Dimensional Analysis

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``` Page 1

Introduction
Dimensional analysis is a mathematical technique in which
study of dimensions is made use of in solving engineer-
ing problems. A physical phenomenon consists of diff erent
quantities which can be expressed in terms of fundamen-
tal or primary quantities. The fundamental quantities mass,
length, time and temperature are designated by M, L, T and
q respectively.
Quantities or variables such as area, velocity, accelera-
tion, viscosity, force, torque, etc., are formed using funda-
mental quantities. These are called derived quantities. The
derived quantities can be expressed in terms of fundamen-
tal quantities. These expressions are called dimensions of
the derived quantities. In general, the quantities can be
classifi ed as fundamental quantities. geometric quantities,
kinematic quantities, dynamic quantities, thermodynamic
quantities, etc.
Sometimes force (dimension F) is taken as a fundamen-
tal quantity instead of mass (dimension M). Accordingly the
system becomes FLT instead of MLT.
Dimension of mass in MLT system is M and in FLT sys-
tem is FL
–1
T
2
.
Dimension of force in MLT system is MLT
–2
and in FLT
system is F.
Some important quantities with their units, symbol and
dimensions in MLT system are given below.
Quantity Unit Symbol
Dimensions
(MLT System)
1. Fundamental
quantities:
Mass
Length
Time
Temperature
kg
m (metre)
s (second)
K
m
l
t
T
M
L
T
q 2. Geometric
quantities:
Area
Volume
m
2
m
3
A, a
V
L
2
L
3
3. Kinematic
quantities:
Velocity
Angular velocity
Acceleration
Angular
acceleration
Discharge
Kinematic
viscosity
m/s
m/s
2
2
m
3
/s
m
2
/s
V, v, u
?
f, a
a
Q
n
LT
–1
T
–1
LT
–2
T
–2
L
3
T
–1
L
2
T
–1
4. Dynamic
quantities:
Force
Density
Speci? c weight
N
kg/m
3
N/m
3
F
r w
MLT
–2
ML
–3
ML
–2
T
–2

Dimensional Analysis
Part III_Unit 8_Chapter 07.indd   728 5/20/2017   6:47:54 PM
Page 2

Introduction
Dimensional analysis is a mathematical technique in which
study of dimensions is made use of in solving engineer-
ing problems. A physical phenomenon consists of diff erent
quantities which can be expressed in terms of fundamen-
tal or primary quantities. The fundamental quantities mass,
length, time and temperature are designated by M, L, T and
q respectively.
Quantities or variables such as area, velocity, accelera-
tion, viscosity, force, torque, etc., are formed using funda-
mental quantities. These are called derived quantities. The
derived quantities can be expressed in terms of fundamen-
tal quantities. These expressions are called dimensions of
the derived quantities. In general, the quantities can be
classifi ed as fundamental quantities. geometric quantities,
kinematic quantities, dynamic quantities, thermodynamic
quantities, etc.
Sometimes force (dimension F) is taken as a fundamen-
tal quantity instead of mass (dimension M). Accordingly the
system becomes FLT instead of MLT.
Dimension of mass in MLT system is M and in FLT sys-
tem is FL
–1
T
2
.
Dimension of force in MLT system is MLT
–2
and in FLT
system is F.
Some important quantities with their units, symbol and
dimensions in MLT system are given below.
Quantity Unit Symbol
Dimensions
(MLT System)
1. Fundamental
quantities:
Mass
Length
Time
Temperature
kg
m (metre)
s (second)
K
m
l
t
T
M
L
T
q 2. Geometric
quantities:
Area
Volume
m
2
m
3
A, a
V
L
2
L
3
3. Kinematic
quantities:
Velocity
Angular velocity
Acceleration
Angular
acceleration
Discharge
Kinematic
viscosity
m/s
m/s
2
2
m
3
/s
m
2
/s
V, v, u
?
f, a
a
Q
n
LT
–1
T
–1
LT
–2
T
–2
L
3
T
–1
L
2
T
–1
4. Dynamic
quantities:
Force
Density
Speci? c weight
N
kg/m
3
N/m
3
F
r w
MLT
–2
ML
–3
ML
–2
T
–2

