hydraulic similitude and model analysis

DIMENSIONAL ANALYSIS,

SIMILITUDE AND MODEL

ANALYSIS

DR. MOHSIN SIDDIQUE

ASSISTANT PROFESSOR

BACKGROUND

Although many practical engineering problems involving fluidmechanics can be solved by

• Equations and Analytical procedures

yet a large number of problems rely on experimental data for theirsolution

In fact, very few problems involving real fluids can be solved byanalytical analysis alone

In general, solution is obtained through the use of a combinationof analysis and experimental data

An obvious goal of any experiment is to make the results aswidely applicable as possible

To achieve this goal, the concept of similitude is often used sothat measurements made on one system (Laboratory) can be usedto describe the behavior of other systems (Outside of laboratory)

2

BACKGROUND

The laboratory systems are usually thought of as modelsand are used to study the phenomenon of interest under carefully controlled conditions

From these model studies, empirical formulations can be developed, or specific predictions of one or more characteristics of some other similar system can be made

However, to do this, it is necessary to establish the relationship between the laboratory model and the “other” system

In present topic, we will learn how to achieve this in a systematic manner

3

BACKGROUND

Dimensional Analysis to predict the physical parameters thatwill significantly influence phenomenon under study

Similitude and Model analysis to investigate the complexityof phenomenon in details

Application of knowledge on actual/prototype model

Wind turbine Dam/spillwaySudden contraction in pipes

4

DIMENSIONAL ANALYSIS

Introduction:

Dimensional Analysis is a mathematical technique makinguse of study of dimensions

It deals with the dimensions of physical quantities involvedin the phenomenon

In dimensional analysis, one first predicts the physicalparameters that will influence the flow, and then by,grouping these parameters in dimensionless combinationsa better understanding of the flow phenomenon is madepossible

It is particularly helpful in experimental work because itprovides a guide to those things that significantly influencethe phenomena; thus it indicates the direction in which theexperimental work should go

This mathematical technique is used in research work fordesign and for conducting model tests

5

TYPES OF DIMENSIONS

There are two types of dimensions

• Fundamental Dimensions or Fundamental Quantities

• Secondary Dimensions or Derived Quantities

Fundamental Dimensions or Fundamental Quantities:These are basic quantities. For Example;

• Time, T

• Distance, L

• Mass, M

Time, TDistance, L, Force, F

Force=Mass x Acceleration = MLT-2

6

TYPES OF DIMENSIONS

Secondary Dimensions or Derived Quantities

The are those quantities which posses more than onefundamental dimensions.

For example;

• Velocity is denoted by distance per unit time L/T

• Acceleration is denoted by distance per unit time square L/T2

• Density is denoted by mass per unit volume M/L3

Since velocity, density and acceleration involve morethan one fundamental quantities so these are calledderived quantities.

7

Characteristics Unit (SI) Dimension

(MLT)

Dimension

(FLT)

Geometry Length m L

Area m2 L2

Volume m3 L3

Kinematic Time S T

Velocity m/S L/T

Acceleration m/S2 L/T2

Discharge m3/S L3/T

Dynamic Mass Kg M FL-1T2

Force N (Kg-m/S2) MLT-2 F

Pressure Pa (N/m2) ML-1T-2 FL-2

Energy J (N-m) ML2T-2 FL

Power Watt (N-m/S) ML2T-3 FLT-1

Table: Flow Characteristics, units and Dimensions

8

METHODOLOGY OF DIMENSIONAL ANALYSIS

The basic principle is Dimensional Homogeneity, whichmeans the dimensions of each terms in an equation on bothsides are equal.

So such an equation, in which dimensions of each term onboth sides of equation are same, is known asDimensionally Homogeneous Equation. Such equations areindependent 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 homogeneousequation.

9

METHODS OF DIMENSIONAL ANALYSIS

If the number of variables involved in a physical phenomenon areknown, then the relation among the variables can be determined by thefollowing two methods;

• Rayleigh’s Method• Buckingham’s π-Theorem

Rayleigh’s Method:

It is used for determining expression for a variable (dependent) whichdepends upon maximum three to four variables (Independent) only.

