Introduction.

Compared to most engineering subjects, the application of fluid mechanics in design depends
more on the use of empirical results built up from extensive experimental research. In most cases
these, these experimental data are presented in the form of tables and charts that a designer may
apply directly. A vivid example is the values of friction factor for pipe flow and separation loss
coefficient for duct and pipe fittings. However, the tables and the underlying experimental work
too bulky and time consuming. It id the organization of experimental work and the result
presentation that dimensional analysis plays such an important role. Dimensional analysis
technique commences with a survey of all the likely variables affecting a phenomenon, the
experimental researcher then suggests the formation of groupings of more than one variable.
Experimental work may be based on these groups rather than on individual variables. This thus
helps to considerably reduce the testing program and leading to simplified design, guides e.g
Moody charts. Similarly on the other hand is inherent in the formulation of relationship such as
Moody charts. It is more commonly associated with the use of models and model testing
technique. Examples of such application is in wind tunnel test.

Note: Mathematics alone is not sufficient in the application of the similarity laws; in many cases
it will be found that fatal equivalence of all the dimensionless groupings will be mutually
impossible. The experience of the researcher is called upon in such cases, examples being found
in the cases of ships model testing and pumps or turbine modelling techniques utilizing gas in
place of H2O. Dimensional analysis and similarity together allow the design engineer to predict
accurately and economically the performance of the prototype system; whether it is an aircraft
wing, ship hull, dam spillway or harbor construction.

Dimensional Analysis

The foundation of fluid mechanics as a subject lies in the exponential investigations of fluid
flow. In order to determine the form dependence of one variable on a range of other controlling
parameters in the absence of an analytical solution, it is necessary to undertake an experimental
investigation, however, simply recording the effect of one variable on another with all others
held constants and operating until all the possible combinations are exhausted is not an option in
terms neither of time nor the utility of the outcome. Dimensional analysis offers an escape from
this problem by allowing the identification of groups of variables whose interrelationships may
be determined experimentally.

In engineering, especially fluid mechanics the parametric relationship needs to be verified.

Dimensions and dimensional reasoning help to formulate and cross check this. If it is possible to
know the factors involved in a physical situation, dimensional analysis can be used to obtain a
form of dimensional relationships between them. Dimensional analysis is based on the principle
of dimensional homogeneity and using the dimensions of relevant variables affects the
phenomenon.

Dimension

In general, the physical situations have some physical quantities (derived quantities) which are
expressed in form of basic (primary) or fundamental quantities. These basic quantities have the
S.I unit dimension of Mass, length, and Time which are given the symbols M , L∧T respectively.

Other quantities are obtained by expressing them in terms of their fundamental quantities and are
called derived or secondary quantities. Examples are velocity, area and density. The expression
for a derived quantity in terms of the primary quantities is called the dimension of the physical
quantity.

Dimensions are group into Geometric, Kinematics, Dynamics and Thermodynamics properties.
Examples are as given below

Quantity Dimensions, MLT System

Fundamental Quantities

Mass M

Length L

Time T
Geometric Quantities

Area L2

Volume L3

Moment Of Inertia L4

Strain L0

Kinematic Quantities

Linear Velocity ¿−1

Linear Acceleration ¿−2

Angular Velocity T −1

Angular Acceleration T −2

Volume Flow Rate/Discharge L3 T −1

Gravity ¿−2

Kinematic Viscosity L2 T −1

Dynamics Quantities

Force MLT −2

Weight MLT −2

Mass Density M L−3

Specific Weight ML−2 T −2

Specific Gravity M 0 L0 T 0
Impulse ML T −1

Linear Momentum ML T −1

Angular Momentum ML2 T −1

Pressure ML−1 T −2

Dynamic Viscosity ML−1 T −1

Work/Energy ML2 T −2
Torque ML2 T −2
Power
ML−2 T −3

Surface Tension M T −2

Thermodynamic Quantities

Temperature θ

Thermal Conductivity ML T −3 θ−1

Enthalpy M L2 T −2

Entropy ML2 T −2 θ−1

Dimensional Reasoning and Homogeneity

Dimensional reasoning is based on the proposition that for an equation to be true,

then both sides of the equation must be numerically and dimensionally identical.
For example, given the expression:
a+ b=c , where a=2, b=6 ,∧c=8