Dimensional Analysis
Part III_Unit 8_Chapter 07.indd   728 5/20/2017   6:47:54 PM

Quantity Unit Symbol
Dimensions
(MLT System)
Dynamic
viscosity
Pressure
Torque
Ns/m
2
N/m
2
N-m
m p
T
ML
–1
T
–1
ML
–1
T
–2
ML
2
T
–2
5. Thermodynamic
properties:
Thermal
conductivity
Enthalpy/unit
mass
Entropy
Internal energy
W/mK
J/kg
J/K
J
k
h
f , s
E, u
MLT
–3
q –1
L
2
T
–2
ML
2
T
2
q –1
L
2
T
–2
Dimensional Homogeneity
Concept of dimensional homogeneity is applicable to phys-
ical equations. An equation is formed using two or more
quantities. An equation is said to be dimensionally homog-
enous when left hand side of the equation is dimensionally
equal to the right hand side of the equation. When each term
in the equation is reduced to their dimensions, fundamental
dimensions in each side of the equation will have identical
powers. According to Fourier’s principle of homogeneity, a
correct equation expressing a physical relationship between
quantities must be dimensionally homogenous and numeri-
cally equivalent.
For example, consider the equation vgH = 2
Dimensions of left hand side = LT
–1
Dimensions of right hand side LT L
-2
= LT
–1
Therefore the equation is dimensionally homogenous.
If the value of g = 9.81 m/s
2
is substituted, the equation
becomes,
V = 29 81 × .H = 4.429 H
Here, RHS =
L
1
2
Therefore the equation is not dimensionally homogenous
and the equation cannot be applied for other unit systems
such as FPS or CGS.
Methods of Dimensional
Analysis
Two important methods used for dimensional analysis are:
1. Rayleighs method
2. Buckinghams p -theorem method
Rayleighs Method
Rayleighs method is suitable when the number of independ-
ent variables in the phenomenon is only 3 or 4.
The interrelated variables or quantities having different
dimensions are expressed in the form of an exponential
equation which must be dimensionally homogenous.
Let  x = f (x
1
, x
2
, x
3
, …, x
n
)
Then an exponential equation can be formed as follows:
x = C(x
1
a
, x
2
b
, x
2
c
…)
Where, C is a non-dimensional factor. Values of a, b, c, …
are obtained by comparing left hand side to right hand side
of the equation after writing dimensions of the variables.
SOLVED EXAMPLE
Example 1
A fluid of density r and viscosity m flows through a pipe of
diameter d. Derive an expression in terms of Reynolds num-
ber for resistance per unit area of surface using Rayleigh’s
method of dimensional analysis.
Solution
Let F be the resistance per unit area:
F is a function of the independent variables. r , m , v and
d or
F = f (r , v, d, m )
This can be written as
F = C r a
v
b
· d
c
m d
Writing in dimensional from
ML
–1
T
–2
= [ML
–3
]
a
[LT
–1
]
b
L
c
[ML
–1
T
–1
]
d
Comparing the powers
For M,
1 = a + d,  (1)
For L,
–1 = –3a + b + c – d (2)
For T,
–2 = –b – d (3)
There are four unknowns and only three equations. Values
cannot be found out. But 3 unknowns can be expressed in
terms of the other.
Writing in terms of ‘a’,
d = 1 – a
b = –d + 2
= –1 + a + 2
= 1 + a
c = –3a – b + d
= –1 + 3a – 1 + 1 – a
= a – 1
Part III_Unit 8_Chapter 07.indd   729 5/20/2017   6:47:54 PM
Page 3

