54424673 Chapter 5 Dimensional Analysis

CHAPTER 5

DIMENSIONAL ANALYSIS AND SIMILITUDE

5.1 DIMENSIONAL ANALYSIS

One approach to solve fluid mechanics problems is by using dimensional

analysis, a mathematical technique that makes use of the study of dimensions.

Dimensional analysis is related to similitude; however, the approach is quite

different. In dimensional analysis, the prediction of physical parameters will

influence the flow, and then group these parameters into dimensionless

combinations for better understanding of flow phenomena. Dimensional

analysis is particularly helpful in experimental work because it provides a guide

to the things that significantly influence the phenomena; thus it indicates the

direction in which experiment work is important.

The significant of dimensional analysis are;

1. Useful for research study especially in design work by reducing the

number of variables.

2. To express in dimensionless equation to find the significant of each

parameters.

3. To simplify the analysis of complex phenomenon in systematic order.

5.2 UNITS AND DIMENSIONS All physical phenomena are expressible in terms of a set of basic or

fundamental dimensions as shown in Table 6.1.

HYDRAULICS

2

Table 5.1

Dimension SI US

Mass (M) Kg Lb

Length (L) M ft

Time (T) S S

Temperature (θ) oC oF

All equations related to a physical phenomenon must be dimensionally

homogeneous. This is known as Principle of Dimensional Homogeneity.

5.3 DIMENSIONAL HOMOGENEITY

An equation which expresses the proper relationship between the variables in a

physical phenomenon will be dimensionally homogenous. This means that

each of additive terms in an equation should have the same dimension. For

example; P (kg/ms) = ρgh (kg/ms), where both sides are in same units.

There are 2 types of system which are MLTθ and FLTθ. In a MLTθ system, the

fundamental dimensions are mass M, length L, time T, and temperatureθ .

While for FLTθ system, mass M is replaced by force F. For instance; Area for

rectangular, A = Length, L x width, b = m2 (SI unit). But in dimensional analysis

principle, value is not important. Thus, Area = Length, L x width, L = L2. Table

6.2 shows quantities of fluid mechanics and hydraulic in MLTθ system.

DIMENSIONAL ANALYSIS AND SIMILARITY

3

Table 5.2 QUANTITY SYMBOL SI UNIT DIMENSION

FUNDAMENTAL Mass Length Time GEOMETRIC Area Volume Angle Strain Moment of Inertia DYNAMIC Force Weight Specific weight Density Pressure Shear stress Modulus of elasticity Momentum Angular momentum Moment of momentum Torsion Energy Work Power Dynamic viscosity Surface tension KINEMATIC Linear velocity Angular velocity Rotational speed Acceleration Angular acceleration Gravity Discharge Kinematics viscosity Stream function Circulation Vorticity

m L t

A V θ e I

F W γ ρ P ι

E, K M

T E W P μ σ

U,v,u ω N a α g Q υ Ψ Γ Ω

kg m s

m2 m3

M4

N (kgm/s2) N

N/m3 kg/m3

Pa (N/m2) Pa

kgm/s

Nm J

W (J/s) Pa.s

(N.s/m2) (kg/ms)

N/m

m/s rad/s (s-1)

s-1 m/s2 s-2

ms-2 m3/s m2/s m2/s m2/s s-1

M L T

L2 L3

M0L0T0 L0 L4

MLT-2 MLT-2

ML-2T-2 ML-3

ML-1T-2 ML-1T-2 ML-1T-2 MLT-1 ML2T-1 ML2T-1 ML2T-2 ML2T-2 ML2T-2 ML2T-3 ML-1T-1

MT-2

LT-1 T-1 T-1

LT-2 T-2

LT-2 L3T-1 L2T-1 L2T-1 L2T-1 T-1

Source: Rajput (1998)

HYDRAULICS

4

5.4 METHODS There are a lot of methods can be used to reduce into a smaller number of

dimensionless parameters such as Bridgman method, Matrix-tenor etc. Two of

the commonly used methods are Raleigh method (basic principle) and

Theorem Pi Buckingham.

