This document discusses two methods of dimensional analysis: Rayleigh's method and Buckingham π-theorem. Rayleigh's method expresses dependent variables as an exponential function of independent variables. It is useful for problems with 3-4 variables but difficult above that. Buckingham π-theorem states that variables can be grouped into dimensionless π terms, with the number of terms equal to the total variables minus the number of fundamental dimensions. This allows problems with many variables to be solved through grouping and repeating variables. Both methods involve setting up equations using dimensional homogeneity to solve for exponent values.

2.  FLUIDMECHANICS
 GROUPNO: 9
 PREPAREDBY:
ENROLLMENTNO:141080119050
141080119052
141080119059
151083119013.
 MECHANICALBRANCH.
 Semester:4th(2nd year)
 TOPICNAME: REYLEIGH’SMETHOD,BUCKINGHAM π-
THEOREM.

3. INTRODUCTION
 It is method of dimensional analysis.
 This two methods is very important for
dimensional analysis.
(1).Rayleigh`s method
(2).Buckingham`s - theorem.

4. Rayleigh'sMethod
 In this method, the expression for the
variables in form of exponential
equation and dimensionally
homogeneous.
 Let, Y is a variable, which depends on
variables, then functional
relationship may be written as:
.............321 ,, xxx
 ,.....,, 321 xxxfY 

5.  Where , Y=dependent variable,
=independent variables,
f=function.
 This method is used for determining
expressions for a variables which
depends upon maximum three or
four variables only..
 If the number of independent
variables becomes more than four,
then it is very difficulty to find
expression for the dependent
variables.
.............321 ,, xxx

6. Method involves the
following steps
 (1).Gather all the independent variables
which govern variation of dependent
variables.
 (2).write the functional relationship with
the given data
 (3).write the equation in terms of a
constant with exponents(power)
a,b,c.....
 ,.....,, 321 xxxfY 
 ,......,, 321
cba
xxxKY 

7. where , K is a dimensionless co-
efficient and a,b,c....are the arbitrary
powers.
 (4).Apply principal of dimensional
homogeneity, and put the
dimensions(M,L,T) of variables on
both sides of equation.
 (5).find out the values of exponents
(a,b,c,...) by obtaining simultaneous
equation.

8.  (6).put the value of exponents (a,b,c...)
in the main equation and form the
dimensionless parameter by
grouping the variables with similar
exponents..

9. Buckingham’s π-Theorem
 This method is minimized difficulties
of Rayleigh's theorem....
 It states, "If there are n numbers of
variables (dependent and
independent variables) in the physical
phenomenon and if these variables m
numbers of fundamental dimensions
(M,L,T), then the variables may be
grouped into (n-m) dimensionless
terms”

10.  This dimensionless term is known as π.
 Let us consider a variable depends
upon independent variables
then the functional equation can be
written as
 The equation may be written in general
form as
1x
nxxx ,......, 32
),.....,,( 321 nxxxkx 
cxxxxf n ),....,,,( 321

11.  Where , c is a constant and f is a
function.
 If there are n variables and m
fundamental dimensions, then
according to Buckingham’s π-theorem
=constant.
 The π-term is dimensionless and
independent of the system.
 mnF  ,.....,,, 3211

12. Buckingham’s method
involved following steps:
 (1).write the functional relationship of
given data,
 (2).write the equation in its general form
 (3).Find the numbers of π-terms . If there
are n variables and m is fundamental
dimensions , numbers of π-terms=n-m
),.....,,( 321 nxxxfx 
0),.....,,( 32 nxxxf

13.  (4).select m number of repeating
variables and write separate
equation for each π-term. Each π-
terms contain the repeating variables.
The repeating variables are written in
exponential form...
.......................................
...........................
43211
111
xxxx
cba

53212
222
xxxx
cba

6321
333
3
xxxx
cba

n
cba
mn xxxx mnmnmn 
 321

14. Where , are repeating variables.
 (5).Each π-term solve by the principle of
dimensional homogeneity, put the
dimensions of variables in each π-term
and find out the value of a,b,c,,, by solving
simultaneous equations..
 (6).Now put the values of a,b,c,... In the π-
terms.
 (7).write the functional relation in the
required form
321 ,, xxx
  0,.....,,, 3211 mnF 

15. Procedure for selection of
Repeating variables:
 Number of repeating variables= no .
of fundamental dimensions=m
 The repeating variables should not be
dependent variable.
 It should not be dimensionless.
 No two variables should have the
same dimensions.

16.  The repeating variables together
must have the dimensions as MLT.
 The repeating variables should be
selected in such a way that
(1).one variables contains geometric
property as length, diameter, height,
width, etc.
(2).other variables contains flow
property as velocity ,acceleration etc.
(3).Third variables contains fluid
property as dynamic viscosity,
density, etc.

17.  In most of common problems of fluid
mechanics , the pair of repeating
variables as (1).
(2).
(3).
(4).
,,Vl
,,Vl
,,Vd
,,Vd