Introduction
Dimensional analysis is a mathematical technique in which
study of dimensions is made use of in solving engineer-
ing problems. A physical phenomenon consists of diff erent
quantities which can be expressed in terms of fundamen-
tal or primary quantities. The fundamental quantities mass,
length, time and temperature are designated by M, L, T and
q respectively.
Quantities or variables such as area, velocity, accelera-
tion, viscosity, force, torque, etc., are formed using funda-
mental quantities. These are called derived quantities. The
derived quantities can be expressed in terms of fundamen-
tal quantities. These expressions are called dimensions of
the derived quantities. In general, the quantities can be
classifi ed as fundamental quantities. geometric quantities,
kinematic quantities, dynamic quantities, thermodynamic
quantities, etc.
Sometimes force (dimension F) is taken as a fundamen-
tal quantity instead of mass (dimension M). Accordingly the
system becomes FLT instead of MLT.
Dimension of mass in MLT system is M and in FLT sys-
tem is FL
–1
T
2
.
Dimension of force in MLT system is MLT
–2
and in FLT
system is F.
Some important quantities with their units, symbol and
dimensions in MLT system are given below.
Quantity Unit Symbol
Dimensions
(MLT System)
1. Fundamental
quantities:
Mass
Length
Time
Temperature
kg
m (metre)
s (second)
K
m
l
t
T
M
L
T
q 2. Geometric
quantities:
Area
Volume
m
2
m
3
A, a
V
L
2
L
3
3. Kinematic
quantities:
Velocity
Angular velocity
Acceleration
Angular
acceleration
Discharge
Kinematic
viscosity
m/s
m/s
2
2
m
3
/s
m
2
/s
V, v, u
?
f, a
a
Q
n
LT
–1
T
–1
LT
–2
T
–2
L
3
T
–1
L
2
T
–1
4. Dynamic
quantities:
Force
Density
Speci? c weight
N
kg/m
3
N/m
3
F
r w
MLT
–2
ML
–3
ML
–2
T
–2

Dimensional Analysis
Part III_Unit 8_Chapter 07.indd   728 5/20/2017   6:47:54 PM

Quantity Unit Symbol
Dimensions
(MLT System)
Dynamic
viscosity
Pressure
Torque
Ns/m
2
N/m
2
N-m
m p
T
ML
–1
T
–1
ML
–1
T
–2
ML
2
T
–2
5. Thermodynamic
properties:
Thermal
conductivity
Enthalpy/unit
mass
Entropy
Internal energy
W/mK
J/kg
J/K
J
k
h
f , s
E, u
MLT
–3
q –1
L
2
T
–2
ML
2
T
2
q –1
L
2
T
–2
Dimensional Homogeneity
Concept of dimensional homogeneity is applicable to phys-
ical equations. An equation is formed using two or more
quantities. An equation is said to be dimensionally homog-
enous when left hand side of the equation is dimensionally
equal to the right hand side of the equation. When each term
in the equation is reduced to their dimensions, fundamental
dimensions in each side of the equation will have identical
powers. According to Fourier’s principle of homogeneity, a
correct equation expressing a physical relationship between
quantities must be dimensionally homogenous and numeri-
cally equivalent.
For example, consider the equation vgH = 2
Dimensions of left hand side = LT
–1
Dimensions of right hand side LT L
-2
= LT
–1
Therefore the equation is dimensionally homogenous.
If the value of g = 9.81 m/s
2
is substituted, the equation
becomes,
V = 29 81 × .H = 4.429 H
Here, RHS =
L
1
2
Therefore the equation is not dimensionally homogenous
and the equation cannot be applied for other unit systems
such as FPS or CGS.
Methods of Dimensional
Analysis
Two important methods used for dimensional analysis are:
1. Rayleighs method
2. Buckinghams p -theorem method
Rayleighs Method
Rayleighs method is suitable when the number of independ-
ent variables in the phenomenon is only 3 or 4.
The interrelated variables or quantities having different
dimensions are expressed in the form of an exponential
equation which must be dimensionally homogenous.
Let  x = f (x
1
, x
2
, x
3
, …, x
n
)
Then an exponential equation can be formed as follows:
x = C(x
1
a
, x
2
b
, x
2
c
…)
Where, C is a non-dimensional factor. Values of a, b, c, …
are obtained by comparing left hand side to right hand side
of the equation after writing dimensions of the variables.
SOLVED EXAMPLE
Example 1
A fluid of density r and viscosity m flows through a pipe of
diameter d. Derive an expression in terms of Reynolds num-
ber for resistance per unit area of surface using Rayleigh’s
method of dimensional analysis.
Solution
Let F be the resistance per unit area:
F is a function of the independent variables. r , m , v and
d or
F = f (r , v, d, m )
This can be written as
F = C r a
v
b
· d
c
m d
Writing in dimensional from
ML
–1
T
–2
= [ML
–3
]
a
[LT
–1
]
b
L
c
[ML
–1
T
–1
]
d
Comparing the powers
For M,
1 = a + d,  (1)
For L,
–1 = –3a + b + c – d (2)
For T,
–2 = –b – d (3)
There are four unknowns and only three equations. Values
cannot be found out. But 3 unknowns can be expressed in
terms of the other.
Writing in terms of ‘a’,
d = 1 – a
b = –d + 2
= –1 + a + 2
= 1 + a
c = –3a – b + d
= –1 + 3a – 1 + 1 – a
= a – 1
Part III_Unit 8_Chapter 07.indd   729 5/20/2017   6:47:54 PM