5.4.1 RAYLEIGH’s METHOD This method is a basic for a small number of parameters; it becomes rather

cumbersome when a large number of parameters are involved.

ncba AAACAA .....4321 =

Where

A1 = dependent variables

nAAAA .....432 = independent variables

C = a dimensionless constant

The dimensions of each quantities, nAAAA .....432 are written and the sum

exponents of each, which are M, L and T on both sides are equated. Solution of

equations on simplification yields dimensionless groups controlling the

phenomenon.

Example 1: Express dimensionless equation for the speed V with a wave pressure travels

through a fluid. Consider the physical factors probably influence the speed are

compressibility, K density, ρ dan kinematics viscosity, ν.

Answer: 1. Write the fundamental dimension for all dimensions given using MLTθ.

⎥⎦

⎤⎢⎣

⎡=⎥⎦

⎤⎢⎣⎡=⎥⎦

⎤⎢⎣⎡=⎥⎦

⎤⎢⎣⎡=

TL

LM

LTMK

TLV

2

32 ,,, νρ


5

2. Let write the equation like this:

dbaCKV νρ=

3. Insert fundamental dimension into the equation while C is dimensionless

constant.

dba

TL

LM

LTM

TL

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛=

2

32

4. To satisfy dimensional homogeneity, net power of each dimension must be

identical on both sides of this equation. Thus,

For M : 0 = a + b

For L : 1 = -a + (-3b) + 2d

For T : -1 = -2a + (-d)

5. Solve those three equations, we obtain a = ½, b = -½ dan d = 0,

6. So the equation will be,

02/12/1 νρ −= CKV or ρ

KCV =

∴ Wave speed is not affected by the fluid’s kinematics viscosity, v

Example 2:

The drag force DF on a sphere in laminar flow is known to depend on its

diameter D, velocity of flow V, density of fluid ρ and coefficient of viscosityμ .

Obtain an expression for DF using Rayleigh’s method.

Answer: Using M, L, and T as primary units,

Since

dcbaD VCDF μρ= where, C = dimensionless constant.

Thus, [ ] [ ] [ ] [ ]dcba TMLMLLTLMLT 11312 −−−−− =

HYDRAULICS

6

Equating the powers of M, L, and T on both sides,

For M: 1 = c + d (1)

For L: 1 = a + b + (-3c) + (-d) (2)

For T: -2 = -b + (-d) (3)

Since there are three equations and four unknowns, three variables can be

expressed in terms of the fourth using substitution method.

From (1) c = 1 - d

From (3) b = 2 - d = 2 – d

From (2) a = 1 - b + 3c + d = 1 - (2 - d) + 3 (1 - d) + d = 2 – d

∴ ddddD VDF μρ ⋅⋅= −−− 122

d

VDVD ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

ρμρ 22

⎟⎟⎠

⎞⎜⎜⎝

⎛ρ

μρ=

VDfnVDF 22

D

5.4.2 Pi Buckingham Theorem When a large number of variables are involved, Raleigh’s method becomes

lengthy. In such circumstances, the Buckingham's method is useful. This

method expressed the variables related to a dimensional homogenous

equation as:

( )nXXXfX ....., 321 =

where, the dimension at each section is the same.

The Buckingham Phi Theorem can also be expressed in terms of ∏ as shown

in on the right.

)....,,()( 321 knphi −ΠΠΠΠ=φ

where, m = the primary dimensions which are M, L, T and θ

n = dimensional variables such as velocity, discharge and

density.

k = reduction


7

Example 3: Drag force FD exerted on a submerged sphere as it moves through a viscous

fluid. Certainly parameters involve are diameter, D, velocity, V, dynamic

viscosity, μ and density of fluid, ρ. Express dimensionless equation by using Pi

Buckingham theorem.

Answer:

1. List the influence factors and count n.

FD = fn (D, V, μ, ρ)

where FD= dependent variable

D,V, μ and ρ = independent variables

n = 5 (FD, D, V, μ, ρ)

2. Choose dimensional system (MLTθ or FLTθ) & list the dimensions of each

variable. Find m.

