6.1 Dimensional Analysis

6.1.1 Fundamental Dimension, System of Units

and Hydraulics Variable

6.1.2 Method of Dimensional Analysis

6.2 Hydraulic Similarity

6.2.1 Types of Similarity

Geometric Similarity

Kinematic Similarity

Dynamic Similarity

CHAPTER 6 : DIMENSIONAL ANALYSIS

& HYDRAULICS SIMILARITY

Learning outcomes

Able to apply and analyze the DIMENSIONAL ANALYSIS

Rayleigh‟s method

Buckingham‟s method

Hydraulic similarity

Geometric similarity

Kinematic similarity

Dynamic similarity

6.1 : DIMENSIONAL ANALYSIS

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.

Types of Dimensions:

Fundamental Dimensions or Fundamental Quantities

Secondary Dimensions or Derived Quantities

6.1.1 Fundamental Dimension, System of Units and

Hydraulics Variable

Dimensionless analysis system involved qualitative and quantitative aspect.

Qualitative aspect ;

Serves to identify the nature, or type, of the characteristics (such as

length, time, stress and velocity)

Quantitative aspect;

provides a numerical measure of the characteristics (such as length,

time, stress and velocity)

Requires both a number and a standard by which various quantities can

be compared.

A standard (unit) for length might be a meter or foot, for time might be

an hour or second, and for mass a slug or kilogram.

The qualitative description is conveniently given in terms of certain primary

quantities, such as length (L), time (T), mass (M) and temperature (θ).

Primary quantities can be used to provide a secondary quantities, for

examples area =L2, velocity =MT-1 and density ML-3.

Alternatively, L, T, and F could also be used, where F is the basic

dimensions for force.

Newton‟s law states that force is equal to mass time acceleration, it follows

that F=MLT-2 or M=FL-1T2.

Secondary quantities expressed in terms of M can be expressed in term of

F. For example; stress, ς, is a force per unit area, so that ς= FL-2, but an

equivalent dimensional equation is ς = ML-1T-2.

Dimensions Associated with Common Physical Quantities

No. Quantity MLT FLT

1 Length, L L L

2 Area, A L2 L2

3 Volume, V L3 L3

4 Time , t T T

5 Velocity, v LT-1 LT-1

6 Acceleration, a LT-2 LT-2

7 Gravitational acceleration, g LT-2 LT-2

8 Frequency, N T-1 T-1

9 Discharge, Q L3T-1 L3T-1

10 Force, F or Weight, W MLT-2 F

11 Power, P ML2T-3 FLT-1

Quantity MLT FLT

12 Work or Energy, E ML2T-2 FL

13 Pressure, p ML-1T-2 FL-2

14 Mass, m M FT2L-1

15 Mass density, ρ ML-3 FT2L-4

16 Specific weight, w ML-2T-2 FL-3

17 Dynamic viscosity, µ ML-1T-1 FTL-2

18 Kinematic viscosity, v L2T-1 L2T-1

19 Surface Tension, ς MT-2 FL-1

20 Shear stress, τ ML-1T-2 FL-2

21 Bulk Modulus, K ML-1T-2 FL-2

Exercise :

What are the dimensions of the following variables?

Shear stress

Dynamic viscosity

Work

Modulus of elasticity

Power

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.

Example MLT and FLT

Determine the dimensions of force, pressure, power, specific weight and

surface tension in MLT system

Determine the dimensions of discharge, torque and momentum FLT system

NON- DIMENSIONAL GROUP

Non-dimensional/dimensionless group can be defined as ratio of two same

quantities or a group of parameters that are not produce any dimension.

For example, main component of force that act on fluid element is

influenced by viscosity, gravity, pressure, surface tension and elasticity.

Combination of these force is named as inertia force.

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.

1. Reynold‟s Number

2. Froude‟s Number

3. Euler‟s Number

Some common variables and Dimensionless

Groups

Reynolds 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:

1. Pipe Flow

2. Resistance experienced by submarines, airplanes, fully immersed

bodies etc.