? F = C r a
v
1+a
d
a–1
m 1–a
= C
vd v
d
v
v
a
?
µ
µ?
?
?
?
?
?
?
?
= C
vd
vd
v
a
?
µ
µ
?
?
?
?
?
?
?
?
?
?
?
?
?
?
2
= C
vd
v
a
?
µ
?
?
?
?
?
?
?
-1
2
= CR v
e
a
()
-12
?
= Cv R
e
?f
2
()
Buckingham’s p -theorem Method
From a set of variables, dimensionless groups are formed
using this method. According to this method, if there are
n variables (dependent and independent) influencing a
phenomenon, which can be fully expressed in terms of m
fundamental units, then these n variables can be grouped as
(n – m) dimensionless terms called p terms.
Method of Forming Dimensionless
Constants
1. Number of fundamental units involved (m) is found
out and thus number of dimensionless group (n – m)
is found out.
2. Repeating variables are selected. Number of
repeating variables is equal to number of fundamental
units involved, i.e., m number. The repeating
variables should together contain all the fundamental
dimensions. One geometric characteristic (like l,
d) one fluid characteristic (like r , m ) and one flow
characteristic (like v) are selected as repeating
variables. l or d, v and r would be the best choice in
most cases. As far as possible dependent variable is
not selected as a repeating variable.
3. p terms are formed using the repeating variables
and one of the non-repeating variables. Repeating
variables are raised to indexes.
For example,
Let x
1
be the dependent variable and x
2
, x
3
, x
4
, …, x
n
be
the independent variables.
Then,
x
1
= f (x
2
, x
3
, x
4
, …, x
n
)
This can be written as
f (x
1
, x
2
, x
3
, …, x
n
) = 0
If there are m fundamental units involved,
Number of p terms = (n – m)
If m = 3,
Let x
2
, x
3
and x
4
be the repeating variables selected.
Then,
p
1 23 4 1
11 1
= xx xx
ab c
.
p
2 23 4 5
22 3
= xx xx
ab c
.
p
3 23 4 6
33 3
= xx xx
ab c
.
p
nm
ab c
n
xx xx
nm nm nm
-
=
-- -
23 4
.
The p terms are written in the dimensional form and val-
ues of powers are found out as in the case of Rayleigh’s
method. Thus the dimensionless constants are identified.
Using the p terms relation between variables are obtained as
follows. For example,
p 1
= f(p 2
, p 3
, …)
or
f
1
(p 1
, p 2
, p 3
, …) = 0
Example 2
Using Buckinghams p -theorem show that the velocity
through a circular orifice is given by:
V = 2gH
D
HvH
f
µ
?
,
?
?
?
?
?
?
Where
D = Diameters of orifice
m = Coefficient of viscosity
r = Mass density
g = Acceleration due to gravity
Solution
V is a function of H, D, m , r and g
or V = f (H, D, m , r , g)
or f
1
(V, H, D, m , r , g) = 0
Total number of variables, n = 6
Writing the variables in dimensions
V = LT
–1
H = L
D = L
m = ML
–1
T
–1
r = ML
–3
g = LT
–2
From the above, number of fundamental dimensions
involved,
m = 3
Part III_Unit 8_Chapter 07.indd   730 5/20/2017   6:47:55 PM
Page 4