The above equation is numerically true. But it does not end here, the equation must
also satisfy it dimensional equality. For example,

1 orange+ 2mangoes=3 pears

The expression above makes no dimensional sense as different objects

(dimensions) cannot be added together. Examples are shown below:

1 orange+ 2oranges =3 oranges

6 m+11 m=17 m

An equation is only dimensionally homogeneous if all the terms have the same
dimensions. In general any equation of the form

Example 1

Prove that the formula F=ρ v 2 l 2 is dimensionally correct.

M × a=ρ v 2 l 2

2
M x LT −2= M L−3(( L T −1) ¿∗L2

M∗L T −2=M L−3∗L2 T −2∗L2

MLT −2=MLT −2

Example 2

Determine the dimension of μ in the dimensional homogeneous equation.

μ= ρvl

ρ=M L−3
 ν= LT −1

l=L

μ= M L−3 × L T −1 × L

μ= M L−1 T −1

Dimensions for derivatives and Integrals

For derivatives, the dimensions are obtained as follows,

dy δy y
[ ][ ]
dx
=
δx
=
x

Similarly,

dy
Increment of
d2 y d dy dx
[ ] ( )
dx 2
= =
dx dx Increment of ( x)

Which gives,

dy y

[ ][
d2 y
dx 2
=
dx
x
= ][ ]
x
x
y
= 2
x

Or in general,

dn y y
[ ][ ]
dx n
= n
x

Dimensions for integrals can also be obtained in a similar way. Recall the integral
term below:
p

∫ y dx
q
The term means the limit of the sum of all the products of yδx between x= p∧x=q.
The dimensions of the integrals will be the same as those of yδx .the dimension for
the integral thus becomes:

[∫ ]
q
y dx =[ yx ]

Similarly for double integral,

[∬ ad z 1 d z 2 ]= [ a z1 z 2 ]

METHODS OF DIMENSIONAL ANALYSIS

The methods of dimensional analysis are based on the Fourier’s principle of homogeneity. The
methods include:

I. Raleigh’s Method
II. Buckingham’s π−method
III. Matrix-tensor method
IV. Bridgman’s method.
V. By visual inspection of the variables involved.
VI. Rearrangement of differential equations.
Only the first two will be considered.

(i) Rayleigh’s Indicial Method of Dimensional Analysis

In dimensional analysis, some variables are selected (repeating variables)

 Geometric Properties (e.g. length, l ; diameter, d; Height, H; etc.)

 Flow Properties (e.g. velocity, v; acceleration, a; etc.)
  Fluid Properties (e.g. mass density ρ ,weight density, w; and dynamic
viscosity μ etc)
The three selected variables are the parameters of which the subject of the formula
is to be made.

Example 1

The drag force on a smooth sphere of diameter d, moving with a uniform velocity
V in a fluid density ρ and dynamic viscosity μ.Find the expression for F in terms of
these quantities.

Solution

The general relationship is mathematically represented as

F=Φ ( D ,V , ρ , μ) ¿

The dimensions for the variables are given as:

F = Force, ML T −2

D= Diameter L

V= Velocity L T −2

ρ = Density ML−3

μ=Dynamic viscosity ML−1 T −1

F=C ( Da V b ρ c μd ), where C is a constant.

Using the M-L-T system, the corresponding equations for dimensions is:

ML T −2=C ¿
For dimension homogeneity, the exponents of each dimension on both sides of the
equation must be identical, that is;

Equating the exponents

For M : 1=c+ d (i)

For L : 1=a+b−3 c−d ( ii )

For T : −2=−b−d ( iii )

There are four unknowns but 3 given equations. It is not possible to find values of a , b , c∧d
under the given condition. However, three of the unknown variables can be expressed in terms of
the 4th, considered to be the most important. The role of viscosity in this problem is a vital one,
hence, a , b , c are expressed in terms of d.

