Dimensional-Analysis UnitIV

DIMENSIONAL ANALYSIS, ,HYDRAULIC SIMILITUDE AND

MODEL INVESTIGATIONMODEL INVESTIGATION

Dr. P. JAGADEESH

DIMENSIONAL ANALYSIS

Dimensional analysis is a mathematical technique which make use of the study of dimensions as an aid to solution of several engineering problems.

Uses:

1. Testing the dimensional homogeneity of any equation of fluid motionmotion.

2. Deriving equations expressed in terms of non-dimensional parameters to show the relative significance of each parameter.

3. Planning model tests and presenting experimental results in asystematic manner in terms of non-dimensional parameters; thusmaking it possible to analyze the complex fluid phenomenon.g p y p p

Dimensional Homogeneity

Fourier’s principle of dimensional homogeneity states that an equation which ou e s p c p e o d e s o a o oge e ty states t at a equat o cexpress a physical phenomenon of fluid flow must be algebraically correct and dimensionally homogeneous.

Ex: lEx:

L.H.S = R.H.S

2 ltg

= ∏

L.H.S R.H.S

Method of Dimensional Analysis

1.Rayleigh Methody g

2.Buckingham Π-MethodRayleigh Method: In this method a functional relationship of some variable is expressed i th f f ti l ti hi h t b di i ll hin the form of an exponential equation which must be dimensionally homogeneous.

( )( ) ( )1 2, ,.... 1nx f x x x=

( ) ( )1 2 .... 2a b nnx c x x x=

Discharge Through Orifice ……Q = f(µ, ρ, d, H, g)

By Rayleigh method may be expressed as

Q C( a b dc Hc e)Q = C(µa ρb dc Hc ge)

Substituting proper dimensions for each variable in the above eqn.

(L3/T) = (M0L0T0) (M/LT)a (M/L3)b (L)c (L)d (L/T2)e

Comparing both sides, Solving eqn 1, 2, 3M: 0 = a+b (1) b = -aL: 3=-a-3b+c+d+e (2) e = ½ - a/2T: -1=-a-2e (3) c = 5/2 – 3a/2 –dT: 1 a 2e (3) c 5/2 3a/2 d

( ) 3 31 12 1 1, , ,dH HQ a gH f C fd d

µ µ⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟= =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠( ) 3 31 1

2 2 2 2, , ,dg

d dd g d gρ ρ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦

Buckingham Π-Method:It t t th t if th di i l i bl i l d iIt states that if there are n dimensional variables involved in aphenomenon, which can be completely described by m fundamentalquantities or dimensions (such as mass, length, time etc.,), and are relatedby a dimensionally homogeneous equation then the relationship amongby a dimensionally homogeneous equation, then the relationship amongthe n quantities can always be expressed in terms of exactly (n-m)dimensionless and independent Π terms.

( )1 2 3, ,.... nx f x x x=

( )1 1 2 3, .... nf x x x x C=

( )( )2 1 2 1, .... n mf Cπ π π − =

Selection of m variables out of the existing variables as non repeating:♦These (m variables) variables should such that none of them is dimensionless♦These (m variables) variables should such that none of them is dimensionless

♦They themselves do not form a dimensionless parameters

♦All the m fundamental dimensions are included collectively in them♦All the m fundamental dimensions are included collectively in them

♦As far as possible the dependent variable should not be taken as repeating variable as otherwise it will not possible to obtain an explicit relationship.

♦For example in fluid problems, usually a characteristic linear dimension (ex. Length, breadth & depth), a characteristic velocity and a characteristic fluid property (generally fluid density) are chosen as repeating variables.

Problem: The resistance force F of a ship is a function of its length L, velocity V, acceleration due to gravity g and fluid properties like density ρand viscosity Write this relationship in a dimensionless form as belowand viscosity µ. Write this relationship in a dimensionless form as below.