Choose MLTθ, )LM

LTM,

TL(L,fn

TML

32 =

3. Find k. It is usually equal to m which is cannot exceed but rarely less than m.

Then find n-k or n-m (the number of dimensionless Π groups needed).

n - m = 5 – 3 = 2 so we can write f (Π1, Π2) = 0

4. Choose repeating variables which is 3 (number must be same as m). These

variables must contain the entire fundamental dimensions which are L, T and

M. In this question, we choose

D (L), V (L/T), ρ (M/L3)

Notes:

Number of repeating variables must same with number of fundamental

dimension and choose from independent variables (in this example choose

3)

m = 3 (M, L and T)

HYDRAULICS

8

To choose these variables must relate to mass, geometry and kinematics.

Make sure that the entire fundamental dimensions (M, L & T) include in the

chosen variable (at least 1).

5. Form the phi group, which are;

π 1 = ρa1 Db1 Vc1 FD

π 2 = ρa2 Db2 Vc2 μ

The repeating variables need to be inserted in the phi group (π 1 and π 2).

Equate the exponents on both sides, solve and form dimensionless groups.

6. Since the π s are dimensionless, we can replace with M0L0T0 in π 1

π 1 = ρa1 Db1 Vc1 FD

M0 L0 T0 = ⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

2

11

1

3 TML

TLL

LM c

ba

M: 0 = a1 + 1

L : 0 = -3a1 + b1 + c1 + 1

T : 0 = - c1 – 2

Solve a1, b1 & c1 a1 = -1, b1 = -2, c1 = -2

7. So π 1 = ρa1 Db1 Vc1 FD

π 1= ρ-1 D-2 V-2 FD

8. Then solve π 2

π 2 = ρa1 Db1 Vc1 μ

M0 L0 T0 = ⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

LTM

TLL

LM c

ba 1

11

3

M: 0 = a1 + 1

L : 0 = -3a1 + b1 + c1 - 1

T : 0 = - c1 – 1


9

Solve a1, b1 & c1 a1 = -1, b1 = -1, c1 = -1

9. So π 2 = ρa1 Db1 Vc1 μ

π 2 = ρ-1 D-1 V-1 μ

10. Rearrange the pi groups as desired and expressed as

( )nf ππππ ......., 321 =

)( 22VDFf

DVD

ρρμ

=

Example 4: Show that

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛=

2/32/3

HgPHgq φ

when flow rate, q (m3/s/m) is over the spillway and assume that would affected

by height of water over weir, H (m), height of weir, P (m) and gravity, g (m/s2)

Answer:

1. List the influence factors and count n.

q = fn (H, P, g) n = 4

2. Choose dimensional system (MLTθ or FLTθ) & list the dimensions of each

variable. Find m.

),,(2

3

TLLLf

TLL

=

3. Find k and then find n-k

So k = 2 same as m

m = 2 (L and T)

HYDRAULICS

10

n - m = 4 – 2 = 2 so there are 2 groups of phi; π 1, π 2 = 0

4. Choose repeating variables which is 2 (same as k). These variables must

contain L and T.

Choose: H and g

In this example, the repeating variables can be found easily in the equation

given.

⎟⎟⎠

⎞⎜⎜⎝

⎛=

2/32/3

HgPHgq φ ⎟

⎟⎠

⎞⎜⎜⎝

⎛=

2/32/3 HgH

Hgq φ

5. Form phi group which is 2 groups:

π 1 = Ha1 gb1 q

π 2 = Ha2 gb2 P

Equate the exponents on both sides, solve its and form dimensionless groups.