Influence of viscosity

Inertia force/Viscous force

Reynolds number (Re)

2

23

2

./

/

/

LLv

LvL

dyAdv

t

vV

tA

ma

vLvL

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

Influence of gravity

Inertia force/Gravitational force

Froude number (Fr)

gL

vL

mg

ma3

22

m

p

r

m

p

r

r

r

m

p

m

p

m

m

p

p

mm

m

pp

p

mrpr

L

LL

v

vv

L

v

L

Lv

v

L

v

L

v

Lg

v

Lg

vFF

,

where

1

or or or

Froude’s model Law

The various ratio for Reynold‟s Law are obtained as :

273

23222

222

2522

m

p

r

r

: ratiopower

: ratio force

: ratio discharge

1a

aa : ratio naccelratio

T : ratio time

: ratiovelocity

since

rrrrrrrrrrrrrrr

rrrrrrrrrrrrr

rrrrr

mm

pp

r

r

r

r

r

mm

pp

r

r

r

mm

pp

m

p

r

m

p

m

p

r

m

m

p

p

LLLvLvvLvFP

vLvvLvQamF

LLLvLvA

vAQ

L

L

T

v

Tv

Tv

LL

L

vL

vL

T

T

LL

L

v

vv

L

v

L

v

Exercise :

Prove that under the influence of pressure, Euler number (Eu) is

dimensionless.

Prove that under the influence of surface tension, Weber number (We) is

dimensionless.

Prove that under the influence of elasticity, Mach number (Ma) is

dimensionless.

NON-DIMENSION ANALYSIS

Rayleigh „s method

Buckingham Pi (π) Theorem

6.1.2 : Methods of Dimensional Analysis

Three methods available:

Rayleigh‟s method

Buckingham‟s method

Hydraulic similarity

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.

6.1.2(1) : Rayleigh’s method

It is used to relate variables or develop dimensionless groups when there is

a limited number of variables (generally five or six)

If the variables are more than four or five this method becomes difficult to

solve

STEPS

Write the functional relationship with the given data

Write the equation in terms of a constant with exponents a, b, c…

Find out the values of a, b, c… by obtaining simultaneous equation.

Substitute the values of these exponents in the main equation and simplify

it.

Example 1 :

Using the Rayleigh method of dimensional analysis, develop an equation

for the power delivered by a pump to lift a fluid of a specific weight with

a rate of Q to a static level of H

Solution:

From the MLT and FLT table:

cba-

cba

-

LTML

HQP

TMLP

QH

1-32-2-32

32

TLTML

: writebecan equation following thethus,

table;MLT/FLT from and

isPower

Cont.. Example 1

Fundamental equations are then written as:

1

M

a

Ma

1322

thus,

)1(3)1(22

1322

LL 32-2

c

c

cba

LLcba

123

thus,

1)1(23

123

TT 1-2-3

b

b

ba

Tba-

QHP

HQPcba

: writebecan

thus,

Example 2 :

Assuming that the distance (z) traveled by a freely falling body is a

function of time, the weight of the body and the acceleration due to

gravity, find the relation between z and other variables.

Solution: From the given information

cba

cba

LTFTL

gWTz

2

:form ldimensionain

: writebecan equation following thethus,

Cont… example 2

1

thus;

1

:as written are equations lfundamenta the

c

cL

2

thus,

)1(20

210

0

:form ldimensionain

2

a

a

ca

LTTca

0

10

0

b

b

Fb

6.1.2 (2) : Buckingham Pi (π) Theorem

Since Rayleigh‟s Method becomes laborious if variables are more than

fundamental dimensions (MLT), so the difficulty is overcome by Buckingham‟s

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”.

x1= f(x2 , x3, x4 ,… xn )

f1(x1, x2 ,x3, … xn ) = 0

f (π1, π2 ,π3, … πn-m ) = 0

π term can be expressed as

π1 = x2a1 x3b1 x4c1

π2 = x2a2 x3b2 x4c2

πn-m = x2a(n-m) x3b(n-m) x4c(n-m)

Repeating variables

Geometrical variables – length, diameter etc

Flow property – velocity, acceleration etc

Fluid property – density, viscosity etc

Steps

List all the variables that are involved in the problems.

Express each of the variables in terms of basic/primary dimension.

Determine the required number of π-terms by k – r where k is the number

of variables in the problem and r is the number of reference dimensions.

Π=n-m (MLT/FLT) (m=2 or 3(usually taken value))

Select a number of repeating variables, where the number required is

equal to the number of reference dimensions.

Form a π-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.

Repeat step 5 for each of the remaining non repeating variables.

Check all the resulting π-terms to make sure they are dimensionless.

Express the final form as a relationship among the π-terms .