,F gLfn µ⎡ ⎤= ⎢ ⎥2 2 2 ,fn

v L V VLρ ρ⎢ ⎥⎣ ⎦

Using Buckingham Π-Method,

F = fn (L, V, g, ρ, µ)

List of dimensions for each variable,

F L V g ρ µ

[MLT-2] [L] [LT-1] [LT-2] [ML-3] [ML-1T-1]

There are n m = 3 dimensionless π terms Take L V and ρ as repeatingThere are n-m = 3 dimensionless π terms. Take L, V and ρ as repeating variables.

: a b cterm FL Vπ π ρ=1 1:term FL Vπ π ρ=

2 2: a b cterm gL Vπ π ρ=2 2:term gL Vπ π ρ

3 3: a b cterm L Vπ π µ ρ=3 3 µ ρ

0 0 0 a b c⎡ ⎤0 0 01 1: a b cterm M L T FL Vπ π ρ⎡ ⎤= =⎣ ⎦

[ ]0 0 0 2 1 3b caM L T MLT L LT ML− − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤[ ]0 0 0 2 1 3M L T MLT L LT ML− − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

Hence by equating powers of M L and T on both sides we getHence by equating powers of M, L and T on both sides, we get

1 + c =0

1+ a + b – 3c=01+ a + b 3c 0

- 2 – b = 0c = -1, b = -2, a = -2

1 2 2

FV L

πρ

=V Lρ

2 2: a b cterm gL Vπ π ρ=

[ ]0 0 0 2 1 3b caM L T LT L LT ML− − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

c=0

1 + a + b = 0

- 2 – b = 0

a = 1, b = -2, c=0

2 2

gLV

π =V

3 3: a b cterm L Vπ π µ ρ=

[ ]0 0 0 1 1 1 3b caM L T ML T L LT ML− − − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦1 + c = 0

- 1 + a+ b – 3c = 0

1 b 0-1 – b = 0

a = -1, b = -1, c = -1

3 LVµπρ

=

2 2 2, ,F gLHence fn orv L V VL

µρ ρ

⎡ ⎤= ⎢ ⎥

⎣ ⎦[ ]2 2 ,ReF Fr

v Lφ

ρ=

v L V VLρ ρ⎣ ⎦ v Lρ

Prob. The pressure drop ∆p in a pipe of diameter D and length l depends on mass density, ρ and viscosity, µ of the flowing fluid, mean velocity flow V and

h i ht K f h j ti th i f U i B ki haverage height K of roughness projections on the pipe surface. Using Buckingham Π-Method, obtain the dimensionless expression for ∆p as

2flVhWhere hf is the loss of head due to friction (= ∆p/ γ γ ), γ is the specific weight

f th fl id d f i th ffi i t f f i ti

2fhgD

=

of the fluid and f is the coefficient of friction

∆p = fn (V, l, D, u, k, ρ)

fn (∆p V l D u k ρ)=Cfn (∆p , V, l, D, u, k, ρ)=C

1 1 11

a b c

b

V D pπ ρ= ∆2 2 2

23 3 3

3

a b c

a b c

V D lV D k

π ρ

π ρ

=

=4 4 4

4a b cV Dπ ρ µ=

p l k µ⎛ ⎞ ⎛ ⎞∆ ⎛ ⎞ ⎛ ⎞1 2 3 42 ; ; ;p l k

V D D VD

p l k

µπ π π πρ ρ

µ

⎛ ⎞ ⎛ ⎞∆ ⎛ ⎞ ⎛ ⎞= = = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

⎛ ⎞∆2 , ,p l kfn

V D D VDwhich may also be written as

µρ ρ

⎛ ⎞∆= ⎜ ⎟

⎝ ⎠

2

,

, ,

which may also be written as

l k VDp V fnD D

ρρ ⎛ ⎞∆ = × ⎜ ⎟

⎝ ⎠2

pD D

p V l k VDfn

ρµ

ρ ρ

⎜ ⎟⎝ ⎠

⎛ ⎞ ⎛ ⎞∆= ×⎜ ⎟ ⎜ ⎟

2

,

; Re

fnD D

p flV kh where f fn

γ γ µ×⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠∆ ⎛ ⎞= = = ⎜ ⎟; ,Re

2fh where f fngD Dγ

= = = ⎜ ⎟⎝ ⎠

SIMILITUDE – TYPES OF SIMILARITIES

Th th t f i il iti t b t bli h d f l t i il itThere are three types of similarities to be established for complete similarity to exist between the model and its prototype.

