1. Dimensional Analysis definition necessity 2. Fundamental quantities- Primary Dimensions 3. Non Primary Dimensions 4. Secondary derived Dimensions 5. Dimensionless quantities 6. Necessity and objectives of Dimensional Analysis 7. Dimensional Homogeneity 8.Methods of Dimensional Analysis 9. Rayleigh's Method of dimensional analysis 10. Numerical Problem on Rayleigh's Method of dimensional analysis

1. DIMENSIONAL ANALYSIS
PART-I
Dr. RAMESH B R
ramesh.burlir@gmail.com

2. Dimensional Analysis
• It is a pure mathematical technique to establish a relationship
between physical quantities involved in a fluid phenomenon by
considering their dimensions.
• In dimensional analysis, from a general understanding of fluid
phenomena, we first predict the physical parameters that will
influence the flow
• then we group these parameters into dimensionless combinations which
enable a better understanding of the flow phenomena.
• Dimensional analysis is particularly helpful in experimental work because it
provides a guide to those things that significantly influence the phenomena
• thus it indicates the direction in which experimental work should go.

3. Fundamental quantities- (Primary dimensions)
The physical quantities which can be treated as independent of other physical
quantities and are not usually defined in terms of other physical quantities, are
called fundamental quantities.(primary dimensions. Primary (sometimes
called basic) dimensions are defined as independent or fundamental dimensions,
from which other dimensions can be obtained.
Ex.: Mass, Length, Time etc.,
• Units are the standard elements we use to quantify these dimensions.
Ex.: Kg, Metre, Seconds etc.,
• For example, length is a dimension that is measured in units such as microns (m),
feet (ft), centimeters (cm), meters (m), kilometers (km), etc.There are seven
primary dimensions (also called fundamental or basic dimensions. They are
mass, length, time, temperature, electric current, amount of light, and
amount of matter

4. Non primary dimensions
• All nonprimary dimensions can be formed by some combination of
the seven primary dimensions
For example, force has the same dimensions as mass times acceleration (by
Newton’s second law). Thus, in terms of primary dimensions,
Dimensions of force:

5. EXAMPLE :
Primary Dimensions of Surface Tension
SOLUTION
The primary dimensions of surface tension are to be determined.
Analysis
Force has dimensions of mass times acceleration, or {mL/t2 }. Thus,

6. Geometric Units Dimensions
Area 𝑚2 𝐿2
Volume 𝑚3 𝐿3
Kinetic
Velocity m/s L/T (L𝑇−1)
Acceleration m/𝑠2 L𝑇2 (L𝑇−2)
Discharge m3 /s L3/T(L3T-1)
Dynamic
Force N ML/T (ML𝑇−1)
Density g/𝑚3 M/𝐿3 (M𝐿−3)
The physical quantities whose defining operations are based on other physical
quantities, are called derived quantities.
All physical quantities other than the seven base quantities are derived
quantities
Swcondary dimensions are those quantities which posses more than one
fundamental dimensions
Secondary or Derived Dimensions:

7. Dimensionless quantity
Physical quantities which do not possess dimensions are called
dimensionless quantities.
Example: Angle, specific gravity, strain.
In general, physical quantity which is a ratio of two quantities of same
dimension will be dimensionless.
Example:
Reynold’s number 𝑅 𝑒 =
𝜌𝑈𝐿
𝜇
is a dimensionless number
Therefore the Dimensions of numerator 𝜌𝑈𝐿 = (M𝐿−3
) (L𝑇−1
) (L)
Simplifying we get 𝑀𝐿−1
𝑇−1
Dimensions of denominator 𝜇 = 𝑀𝐿−1
𝑇−1
Therefore 𝑅 𝑒 =
𝜌𝑈𝐿
𝜇
=
𝑀𝐿−1 𝑇−1
𝑀𝐿−1 𝑇−1 hence dimensionless.

8. Necessity of Dimensional analysis
• In some of the practical real flow problems in fluid mechanics can be
solved by using equations and analytical procedures.
• Solutions of some real flow problems depend heavily on experimental
data.
• Sometimes, the experimental work in the laboratory is not only time
consuming, but also expensive.
• So, the main goal is to extract maximum information from fewest
experiments.
• In this regard, dimensional analysis is an important tool that helps in
correlating analytical results with experimental data and to predict the
prototype behaviour from the measurements on the model.