Thus from equation (i), 1=c+ d , therefore , c=1−d

From equation( iii ), −2=−b−d , therefore , b=2−d

An expression for ' a ' can be found by substituting values of b∧c obtained into equation (ii)

1=a+ ( 2−d ) −3 ( 1−d )−d

1=a+2−d−3+3 d−d

a−1+d=1

a=2−d

Substituting values of a , b , c obtained into equation ¿will give:

F=C ( D2−d V 2−d ρ1−d μd )

2 2 μ d
F=C [ ρ v D ( )]
ρvD

F=ρ v 2 D 2 Φ ¿ )
Example 2

A partially submerged body is towed in water. The resistance R to its motion suspends on the
density ρ ,the viscosity μ of the body, the velocity V of the body and the acceleration due to
gravity, g .show that the resistance to motion can be expressed in the form.

μ lg
2 2
R= ρ L v Φ ( [ ) ,( 2 )
ρLv v ]
The mathematical relationship is defined mathematically as

R= Φ ( ρ , μ , L , v , g)

R= Resistance, ML T −2

ρ=¿Density, ML−3

μ=Dynamic Viscosity ML−1 T −1

L= Length, L

v= Velocity, LT −1

g= Acceleration due to gravity, LT −2

Introducing exponents to the variables, gives:

R=C(ρ a μb Lc v d ge ) (i)

Using the M-L-T system, the corresponding equations for dimensions is:
a b d
ML T −2=C [ ( M L−3 ) ∗( M L−1 T −1 ) ∗Lc∗( LT −1 ) ∗( L T −2)e ]

Equating the exponents

For M : 1=a+b ( ii )

For L : 1=−3 a−b+ c +d +e ( iii )

 For T : −2=−b−d−2 e ( iv )

Since there are 5 variables and 3 equations, evaluating of ρ , L, v (i. e . a , c , d)in terms
of μ∧g ( withexponents b∧e ) will give the desired result;

Thus from equation (i), 1=a+b , therefore , a=1−b

From equation( iii ), −2=−b−d−2 e , therefore , d=2−b−2 e

Then to find an expression for' c ', from equation (iii);

1=−3(1−b)−b+c +(2−b−2 e)+e

c=1+3−3 b+b−2+b+ 2e−e=2−b+ e

Now substituting the known parameters into equation (i), get:

R=C ( ρ1−b μ b L2−b +e v 2−b−2 e g e )

2 2 μ b Lg e
R=C[ ρ L v ( ) ( 2)]
ρvL v

μ Lg
R=ρ L2 v 2 Φ ¿( ρνL ¿ ,( 2 )
v

(ii) Buckingham’s Pi’s Theorem

If there are large number of variables involved, the Indicial or Raleigh’s method can become
lengthy and difficult to solve. It was therefore necessary to develop a more generalized
methodology which can lead directly to a set of dimensionless groups whose numbers can be
determined in advance by a scrutinizing the matrix formed from the variables considered to be
relevant to the investigation. Such technique is known as the Buckingham π−method, also
known as the group method.

This theorem states that if there are n variables in a problem and these variables contain
in primary dimension (For example M, L, T), the equation relating all the variables will have (n-
m) dimensionless groups.
Assume, a physical phenomenon is a described by a number of independent variables like
x 1 , x2 , x 3 , ........... , x m.The phenomenon may be expressed analytically by an implicit functional
relationship of the controlling variables:

F=( x 1 , x 2 , x3 , ........ x m )=0

The analytical version of the phenomenon can be reduced to

F=( π 1 , π 2, π 3 .......... π m −n )=0

The dimensional group are called πgroups.