1. Geometric Similarity

2. Kinematic Similarity

3. Dynamic Similarity

Geometric Similarity: Geometric similarity exists between the model and prototype if the ratios of corresponding length dimensions in the model and the prototype are equal.

L b dLength scale ratio =

m m mr

P P P

L b dLL b d

= = =

2m m mr r

A L xbA L= = =Area scale ratio =

Volume scale ratio =

r rP p pA L xbArea scale ratio

2m m m mr r

P

V L xb xdV LV L xb xd

= = =P p p pV L xb xd

Kinematic similarity: Kinematic similarity exists between the model and prototype if

i the paths of the homologous moving particles are geometrically similar andi. the paths of the homologous moving particles are geometrically similar, and

ii. If the ratios of the velocities as well as acceleration of the homologous particles are equal.

TTime scale ratio =

mm m r

LV LTV

mr

P

TTT

=

V l it l ti m m rr

pP r

p

V LTV LV TT

= = =Velocity scale ratio =

L

Acceleration scale ratio =( )

( )( )

2

2

2

m

mm rr

pP r

LTa La La TT

= = =

( )2PT

Dynamic similarity: Dynamic similarity exists between the model and the prototype which are geometrically and kinematically similar if the ratio of all the forces p yp g y yacting at homologous points in the two systems viz., the model and the prototype are equal.

tanGm mIm Em TmF FF F F Cons tF F F F F

ν= = = = =Gp Ip p Ep TpF F F F Fν

TYPES OF MODELS

1 U di t t d M d l1.Undistorted Models

2.Distorted Models

U di t t d M d l A di t t d d l i th t hi h iUndistorted Models: An undistorted model is that which isgeometrically similar to it prototype by satisfying the basic condition ofperfects similitude. The predictions in the case of such models isp prelatively easy and many of the results obtained from the model testscan be transferred directly to the prototype.

Undistorted Models: Distorted models are those in which one or moreterms of the model are not identical with their counter-parts in theproto type Since the basic condition of perfect similitude is notproto-type. Since the basic condition of perfect similitude is notsatisfied, the results obtained with the help of a distorted model areliable to distortion and have more quantitative value only.Ex: Rivers, Dam across wide river, harbours, estuaries etc.,

REASONS FOR ADOPTING DISTORTED MODELS

• To maintain accuracy in vertical measurements• To maintain accuracy in vertical measurements.

• To maintain turbulent flow.

• To obtain suitable bed material and its adequate movement.To obtain suitable bed material and its adequate movement.

• To obtain suitable roughness condition.

• To accommodate the available facilities such as space, money, water

supply and time.

ADVANTAGES

The vertical exaggeration results in steeper water surface slopes andmodification of wave heights in models, which cab be measured easilyaccurately.y

Due to exaggarated slopes, the Reynolds numbers of a model isconsiderably increased and the surface resistance is lowered. This assist insimulation of the flow conditions in the model and prototypesimulation of the flow conditions in the model and prototype.

Model size can be sufficiently reduced by its distortion, thereby its operation simplified and also cost is lowered automatically.

In case of distorted models sufficient tractive force can be developed to produce adequate bed movement with a reasonably small model.

DIS ADVANTAGESDIS-ADVANTAGES

The magnitude and distribution of velocities are incorrectly re-produced because of vertical exaggeration of measurements.

The pressure may not be correctly reproduced in magnitude and direction.

Slope of river bends, earth cuts and dikes are often too steep that they cannot be moulded satisfactorily in sand or other erodible material.

A model wave may differ in type and possibly in action from that of the y yp p yprototype.

There is an unfavorable psychological effect on the observer.

NON-DIMENSIONAL NUMBERS

Since Inertial force always exists when any mass is in motion the conditionSince Inertial force always exists when any mass is in motion, the condition for dynamic similarity are developed by considering the ratio of the and any one of the remaining force.