If the number of independent variables are more than 4 then it is verydifficult to obtain expression for dependent variable.

Let X is a dependent variable which depends upon X1, X2, and X3 asindependent variables. Then according to Rayleigh’s Method

X=f(X1, X2, X3) which can be written as

X=K X1a, X2

b, X3c

Where K is a non-dimensional constant and a, b, c are arbitrarypowers which are obtained by comparing the powers offundamental dimensions (Dimensional Homogeneity).

10

RAYLEIGH’S METHODQ. The resisting force R of a supersonic plane during flight can beconsidered as dependent upon the length of the aircraft l, velocityV, air viscosity µ, air density ρ, and bulk modulus of air k. Expressthe functional relationship between the variables and the resistingforce.

-2 1 1 1 3 1 2

( , , , , ) , , , , (1)

Where: A = Non dimensional constant

Substituting the powers on both sides of the equation

( ) ( ) ( ) ( )

Equating the powers of MLT on both

a b c d e

a b c d e

R f l V K R Al V K

MLT AL LT ML T ML ML T

µ ρ µ ρ

− − − − − −

= ⇒ =

=

sides

Power of M 1

Power of L 1 - -3 -

Power of T 2 - - - 2

c d e

a b c d e

b c e

⇒ = + +

⇒ = +

⇒ − =

� Solution:

11

RAYLEIGH’S METHOD

Since the unkown(5) are more than number of equations(3). So expressing

a, b & c in terms of d & e

1- -

2 - - 2

1- 3 1- (2 - - 2 ) 3(1- - )

1- 2 2 3-3 -3 2 -

Substituting the values

d c e

b c e

a b c d e c e c c e e

c e c c e e c

=

=

= + + + = + + +

= + + + + + =

2 2 2 1 2 2 2

2 2

2

2 2

2

in (1), we get

( )( )c c e c c e e c c c c e e e

c e

R Al V K Al V l V V K

KR A l V

Vl V

KR A l V

Vl V

µ ρ ρ µ ρ ρ

µρ

ρ ρ

µρ φ

ρ ρ

− − − − − − − − − −= =

=

=

OR

12

RAYLEIGH’S METHOD

( )ρ,Kfu = ( )baKCu ρ,=

13

RAYLEIGH’S METHOD

14

RAYLEIGH’S METHOD

15

BUCKINGHAM’S Π-THEOREM:

Buckingham’s π-Theorem: Since Rayleigh’s Method becomeslaborious if variables are more than fundamental dimensions (MLT), sothe difficulty is overcome by Buckingham’s π-Theorem which statesthat

“If there are n variables (Independent and Dependent) in a physicalphenomenon and if these variables contain m fundamental dimensionsthen the variables are arranged into (n-m) dimensionless terms whichare called π-terms.”

Let X1, X2, X3,…,X4, Xn are the variables involved in a physicalproblem. Let X1 be the dependent variable and X2, X3, X4,…,Xn arethe independent variables on which X1 depends. Mathematically it canbe written as

X1=f(X2 ,X3 ,X4 ,Xn) which can be rewritten as

f(X1X2 ,X3 ,X4 ,Xn)=0

Above equation is dimensionally homogenous.

It contain n variables and if there are m fundamental dimensions then itcan be written in terms of dimensions groups called π-terms which areequal to (n-m)

Hence; f(π1, π2 π3,… πn-m)=0

16


Properties of π-terms:

• Each π-term is dimensionless and is independent of system ofunits.

• Division or multiplication by a constant does not change thecharacter of the π-terms.

• Each π-term contains m+1 variables, where m is the number offundamental dimensions and also called repeating variable.

Let in the above case X2, X3, X4 are repeating variables and iffundamental dimensions m = 3 then each π-term is written as

Π1=X2a1. X3

b1. X4c1 .X1

Π2=X2a2. X3

b2. X4c2 .X5

.