6. Since the phi is dimensionless, we can replace with M0L0T0 in π1

π1 = Ha1 gb1 q

L0 T0 = ( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

LTL

TLL

ba

31

21

L : 0 = a1 + b1 + 2 T : 0 = - 2b1 - 1

Solve a1, b1 & c1 a1 = -3/2, b1 = -1/2

7. So π1 = Ha1 gb1 q π1= H-3/2 g-1/2 q

8. Repeat step 6 – 7 for π2

9. Rearrange the pi groups as desired and expressed as

( )21 ππ fn=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

gHPfn

gHq

2/32/3

⎟⎟⎠

⎞⎜⎜⎝

⎛=

gHPfngHq2/3

2/3


11

5.5 SIMILARITY In hydraulic and aeronautical engineering valuable results are obtained at

relatively small cost by performing tests on small scale models of full size

systems (prototypes). Similarity laws help us to interpret the results of model

studies. The relation between model and prototype is classified into three:

1. Geometry Similarity

2. Kinematics Similarity

3. Dynamic Similarity

5.5.1 MODEL (m) Model is similar with object/structure required in certain scale ratio. It is need to

be tested in laboratory with similar condition in real phenomenon. The size of

model is not necessary smaller than prototype.

5.5.2 PROTOTYPE (p) Prototype is an object/actual structure in full size. It is need properly tested in

actual phenomenon, example: spillway structure in open channel, ship etc

5.5.3 GEOMETRY SIMILARITY The prototype and model have identical shapes but differ only in size. Ratio of

corresponding length in prototype and model show as,

Length, rm

p LLL

=

Area, 2

2

2

rm

p

m

p LL

LAA

==

Volume, 3

3

3

rm

p

m

p LLL

VV

==

HYDRAULICS

12

5.5.4 KINEMATICS SIMILARITY In addition to geometric similarity, ratio of velocities at all corresponding points

in flow are the same.

Velocity scale ratio, m

pr V

VV =

As time, T is dimensionally L/V. By that

Time scale ratio, rm

p TTT

= and r

rr V

LT =

While, for discharge and acceleration scale ratio

Discharge scale ratio, r

r

m

p

m

p

m

m

p

p

m

pr T

L

TTL

L

TL

TL

QQ

Q33

3

3

3

====

Acceleration scale ratio, 2

2

2

2

2

r

r

m

p

m

p

m

m

p

p

m

pr T

L

T

TLL

TLT

L

aa

a ====

5.5.5 DYNAMIC SIMILARITY Two systems have dynamic similarity if, in addition to dynamic similarity,

corresponding forces are in the same ratio in both. The force scale ratio is

rm

p FFF

=

Basically, if the geometric and kinematics similarities exist, it shows two

systems are dynamically similar. The ratios of these systems of all

corresponding forces are the same. The respective forces includes;

a) Gravity = GF b) Viscosity = vF c) Elasticity = EF d) Surface tension = TF e) Inertia = IF


13

Resultant=++++=∑ TEVPG FFFFFF and FI = - Resultant

Then, strict dynamic similarity means;

=====Ip

Im

Tp

Tm

Ep

Em

vp

vm

Gp

Gm

FF

FF

FF

FF

FF

Constant

5.5.6 ADVANTAGES USING SIMILARITY The advantages are;

1. Performances of object can be predicted.

2. Economy and easy to build, where design of model can be done many

times until reach a certain values.

3. Nonfunctional structure also can be measured such as dam.

5.5.7 NON-DIMENSIONAL PARAMETER

By using Raleigh’s Method or the Buckingham Phi Theorem, the number of

dimensional variables such as mass, length and time used in an analysis of

flow is reduced to a few non-dimensional variables.

There are the five non-dimensional parameters that represent the ratio of

forces per unit volume.

1. Reynolds Number

2. Froude Number

3. Mach Number

4. Euler Number

5. Weber Number

thus 0=+++++ ITEVPG FFFFFF

HYDRAULICS

14

Table 5.2

5.5.7.1 Reynolds Number Reynolds number is a non-dimensional parameter that is used when viscous

force is dominant. Reynold’s number represents the ratio between inertia force

FI and viscosity force FV.

