FLUID MECHANICS-II MENG 315
3. DIMENSIONAL ANALYSIS AND SIMILARITY
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DIMENSIONAL ANALYSIS AND SIMILARITY
1. Dimensional Analysis and Dimensional Homogeneity
Dimensional analysis
❖is a computational technique which makes use of the study of the
dimensions of quantities to analyze and solve diverse engineering
problems.
❖To do this, we leverage on the fact that; each physical phenomenon can
be modelled (expressed) in terms of relevant quantities that influence
the cause, effect and behaviour of a system by an equation.
❖The relevant quantities can be dimensional and non-dimensional as the
case maybe.
❖In dimensional analysis, we find a systematic arrangement of the
variables to give us a set of dimensional quantities.
❖This technique is widely used in analyzing and modelling fluid systems.
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DIMENSIONAL ANALYSIS AND SIMILARITY
Fig. Fluids in the design and analysis of aircrafts
Fig. Fluids in the design and analysis of marine crafts
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DIMENSIONAL ANALYSIS AND SIMILARITY
Fig. Fluids in hydropower systems-turbomachinery
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DIMENSIONAL ANALYSIS AND SIMILARITY
Dimensional analysis is very useful in the following ways:
❖ Can be useful in testing for the dimensional homogeneity of any equation of
a fluid phenomenon.
❖ To derive functional relationships for a fluid phenomenon.
❖ To derive equations expressed in terms of non-dimensional parameters and
to show the relative significance of each parameter.
❖ For planning model tests and presenting experimental results in an orderly
manner, thus making the analysis of complex phenomena simple.
❖ Dimensional analysis therefore offers a qualitative route to the understanding
of fluid flow mechanisms.
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DIMENSIONAL ANALYSIS AND SIMILARITY
Advantages of dimensional analysis
❖ It expresses the functional relationship between variables in dimensionless
terms.
❖ In hydraulic model studies, it reduces the number of variables in a physical
phenomenon (generally by three).
❖ By careful selection of variables, the dimensionless parameters can be used to
make certain logical deductions about the problem.
❖ Design curves, by the use of dimensional analysis, can be developed from
experimental data or developed from direct solution of the problem.
❖ It enables getting up a theoretical equation in a simplified dimensional form.
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DIMENSIONAL ANALYSIS AND SIMILARITY
❖ It enables setting up a theoretical equation in a simplified dimensional form.
❖ Dimensional analysis provides partial solutions to the problems that are too
complex to be dealt with mathematically.
❖ The conversion of units of quantities from one system to another is
facilitated.
Dimensions (basic and derived)
❖ All physical quantities used in fluid phenomenon can be expressed in terms of
fundamental/basic/primary quantities.
❖ The fundamental quantities we will encounter are; mass, length, time,
temperature.
❖ Other in other phenomena are; luminous intensity, molarity, and current.
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DIMENSIONAL ANALYSIS AND SIMILARITY
Physical quantities
Mass
Length
SI –units (abbreviations)
Kilogram (kg)
Metre (m)
Dimensions
M
L
Time
Temperature
Quantity of a substance
(mole)
Second (s)
Kelvin (K)
Mole (mol)
T
/K
Mol
Current
Luminous Intensity.
Ampere (A)
Candela (cd)
A
cd
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DIMENSIONAL ANALYSIS AND SIMILARITY
❖ Other quantities which are expressed in terms of the fundamental ones are
called derived/ secondary quantities.
❖ The dimensions of derived quantities are expressed in terms of the dimensions
of the fundamental quantities.
❖ Some commonly encountered derived quantities are: area, velocity, volume,
density, pressure etc.
❖ The dimensions of derived physical quantities can be expressed in the MassLength-Time (M-L-T) system,
❖ or the Force-Length-Time (F-L-T) system.
❖ Some engineers prefer the use of force to mass as a fundamental quantity
because the force can be easily measured.