Introduction
Dimensional analysis is a mathematical technique in which
study of dimensions is made use of in solving engineer-
ing problems. A physical phenomenon consists of diff erent
quantities which can be expressed in terms of fundamen-
tal or primary quantities. The fundamental quantities mass,
length, time and temperature are designated by M, L, T and
q respectively.
Quantities or variables such as area, velocity, accelera-
tion, viscosity, force, torque, etc., are formed using funda-
mental quantities. These are called derived quantities. The
derived quantities can be expressed in terms of fundamen-
tal quantities. These expressions are called dimensions of
the derived quantities. In general, the quantities can be
classifi ed as fundamental quantities. geometric quantities,
kinematic quantities, dynamic quantities, thermodynamic
quantities, etc.
Sometimes force (dimension F) is taken as a fundamen-
tal quantity instead of mass (dimension M). Accordingly the
system becomes FLT instead of MLT.
Dimension of mass in MLT system is M and in FLT sys-
tem is FL
–1
T
2
.
Dimension of force in MLT system is MLT
–2
and in FLT
system is F.
Some important quantities with their units, symbol and
dimensions in MLT system are given below.
Quantity Unit Symbol
Dimensions
(MLT System)
1. Fundamental
quantities:
Mass
Length
Time
Temperature
kg
m (metre)
s (second)
K
m
l
t
T
M
L
T
q 2. Geometric
quantities:
Area
Volume
m
2
m
3
A, a
V
L
2
L
3
3. Kinematic
quantities:
Velocity
Angular velocity
Acceleration
Angular
acceleration
Discharge
Kinematic
viscosity
m/s
m/s
2
2
m
3
/s
m
2
/s
V, v, u
?
f, a
a
Q
n
LT
–1
T
–1
LT
–2
T
–2
L
3
T
–1
L
2
T
–1
4. Dynamic
quantities:
Force
Density
Speci? c weight
N
kg/m
3
N/m
3
F
r w
MLT
–2
ML
–3
ML
–2
T
–2

Dimensional Analysis
Part III_Unit 8_Chapter 07.indd   728 5/20/2017   6:47:54 PM