Exercise :

The pressure drop (∆p) in a pipe depends upon the mean velocity of

flow(v), length of pipe (l), diameter of pipe (d), viscosity of fluid (µ),

average height of roughness projections on the inside surface (k) mass

density of fluid (ρ). By using Buckingham‟s π- theorem, obtain a

dimensionless expression for ∆p. Also show that hf = 4flv2/ 2gd

where the hf is due to friction (∆p/w) and w is the specific weight of the

fluid and f is the coefficient of friction

Buckingham- method

Using the Buckingham- method, derive an expression for the shear stress,

, in fluid flowing in a pipe assuming that it is a function of the diameter, D

pipe roughness e, fluid density, , dynamic viscosity and fluid velocity v.

Solution:

Variables are : , , D, e, , , v. (6 variables (n) contain M and F,

the m=3; thus repeating =6-3).

6.2 : HYDRAULICS SIMILARITY

Hydraulics similarity or similitude is the science of study of the flow by

performing experiment over a model, which has been designed so that the

experimentation may be carried out on it in a laboratory, and the

performance of the prototype be predicted from the results obtained on

the model.

The actual prototype may be very big or small in size, or maybe using a

fluid that may be replaced by a more convenient fluid during the

experimentation.

This kind of study requires that the model and the flow pattern over it

should be exactly similar to that for the prototype. The two flow patterns

will be similar when these have geometric, kinematic, and dynamic

similarity.

6.2.1 : Type of Similarities

Similitude is a concept used in testing of Engineering Models and also

known as model studies

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 into

wing basins, air planes in wind tunnel, hydraulic

turbines, centrifugal pumps, spillways of dams, river channels etc and to stu

dy such phenomenon as the action of waves and tides on beaches, soil

erosion, and transportation of sediment etc.

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.

Similarity between hydraulic model and prototype may be achieved in 3

basic forms.

• 1. Geometric Similarity

• 2. Kinematic Similarity

• 3. Dynamic Similarity

6.2.1(1) : Geometric Similarity

Geometric similarity exists between a model and a prototype when the

linear dimension between a model and the prototype satisfy the

corresponding linear dimension ratio.

Mathematically,

where Lr = length ratio and L, b, D represent length, width

and depth. Lp=length of prototype and so on.

Same for ,

LrD

D

b

b

L

L

m

p

m

p

m

p

2

r

m

pL

Area

Area

3

r

m

pL

Volume

Volume r

m

pL

U

U

Example : Geometric Similarity

The ratio of the corresponding linear dimensions of a model and a

prototype are equal

Lr = Lp / Lm

where Lr = length ratio

Lp = length prototype

Lm = length model

6.2.1(2) : Kinematic Similarity

Kinematic similarities mean similarities of motion of fluid between a model

and prototype.

Mathematically,

where Vp1 and Vp2 are velocities in prototype and so on and Vr is

velocity ratio

Similarly, ar=acceleration ratio=

r

m

p

m

pV

V

V

V

V

2

2

1

1

r

m

p

m

pa

a

a

a

a

2

2

1

1

Example : Kinematic Similarity

Similarity of motion in a model and prototype

Vr = Vp / Vm

where Vr = velocity ratio

Vp = velocity prototype

Vm = velocity model

6.2.1(3) : Dynamic Similarity

Dynamic similarities means the similarity of forces between a model and a

prototype.

Let Fi = Inertia force

Fv = Viscous force

Fg = Acceleration due to gravity force

Then, (Fr = force ratio)

r

mg

pg

mv

pv

mi

piF

F

F

F

F

F

F ......

Example : Dynamic Similarity

Similarity between forces in a model and prototype

Fr = Fp / Fm

where Fr = force ratio

Fp = force prototype

Fm = force model

Example :

In a model test of a spillway, the discharge and velocity of flow over the

model were 1.5 m3/s and 1.2 m/s respectively. Calculate the discharge and

velocity over the prototype, if the prototype is 50 times larger than the model

size.

Example 2:

The characteristics of a spillway are to be studied by means of a

geometrically similar model constructed to the scale ratio of 1:10.

If the maximum rate of flow in the prototype is 28.3 m3/s, what will be

the corresponding flow in the model?

If the measured velocity in the model at a point on the spillway is 2.4 m/s,

what will be the corresponding velocity in the prototype?

If the hydraulic jump at the floor of the model is 50 mm high, what will be

the height of jump in the prototype?

If the energy dissipated per second in the model is 3.5 J, what energy is

dissipated in the prototype?

Example 3:

A 7.2 m high and 15 m long spillway discharges 94 m3/s discharge under

a head of 2.0 m. If a 1:9 scale model is to be made

Determine :

The model dimensions, head over spillway and the model discharge.

If the model experiences a force of 7.5 kN, determine force on the

prototype.