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DIMENSIONAL ANALYSIS AND SIMILARITY
❖ Some quantities (fundamental and derived) used in thermo-fluids phenomena
Quantity and symbol
SI-Unit
Dimensions
M-L-T System
F-L-T System
(A) Fundamental quantities
Length (𝐿, 𝑙)
𝑚
L
L
Mass (𝑀, 𝑚)
𝐾𝑔
M
F
Time (𝑇, 𝑡)
𝑠
T
T
(B) Geometric quantities
Area (𝐴, 𝑎)
𝑚2
L2
L2
Volume (𝑉)
𝑚3
L3
L3
Moment of inertia (𝐼)
𝑚4
L4
L4
LT −1
LT −1
(C) Kinematic quantities
Linear velocity (𝑢, 𝑉, 𝑈)
𝑚𝑠 −1
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DIMENSIONAL ANALYSIS AND SIMILARITY
Angular velocity (), rotational speed N
𝑟𝑎𝑑𝑠 −1
𝑟𝑝𝑚
T −1
T −1
Acceleration (a)
𝑚𝑠 −2
LT −2
LT −2
Angular acceleration ()
𝑟𝑎𝑑𝑠 −2
T −2
T −2
Volume flow rate/discharge (Q)
𝑚3 𝑠 −1
L3 T −1
L3 T −1
Gravitational intensity/
gravity (g)
𝑚𝑠 −2
LT −2
LT −2
Kinematic viscosity ( )
𝑚2 𝑠 −1
L2 T −1
L2 T −1
Stream function (), circulation ()
𝑚2 𝑠 −1
L2 T −1
L2 T −1
Vorticity ()
𝑠 −1
T −1
T −1
MLT −2
𝐹
acceleration due to
(D) Dynamic quantities
Force (F)
𝑘𝑔𝑚𝑠 −2
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DIMENSIONAL ANALYSIS AND SIMILARITY
Density ()
𝑘𝑔𝑚−3
ML−3
FL−4 T 2
Acceleration (a)
𝑚𝑠 −2
LT −2
LT −2
Dynamic viscosity()
𝑘𝑔𝑚−1 𝑠 −1
ML−1 T −1
FL−2
Specific weight (w)
𝑘𝑔𝑚−2 𝑠 −2
ML−2 T −2
FL−3
Pressure (p); shear stress ()
𝑘𝑔𝑚−1 𝑠 −2
ML−1 T −2
FL−2
Modulus of elasticity, (E,K)
𝑘𝑔𝑚−1 𝑠 −2
ML−1 T −2
FL−2
Momentum
𝑘𝑔𝑚𝑠 −1
MLT −1
FT
Angular momentum/moment of momentum
𝑘𝑔𝑚2 𝑠 −1
ML2 T −1
FLT
Work (W); energy, E
𝑘𝑔𝑚2 𝑠 −2
ML2 T −2
FL
Torque (T)
𝑘𝑔𝑚2 𝑠 −2
ML2 T −2
FL
Power (P)
𝑘𝑔𝑚2 𝑠 −3
ML2 T −3
FLT −1
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FLUID MECHANICS 2-MENG 315/FBC/USL/STN
DIMENSIONAL ANALYSIS AND SIMILARITY
(E) Thermodynamic quantities
𝐾


𝑘𝑔𝑚−1 𝑠 −3 𝐾 −1
ML−1 T −3  −1
FT −1  −1
Specific enthalpy
𝑚2 𝑠 −2
L2 T −2
L2 T −2
Gas constant (R)
𝑚2 𝑠 −2 𝐾 −1
L2 T −2  −1
L2 T −2  −1
Entropy (S)
𝑘𝑔𝑚2 𝑠 −2 𝐾 −1
ML2 T −2  −1
FL −1
Specific internal energy
𝑚2 𝑠 −2
L2 T −2
L2 T −2
Rate of Heat transfer
𝑘𝑔𝑚2 𝑠 −3
ML2 T −3
FLT −1
Temperature (a)
Thermal conductivity (k)
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DIMENSIONAL ANALYSIS AND SIMILARITY
Dimensional homogeneity
❖ The correctness of any physical equation used to model the relationship
between any set of quantities
❖ can be assessed by establishing a case of the dimensional homogeneity of every
term in the equation.