Quantity Unit Symbol
Dimensions
(MLT System)
Dynamic
viscosity
Pressure
Torque
Ns/m
2
N/m
2
N-m
m p
T
ML
–1
T
–1
ML
–1
T
–2
ML
2
T
–2
5. Thermodynamic
properties:
Thermal
conductivity
Enthalpy/unit
mass
Entropy
Internal energy
W/mK
J/kg
J/K
J
k
h
f , s
E, u
MLT
–3
q –1
L
2
T
–2
ML
2
T
2
q –1
L
2
T
–2
Dimensional Homogeneity
Concept of dimensional homogeneity is applicable to phys-
ical equations. An equation is formed using two or more
quantities. An equation is said to be dimensionally homog-
enous when left hand side of the equation is dimensionally
equal to the right hand side of the equation. When each term
in the equation is reduced to their dimensions, fundamental
dimensions in each side of the equation will have identical
powers. According to Fourier’s principle of homogeneity, a
correct equation expressing a physical relationship between
quantities must be dimensionally homogenous and numeri-
cally equivalent.
For example, consider the equation vgH = 2
Dimensions of left hand side = LT
–1
Dimensions of right hand side LT L
-2
= LT
–1
Therefore the equation is dimensionally homogenous.
If the value of g = 9.81 m/s
2
is substituted, the equation
becomes,
V = 29 81 × .H = 4.429 H
Here, RHS =
L
1
2
Therefore the equation is not dimensionally homogenous
and the equation cannot be applied for other unit systems
such as FPS or CGS.
Methods of Dimensional
Analysis
Two important methods used for dimensional analysis are:
1. Rayleighs method
2. Buckinghams p -theorem method
Rayleighs Method
Rayleighs method is suitable when the number of independ-
ent variables in the phenomenon is only 3 or 4.
The interrelated variables or quantities having different
dimensions are expressed in the form of an exponential
equation which must be dimensionally homogenous.
Let  x = f (x
1
, x
2
, x
3
, …, x
n
)
Then an exponential equation can be formed as follows:
x = C(x
1
a
, x
2
b
, x
2
c
…)
Where, C is a non-dimensional factor. Values of a, b, c, …
are obtained by comparing left hand side to right hand side
of the equation after writing dimensions of the variables.
SOLVED EXAMPLE
Example 1
A fluid of density r and viscosity m flows through a pipe of
diameter d. Derive an expression in terms of Reynolds num-
ber for resistance per unit area of surface using Rayleigh’s
method of dimensional analysis.
Solution
Let F be the resistance per unit area:
F is a function of the independent variables. r , m , v and
d or
F = f (r , v, d, m )
This can be written as
F = C r a
v
b
· d
c
m d
Writing in dimensional from
ML
–1
T
–2
= [ML
–3
]
a
[LT
–1
]
b
L
c
[ML
–1
T
–1
]
d
Comparing the powers
For M,
1 = a + d,  (1)
For L,
–1 = –3a + b + c – d (2)
For T,
–2 = –b – d (3)
There are four unknowns and only three equations. Values
cannot be found out. But 3 unknowns can be expressed in
terms of the other.
Writing in terms of ‘a’,
d = 1 – a
b = –d + 2
= –1 + a + 2
= 1 + a
c = –3a – b + d
= –1 + 3a – 1 + 1 – a
= a – 1
Part III_Unit 8_Chapter 07.indd   729 5/20/2017   6:47:54 PM

? F = C r a
v
1+a
d
a–1
m 1–a
= C
vd v
d
v
v
a
?
µ
µ?
?
?
?
?
?
?
?
= C
vd
vd
v
a
?
µ
µ
?
?
?
?
?
?
?
?
?
?
?
?
?
?
2
= C
vd
v
a
?
µ
?
?
?
?
?
?
?
-1
2
= CR v
e
a
()
-12
?
= Cv R
e
?f
2
()
Buckingham’s p -theorem Method
From a set of variables, dimensionless groups are formed
using this method. According to this method, if there are
n variables (dependent and independent) influencing a
phenomenon, which can be fully expressed in terms of m
fundamental units, then these n variables can be grouped as
(n – m) dimensionless terms called p terms.
Method of Forming Dimensionless
Constants
1. Number of fundamental units involved (m) is found
out and thus number of dimensionless group (n – m)
is found out.
2. Repeating variables are selected. Number of
repeating variables is equal to number of fundamental
units involved, i.e., m number. The repeating
variables should together contain all the fundamental
dimensions. One geometric characteristic (like l,
d) one fluid characteristic (like r , m ) and one flow
characteristic (like v) are selected as repeating
variables. l or d, v and r would be the best choice in
most cases. As far as possible dependent variable is
not selected as a repeating variable.
3. p terms are formed using the repeating variables
and one of the non-repeating variables. Repeating
variables are raised to indexes.
For example,
Let x
1
be the dependent variable and x
2
, x
3
, x
4
, …, x
n
be
the independent variables.
Then,
x
1
= f (x
2
, x
3
, x
4
, …, x
n
)
This can be written as
f (x
1
, x
2
, x
3
, …, x
n
) = 0
If there are m fundamental units involved,
Number of p terms = (n – m)
If m = 3,
Let x
2
, x
3
and x
4
be the repeating variables selected.
Then,
p
1 23 4 1
11 1
= xx xx
ab c
.
p
2 23 4 5
22 3
= xx xx
ab c
.
p
3 23 4 6
33 3
= xx xx
ab c
.
p
nm
ab c
n
xx xx
nm nm nm
-
=
-- -
23 4
.
The p terms are written in the dimensional form and val-
ues of powers are found out as in the case of Rayleigh’s
method. Thus the dimensionless constants are identified.
Using the p terms relation between variables are obtained as
follows. For example,
p 1
= f(p 2
, p 3
, …)
or
f
1
(p 1
, p 2
, p 3
, …) = 0
Example 2
Using Buckinghams p -theorem show that the velocity
through a circular orifice is given by:
V = 2gH
D
HvH
f
µ
?
,
?
?
?
?
?
?
Where
D = Diameters of orifice
m = Coefficient of viscosity
r = Mass density
g = Acceleration due to gravity
Solution
V is a function of H, D, m , r and g
or V = f (H, D, m , r , g)
or f
1
(V, H, D, m , r , g) = 0
Total number of variables, n = 6
Writing the variables in dimensions
V = LT
–1
H = L
D = L
m = ML
–1
T
–1
r = ML
–3
g = LT
–2
From the above, number of fundamental dimensions
involved,
m = 3
Part III_Unit 8_Chapter 07.indd   730 5/20/2017   6:47:55 PM

