems it is seen that one force is most significant
compared to others 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
I. • Reynold’s Model Law
II. • Froude’s Model Law
III. • Euler’s Model Law
IV. • Weber’s Model Law
V. • Mach Model Law

52. REYNOLD’S MODEL LAW
It is based on Reynold’s number which 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

54. The Various Ratios for Reynolds’s Law are obtained as

56. Example 1

58. Example 2

61. 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

63. (ρr = 1)
= Lr3
= Lr3.5

66. Example 3

67. Example 4

70. Euler's Model Law: states that Euler's number for model must be equal
to the Euler's number for prototype.
When pressure forces alone are predominant, a model may be taken to be
dynamically similar to the prototype when the ratio of the inertia to the
pressure forces is the same in the model and the prototype.
[Eu]p = [Eu]m

72. Weber’s Model Law : states that Weber's number for model
must be equal to the Weber's number for prototype.
[We]p = [We]m

74. Mach Model Law : states that Mach number for model must be
equal to the Mach number for prototype.
[M]p = [M]m

75. Question : Types of Models
The hydraulic models basically two types as,
1. Undistorted models
2. Distorted models
1.Undistorted model:
This model is geometrical similar to its prototype.
The scale ratio for corresponding linear dimension of the model and its
prototype are same.
The behavior of the prototype can be easily predicted from the result of
these type of model.

76. Advantages of undistorted model
1. The basic condition of perfect geometrical similarity is satisfied.
2. Predication of model is relatively easy.
3. Results obtained from the model tests can be transferred to directly to
the prototype.
Limitations of undistorted models
1. The small vertical dimension of model can not measured accurately.
2. The cost of model may increases due to long horizontal dimension to
obtain geometric similarity.

77. 2.Distorted Models:
This model is not geometrical similar to its prototype the different scale
ratio for linear dimension are adopted.
Distorted models may have following distortions:
Different hydraulic quantities such as velocity, discharge etc
Different materials for the model and prototype.
The main reason for adopting distorted models
To maintain turbulent flow
To minimize cost of models

78. Advantages of distorted models
• Accurate and precise measurement are made possible due to increase
vertical dimension of models.
• Model size can be reduced so its operation is simplified and hence the
cost of model is reduced
Disadvantage of distorted Models
• Depth or height distortion changes wave patterns.
• Slopes, bands and cuts may not properly reproduced in model.

79. Scale Ratios for Distorted Models

80. Example 1.
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.

82. 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/s, the average width and
depth of flow are 60m and 4.2m respectively. Determine the corresponding
discharge in model.
Example 2.

84. Summery of Chapter
Rayleigh's method- Theory and Examples
Buckingham Pi Theorem- Theory and Examples
Model and Similitude
Forces on Fluid
Dimensionless Numbers
Model laws
Distorted models