❖ The situation is summed up in an axiom referred to as the principle of
dimensional homogeneity;
❖ This principle states that “ every term in an equation when reduced to
fundamental dimensions must contain identical powers of each dimension”.
❖ It must be noted that dimensionally homogeneous equations are applicable to
all systems of unit.
❖ Only quantities having the same dimensions can be added in any equation
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DIMENSIONAL ANALYSIS AND SIMILARITY
Example 1.0. Establish that the equation 𝒗 = 𝒖 + 𝒂𝒕 is a correct
relationship between the kinematic properties of a moving particle
(𝒖=initial velocity, 𝒗 = final velocity, 𝒂 =acceleration, and 𝒕 =time).
Solution
Dimensions of 𝐯 = LT −1 , 𝐮 = LT −1 , 𝐚 = LT −2 andm 𝐭 = T
So from 𝒗 = 𝒖 + 𝒂𝒕 ;
𝐷𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑓 𝑣 = 𝐷𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑓 𝑢 + 𝐷𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑓 𝑎 𝐷𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑓 𝑡
LT −1 = LT −1 + (LT −2 )(T)
LT −1 = LT −1 + LT −1
Since all the terms reduced to fundamental dimensions are the same
in the Equation, it implies that the equation is correct.
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DIMENSIONAL ANALYSIS AND SIMILARITY
Example 2.0. Determine the dimensions of 𝑬 in the dimensionally
homogeneous Einstein’s equation.
1
2
𝐸 = 𝑚𝑐
−1
 2
2−
𝑐
𝒎 =mass, 𝒄 =speed of sound,  =speed of a body
Solution
Dimensions of 𝐦 in terms of basic dimensions = M
Dimensions of 𝐜 and  in terms of basic dimensions = LT −1
Since 𝒄 and  have the same dimensions, it implies that
1

2
− 1 is
2− 𝑐
dimensionless, so;
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DIMENSIONAL ANALYSIS AND SIMILARITY
Dimension of E = Dimension of m dimension of c 2
= M LT −1 2
= ML2 T −2
Hence, 𝑬 has the dimension of energy.
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DIMENSIONAL ANALYSIS AND SIMILARITY
2. Methods for Dimensional analysis
❖ With the aid of dimensional analysis, the equation relating the relevant
quantities
❖ that influence a physical phenomenon can be derived in terms of
dimensionless groups.
❖ This can then result in the reduction of the number of variables.
❖ The methods of dimensional analysis are based on principle of dimensional
homogeneity.
❖ The methods for dimensional analysis developed thus far are as follows:
i.
Rayleigh’s method
ii. Buckingham-
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DIMENSIONAL ANALYSIS AND SIMILARITY
iii. Bridgman method
iv. Matrix-tensor method
v.
Visual inspection of variables method
vi. Rearrangement of differential equation method.
I. Rayleigh’s method of dimensional analysis
❖ This method gives a special form of relationship between the dimensionless
group.
❖ It however has one shortcoming; that is, it does not give the number of
dimensionless groups to be derived before the analysis.
❖ It can be used to determine the expression for variables which depend upon
a maximum of three to four variables only corresponding to the three to
four relevant basic dimensions.
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DIMENSIONAL ANALYSIS AND SIMILARITY
❖ When the number of independent variables exceeds four,
❖ then it becomes difficult to find the relation for the independent variable.
❖ This method is based on the following principles:
❖ Given a variable 𝒀 which is a function of:
𝑿𝟏 , 𝑿𝟐 , 𝑿𝟑 , … 𝑿𝒏 ;
❖ a functional relationship between these variables can be written as:
𝒀 = 𝒇(𝑿𝟏 , 𝑿𝟐 , 𝑿𝟑 , … 𝑿𝒏 )
❖ Where 𝒀 is the dependent variable, while 𝑿𝟏 , 𝑿𝟐 , 𝑿𝟑 , … 𝑿𝒏 are the independent
variables.