The Reynolds number can be expressed as:

Re =V

I

FF

=ForceViscosity

Force Inertia

νμ

ρμρ LVLV

LVVL

===22


15

Below is the equation for dynamic similarity where viscous forces are

predominant.

ppm

m

LVLV⎟⎠⎞

⎜⎝⎛===⎟

⎠⎞

⎜⎝⎛

ννReRe

Reynolds number is used for the following types of flow:

Completely submerged flow

Completely enclosed flow through pipes and plates

Viscous flow as in settling of particles in fluids

Flow in flow meter in pipes, venturi meter, or orifice meter

Example 5: An oil (density = 917 kg/m3, dynamic viscosity = 0.29 Pa.s) flows in a 15 cm

diameter pipe at a velocity of 2 m/s. What would be the velocity of water flow in

a 1 cm diameter pipe, to make the two flows dynamically similar? The density

and viscosity of water can be taken as 998 kg/m3 and 1.31 x 10-3 Pa

respectively.

Answer: Reynolds similarity law is applicable,

p

pmm

LVLV⎟⎠⎞

⎜⎝⎛===⎟

⎠⎞

⎜⎝⎛

ννReRe

rr

r

r

r

p

mr LLV

VVρμν

===

0623.0

917998

151

1.29.01031.11 3

=⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛

=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

−x

LLV

V

p

m

p

mp

m

p

m

ρρμ

μ

smxVm /1246.00623.02 ==

HYDRAULICS

16

Example 6: A 1:6 scale model of a passenger car is tested in a wind tunnel. The prototype

velocity is 60 km/h. If the model drag is 250 N, what is the drag force and

power are required for prototype. The air in the model and prototype can be

assumed to have the same properties.

Answer: Reynolds similarity law is applicable

p

pmm

LVRRLV⎟⎠⎞

⎜⎝⎛===⎟

⎠⎞

⎜⎝⎛

νν then

r

rr L

V ν=

If ),..(1 pmpmr ei μμρρν === then r

r LV 1

=

100m/skm/h3606/1

60====

r

pm L

VV

5.5.7.2 Froude Number

The Froude number represents the ratio between inertia force and gravity

force. Froude number is expressed as:

Fr = gLV

The Froude number is applied where gravity forces are predominant. The

number is used in the analysis of:

Wave action such as breakwaters and ships

Free surface flow in open-channels

Hydraulic structures, such as spillways, stilling basins, weirs, and notches

Forces on bridge piers and offshore structures


17

The equation for dynamic similarity where gravity forces is predominant is given

below.

ppm

mgLVFrFr

gLV

⎟⎟⎠

⎞⎜⎜⎝

⎛===⎟

⎟⎠

⎞⎜⎜⎝

⎛)()(

Example 7: A spillways model with 1:50 scale can manage discharge of 1.25 m3/s, find the

discharge of prototype? If flood scenario takes only 12 hours in prototype, how

long should it take in the model?

Answer:

( ) ( )m

mm

p

pp gL

VFr

gL

VFr ===

Graviti g is same for model and prototype.

The length scale ratio, Lr =Lp/Lm, when rr LV =

Discharge scale ratio, Qr = VrLr2 = Lr

5/2

As Lr =1/50 , 5.2

150

25.1⎟⎠⎞

⎜⎝⎛== p

m

p QQQ

smQp /22097 3=

Time ratio, rr

rr L

VLT ==

50112

== m

p

m TTT

hours697.150

12==mT

Example 8: A model boat 1/100 size of its prototype has 0.12 N of resistance when

simulating a speed of 5 m/s of the prototype. What is the corresponding

HYDRAULICS

18

resistance in the prototype? Water is the fluid in both cases and frictional forces

can be neglected.

Answer:

( ) ( )m

mm

p

pp gL

VFrgL

VFr ===

If Lp/Lm = Lr , rr LV = and ( ) ( )322

)()(

rrrm

pr LVL

DayaDaya

F ρρ ===

The density of fluid is same, by that,

ρp = ρm and ρr = 1,

Maka, 3r

LFF

m

p =

Fp = Prototype force = (100/1)3 x (0.12) = 120 kN 5.5.7.3 Mach Number The Mach number represents the ratio between inertia force and

compressibility or elastic force. The ratio is mathematically presented as

below.