.Πn-m=X2

a(n-m). X3b(n-m). X4

a(n-m) .Xn

Each equation is solved by principle of dimensionlesshomogeneity and values of a1, b1 & c1 etc are obtained. Final resultis in the form of

Π1=ϕ’(Π2, Π3, Π4 ,…, Π(n-m))

Π2=ϕ”(Π1, Π3, Π4 ,…, Π(n-m))17

METHODS OF SELECTING REPEATING VARIABLES

The number of repeating variables are equal to number of

fundamental dimensions of the problem. The choice of repeatingvariables is governed by following considerations;

• As far as possible, dependent variable should not be selected asrepeating variable

• The repeating variables should be chosen in such a way that onevariable contains geometric property, other contains flow propertyand third contains fluid property

• The repeating variables selected should form a dimensionlessgroup

• The repeating variables together must contain all threefundamental dimension i.e., MLT

• 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.

18


Q 1. The resisting force R of a supersonic plane during flight canbe 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 theresisting 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

µ ρ µ ρ

π

π π π

= ⇒ =

=

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

ρ

π ρ

π ρ µ

π ρ

=

=

=19


Now each Pi-term is solved by the principle of dimensionalhomogeneity

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 cterm M L T L LT ML MLT

a

π − − −− ⇒ =

⇒

⇒ = −

⇒

∴ -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

Rl V R

L V

term M L T L LT ML ML T

c c

a b c a

π ρ πρ

π − − − −

= ⇒ =

− ⇒ =

= + ⇒ =

= + ⇒ = −

2 2

-1 -1 -1

2 2

er of T: 0 - -1 -1

b b

l VlV

µπ ρ µ π

ρ

= ⇒ =

∴ = ⇒ =

20


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 cterm M L T L LT ML ML T

c c

a b c a

b b

π − − − −− ⇒ =

= + ⇒ =

= + ⇒ = −

= ⇒ =

∴ 0 -2 -1

3 2 2

1 2 3 2 2 2

2 2

2 2 2 2

( ) , , 0

, ,

Kl V K

V

Hence

R Kf f or

l V lV V

R K KR l V

l V lV V lV V

π ρ πρ

µπ π π

ρ ρ ρ

µ µφ ρ φ

ρ ρ ρ ρ ρ

= ⇒ =

= =

= ⇒ =

21


Q 2. A thin rectangular plate having a width, w, and height, h, is locatedso that it is normal to a moving stream of fluid. Assume the drag D, thatthe fluid exerts on the plate is a function of w and h, the fluid viscosityand density µ, and ρ, respectively, and velocity V of the fluidapproaching the plate. Determine a suitable set of pi terms to study thisproblem experimentally

( )Vhwf ,,,,D

:can write weproblem, ofstatement theFrom

ρµ=

33-6m-n terms,pi of No.

3mdimension, lfundamenta of No.

6n, variablesofnumber Total

==

=

=

V,w,

variablesRepeating

ρ

1

11

2 ,,D

are system MLT using variablesofdimension The

−

−−

=

=

===

LTV

TML

LhLwMLT-

µ

22


333

3

222

2

111

1 ,,

as written becan termspi theNow

cbacbacbaVwVhwVDw ρµπρπρπ ===

( )( ) ( ) ( )

ρπ

ρπ

ρπ

π

221

122

1

111

1

111

1

131112000

111

1

1

D

D

-1c and -2b-2,a Therefore,

b--20:TFor

3c-ba10:LFor

10:MFor

D

:For

Vw

Vw

c

MLLTLMLTTLM

Vw

cba

cba

=

=

===

=

++=

+=

=

=

−−−

−−−

23


333

3

222

2

111

1 ,,



( )( ) ( ) ( )

w

h

Vhw

c

MLLTLLTLM

Vhw

cba

cba

=

=

===

=

++=

=

=

=

−

−−

2

001

2

222

2

222

2

23212000

222

2

2

0c and 0b-1,a Therefore,

-b0:TFor

3c-ba10:LFor

0:MFor

:For

π

ρπ

ρπ

π

24


333

3

222

2

111

1 ,,



( )( ) ( ) ( )