The steps to solving with this method is as follows:

The steps involved in this method include:

1. List the variables considered to be significant and form a matrix with their dimensions (optional)
2. Then determine the number of dimensionless groups into which these variables can be
combined. This is obtained from the expression
n−m , where ' n ' is thenumber of variables∧' m' is thenumber of dimensions
3. Combine the repeating variables with the remaining independent variables to form the required
number of group. Repeating variables are guided by the following rules:
 The repeating variables must include all the dimensions taken to describe the system or
phenomenon. Usually the fundamental dimensions are M , L∧T . If only 2 of the dimensions are
involved, however, there will be 2 repeating variables and they must contain together the 2
dimensions involved.
 The repeating variables must not form the non-dimensional parameter
 Dependent variable should not be selected as repeating variables.
 No 2 repeating variables should have the same dimensions
 The repeating variables should be chosen with some regard for the practicality of any
experimental investigation, they should be easily measured or set by the investigator. Also,
where the result of the dimensional analysis are to be the basis for a later design methodology,
the repeating variable should be of prime interest to the designer. For example, it is more
sensible to define pipe type in terms of pipe diameter than surface roughness as a repeating
parameter, and density is perhaps better than viscosity as a descriptor of fluid type.
 The repeating variables should be chosen in such a way that one variable contains
I. Geometric property (e.g. length, l; diameter, d; height, H etc.),
II. Flow property (e.g. velocity, V; acceleration, a etc.)
III. The third variable contains fluid property (e.g. mass density, ρ; weight density, w
dynamic viscosity, µ etc.).
 4. A variable that is considered to be of minor significance will as a result of the points considered
above only appear in one group. The influence of this group will be negligible if this variable is
truly inconsequential.
5. There are no right or wrong answers in dimensional analysis. Some answers are only more
useful than others.

It should be noted that for the Buckingham Pi-method, the following rules applies for equation of
the form:

∅ ( π 1 , π 2 , π 3 … … .., π n−1 )=0

I. Any number of dimensionless groups may be combined by multiplication or division to form a

' π1
new valid group. Thus π 1∧π 2 may be combined to form π 1= , and the defining equation
π2
becomes.
II. The reciprocal of any dimensional group remains valid. An example of this will be met in the
later treatment of fans and pumps where a reciprocal form of the Reynolds number will be
recognizable.
III. Any dimensional group may be raised to any power and remain valid.
IV. Any dimensional group may be multiplied by a constant and remain valid. This is useful in
relating a particular group to an easily measured quantity, e.g. pressure coefficients.

Example 1

The drag force on a smooth sphere of diameter D moving with a uniform velocity v in a fluid
density ρ and a dynamic viscosity μ find the expression for F in terms of these quantities.

Solution

Here, the drag force,F is a function of:

Diameter , D, velocity ,V , density , ρ ,∧dynamic viscosity , μ

Mathematically, this is written as: F=f ( D , V , ρ , μ )

This can also be written as f (F , D ,V , ρ , μ)=0

Thus,total number of variable ,=n=5.

To get the number of dimensions in this phenomenon, we have to write out the dimensions for
each variable, i.e.
 F=¿MLT −2, D=L , V =L T −1 , ρ=M L−3 ,∧μ=M L−1 T −1

The fundamental dimension; m=3 (M, L, and T)

The number of pi- groups, π=n−m = 5 - 3

π=2

The equation can thus be written as f ( π 1 , π 2)

Repeating variables.

Since m = 3, it means 3 variables can be selected as repeating variables

F cannot be selected as a repeating variable because it is a dependent variable. For the remaining
4, we are guided in picking the variables such that one should:

Geometric Property----------- D (L)

Flow Property-------------------V (LT −1)

Fluid Property------------------ ρ(M L−3 )

Each of the π−term , canbe written as follows :

π 1 = ( D a . V b . ρ .c . F )
1 1 1

π 2=( D a . V b . ρ .c . μ )
2 2 2

Using the principle of dimensional homogeneity, the π−terms can be resolved as follows :

For the first one,

π 1 = ( D a . V b . ρ .c . F )
1 1 1

ci
π i=La ¿( LT −1 )b ∗( M L−3 ) ∗M 1 L1 T −2
i i

ci
π i=M 0 L0 T 0=La ¿( LT −1 )b ∗( M L−3 ) ∗M 1 L1 T −2
i i
Equating the exponents of M, L and T, we get,