Mach Number = 1/2

1/2

force)bility (Compressi

force) (Inertia

The Mach number is used where compressibility effects are predominant in the

flow and can be expressed as:

M = CV

/EV

=ρ

Where:

C = velocity of sound in medium


19

The equation for dynamic similarity where compressibility effects are

predominant is given on the right.

ppm

m CVMM

CV

⎟⎠⎞

⎜⎝⎛===⎟

⎠⎞

⎜⎝⎛

Example 9: An airfoil moves at 650 km/h through still air at 20oC. If the elastic stress and

density of air at this temperature is 21 kg/cm2 and 0.216 kg/m3, find Mach’s

number.

Answer: V = 650 km/h = 180.6 m/s K = 21 kg/cm2 = 21 x 104 kg/m2 ρ = 0.216 kg/m3

14.0

126.010216.180

4===

xKVMρ

5.5.7.4 Euler Number

The Euler number represents the ratio between inertia force and pressure force

as shown on the right. When pressure dominates flow, dynamic similarity is

obtained using the Euler number for both prototype and model.

Example 10: A prototype spillway has a characteristic velocity of 2 m/s, 996 kg/m3 of density

and surface tension is 0.0712. What is corresponding length of prototype if

Weber number of model is 5.03 x 105 ?

HYDRAULICS

20

Answer:

mmp

m

LVWWLV⎟⎟⎠

⎞⎜⎜⎝

⎛===⎟

⎟⎠

⎞⎜⎜⎝

⎛

σρ

σρ 22

522

1003.50712.0

)2(996 xLLVWp ===σ

ρ

5.5.7.5 Weber Number

The Weber number represents the ratio between inertia force and surface

tension as displayed on the right. When surface tension dominates flow,

dynamic similarity is obtained using the Weber number for both prototype and

model.

m

mp

m

LVWWLV⎟⎠⎞

⎜⎝⎛===⎟

⎠⎞

⎜⎝⎛

σρ

σρ 22

mLp 9=


21

EXERCISE 1. The capillary rise, h of a fluid of density, ρ and surface tension σ in a

tube of diameter D depends upon the contact angle θ and gravity g.

Obtain an expression for h by Rayleigh’s method.

2. The stagnation pressure, ps in an air flow depends upon the static

pressure, po , the velocity, V of the free stream and density, ρ of the air.

Derive a dimensionless expression for ps by Rayleigh’s method.

3. The velocity of flow, u is very near to the rotating disk depends on the

angular velocity, ω of the disk, the radial distance r, vertical distance z

and kinematics viscosity of the fluid, ν. Derive a relationship for u in

dimensionless form by using Pi Theorem.

4. The shear stress, ιo at the bed of a rough channel depends upon the

depth of flow, y, velocity of the fluid, V, roughness of the bed, ε and fluid

density, ρ and viscosity, μ. Derive an expression for ιo in dimensionless

form by using Pi Theorem.

5. Obtain expressions for the velocity and force ratio similitude for a model

which obeys Mach’s law similarity.

6. If a 1.0 m diameter of pipe carrying air at a velocity 3.8 m/s is to be

modeled for dynamic similarity by 10 cm diameter of water pipe, what

would be the velocity of water?

7. The resistance offered to the movement of a 2.0 m long ship model in a

towing tank full of fresh water while moving with a speed of 1.5 m/s was

450 N.

(a) If the prototype is 60 m in length, what will be the corresponding

velocity?

(b) What would be the force required to drive at a corresponding

velocity for a prototype of 80 m length in seawater (relative

density 1.025)?

REFERENCES:

HYDRAULICS

22

1. Franzini J.B. and Finnermore E.J. 2006. Fluid Mechanics. Mc Graw Hill,

10th Edition.

2. Subramanya, K (1993). Theory and Application of Fluid Mechanics. Mc

Graw Hill. New Delhi.

3. Zarina Md Ali and Ishak Baba (2003) E-module Hidraulics. UTHM

Documents

54424673 Chapter 5 Dimensional Analysis