ρ

µπ

ρµπ

ρµπ

π

Vw

Vw

c

MLLTLTMLTLM

Vw

cba

cba

=

=

=−=−=

=

++−=

+=

=

=

−−−

−−−−

3

111

3

333

3

333

3

3331311000

333

3

3

-1c and 1b1,a Therefore,

b--10:TFor

3c-ba10:LFor

10:MFor

:For

25


( )

( )

=

=

=

−

−

ρ

µφ

ρ

ππππφπ

πππππφ

wVh

w

Vw

mn

mn

,D

,...,,,

0,...,,,,

form in the drepresente

becan analysis ldimensiona of results eFinally th

22

4321

4321

26


27

Let’s solve it now


28Let’s solve it now


31

For more worked problems refer to

Civil Engineering Hydraulics by Nalluri and Featherstone

(Worked problems 9.1 to 9.11)

SIMILITUDE AND MODEL ANALYSIS

Similitude is a concept used in testing of EngineeringModels.

Usually, it is impossible to obtain a pure theoretical solutionof hydraulic phenomenon.

Therefore, experimental investigations are often performedon small scale models, called model analysis.

A few examples, where models may be used are ships intowing basins, air planes in wind tunnel, hydraulic turbines,centrifugal pumps, spillways of dams, river channels etcand to study such phenomenon as the action of waves andtides on beaches, soil erosion, and transportation ofsediment etc.

33

MODEL ANALYSIS

Model: is a small scale replica of the actual structure

Prototype: the actual structure or machine

Prototype Model

Lp3

Lp1

Lp2

Fp1

Fp3

Fp2

Lm3

Lm1

Lm2Fm1

Fm3

Fm2

34

Note: It is not necessary that the models should be smaller that theprototype, they may be larger than prototype

MODEL ANALYSIS

Model Analysis is actually an experimental method offinding 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 forobtaining, in advance, useful information about theperformance of the prototype only if a complete similarityexits between the model and the prototype

35

SIMILITUDE-TYPE OF SIMILARITIES

Similitude: is defined as similarity between the modeland prototype in every respect, which mean model andprototype have similar properties or model andprototype are completely similar.

Three types of similarities must exist between modeland prototype.

• Geometric Similarity

• Kinematic Similarity

• Dynamic Similarity

36


Geometric Similarity: is the similarity of shape. It is said to existbetween model and prototype if ratio of all the correspondinglinear dimensions in the model and prototype are equal. e.g.

p p p

r

m m m

L B DL

L B D= = =

� Where: Lp, Bp and Dp are Length, Breadth, and diameter of prototypeand Lm, Bm, Dm are Length, Breadth, and diameter of model.

� Lr= Scale ratio

� Note: Models are generally prepared with same scale ratios in everydirection. Such models are called true models. However, sometimesit is not possible to do so and different convenient scales are used indifferent directions. Thus, such models are called distorted model

37


Kinematic Similarity: is the similarity of motion. It is said to existbetween model and prototype if ratio of velocities and accelerationat 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 aV a

V V a a= = = =

� Where: Vp1& Vp2 and ap1 & ap2 are velocity and accelerations at point1 & 2 in prototype and Vm1& Vm2 and am1 & am2 are velocity andaccelerations at point 1 & 2 in model.

� Vr and ar are the velocity ratio and acceleration ratio

� Note: Since velocity and acceleration are vector quantities, hencenot only the ratio of magnitude of velocity and acceleration at thecorresponding points in model and prototype should be same; butthe direction of velocity and acceleration at the corresponding pointsin model and prototype should also be parallel.

38


Dynamic Similarity: is the similarity of forces. It is said to existbetween model and prototype if ratio of forces at thecorresponding points in the model and prototype are equal. e.g.

( )

( )

( )

( )

( )

( )gi vp p p

r

i v gm m m

FF FF

F F F= = =

� Where: (Fi)p, (Fv)p and (Fg)p are inertia, viscous and gravitationalforces in prototype and (Fi)m, (Fv)m and (Fg)m are inertia, viscous andgravitational forces in model.

� Fr is the Force ratio

� Note: The direction of forces at the corresponding points in modeland prototype should also be parallel.