I I ti Vi f ti R ld b (R )I. Inertia – Viscous force ratio – Reynolds number (Re)

Inertial force = ma=ρxV xv/t = ρxV/t xv = ρ x Av x v = ρL2v2

Viscous force = τ=(µdv/dy)A = µ(v/L ) L2= µvLViscous force = τ=(µdv/dy)A = µ(v/L ) L µvL2 2

i

v

F L v vLF vL

ρµ ν

= =

II. Inertial – Gravitational Force ratio – Froude number (Fn)

2iF vi

vF Lg=

III. Inertia – Pressure force ratio – Euler number (Eu)2F v

1/Eu = Newton Number

i

v

F vpFρ

=

IV. Inertial – Elasticity Force ratio – Mach number (Ma)

2F 2

2

2

;i

v

F v c kF cv Cauchy Number

ρ= =

2

11

/

Cauchy NumbercMa v c Supersonic

Subsonic

=

= ><11

SubsonicSonic

<=

Inertial – Surface tension Force ratio – Weber number (We or Nw)

2i

v

F vF L

σρ

=

2v v

L Lσ σρ ρ

=

Prob.1. 1/10 model of an airplane is tested in a variable density wind tunnel. Theprototype plane is to fly at 400 km/h speed under atmospheric conditions. Theprototype plane is to fly at 400 km/h speed under atmospheric conditions. Thepressure used in the wind tunnel is 10 times the atmospheric pressure. Calculatethe velocity of air in the model. To what prototype value would a measured drag of500 N in the model correspond?

Sol. Reynolds similarity law is applicable.

( ) ( )Re ; Re p p pm m m v Lv L ρρ( ) ( )

110

Re ; Re p p pm m mm p

m p

mr

L LL

µ µ= =

= =

Since pressure does not affect the viscosity appreciably, µm= µp... Further at constant temperature p/ρ=constant.

Pressure ratio

10pL

Pressure ratio, 10

400 400 /

mr r

p

r

p pp

v v km h

ρ

µ

= = =

= × = = 400110 10/m p

r r

v v km hLρ

= × = =×

Model velocity is the same as prototype velocity, i.e. vr = 1.0

Force ratio Fr = ( )22 2 21 110 1F L v × ×Force ratio, Fr = ( )2 2 21 110 110 10

500 10 5000

r r r r

mp

r

F L v

FF NF

ρ= = × × =

= = × =

Prob.2. A model boat, 1/100 size of its prototype has 0.12N of resistance when simulating a speed of 5 m/s of the prototype. Water is the fluid in both cases. What is the corresponding resistance in the prototype? Assume frictional forces

r

are neglected.

Sol. The resistance offered at the free surface is the significant force and as suchFroude model law is appropriate.

pmm

m p

vvFrgL gL

LIf L V L

= =

( ) 2 2 3

;mr r r

p

m

LIf L V LL

ForceL v Lρ ρ

= =

= =( ) r r r r r

p

L v LForce

ρ ρ

Since the same fluid is used in the model and prototype, ρm=ρr,and ρr=1.

3mr

F LF

=

3

310 12 120000 120

100.

p

mp

F

FF N kNL⎛ ⎞= = = =⎜ ⎟⎝ ⎠100p

rL ⎜ ⎟⎝ ⎠

Prob.3. A 1:6 scale model of a passenger car is tested in a wind tunnel. Theprototype velocity is 60 km/h. If the model drag is 250 N what is the drag and the

i d t th d i th t t Th i i th d l dpower required to overcome the drag in the prototype. The air in the model andprototype can be assumed to have the same properties.

