Similitude & Model Analysis

Model Analysis Model: is a small scale replica of the actual structure. Prototype: the actual structure or machine.

Prototype Model

Lp3

Lp1

Lp2Fp1

Fp3

Fp2

Lm3

Lm1

Lm2Fm1

Fm3

Fm2

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 helps in conducting tests.

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.

Similitude-Type of Similarities Similitude: is defined as similarity between the model

and prototype in every respect, which means model andprototype have similar properties or model and prototypeare completely similar.

Three types of similarities must exist between model andprototype.

Geometric SimilarityKinematic SimilarityDynamic Similarity

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 pr

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

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 pr 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.

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 pr

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.

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 toviscosity 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 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.

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 one force to other, it will be adimensionless number. These are also called non-dimensional parameters.

The following are most important dimensionlessnumbers. Reynold’s Number Froude’s Number Euler’s Number Weber’s Number Mach’s Number

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

flowing fluid.

. .Re

. .. . .

. . .

Velocity VolumeMass VelocityFi Time TimeFv Shear Stress Area Shear Stress Area

QV AV V AV V VL VLdu VA A Ady L

ρ

ρ ρ ρ ρτ µ υµ µ

= = =

= = = = =

2

. .

. .

. .. .

Velocity VolumeMass VelocityFi Time TimeFeFg Mass Gavitational Acceleraion Mass Gavitational Acceleraion

QV AV V V VVolume g AL g gL gL

ρ

ρ ρρ ρ

= = =

= = = =

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

Dimensionless Numbers Eulers’s Number, Re: It is the ratio of inertia force to the pressure force of

flowing fluid.

2

. .

Pr . Pr .

. .. . / /

u

Velocity VolumeMass VelocityFi Time TimeEFp essure Area essure Area

QV AV V V VP A P A P P

ρ

ρ ρρ ρ

= = =

= = = =

2 2

. .

. .

. .. . .

Velocity VolumeMass VelocityFi Time TimeWeFg Surface Tensionper Length Surface Tensionper Length

QV AV V L V VL L L

L

ρ

ρ ρ ρσ σ σ σ

ρ

= = =

= = = =

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

Dimensionless Numbers Mach’s Number, Re: It is the ratio of inertia force to the elastic force of

flowing fluid.

2 2

2

. .

. .

. .. . /

: /

Velocity VolumeMass VelocityFi Time TimeMFe Elastic Stress Area Elastic Stress Area

QV AV V L V V VK A K A KL CK

Where C K

ρ

ρ ρ ρρ

ρ

= = =

= = = = =

=

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 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 thatv 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 inertia 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

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 Reynold’s Model Law Froude’s Model Law Euler’s Model Law Weber’s Model Law mach’s Model Law

Reynold’s Model Law It is based on Reynold’s number and states that

Reynold’s number for model must be equal to theReynolds number for prototype.

Reynolds Model Law is used in problems where viscousforces are dominant. These problems include: Pipe Flow Resistance experienced by submarines, airplanes, fully immersed

bodies etc.

( ) ( )Re Re

1

: , ,

m mP PP 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 LV L

υ υ

υυυ

υυυ

= =

= =

= = =

Reynold’s Model Law The Various Ratios for Reynolds’s Law are obtained as

rr

r

P P P r

m m m r

P Pr

m m

2r

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

LVV L

a Ta T

A V L VA V

m

υ µ ρυ υ

υ υυ

= =

= =

= =

= =

= =

2 2 2

2 2 2 3r 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 VL V V L V

ρ ρ ρ

ρ ρ

= = =

=

Reynold’s Model Law Q. A pipe of diameter 1.5 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 DD

900 1.5 1 10 3.01000 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

VV

QA

m sV V m s

and Q V A

ρ ρ µρµ µ ρ µ

π

−

−

= ⇒ =

× ×= =

× ×

= =

=∴ = =

= = 2

3

091 / 4(0.15)

0.0899 /m sπ×

=

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/m3n Velocity of flow Vm=? Discharge Qm=?

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 numberfor prototype.