39

TYPES OF FORCES ENCOUNTERED IN FLUID

PHENOMENON

Inertia Force, Fi: It is equal to product of mass and acceleration in theflowing fluid.

Viscous Force, Fv: It is equal to the product of shear stress due toviscosity and surface area of flow.

Gravity Force, Fg: It is equal to product of mass and acceleration dueto gravity.

Pressure Force, Fp: it is equal to product of pressure intensity andcross-sectional area of flowing fluid.

Surface Tension Force, Fs: It is equal to product of surface tensionand length of surface of flowing fluid.

Elastic Force, Fe: It is equal to product of elastic stress and area offlowing fluid.

40

DIMENSIONLESS NUMBERS

Dimensionless numbers are the numbers which areobtained by dividing the inertia force by viscous force orgravity force or pressure force or surface tension force orelastic force.

As this is ratio of once force to other, it will be adimensionless 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

41


Reynold’s Number, Re: It is the ratio of inertia force to the viscous force

of flowing fluid.

. .

Re. .

. . .

.. .

Velocity VolumeMass Velocity

Fi Time Time

Fv Shear Stress Area Shear Stress Area

QV AV V AV V VL VL

du VAA A

dy L

ρ

ρ ρ ρ ρ

τ µ υµ µ

= = =

= = = = =

2

. .

. .

. .

. .


Fi Time TimeFeFg Mass Gavitational Acceleraion Mass Gavitational Acceleraion

QV AV V V V

Volume g AL g gL gL

ρ

ρ ρ

ρ ρ

= = =

= = = =

� Froude’s Number, Fe: It is the ratio of inertia force to the gravity force

of flowing fluid.

42


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

flowing fluid.

2

. .

Pr . Pr .

. .

. . / /

u


Fi Time TimeEFp essure Area essure Area

QV AV V V V

P A P A P P

ρ

ρ ρ

ρ ρ

= = =

= = = =

2 2

. .

. .

. .

. . .


Fi Time TimeWeFg Surface Tensionper Length Surface Tensionper Length

QV AV V L V V

L L L

L

ρ

ρ ρ ρ

σ σ σ σ

ρ

= = =

= = = =

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

43


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

flowing fluid.

2 2

2

. .

. .

. .

. . /

: /


Fi Time TimeMFe Elastic Stress Area Elastic Stress Area

QV AV V L V V V

K A K A KL CK

Where C K

ρ

ρ ρ ρ

ρ

ρ

= = =

= = = = =

=

44

MODEL LAWS OR SIMILARITY LAWS

We have already learned that for dynamic similarity, ratio ofcorresponding forces acting on prototype and model should beequal i.e.

( )

( )

( )

( )

( )

( )( )

( )

( )

( )

( )

( )

g pv s e Ip p p p p p

v s e Ig pm m m mm m

F FF F F F

F F F FF F= = = = =

( ) ( )

( )

( )( )

( )

Thus dynamic similarity require that

v g p s e I

v g p s e Ip p

Iv g p s e mm

F F F F F F

F F F F F F

FF F F F F

+ + + + =

+ + + +=

+ + + +

� Force of inertial comes in play when sum of all other forces is notequal to zero which mean:

� In case all the forces are equally important, the above two equationscannot be satisfied for model analysis

45

MODEL LAWS OR SIMILARITY LAWS

However, for practical problems it is seen that one force is mostsignificant compared to others and is called predominant force ormost significant force.

Thus, for practical problem only the most significant force isconsidered for dynamic similarity. Hence, models are designed onthe basis of ratio of force, which is dominating in thephenomenon.

Finally, the laws on which models are designed for dynamicsimilarity are called models laws or laws of similarity. Thefollowings are these laws

• Reynold’s Model Law

• Froude’s Model Law

• Euler’s Model Law

• Weber’s Model Law

• Mach’s Model Law

46

REYNOLD’S MODEL LAW

It is based on Reynold’s number and states that Reynold’s numberfor model must be equal to the Reynolds number for prototype.