9. Objectives of Dimensional Analysis
1) Checking the dimensional homogeneity of any fluid flow equation.
2) Deriving fluid mechanics equations expressed in terms of non-dimensional
parameters to show the relative significance of each parameter.
3) Planning tests and presenting experimental results in a systematic manner.
4) Analysing complex flow phenomena by use of scale models (model similitude).
5) Conversion from one dimensional unit to another
6)Checking units of equations (Dimensional Homogeneity)
7) Defining dimensionless relationship using
a) Rayleigh’s Method
b) Buckingham’s π-Theorem
8) Model Analysis

10. DIMENSIONAL HOMOGENEITY
Principle of dimensional homogeneity states that an equation which
expresses a physical phenomenon of fluid flow must be algebraically correct
and dimensionally homogeneous
An equation is said to be dimensionally homogeneous, if the dimensions on
its left hand side are same as the dimensions pf the terms on the left hand
side
Example : Consider the equation V= 2𝑔ℎ
Dimensions of LHS V = L/T =LT-1
Dimensions of RHS = 2𝑔ℎ =
𝐿
𝑇2xL =
𝐿2
𝑇2 =
𝐿
𝑇
= LT-1
Dimensions of LHS= Dimensions of RHS
Hence the equation is homogeneous

11. Example : Consider the equation s= ut+
1
2
a𝑡2
Dimensions of LHS S= distance = m= L1
Dimensions of RHS =ut=m/s xs = L/T xT = L1
Dimensions of RHS= ½ at2 =m/s2 x s2 = L/T2 xT2 = L1
Dimensions of LHS= Dimensions of RHS
Hence the equation is homogeneous

12. Check the dimensional homogeneity of Bernoulli’s equation of energy
Bernoulli’s equation is P+1/2 𝜌𝑉2
+𝜌𝑔ℎ = 𝑐
Solution :
Dimension of P =
𝐹𝑜𝑟𝑐𝑒
𝐴𝑟𝑒𝑎
=
𝑀𝑎𝑠𝑠 𝑥
𝐿𝑒𝑛𝑔𝑡ℎ
𝑇𝑖𝑚𝑒2
𝐿𝑒𝑛𝑔𝑡ℎ2 =
𝑀
𝐿
𝑇2
𝐿2 =
𝑀
𝑇2 𝐿
Dimensions of 1/2 𝜌𝑉2
=
𝑀𝑎𝑠𝑠
𝑉𝑜𝑢𝑚𝑒
𝐿𝑒𝑛𝑔𝑡ℎ
𝑇𝑖𝑚𝑒
2
=
𝑀
𝐿3
𝐿
𝑇
2
=
𝑀
𝑇2 𝐿
Dimensions of 𝜌𝑔ℎ =
𝑀𝑎𝑠𝑠
𝑉𝑜𝑢𝑚𝑒
𝐿𝑒𝑛𝑔𝑡ℎ
𝑇𝑖𝑚𝑒2 𝐿𝑒𝑛𝑔𝑡ℎ =
𝑀
𝐿3
𝐿
𝑇2 𝐿 =
𝑀
𝑇2 𝐿
Therefore all the additive terms of the Bernoulli’s equation are having the same
dimensions
Therefore from the law of homogeneity the dimensions of the constant shall also
have the dimensions on the left hand side
Therefore the dimension of the right hand side C is
𝑀
𝑇2 𝐿

13. METHODS OF DIMENSIONAL ANALYSIS
There are two methods of dimensional analysis used.
(i) Rayleigh's method
(ii) Buckingham π Theorem
RAYLEIGH'S METHOD
In this method, the expression is determined for a variable for
maximum three or four variables only.
If the number of independent variables becomes more than four, it is
very difficult to find the expression for the dependent variable

14. Steps involved in Rayleigh's method
1. First, the functional relationship is written with the given data.
Consider X as a variable which depends on X1, X2, X3,… Xn
So, the functional equation is written X=f(X1, X2, X3,…Xn)
2. Then the equation is expressed in terms of a constant with exponents like
powers of a, b, c ... Therefore, the equation is again written as:
X=ϕ(X1
a,X2
b,X3
c, ...Xn
z)
Here, (ϕ)= Constant a, b, c, ... z = Arbitrary power
3. The values of a, b, c, ... z are determined with the help of dimensional
homogeneity. It means, the powers of the fundamental dimensions on both
sides are compared to obtain the values of exponents.
4. Finally, these exponents/power values are substituted in the functional
equation and simplified to obtain the suitable form.