For M: 0=c 1+ 1=0 ,∧c 1=−1 ………….( i )

For L: 0=a1 +b 1−3 c 1+1=0 , … ……..( ii )

For T: 0=−b1 −2=0 ,∧b1=−2

substitute known values into equation (ii ) , get

a 1+ (−2 )−3 (−1 )=−1 , where a1=−2

Therefore,

F
π 1=( D −2 . V −2 . ρ .−1 . F ) =
ρ D2 V 2

Similarly,

π 2 = ( D a . V b . ρ .c . μ )
2 2 2

c2
π 2=La ¿(L T −1 )b ∗( M L−3 ) ∗M 1 L−1 T −1
2 2

c2
π 2=M 0 L0 T 0=La ¿( L T −1 )b ∗( M L−3 ) ∗M 1 L−1 T −1
2 2

For M: 0=c 2+ 1=0 ,∧c 2=−1 ………….( iv )

For L: 0=a2 +b 2−3 c 2−1=0 , … ……..( v )

For T: 0=−b2 −1=0 ,∧b1=−1……….( vi )

substitute known values into equation ( v ) , get

a 2+ (−1 ) −3 (−1 )=1 , where a2=−1Therefore,

μ
π 2=( D −1 . V −1 . ρ .−1 . μ )=
ρDV

f (π 1 , π 2)=0
f
( ρ DF V , ρDVμ )
2 2

μ
F=ρ v 2 D2 Φ ( )
ρV D

Example 2

Prove that the shear stress τ in a fluid flowing through a pipe can be expressed by the equation

μ K
τ =ρ v 2 D2 Φ ( , )
ρDv D

Where;

D= Diameter of pipe

V= Velocity

K =Height of Roughness Projection

ρ= Mass Density

μ= Dynamic Viscosity

f (τ , D , v , ρ , μ , κ)=0

τ =M L−1 T −2 , D = L, ν=LT −1, ρ=M L−3, μ= M L−1 T −1 , K = L

Number of Variables, n=6

Number of |Fundamental Dimensions = 3

Number of Pi groups =6−3=3

Repeating Variables
Geometric Property D

Flow Property v

Fluid Property ρ

π i=ρ a D b v c . B
i i i

π i=( M L−3 )a Lb ( L T −1 )C M x L y T z
i i i

π i=M a L−3 a Lb LC T −C M x L y T z
i i i i i

M 0 L0 T 0=M a + x L−3 a +b +c + y T −C +Z
i i i i i

a i+ x=0

−3 ai +b i+ Ci + y=0

−C i+ Z=0

For 1: τ =M L−1 T −2

x=1 y=−1 Z =−2

a 1+1=0 (i)

−3 ai +b i+ c i−1=0 (ii)

−C i+−2=0 (iii)

From eqn (i)

a 1+1=0 a 1=−1

From eqn (iii)

−C i−2=0 C i=−2

From eqn (ii)

−3 ai +b i+ c i−1=0

−3(−1)+b i−2−1=0

−3+b i−3=0

b i=0

π 1=ρ−1 D0 v −2 τ

τ
π 1=
ρ v2

For 2: μ= M L−1 T −1

x=1 , y=−1∧Z=−1

The equation becomes

a 2+1=0

−3 a2 +b 2+C 2−1=0

−C 2−1=0

From (i)

a 2+1=0 a 2=−1

From (iii)

−C 2−1=0 C 2=−1
From (ii)

−3 a2 +b 2+C 2−1=0

−3(−1)+b 2−1−1=0

−3+b 2−2=0

b 2=−1

π 1=ρ−1 D−1 v −1 μ

μ
π 2=
ρDv

For 3: k=L

x=0 , y=1 ,∧z=0

From (i)

a 3+ 0=0

−3 a3 +b 3+ C3 +1=0

−C 3+ 0=0

From (i) a 3=0

From (iii) −C 3=0

From (ii)