? Number of p terms
= n – m
= 6 – 3 = 3
? The equations in terms of p terms is f
1
(p 1
, p 2
, p 3
) = 0.
Choosing H, g, r as repeating variables, the p terms are
as follows.
p?
1
11 1
= Hg v
ab c
p?
2
22 2
=Hg D
ab c
p? µ
3
33 3
= Hg
ab c
Writing dimensions:
p 1
term:
ML TL LT ML LT
ab c
°° °=
-- -
11 1
23 1
() ()
Equating exponents of M, L and T
M: 0 = C
1
L: 0 = a
1
+ b
1
+ –3c
1
+ 1
T: 0 = –2b
1
–1
From the above,
c
1
= 0
b
1
=
-1
2
a
1
= –b
1
+ 3c
1
– 1
=
1
2
+ 0 –1
=
-1
2
? p?
1
1
2
1
2
=
--
Hg v

=
v
gH
p 2
term:
ML T °° °=
LLTMLL
ab c
22 2
23
() ()
--
M: 0 = c
2
L: 0 = a
2
+ b
2
– 3c
2
+ 1
T: 0 = –2b
2
? c
2
= 0
b
2
= 0
a
2
= –b
2
+ 3c
2
– 1 = –1
?  p 2
= H
–1
g° r ° D
=
D
H
p 3
term:

· °° ° ML T
=
LLTML
ab c
33 3
23
() ()
--
ML
–1
T
–1
M: 0 = c
3
+ 1
L: 0 = a
3
+ b
3
+ –3c
3
– 1
T: 0 = –2b
3
– 1
? c
3
= –1
b
3
=
-1
2
a
3
= –b
3
+ 3c
3
+ 1
=
1
2
31 -+ =
-3
2
?=
-
-
-
p? µ
3
3
2
1
2 1
Hg
=
µ
? gH
3
2
=
µ
?
v
HV gH
=
µ
? HV
p 1
? Equation can be written as,
f
1
V
gH
D
HvH
,,
µ
?
p
1
0
?
?
?
?
?
?
?
?
=
or
V
gH
D
HvH
=
?
?
?
?
?
?
f
µ
?
p ,
1
or VgH
D
HvH
=
?
?
?
?
?
?
2 f
µ
?
,
Because, multiplying or dividing by a constant does not
change the character of p terms.

Part III_Unit 8_Chapter 07.indd   731 5/20/2017   6:47:57 PM
```

## Irrigation Engineering

7 videos|35 docs|31 tests

## Irrigation Engineering

7 videos|35 docs|31 tests

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