Reynolds Model Law is used in problems where viscous forces aredominant. These problems include:

• Pipe Flow

• Resistance experienced by submarines, airplanes, fully immersedbodies etc

( ) ( )Re Re

1

: , ,

m mP P

P m

P m

P P r r

rPm m

m

P P Pr r r

m m m

V LV Lor

V L V L

V L

V Lwhere V L

V L

υ υ

υυ

υ

υυ

υ

= =

= =

= = =

47


The Various Ratios for Reynolds’s Law are obtained as

rr

r

P P P r

m m m r

P Pr

m m

2

r

r

sin /

Velocity Ratio: V =L

T L /V LTime Ratio: Tr=

T L /V V

V / VrAcceleration Ratio: a =

V / Tr

Discharge Ratio: Q

Force Ratio: F =

P m

mP P

m P m

P

m

P Pr r

m m

VL VLce and

LV

V L

a T

a T

A VL V

A V

m

υ µ ρυ υ

υ υ

υ

= =

= =

= =

= =

= =

2 2 2

2 2 2 3

r r rPower 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

ρ ρ ρ

ρ ρ

= = =

=

48


Q. A pipe of diameter 1.5 m is required to transport an oil of specificgravity 0.90 and viscosity 3x10-2 poise at the rate of 3000litre/s.Tests were conducted on a 15 cm diameter pipe using water at20oC. Find the velocity and rate of flow in the model.

p p p p pm m m

m m

2

2

p 2

For pipe flow,

According to Reynolds' Model Law

V D DV D

D

900 1.5 1 103.0

1000 0.15 3 10

3.0Since 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

ρ ρ µρ

µ µ ρ µ

π

−

−

= ⇒ =

× ×= =

× ×

= =

=

∴ = =

= = 2

3

091 / 4(0.15)

0.0899 /m s

π×

=

� 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

� Velocity of flow Vm=?

� Discharge Qm=?49

REYNOLD’S MODEL LAWQ. A ship 300m long moves in sea water, whose density is 1030kg/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 andthe resistance of the model is 60N.

Determine the velocity of ship in sea water and also the resistanceof ship in sea water. The density of air is given as 1.24kg/m3. Takethe kinematic viscosity of air and sea water as 0.012 stokes and0.018 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

50

( )

( )

4

4

2 2

2 2 2 2

2 2

0.012 10 330 0.2 /

0.018 10 300

Resistance= Mass Acceleration= L V

L V 1030 300 0.2369.17

1.24 3 30L V

369.17 60 22150.2

p mp m

p m m p

p p

mm

p

LVL VLV V

L

Vp m s

Since

FThus

F

F N

υ

υ υ υ

ρ

ρ

ρ

−

−

= ⇒ =

×= =

×

×

= = =

= × =


For dynamic similarity between model and prototype, the Reynolds number for both of them should be equal.

51

0.018

0.012

FROUDE’S MODEL LAW

It is based on Froude’s number and states that Froude’s number formodel must be equal to the Froude’s number for prototype.

Froude’s Model Law is used in problems where gravity forces isonly dominant to control flow in addition to inertia force. Theseproblems 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 mP P

P m

P P m m P m

P P Pr r r r

m mPm

m

V VV VF F or or

g L g L L L

V V LV L where V L

V LLV

L

= = =

= = = =

52


The Various Ratios for Reynolds’s Law are obtained as;

r

P P P r

m m m

P Pr

m m

2 2 5/ 2

r

sin

Velocity Ratio: V

T L /V LTime Ratio: Tr=

T L /V

V / VrAcceleration Ratio: a = 1

V / Tr

Discharge Ratio: Q

Force Ratio: Fr=

mP

P m

pPr

m m

r

r

rP

m r

P Pr r r r r

m m

r r

VVce

L L

LVL

V L

LL

La T

a T L

A VL V L L L

A V

m a

=

= = =

= = =

= = = =

= = = =

=

( )

2 2 2 2 3

32 2 2 3 2 7 / 2Power 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

ρ ρ ρ ρ ρ

ρ ρ ρ ρ

= = = =

= = =

53


Q 1. In the model test of a spillway the discharge and velocity of flowover the model were 2 m3/s and 1.5 m/s respectively. Calculate thevelocity and discharge over the prototype which is 36 times themodel size.