−3 a3 +b 3+ C3 +1=0

−3(1)+b3 +0+ 1=0

 b 3=−1

π 3=D −1 k

π 3= K / D

f (π 1 , π 2, π 3 )

K
f ¿, )=0
D

τ
2
=Φ ¿, K )
ρv D

K
τ =ρ v 2 Φ ¿, )
D

Similitude

In fluid mechanics and hydraulic engineering, a model is sometimes produced to easily describe
the behavior of a too big or a too small prototype. In building such model, there must be three
similarities:

1. Geometric Similarity
2. Kinematic Similarity
3. Dynamic Similarity

Geometric Similarity

For the model and the prototype to be geometrically similar, the ratio of the lengths of all sides
must be the same. Also, the included angles must be the same.
If Lm, Dm, and Hm are Length, Diameter, and Height respectively of a model. If Lp, Dp, and Hp are
Length, Diameter, and Height respectively of its prototype. Then;

For a geometric similarity,

Lm D m H m
, , =Lr
Lp Dp H p

Kinematic Similarity

Kinematics is the motion of a body without considering the force causing the motion. In
Kinematic Similarity, the kinematic parameters (velocity and acceleration) ratios must be
constant.

If Vm and am are velocity and acceleration respectively of a model and V p and ap are velocity and
acceleration respectively of its prototype. Then;

V m V r∧a m
= =ar
Vp ap

Dynamic Similarity

Dynamics is the motion of a body with considering to the force causing the motion. This is much
associated with forces, and these forces which act on corresponding masses in the model and
prototype shall be in the same ratio throughout the area of flow modeled.

Consider a hypothetical flow situation where the pressure change DP between two points is
dependent on mean velocity V, length l, densityρ, viscosity μ, bulk modulus K, surface tension σ
, and gravitational acceleration g.

∆ P=M L−1 T −2

V =L T −1

l=L

ρ=M L−3
 μ= M L−1 T −1

K= M L−1 T −2

σ =M L−1 T −2

g= LT −2

Choosing the repeating variables

1. Geometric property – l (L)

2. Fluid property - ρ(M L−3 )
3. Flow property - V (L T −1)

Number of variables = 8

Number of pi = 8 – 3 = 5

π i=ρ a V b l c B
i i i

π i=( M L−3 )a (LT −1 )b Lc M X LY T Z

i i i

M 0 L0 T 0=M a + x L−3 a +b +c +Y T −b + z
i i i i i

a i+ x=0

−3 ai +b i+ c i+ y=0

−b i+ z =0

This is termed as equation *

For 1: B = ∆ P( M L−1 T −2 ) x = 1, y = -1, and z = -2

Substitute into *

a 1+1=0 1
−3 a1 +b 1+ c1 −1=0 2

−b 1−2=0 3

From 1

a 1+1=0

a 1=−1

From 3

−b 1−2=0

b 1=−2

From 2

−3 a1 +b 1+ c1 −1=0

−3 ×−1−2+ c1 −1=0

c 1=0

∆P
π 1=ρ−1 V −2 ∆ P=
ρV2

Since ∆ P , K ,∧σ have the same dimensions, their relationships will be the same

K σ
π 2= 2
∧π 3 =
ρV ρV2

For 4: B = μ( M L−1 T −1 ) x = 1, y = -1, and z = -1

Substitute into *

a 4 +1=0 1

−3 a 4 +b 4+ c 4−1=0 2
−b 4−1=0 3

From 1

a 4 +1=0

a 4=−1

From 3

−b 4−1=0

b 4=−1

From 2

−3 a 4 +b 4+ c 4−1=0

−3 ×−1−1+ c 4−1=0

c 4 =−1

μ
π 4 =ρ−1 V −1 l −1 μ=
ρVl

For 5: B = g( L T −2) x = 0, y = 1, and z = -2

Substitute into *

a 5=0 1

−3 a5 +b 5+ c 5+ 1=0 2

−b 5−2=0 3

From 1

a 5=0

From 3
−b 5−2=0

b 5=−2

From 2

−3 a5 +b 5+ c 5+ 1=0

−3 ×0−2+c 5 +1=0

c 1=1

lg
π 4 =V −2 lg =
V2

∆P μ σ K lg
ρV 2
=f
ρVl(, , ,
ρ V ρV 2 V 2
2 )
∆P ρVl V ρV 2 V
1
2
ρV 2
=f
( μ
, , ,
K σ √ lg
√
ρ
)
C P =f ( ℜ , Ma ,We , Fr)

Reynold’s Number

Reynold’s number is defined as the ratio of the inertia force to the viscous force.