❖ We can now write the functional relationship in the form below;
𝒀 = 𝑪 𝑿𝒂𝟏 , 𝑿𝒃𝟐 , 𝑿𝒄𝟑 , … . 𝑿𝒏𝒏
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DIMENSIONAL ANALYSIS AND SIMILARITY
❖ Where 𝑪 is a constant and 𝒂, 𝒃, 𝒄 … are the arbitrary powers.
❖ The values of 𝒂, 𝒃, 𝒄 … 𝑛 are obtained by comparing the powers of the
fundamental dimensions on both sides.
❖ Thus the expression is obtained for the dependent variable.
❖ When the number of unknowns is more than the number of fundamental
dimensions,
❖ we can solve for the unknown powers in terms of power of the quantity
that is most significant.
❖ The significance of a quantity is measured in terms the number
fundamental dimensions that it is made up of.
❖ The higher the number of fundamental dimensions the more significant.
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DIMENSIONAL ANALYSIS AND SIMILARITY
Example 3.0.
Find an expression for the drag force 𝑭 on a smooth sphere of diameter 𝑫,
moving with a uniform velocity 𝒗 in a fluid of density 𝝆 and dynamic viscosity
𝝁.
Solution
Consider the table of variables, their symbols, and dimensions below.
S/N
1.
2.
Variables
Drag force
Diameter
symbols
𝑭
𝑫
Dimensions
MLT −2
L
3.
Velocity
𝒗
LT −1
4.
5.
Density
Dynamic viscosity
𝝆
𝝁
ML−3
ML−1 T −1
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DIMENSIONAL ANALYSIS AND SIMILARITY
❖ The dependent variable is 𝑭, and the independent variables are 𝑫, 𝒗, 𝝆, and
.
❖ So the functional relationship between the variables will be.
𝐹 = 𝐶 𝐷𝑎 . 𝑣 𝑏 . 𝜌𝑐 . 𝜇𝑑
❖ Replacing he variables by their dimensions we have;
𝑀𝐿𝑇 −2 = 𝐶 𝐿 𝑎 . 𝐿𝑇 −1 𝑏 . 𝑀𝐿−3 𝑐 . 𝑀𝐿−1 𝑇 −1 𝑑
𝑀𝐿𝑇 −2 = 𝐶 𝑀𝑐 . 𝑀𝑑 . 𝐿𝑎 . 𝐿𝑏 . 𝐿−3𝑐 . 𝐿−𝑑 . 𝑇 −𝑏 . 𝑇 −𝑑
𝑀𝐿𝑇 −2 = 𝐶 𝑀𝑐+𝑑 . 𝐿𝑎+𝑏−3𝑐−𝑑 . 𝑇 −𝑏−𝑑
❖ Equating powers of the fundamental dimensions on both side we get;
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DIMENSIONAL ANALYSIS AND SIMILARITY
For 𝑀, we have;
1 = 𝑐 + 𝑑…………….... 𝑖
For 𝐿, we have;
1 = 𝑎 + 𝑏 − 3𝑐 − 𝑑 … . . (𝑖𝑖)
For 𝑇, we have;
−2 = −𝑏 − 𝑑 … … … … … . 𝑖𝑖𝑖
❖ We have four unknowns to solve for in three equations.
❖ This is not possible without solving for some unknowns in terms of others.
❖ So, from the table of variables and their dimensions, we can see that, of the
four independent variables, 𝝁 is the most important,
❖ since its dimension consists of all three fundamental dimensions (𝑀𝐿 and 𝑇).