Froude’s 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

( ) ( )e e

/ 1; : ,

m mP PP m

P P m m P m

P P Pr r r r

m mPm

m

V VV VF F or org L g L L L

V V LV L where V LV LLV

L

= = =

= = = =

Froude’s Model Law 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/ 2r

sin

Velocity Ratio: V

T L /V LTime Ratio: Tr=T L /V

V / VrAcceleration Ratio: a = 1V / Tr

Discharge Ratio: Q

Force Ratio: Fr=

mP

P m

pPr

m m

rr

rP

m r

P Pr r r r r

m m

r r

VVceL L

LV LV L

LL

La Ta T L

A V L V L L LA 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

= = = =

= = =

Froude’s 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 thevelocity and discharge over the prototype which is 36 times the modelsize.

( ) ( )

( )

2.5 2.5p

m2.5 3

For DischargeQ

36Q

36 2 15552 / sec

r

p

L

Q m

= =

= × =

p

m

For Dynamic Similarity, Froude Model Law is usedV

36 6V

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=?

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

geometrically similar to prototype or in other words if the scale ratiofor linear dimensions of the model and its prototype is same, themodels is called undistorted model. The behavior of prototype canbe easily predicted from the results of undistorted or true model.

Undistorted Models: A model is said to be distorted if it is notgeometrically 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 ratioare taken to be same, then depth of water in the model of river willbe 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.

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 PH

m m

PV

m

Pm r V

m

P Pm

m m

L BL Bhh

ghV L

ghB hAB h

=

=

= =

= = ( ) ( )

( ) ( ) ( ) ( ) ( )3/ 2PScale 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 LA V

= = =

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 themodel=1/50 the horizontal dimension of prototype and verticaldimension of model =1/10 the vertical dimension of prototype.

( )

( )

( ) ( )

3p

r

r

3/ 2P

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 10V

PH

m

PV

m

m r rH

p m

LL

hh

Q L L

Q Q

=

=

=

∴ = ×3

1581.14

1.5 /1581.14 0.000948 /mQ m s

=

⇒ = =

Distorted model 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 anddepth of flow are 60m and 4.2m respectively. Determine the correspondingdischarge in model and check the Reynolds’ Number of the model flow.

( )

( )

( ) ( )

3

r

r

3/ 2r 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 7V

p

p p

PH

m

PV

m

m r rH

p m

Disch e of River Q m sWidth B m and Depth y m

BB

yy

Q L L

Q Q

= =

= = = =

=

=

=

∴ = × =3 3

0710.7

450 /1581.14 6.365 10 /mQ m s−⇒ = = ×

Distorted model

( )( )

mVLReynolds Number, Re =

4/ 60 / 200 0.3

/ 4.2 / 50 0.084

0.3 0.084 0.02522 0.3 2 0.084 0.4680.0252 0.053850.468

Kinematic Viscosity of w

m

m m

m p r H

m p r V

m m m

m m m

m

m

L RWidth B B L m

Depth y y L m

A B y mP B y m

ARP

υ

=

= = = =

= = = =

= = × == + = + × =

= = =

6 2

6

ater = =1 10 / sec4 4 0.253 0.05385Re 54492.31

1 10 >2000

Flow is in turbulent range

m

mVR

υ

υ

−

−

×

× × = = = ×

∴

Assignment Problems: Q.1: The characteristics of the spillway are to be studied by means of a geometrically

similar model constructed to a scale of 1:10.(a) If 28.3 cumecs, is the maximum rate of flow in prototype, what will be the corresponding

flow in model?(b) If 2.4m/sec, 50mm and 3.5 Nm are values of velocity at a point on the spillway, height of

hydraulic jump and energy dissipated per second in model, what will be thecorresponding velocity, height of hydraulic jump and energy dissipation per second inprototype?

Q.2: A 10m wide rectangular headrace channel carries 450 m3/sec at a flow depth of 3.5m. It is intended to construct its model at a model station to a scale of 1:20. What will bethe discharge in the model headrace channel? If the head loss due to friction in themodel is 0.50 m. What is the corresponding power loss (kW) in the prototype channel?

Q.3: A 5m diameter penstock carries 1500 m3/sec design discharge for theturbine. A model of this penstock is developed at a scale of 1:10 at a hydraulicmodel station. If the head loss due to friction in the model is 0.5m what will bethe head loss in the prototype. Also find the velocity, discharge and reynolde’snumber for the model penstock.