Inertia Force
ℜ=
Viscous Force

Inertia force = Mass × acceleration = ρA V 2

V
Viscous force = Shear stress × Area = μ ×A
L

ρVl
ℜ=
μ
Froude’s Number (Fr)

Froude’s number is defined as the square root of the ratio of the inertia force to the gravity force.

Fi
Mathematically, Fr=
√ Fg

Inertia force = Mass × acceleration = ρA V 2

Gravity force = Mass × accelerationdue ¿ gravity =ρALg

ρAV 2
Fr=
√ ρALg

V
Fr=
√ Lg
Euler’s Number (Eu)

Euler’s number is defined as the square root of the ratio of the inertia force to the pressure force.

Fi
Mathematically, Eu=
√ Fp

Inertia force = Mass × acceleration = ρA V 2

Gravity force = Intensity of pressure ×area ¿ p × A

ρA V 2
Eu=
√
p× A

V
Eu=
P
√ ρ

Weber’s Number (We)

Weber’s number is defined as the square root of the ratio of the inertia force to the pressure
force.
 Fi
Mathematically, We=
√ Fs

Inertia force = Mass × acceleration = ρA V 2

Gravity force = SurfaceTension Force=Surface tension ×length=σ × L

ρA V 2
We=
√ σL

V
We=
σ
√ ρL

Mach Number (Ma)

Mach number is defined as the square root of the ratio of the inertia force to the elastic force.

Fi
Mathematically, Ma=
√ Fs

Inertia force = Mass × acceleration = ρA V 2

Elastic force = Elastic stress × Area=K × A

ρA V 2
We=
√ KA

V
We=
K
√ ρ

K
√ ρ
=C=Velocity of sound ∈the field

V
Ma=
C

Examples
 1
1. A geometrically similar model of an air duct is built to scale and tested with water
25
which is 50 times more viscous and 800 times denser than air. When tested under
dynamically similar conditions, the pressure drop is 2 bar in the model. Find the
corresponding pressure drop in the full scale prototype.

Given: Scale ratio,

Lm 1 μ p 1 ρ 1
= , = , and p = and the Pressure drop in the model = 2 bar
LP 25 μ m 50 ρm 800

ρVL
From Reynold: =ℜ
μ

ρ p V p L p ρ m V m Lm
=
μp μm

V p μ p ρ m Lm
= × ×
V m μm ρp Lp

Substituting the given values:

Vp 1 1 16
= ×800 × =
V m 50 50 25

V
From Euler: Eu = P
√ ρ

Vm Vp
=
Pm Pp
√ √
ρm ρp

P p ρ p V 2p
= ×
P m ρm V 2m

Pp 1 16 2
= ×
P m 800 25 ( )
=5.12 ×10
−4
P p=P m × 5.12×10−4

P p=2×5.12 ×10−4=1.024 ×10−3 ¯¿

2. A ship whose hull length is 140m is to travel at 7.6m/s (a) Compute the Froude number
Fr. (b) For dynamic similarity, at what velocity should a 1:30 model be towed through
water?
a. Given: Velocity, V = 7.6m/s; Length, L = 140m

V 7.6
Fr= = =0.205
√ gL √ 9.81× 140
b. When two flow patterns with geometric similar boundaries are influence by inertia and
gravity forces, the Froude number is the significant ratio in model studies. Then:

Froude number of model=Froude number of prototype

Vm Vp
=
√ g m Lm √ g p L p

gm =g p=9.81 m/s 2

7.6 Vp
=
√ 140 √ 140/30
V p=1.39 m/ s