45746431 hydraulic similitude lectures 1

7/30/2019 45746431 Hydraulic Similitude Lectures 1

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Dimensional Analysis And

Similitude & Model Analysis


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Dimensional AnalysisIntroduction:

Dimensional analysis is a mathematical techniquemaking use of study of dimensions.

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

It deals with the dimensions of physical quantitiesinvolved in the phenomenon. All physical quantities aremeasured by comparison, which is made with respect toan arbitrary fixed value.

Types of Dimensions:

Fundamental Dimensions or Fundamental Quantities

Secondary Dimensions or Derived Quantities


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Methodology of Dimensional Analysis The Basic principle is Dimensional Homogeneity, which

means the dimensions of each terms in an equation onboth sides are equal.

So such an equation, in which dimensions of each termon both sides of equation are same, is known as

Dimensionally Homogeneous equation. Such equationsare 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 dimensionallyhomogeneous equation.


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Methods of Dimensional Analysis 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;Rayleighs Method: It is used for determining expression for a variable (dependent)

which depends upon maximum three to four variables (Independent)only.

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

Buckinghams (pie)-Theorem: Since Rayleighs Methodbecomes laborious if variables are more than fundamentaldimensions (MLT), so the difficulty is overcome by Buckinghams-

Theorem which states that Ifthere are n variables (Independent and Dependent) in a physical

phenomenon and if these variables contain m fundamentaldimensions then the variables are arranged into (n-m)dimensionless terms which are called -terms.


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Similitude and Model Analysis Similitude is a concept used in testing of Engineering

Models.

Usually, it is impossible to obtain a pure theoreticalsolution of hydraulic phenomenon.

Therefore experimental investigations are oftenperformed 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, hydraulicturbines, centrifugal pumps, spillways of dams, riverchannels etc and to study such phenomenon as theaction of waves and tides on beaches, soil erosion, andtransportation of sediment etc.
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Model Analysis

Model: is a small scale replica of the actual structure.

Prototype: the actual structure or machine.

Note: It is not necessary that the models should besmaller that the prototype, they may be larger thanprototype (e.g. Water Treatment Plant, Floatation

Columns).

Prototype Model

Lp3

Lp1

Lp2 Fp1

Fp3

Fp2

Lm3

Lm1

Lm2Fm1

Fm3

Fm2


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Model Analysis Model Analysis is actually an experimental method of

finding solutions of complex flow problems.

The followings are the advantages of the model analysis The performance of the hydraulic structure can be predicted in

advance from its model. Using dimensional analysis, a relationship between the variables

influencing a flow problem is obtained which help in conductingtests.

The merits of alternative design can be predicted with the help ofmodel 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 similarity

exits between the model and the prototype.


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Similitude-Type of Similarities

Similitude: is defined as similarity between the modeland 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 SimilarityKinematic Similarity

Dynamic Similarity


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Similitude-Type of Similarities Geometric Similarity: is the similarity of shape. It is said to exist

between model and prototype if ratio of all the corresponding lineardimensions 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 a model is called true model. However, sometimes itis not possible to do so and different convenient scales are used indifferent directions. Such a models is call distorted model


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Similitude-Type of Similarities Kinematic Similarity: is the similarity of motion. It is said to exist

between model and prototype if ratio of velocities and acceleration atthe 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 points

in model and prototype should also be parallel.


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Similitude-Type of Similarities Dynamic Similarity: is the similarity of forces. It is said to exist

between model and prototype if ratio of forces at the correspondingpoints 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.


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Types of forces encountered in fluid Phenomenon

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

Viscous Force, Fv: It is equal to the product of shear stress due to

viscosity and surface area of flow.

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

Pressure Force, Fp: it is equal to product of pressure intensity and

cross-sectional area of flowing fluid.

Surface Tension Force, Fs: It is equal to product of surface tension

and length of surface of flowing fluid.

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

flowing fluid.


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Dimensionless Numbers These are numbers which are obtained by dividing the

inertia force by viscous force or gravity force or pressureforce or surface tension force or elastic 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 dimensionlessnumbers.

Reynolds Number Froudes Number

Eulers Number

Webers Number

Machs Number


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Dimensionless Numbers Reynolds 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 VA A Ady L

2

. .

. .

. .

. .

Velocity VolumeMass VelocityFi Time TimeFeFg Mass Gavitational Acceleraion Mass Gavitational Acceleraion

QV AV V V V

Volume g AL g gL gL

Froudes Number, Re: It is the ratio of inertia force to the gravity force

of flowing fluid.


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Dimensionless Numbers Eulerss Number, Re: 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

2 2

. .

. .

. .

. . .

Velocity Volume

Mass VelocityFi Time TimeWeFg Surface Tensionper Length Surface Tensionper Length

QV AV V L V V

L L L

L

Webers Number, Re: It is the ratio of inertia force to the surface

tension force of flowing fluid.


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Dimensionless Numbers Machs Number, Re: 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


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Model Laws or similarity Laws We have already read that for dynamic similarity ratio of corresponding

forces acting on prototype and model should be equal. i.e

g pv s e I p p p p p p

v s e I g 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 FFF F F F F

Force of inertial comes in play when sum of all other forces is not

equal to zero which mean

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


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Model Laws or similarity Laws However, for practical problems it is seen that one force

is most significant compared to other and is calledpredominant force or most significant force.