15. Example:
Let us consider the frictional resistance of fluid flow per unit area of the
inside surface of the pipe
A reasonable assumption can be made that the resistance which causes
pressure drop of the fluid (Δp)is a function of diameter of pipe (D),
fluid density (𝜌) fluid velocity u and fluid viscosity 𝜇, or
Δp= f[u,D, 𝜌, 𝜇]
Δp = c ua D b 𝜌c 𝜇d
where C is a dimensionless constant. The dimensional equation of the
above expression in fundamental dimensions M, L and T are
𝑀𝐿𝑇−2
𝐿2 = 𝐿𝑇−1 𝑎
𝐿 𝑏
𝑀𝐿−3 𝑐
𝑀𝐿−1
𝑇−1 𝑑
M𝐿−1
𝑇−2
=𝐿 𝑎−𝑏−3𝑐−𝑑
𝑇−𝑎−𝑑
𝑀−𝑐+𝑑

16. For the homogeneity of
M : 1 = c + d,
L : − 1 = a + b – 3 c – d and,
t : − 2 = − a – d.
On solving these equations
we have
b = − d,
c = 1 – d and
a = 2 – d.
∆𝑝 = 𝑐𝑢2−𝑑
𝐷−𝑑
𝜌1−𝑑
𝜇 𝑑
= C𝜌𝑢2 𝜇
𝜌𝑢𝐷
𝑑
=C
𝜌𝑢2
𝑅 𝑒
𝐷
Where 𝑅 𝑒
𝐷
=
𝜌𝑢𝐷
𝜇
= Reynolds number.
The values of constants C and D have to
be determined by experiment

17. Problem: Fluid flow through a small orifice discharging freely into atmosphere under a constant head depends on the
parameters discharge Q diameter d, constant head H. ρ the mass density and µ the dynamic viscosity of the fluid flowing
through the orifice.
Q= f(µ, ρ, d, H, g)
Q = C(μa ,ρb,dc,Hdge) where C is a dimensionless constant.
Substituting the proper dimensions for each variable in this exponential equation in M-L-T system,
(L3/T) = (M°L°T°) (M/LT)a (M/L3)b (L)c(L)d(L/T2)e
For dimensional homogeneity the exponents of each dimension on the both sides of the equation must be identical.
Thus
for M : 0 = a + b
for L : 3 = – a –3b + c + d + e
for T :– 1 = – a – 2e
Since there are five unknowns in three equations, three of the unknowns must be expressed in terms of the other two
b=-a
e=
1
2
-
𝑎
2
c=
5
2
−
3𝑎
2
-d

18. Q=c 𝜇 𝑎
𝜌−𝑎
𝑑
5
2
−
3𝑎
2
−𝑑
𝐻 𝑑
𝑔
1
2
−
𝑎
2
=c 𝑑
5
2 𝑔
1
2 𝜇 𝑎
𝜌−𝑑
𝑑
−3𝑎
2 𝑔
−𝑎
2 , 𝐻 𝑑
𝑑−𝑑
Q=c
𝜇
𝜌𝑑
3
2 𝑔
1
2
𝑎 𝐻
𝑑
𝑑−
1
2
𝑑2
𝐻
1
2 𝑔
1
2 ( multiplying and dividing by
𝜋
4
2)
=
𝐶
𝜋
4
2
𝜇
𝜌𝑑
3
2
𝑔
1
2
𝑎 𝐻
𝑑
𝑑−
1
2 𝜋
4
𝑑2
2𝑔𝐻
= 𝑎 2𝑔𝐻 𝑓1
𝜇
𝜌𝑑
3
2 𝑔
1
2
,
𝐻
𝑑
This expression may be written in usual form i. e., Q=𝐶 𝑑a 2𝑔𝐻
Where 𝐶 𝑑 is the coefficient of discharge of the orifice which can be
expressed as 𝐶 𝑑 =𝑓1
𝜇
𝜌𝑑
3
2 𝑔
1
2
,
𝐻
𝑑
Note: 𝐶 𝑑 is a non dimensional factor

19. Example :Find the equation for the power developed by a pump if it
depends on head H ;discharge Q and specific weight 𝛾 of the fluid
P = f (H, Q, 𝛾 )
P = C HaQb 𝛾 c
[P] = [H]a [Q]b [𝛾]c -----(a)
[L2MT-3] = [LMoTo]a[L3M 0T-1]b[L-2MT-2]c
for M : 1 = c ---(i)
for L : 2 = a + 3b – 2c ---(ii)
for T :– 3 = – b – 2c ---(iii)
(i )in( iii) gives b=1
substituting b and c value in( ii) we get a=1
Substituting a,b and c in (a) we get
[P] = [H]1 [Q]1 [𝛾]1
When K = 1
P= 𝛾𝑄𝐻