So; Solving for 𝑎, 𝑏, 𝑐 in terms of 𝑑 we have;
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DIMENSIONAL ANALYSIS AND SIMILARITY
❖ From equation (𝑖);
1=𝑐+𝑑 𝒄=𝟏−𝒅
❖ From equation (𝑖𝑖𝑖) − 2 = −𝑏 − 𝑑  𝒃 = 𝟐 − 𝒅
❖ From equation (𝑖𝑖)
1 = 𝑎 + 𝑏 − 3𝑐 − 𝑑
1 = 𝑎 + (2 − 𝑑) − 3(1 − 𝑑) − 𝑑
1 = 𝑎 + 2 − 𝑑 − 3 + 3𝑑 − 𝑑
1=𝑎−1+𝑑
𝒂=𝟐−𝒅
❖ Hence; the functional relation now become;
𝐹 = 𝐶 𝐷𝑎 . 𝑣 𝑏 . 𝜌𝑐 . 𝜇𝑑
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DIMENSIONAL ANALYSIS AND SIMILARITY
𝐹 = 𝐶 𝐷𝟐−𝒅 . 𝑣 𝟐−𝒅 . 𝜌𝟏−𝒅 . 𝜇𝑑
𝐹 = 𝐶 𝐷2 . 𝐷 −𝑑 . 𝑣 2 . 𝑣 −𝑑 . 𝜌. 𝜌−𝑑 . 𝜇𝑑
𝐹 = 𝐶 𝜌𝐷2 𝑣 2 . 𝜌−𝑑 𝐷−𝑑 𝑣 −𝑑 𝜇𝑑
𝑭=
𝝆𝑫𝟐 𝒗𝟐 
𝝁
𝝆𝒗𝑫
Example 4.0. The efficiency of a fan  of a fan depends on the density 𝝆, the
dynamic viscosity 𝝁 of the fluid, the angular velocity 𝝎, diameter 𝑫, of the
rotor and the discharge 𝑸. Express  in terms of dimensionless parameters.
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DIMENSIONAL ANALYSIS AND SIMILARITY
Solution
Consider the table of variables involved in fluid phenomenon, their symbols,
and dimensions below.
S/N
1.
2.
3.
4.
5.
6.
Variables
Efficiency
Diameter
Velocity
Density
Dynamic viscosity
Discharge
symbols

𝑫
𝝎
𝝆
𝝁
𝑸
Dimensions
M 0 L0 T 0
L
T −1
ML−3
ML−1 T −1
L3 T −1
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DIMENSIONAL ANALYSIS AND SIMILARITY
The dependent variable is , and the independent variables are 𝑫, 𝝎, 𝝆, . and
𝑸. So the functional relationship between the variables will be.
 = 𝐶 𝐷 𝑎 . 𝜔𝑏 . 𝜌𝑐 . 𝜇 𝑑 . 𝑄 𝑒
Replacing he variables by their dimensions we have;
𝑀0 𝐿0 𝑇 0 = 𝐶 𝐿 𝑎 . 𝑇 −1 𝑏 . 𝑀𝐿−3 𝑐 . 𝑀𝐿−1 𝑇 −1 𝑑 . 𝐿3 𝑇 −1 𝑒
𝑀0 𝐿0 𝑇 0 = 𝐶 𝑀𝑐 . 𝑀𝑑 . 𝐿𝑎 . 𝐿−3𝑐 . 𝐿−𝑑 . 𝐿3𝑒 . 𝑇 −𝑏 . 𝑇 −𝑑 . 𝑇 −𝑒
𝑀0 𝐿0 𝑇 0 = 𝐶 𝑀𝑐+𝑑 . 𝐿𝑎−3𝑐−𝑑+3𝑒 . 𝑇 −𝑏−𝑑−𝑒
Equating powers of the fundamental dimensions on both side we get;
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DIMENSIONAL ANALYSIS AND SIMILARITY
For 𝑀, we have;
0 = 𝑐 + 𝑑……….…..………… 𝑖
For 𝐿, we have;
0 = 𝑎 − 3𝑐 − 𝑑 + 3𝑒 … … … . . (𝑖𝑖)