( ) ( )

( )

2.5 2.5p

m

2.5 3

For Discharge

Q36

Q

36 2 15552 / sec

r

p

L

Q m

= =

= × =

p

m

For Dynamic Similarity,

Froude Model Law is used

V36 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=?

54


Q 2. The characteristics of the spillway are to be studied by means of a geometricallysimilar model constructed to a scale of 1:10.

(i) If 28.3 cumecs, is the maximum rate of flow in prototype, what will be thecorresponding flow in model?

(i) If 2.4m/s, 50mm and 3.5 Nm are values of velocity at a point on the spillway, heightof hydraulic jump and energy dissipated per second in model, what will be thecorresponding velocity, height of hydraulic jump and energy dissipation per second inprototype?

� Solution: Given that

For Model

� Discharge over model, Qm=?

� Velocity over model, Vm = 2.4 m/sec

� Height of hydraulic jump, Hm =50 mm

� Energy dissipation per second, Em =3.5 Nm

� Linear Scale ratio, Lr =10

� For Prototype

� Discharge over model, Qp=28.3 m3/sec

� Velocity over model, Vp =?

� Height of hydraulic jump, Hp =?

� Energy dissipation per second, Ep =? 55


p 2.5 2.5

m

2.5 3

p

m

For Discharge:

Q10

Q

28.3/10 0.0895 / sec

For Velocity:

V10

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:

H10

H

50 10 500

For Energy Dissipation:

E10

E

3.5 10 11067.9 / sec

r

p

r

p

L

H mm

L

E Nm

= =

= × =

= =

= × =

56

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 ofthe model and its prototype is same, the models is called undistortedmodel. The behavior of prototype can be easily predicted from the resultsof undistorted or true model.

Distorted Models: A model is said to be distorted if it is not

geometrically similar to its prototype. For distorted models different scaleratios for linear dimension are used.

For example, if for the river, both horizontal and vertical scale ratio aretaken to be same, then depth of water in the model of river will be veryvery 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 theresults of such models cannot be directly transferred to prototype.

57

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

2Scale 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 Pm

m m

L B

L B

h

h

ghV L

gh

B hA

B h

=

=

= =

= = ( ) ( )

( ) ( ) ( ) ( ) ( )3/ 2

PScale Ratio for discharge: Qr=Q /V

r rH V

P Pm r r r r rH V V H

m m

L L

A VQ L L L L L

A V= = =

58

DISTORTED MODEL

Q 1. The discharge through a weir is 1.5 m3/s. Find the dischargethrough the model of weir if the horizontal dimensions of themodel=1/50 the horizontal dimension of prototype and verticaldimension 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 rH

p m

L

L

h

h

Q L L

Q Q

=

=

=

∴ = ×

3

1581.14

1.5 /1581.14 0.000948 /m

Q m s

=

⇒ = =

59

DISTORTED MODEL

Q 2. A river model is to be constructed to a vertical scale of 1:50 and ahorizontal of 1:200. At the design flood discharge of 450m3/s, the averagewidth and depth of flow are 60m and 4.2m respectively. Determine thecorresponding 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 rH

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−⇒ = = ×

6070710.7

DISTORTED MODEL

( )

( )

m

VLReynolds 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

0.02520.05385

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

AR

P

υ

=

= = = =

= = = =

= = × =

= + = + × =

= = =

6 2

6

ater = =1 10 / sec

4 4 0.253 0.05385Re 54492.31

1 10

>2000

Flow is in turbulent range

m

m

VR

υ

υ

−

−

×

× × = = =

×

∴ 61

PROBLEM


Repeating variables

MODEL ANALYSIS

PROBLEM

MODEL ANALYSIS

THANK YOU

68

WORKED PROBLEMS

69

For Model

• Pipe diameter=3 in

• Fluid=water

• Viscosity=1.21x10-5 ft2/s

For Prototype

• Pipe diameter= 3ft

• Liquid=SAE30

• viscosity=4.5x10-3 ft2/s

For Reynolds number similarity

P1. SAE 30 oil at is pumped through a 3-ft-diameter pipeline at a rate of 6400 gallon/min. A model of this pipeline is to be designed using a 3-in.-diameter pipe and water at as the working fluid. To maintain Reynolds number similarity between these two systems, what fluid velocity will be required in the model?