Thus for practical problem only the most significant forceis considered for dynamic similarity. Hence, models aredesigned on the basis of ratio of force, which isdominating in the phenomenon.

Finally the laws on which models are designed fordynamic similarity are called models laws or laws ofsimilarity. The followings are these laws Reynolds Model Law

Froudes Model Law

Eulers Model Law

Webers Model Law

machs Model Law


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Reynolds Model Law It is based on Reynolds number and states that

Reynolds number for model must be equal to the

Reynolds number for prototype.

Reynolds Model Law is used in problems where viscous

forces are dominant. These problems include:

Pipe Flow

Resistance experienced by submarines, airplanes, fully immersed

bodies etc.

Re Re

1

: , ,

m mP P

P mP m

P P r r

rPm m

m

P P P

r r rm m m

V LV Lor

V L V L

V L

V L

where V LV L


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Reynolds Model Law The Various Ratios forReynoldss 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

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


21/31

Reynolds Model Law Q. A pipe of diameter1.5 m is required to transport an oil of specific

gravity 0.90 and viscosity 3x10

-2

poise at the rate of 3000litre/sec.Tests were conducted on a 15 cm diameter pipe using water at 20oC.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

VV

V

V

Q

A

m s

V V m s

and Q V A

2

3

091 / 4(0.15)

0.0899 /m s

Solution:

Prototype Data: Diameter, Dp= 1.5m

Viscosity of fluid, p= 3x10-2 poise

Discharge, Qp =3000litre/sec

Sp. Gr., Sp=0.9

Density of oil=p=0.9x1000

=900kg/m3

Model Data: Diameter, Dm=15cm =0.15 m

Viscosity of water, m =1x10-2 poise

Density of water, m=1000kg/m3

Velocity of flow Vm=?

Discharge Qm=?


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Froudes Model Law It is based on Froudes number and states that Froudes

number for model must be equal to the Froudes numberfor prototype.

Froudes Model Law is used in problems where gravityforces is only dominant to control flow in addition to inertiaforce. These problems include: Free surface flows such as flow over spillways, weirs, sluices,

channels etc.

Flow of jet from orifice or nozzle

Waves on surface of fluid

Motion of fluids with different viscosities over one another

e e

/ 1; : ,

m mP PP m

P P m m P m

P P Pr r r r

m mP

mm

V VV VF F or or

g L g L L L

V V LV L where V L

V LL

V L


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Froudes Model Law The Various Ratios forReynoldss 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

mr

P Pr r r r r

m m

r r

VVceL 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

3

2 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


24/31

Froudes Model Law Q. In the model test of a spillway the discharge and velocity of flow

over the model were 2 m3/s and 1.5 m/s respectively. Calculate the

velocity and discharge over the prototype which is 36 times the model

size.

2.5 2.5p

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

V6 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=?


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Numerical Problem: Q. The characteristics of the spillway are to be studied by means of a geometrically

similar model constructed to a scale of 1:10.

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

(i) If 2.4m/sec, 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 =?


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Froudes Model Law

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


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Classification of Models Undistorted or True Models: are those which are

geometrically similar to prototype or in other words if the scale ratio

for linear dimensions of the model and its prototype is same, the

models is called undistorted model. The behavior of prototype can

be easily predicted from the results of undistorted or true model.

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

geometrically similar to its prototype. For distorted models differentscale ratios for linear dimension are used.

For example, if for the river, both horizontal and vertical scale ratio

are taken to be same, then depth of water in the model of river will

be very very small which may not be measured accurately.

The followings are the advantages of distorted models The vertical dimension of the model can be accurately measured

The cost of the model can be reduced

Turbulent flow in the model can be maintained

Though there are some advantage of distorted models, however theresults of such models cannot be directly transferred to prototype.


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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

Hm m

P

Vm

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 r H V V H

m m

L L

A VQ L L L L L

A V


29/31

Distorted model Q. The discharge through a weir is 1.5 m3/s. Find the discharge

through the model of weir if the horizontal dimensions of the

model=1/50 the horizontal dimension of prototype and 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

Hm

P

Vm

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


30/31

Distorted model Q. A river model is to be constructed to a vertical scale of 1:50 and a

horizontal of 1:200. At the design flood discharge of 450m3/sec, the average

width and depth of flow are 60m and 4.2m respectively. Determine the

corresponding discharge in model and check the Reynolds Number of the

model flow.

3

r

r

3/ 2

r P

3/ 2

arg 450 /

60 4.2

Horizontal scale ratio= L =200

Vertical scale ratio= L =50

Since Scale Ratio for discharge: Q =Q /

/ 200 50 7

V

p

p p

P

H

m

P

Vm

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 /mQ m s


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Distorted model

m

VLReynolds Number, Re =

4

/ 60 / 200 0.3

/ 4.2 / 50 0.084

0.3 0.084 0.02522 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 mP B y m

AR

P

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

45746431 hydraulic similitude lectures 1

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