For 𝑇, we have;
0 = −𝑏 − 𝑑 − 𝑒 … … … … … . (𝑖𝑖𝑖)
We have five unknowns to solve for in three equations. This is not possible
without solving for some unknowns in terms of others. So from the table of
variables and their dimensions, we can see that, of the five independent
variables, 𝝁 and 𝑸 are the most important, since the dimension of 𝝁 consists of
all three fundamental dimensions (𝑀𝐿 and 𝑇) and 𝑸 has two of three
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DIMENSIONAL ANALYSIS AND SIMILARITY
dimensions and the highest positive index. We can solve for 𝑎, 𝑏, 𝑐 in terms of 𝑑
and 𝑒 we have;
From equation (𝑖);
0 = 𝑐 + 𝑑  𝒄 = −𝒅
From equation (𝑖𝑖𝑖);
0 = −𝑏 − 𝑑 − 𝑒  𝒃 = −𝒅 − 𝒆
From equation (𝑖𝑖);
0 = 𝑎 − 3𝑐 − 𝑑 + 3𝑒  𝑎 = 3𝑐 + 𝑑 − 3𝑒
𝒂 = 3 −𝑑 + 𝑑 − 3𝑒 = −𝟐𝒅 − 𝟑𝒆
Now
 = 𝐶 𝐷 𝑎 . 𝜔𝑏 . 𝜌𝑐 . 𝜇 𝑑 . 𝑄 𝑒
 = 𝐶 𝐷 −2𝑑−3𝑒 . 𝜔−𝑑−𝑒 . 𝜌−𝑑 . 𝜇𝑑 . 𝑄 𝑒
 = 𝐶 𝐷−2𝑑 . 𝜔−𝑑 . 𝜌−𝑑 . 𝜇𝑑 . 𝐷−3𝑒 . 𝜔−𝑒 . 𝑄 𝑒 .
 = 𝐶 𝐷−2 . 𝜔−1 . 𝜌−1 . 𝜇 𝑑 . 𝐷−3 . 𝜔−1 . 𝑄 𝑒
30
DIMENSIONAL ANALYSIS AND SIMILARITY
𝑑
𝜇
𝑄
=
. 3
2
𝐷 𝜌𝜔
𝐷 𝜔
𝝁
𝑸
=
𝝆𝝎𝑫𝟐 𝝎𝑫𝟑
𝑒
Example 5.0. The pressure drop ∆𝒑 in a pipe of diameter 𝑫 and length 𝒍
depends on the density 𝝆 and viscosity 𝝁 of fluid flowing, mean velocity 𝑽 of
flow and average height of protuberance 𝒕. Show that the pressure drop can be
expressed in the form:
𝒍
𝝁
𝒕
𝟐
∆𝒑 = 𝝆𝑽 
𝑫 𝑽𝑫𝝆 𝑫
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DIMENSIONAL ANALYSIS AND SIMILARITY
Solution
Consider the table of variables involved in fluid phenomenon, their symbols, and
dimensions below.
S/N
1.
2.
3.
4.
5.
6.
7.
Variables
Pressure drop
Diameter of pipe
Length of pipe
Density
Dynamic viscosity
Mean velocity
Protuberance
symbols
∆𝒑
𝑫
𝒍
𝝆
𝝁
𝑽
𝒕
Dimensions
ML−1 T −2
L
L
ML−3
ML−1 T −1
LT −1
L
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DIMENSIONAL ANALYSIS AND SIMILARITY
The dependent variable is ∆𝒑, and the independent variables are 𝑫, 𝒍, 𝝆, , 𝑽,
and 𝒕. So the functional relationship between the variables will be.