Pm

VDVD

=

νν

( ) sftAQVp

sftgallonQp

pp /02.234//26.14/

/26.14min/6400

2

3

=×==

=×=

π

p

m

p

p

mm V

D

DV

=

ν

ν

sftVm /104.6 2−×=

70

For Model

For Prototype

P2. Glycerin at 20oC flows with a velocity of 4 m/s through a 30-mm-diameter tube. A model of this system is to be developed using standard air as the model fluid. The air velocity is to be 2 m/s. What tube diameter is required for the model if dynamic similarity is to be maintained between model and prototype?

Pm

VDVD

=

νν

p

m

p

p

mm D

V

VD

=

ν

ν

mDm

310736.0 −×=

71

For Model

• Fluid: water

• Viscosity@20oC =1.004x10-6

m2/s

• Vm=?

For Prototype

• Fluid: Sea water

• [email protected]=1.17x10-6

m2/s

• Vp=30 m/s

P3. The drag characteristics of a torpedo are to be studied in a water tunnel using a 1 : 5 scale model. The tunnel operates with freshwater at 20oc , whereas the prototype torpedo is to be used in seawater at 15.6oc. To correctly simulate the behaviour of the prototype moving with a velocity of 30 m/s, what velocity is required in the water tunnel?

Pm

VDVD

=

νν

p

m

p

p

mm V

D

DV

=

ν

ν

smVm /129=

72

For Model

• dm=0.1m

• Vm=1.2m/s

For Prototype

dp=3m

Vp=?

For Froude No. Similarity

P4. The design of a river model is to be based on Froude number similarity, and a river depth of 3 m is to correspond to a model depth of 100 mm. Under these conditions what is the prototype velocity corresponding to a model velocity of1.2 ms?

Pmgd

V

gd

V

=

( )m

ppgd

VgdV

=

smVp /11=

73

For Model

• lm=0.9m

• Qm?

For Prototype

• lp= 65m

• Qp=40m3/s

For Froude number similarity

P5. Water flows at a rate of 40 m3/s through the spillway of a dam which is 65 m wide. A model spillway, having a width of 0.9 m, is to be constructed and tested in the laboratory. What is the required flowrate in the model?

Pmgd

V

gd

V

=

smQ

lQQ

llVA

VA

Q

Q

m

rpm

rr

mm

Pp

m

p

/1002.9

/

34

5.2

2/12

−×=

=

==

=

m

p

m

p

gd

gd

V

V

( )mpr ddl /=

74

P12. The drag characteristics for a newly designed automobile having a maximumcharacteristic length of 20 ft are to be determined through a model study. Thecharacteristics at both low speed (approximately 20 mph) and high speed (90mph) are of interest. For a series of projected model tests, an unpressurizedwind tunnel that will accommodate a model with a maximum characteristic lengthof 4 ft is to be used. Determine the range of air velocities that would be requiredfor the wind tunnel if Reynolds number similarity is desired. Are the velocitiessuitable? Explain.?

Pm

VDVD

=

ννp

m

p

p

mm V

D

DV

=

ν

ν

Since the wind tunnel is unpressurized, the air properties will be approximately the same for model and prototype

ppp

m

p

m VVVl

lV 5

4

20=

=

=

At low speed At high speed

( ) mphVm 100205 == ( ) mphVm 450905 ==

At high velocity in wind tunnel, compressibility of the air would start tobecome an important factor, whereas compressibility is not important forthe prototype. Thus velocity required for the model would not be suitable

76

For more worked problems refer to

Civil Engineering Hydraulics by Nalluri and Featherstone

(Worked problems 9.1 to 9.11)

THANK YOU

78