Sol. Reynolds similarity law is applicable

( ) ( )Re ; Re p pm mm p

m p

v Lv Lν ν

ν

= =

( )1 . . ,

rr

r

r m p m p

vL

If i e

ν

ν ρ ρ µ µ

=

= = =

1

60 6 360 100/ /

rr

m p r

then v Lv v L km h m s

=

= = × = =2 2 2

2 2 22

p

r r rr r r r r r

r r r

FmForce ratio L v LFp L

ν µ µρ ρ ρρ ρ

⎛ ⎞= = = × = × =⎜ ⎟

⎝ ⎠

1 1 1 0, .r rFmIf andFp

ρ µ= = =

Fp=250 N (Same as in the model)

Power to overcome drag in the prototype:

360 10⎛ ⎞×60 10250 4167 4 1673600

. .p p pP F v W kW⎛ ⎞×

= = × = =⎜ ⎟⎝ ⎠

Unit quantities & Specific quantities:

In order to predict the behavior of a turbine working under varying conditions and to facilitate comparison between the performances of the turbine of the same type but having different outputs and speeds and working under different heads, it is often convenient to express the test results in terms of

t i it titicertain unit quantities.

Performance under unit head - Unit quantities:

From the output of a turbine corresponding to different working heads it is possibleFrom the output of a turbine corresponding to different working heads it is possible

to compute the output which would be developed if the head was reduced to unit (1 meter); the speed being adjusted so that the efficiency remains unaffected.

( )( )

1 1 1 1,uQ Q H Q H Q Q H H

N N H N H N N H H

= = =

( )( )3 2 3 2 3 2

1 1 1 1

3 21 1 1 1

,

,

u

u

N N H N H N N H H

P P H P H P PH H

= = =

= = =

Assumption: The above expressions are based on the assumption that the efficiencies remains constant at all the heads, which is however not correct.

Note: The expression for the various quantities as derived above can be used for comparing the performance of any one turbine only under different conditions of operationconditions of operation.

Unit quantities obtained below would facilitate a comparison between the performance of different turbines of the same type. According to this, the unit speed, unit discharge and unit power are defined respectively as the speed discharge and

ND Q P

unit discharge and unit power are defined respectively as the speed, discharge and power of a turbine having a runner diameter of 1m and operating under a head of 1m.

2 3 22; ;u u u

ND Q PN Q PD HH D H

= = =

Note: The unit quantities for similar turbines will be equal only if their efficiencies are equal.

However, the performance of different types may be compared by considering an imaginary turbine called specific turbineimaginary turbine called specific turbine.

The specific turbine is an imaginary turbine which is identical in shape, geometrical proportions, blade angles, gate settings etc., with the actual turbine but reduced to such size that it will develop one kilowatt power under unit headbut reduced to such size that it will develop one kilowatt power under unit head.

5 4

N PN = 5 4sNH

Problem 1. A Francis turbines working under a head of 5m at a speed of 210r.p.m. develop 75 kW when the rate of flow of water is 1.8 m3/s. The runnerdiameter is 1m. If the head on this turbine is increased to 16m, determine its newspeed, discharge and power.speed, discharge and power.

1 210 16N H⎛ ⎞ ×⎜ ⎟1

1

1 3

210 16 375.7 . . .5

1.8 16 3 22 /

N HN r p m

H

Q HQ

⎛ ⎞ ×= = =⎜ ⎟⎜ ⎟⎝ ⎠⎛ ⎞ ×⎜ ⎟1 3

1

3 2 3 21

1 3 2 3 2

3.22 /5

75 16 429.33

Q m sH

PHP kW

= = =⎜ ⎟⎜ ⎟⎝ ⎠⎛ ⎞ ×

= = =⎜ ⎟1 3 2 3 25H⎜ ⎟⎝ ⎠

Problem 2. A water turbine develops 130 kW at 230 r.p.m. under a head of 16 m. Determine the scale ratio and the speed of a similar machine which will generate 660 kW, when working under a head of 25m.

Solution: 1 22 3 2 2 3 2

1 1 2 2

, ;P PS in ceD H D H

=

( ) ( )

1 1 2 2

3 2 3 22 21 2

130 66016 25D D

D

=

2

1

1 612.DD

=

N D N D

( )

1 1 2 2

1 2

, N D N DAlsoH H

=

( )2 11 1 61223016 25

.N DD ××=

2 178 35. . . .N r p m=