∆𝑝 = 𝐶 𝐷𝑎 . 𝑙 𝑏 . 𝜌𝑐 . 𝜇𝑑 . 𝑉 𝑒 . 𝑡 𝑓
Replacing the variables by their dimensions we have;
𝑀𝐿−1 𝑇 −2 = 𝐶 𝐿 𝑎 . 𝐿 𝑏 . 𝑀𝐿−3 𝑐 . 𝑀𝐿−1 𝑇 −1 𝑑 . 𝐿𝑇 −1 𝑒 . 𝐿 𝑓
𝑀𝐿−1 𝑇 −2 = 𝐶 𝑀𝑐 . 𝑀𝑑 . 𝐿𝑎 . 𝐿𝑏 . 𝐿−3𝑐 . 𝐿−𝑑 . 𝐿𝑒 . 𝐿𝑓 . 𝑇 −𝑑 . 𝑇 −𝑒
𝑀𝐿−1 𝑇 −2 = 𝐶 𝑀𝑐+𝑑 . 𝐿𝑎+𝑏−3𝑐−𝑑+𝑒+𝑓 . 𝑇 −𝑑−𝑒
Equating powers of the fundamental dimensions on both side we get;
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DIMENSIONAL ANALYSIS AND SIMILARITY
For 𝑀, we have;
1 = 𝑐 + 𝑑……….…..………… 𝑖
For 𝐿, we have;
−1 = 𝑎 + 𝑏 − 3𝑐 − 𝑑 + 𝑒 + 𝑓 … . 𝑖𝑖
For 𝑇, we have;
−2 = −𝑑 − 𝑒 … … … . . … … … … . (𝑖𝑖𝑖)
❖ We have six unknowns to solve for in three equations. This is not
possible without solving for some unknowns in terms of others.
❖ So, from the table of variables and their dimensions, we can see
that, of the six independent variables, 𝝁 , 𝒍 and 𝒕 are the most
important,
❖ since the dimension of 𝝁 consists of all three fundamental
dimensions (𝑀𝐿 and 𝑇) and 𝒍 and 𝒕 have the most
34
DIMENSIONAL ANALYSIS AND SIMILARITY
positive index of all the dimensions. We can now solve for 𝑎, 𝑐 and 𝑒 in terms of
𝑏, 𝑑 and 𝑓. We have
From equation (𝑖𝑖𝑖);
−2 = −𝑑 − 𝑒 𝑎𝑛𝑑 𝒆 = 𝟐 − 𝒅
From equation (𝑖); 1 = 𝑐 + 𝑑 𝑎𝑛𝑑 𝒄 = 𝟏 − 𝒅
From equation (𝑖𝑖); −1 = 𝑎 + 𝑏 − 3𝑐 − 𝑑 + 𝑒 + 𝑓
 𝑎 = 𝑑 − 1 − 𝑏 + 3(1 − 𝑑) − (2 − 𝑑) − 𝑓
𝒂 = 𝑑 − 1 − 𝑏 + 3 − 3𝑑 − 2 + 𝑑 − 𝑓 = −𝒃 − 𝒅 − 𝒇
Now;
∆𝑝 = 𝐶 𝐷𝑎 . 𝑙 𝑏 . 𝜌𝑐 . 𝜇𝑑 . 𝑉 𝑒 . 𝑡 𝑓
∆𝑝 = 𝐶 𝐷−𝒃−𝒅−𝒇 . 𝑙 𝑏 . 𝜌𝟏−𝒅 . 𝜇𝑑 . 𝑉 𝟐−𝒅 . 𝑡 𝑓
∆𝑝 = 𝐶 𝜌. 𝑉 𝟐 . 𝐷 −𝑏 . 𝑙 𝑏 . 𝜇𝑑 . 𝐷 −𝑑 . 𝜌−𝒅 . 𝑉 −𝒅 . 𝐷 −𝑓 . 𝑡 𝑓
35
DIMENSIONAL ANALYSIS AND SIMILARITY
𝑙
𝐷
∆𝑝 = 𝜌𝑉 𝟐 𝐶
∆𝒑 =
𝝆𝑽𝟐

𝑏
𝜇
.
𝜌𝑉𝐷
𝑑
𝑡 𝑓
.
𝐷
𝒍
𝝁
𝒕
.
.
𝑫
𝝆𝑽𝑫
𝑫
36