Example of BUCKINGHAM’S PI THEOREM EXAMPLE

1. Government Engineering College
- Bhuj
Subject:- Fluid mechanics
TOPIC:- BUCKINGHAM’S PI THEOREM
EXAMPLE

2. Example No:- 1
 Show that the thrust F of a propeller having
diameter d, moving with a velocity V in a
fluid of viscosity μ and density ρ ,
rotating at a speed of N may be expressed
as,
 Soln.: The relationship between dependant
and independent variable may be expressed
as: F = f(d, v, μ, ρ, N)










Vd
V
Nd
fdVF ,22

3.  Dimension of the variable involved :
Number of variables n = 6
Number of fundamental dimension m = 3
Number of dimensionless л term (n – m) = 3
Sr. No. Variable Symbol Dimension
1. Thrust F
2. Diameter d L
3. Velocity V
4. Viscosity μ
5. Density ρ
6. Speed N
2
MLT
1
LT
11 
TML
3
ML
1
T

4. f1 (л1, л2, л3)= 0 ………….(1)
Selecting the repeating variable as d, V, ρ
• By using Buckingham’s theorem,
• Solve the л equation by the principle of dimension
homogeneity:
• For term:






333
3
222
2
111
1
cba
cba
cba
Vd
VVd
FVd
20:
130:
10:
)()()(
1
111
1
213111000
111
1






bT
bcaL
cM
MLTMLLTLTLM
FVd
cba
cba

1
2
2
1
1
1
1



b
a
c

5. • SO, Substituting the value of a1, b1, c1 in equation,
• For term:
• For term:
1

 22
122
11
Vd
F
FVd  
2
10:
30:
10:
)()()(
2
212
2
223212000
222
2






bT
bcaL
cM
MLTMLLTLTLM
VVd
cba
cba

1
1
0
2
2
2



b
a
c
V
Nd
VVd   011
2 
3
)()()( 233313000
333
3



MLTMLLTLTLM
Vd
cba
cba


6. • Equation the powers of MLT on both sides,
• So, Substituting the value of , , in equation(1),
• Inverse of is also dimensionless term, replace
…..Ans.
10:
130:
10:
3
333
3



bT
bcaL
cM
1
1
1
3
3
3



b
a
c



dV
Vd   111
3
1 2 3
0,,221 







 dVV
Nd
Vd
F
f









 dVV
Nd
f
Vd
F
,22
3 .33 /1  










dVV
Nd
fdVF ,22

7. Example No:- 2
 Show that the lift on airfoil can be expressed
as:
where, ρ = Mass density V = Velocity of flow,
d = Characteristic depth
= Angle of incidence,
μ = Coefficient of viscosity.
• Soln.: The relationship between dependant and
independent variable may be expressed as:
F = f(d, v, μ, ρ, , ,)
LF






 


 ,22 Vd
dVFL


 LF

8.  Dimension of the variable involved :
Number of variables n = 6
Number of fundamental dimension m = 3
Number of dimensionless л term (n – m) = 3
Sr. No. Variable Symbol Dimension
1. Lift
2. Mass density ρ
3. Velocity V
4. Depth d L
5. Angle of
incidence
6. Viscosity μ
LF

2
MLT
3
ML
1
LT
11 
TML
00
TML

9. f1 (л1, л2, л3)= 0 ………….(1)
Selecting the repeating variable as d, V, ρ
Each л term contain m+1 = 3+1 = 4 variables.
• By using Buckingham’s theorem,
• Solve the л equation by the principle of dimension
homogeneity:
• For term:






333
3
222
2
111
1
cba
cba
cba
Vd
Vd
FVd L
1
20:
130:
10:
)()()(
1
111
1
213111000
111
1






bT
bcaL
cM
MLTMLLTLTLM
FVd
cba
cba
L
2
2
1
1
1
1



b
a
c

10.  Substituting the value of a1, b1, c1 in equation,
 For term:
 For term:
1

 22
122
1
Vd
F
FVd L
L  
2
10:
30:
0:
)()()(
2
222
2
00023212000



 
bT
bcaL
cM
TLMMLLTLTLM
cba
  000
2 Vd
3
10:
130:
10:
)()()(
3
333
3
1133313000



 
bT
bcaL
cM
TMLMLLTLTLM
cba
0
0
0
2
2
2



b
a
c



dV
Vd   111
3
1
1
1
3
3
3



b
a
c

11.  Substituting the value of , , in equation(1)
…..Ans.
1 2 3































dV
VdF
dVVd
F
dVVd
F
f
L
L
L
,
,
0,,
22
22
22

12. Example No:- 3
 Fluid of density ρ and viscosity μ flows at an average
velocity V through a circular pipe diameter d. show by
dimensional analysis, that the shear stress of the pipe
wall.
• Soln.: The relationship between dependant and
independent variable may be expressed as:
F = f(d, v, μ, ρ, )
 Dimension of the variable involved :










Vd
fVo
2
o
Sr. No. Variable Symbol Dimension
1. Shear stress
2. Viscosity μ
3. Density ρ
o
11 
TML
21 
TML
3
ML

13. Number of variables n = 5
Number of fundamental dimension m = 3
Number of dimensionless л term (n – m) = 2
f1 (л1, л2)= 0 ………….(1)
Selecting the repeating variable as d, V, ρ
Each л term contain m+1 = 3+1 = 4 variables.
• By using Buckingham’s theorem,
• Solve the л equation by the principle of dimension
homogeneity:
• For term:




222
2
111
1
cba
cba
Vd
Vd o
1
o
cba
Vd   111
1

14. • For term:
20:
130:
10:
)()()(
1
111
1
2113111000



 
bT
bcaL
cM
TMLMLLTLTLM
cba
2
0
1
1
1
1



b
a
c


 2
120
11
V
Vd o
o  
2
10:
130:
10:
)()()(
2
222
2
1123212000



 
bT
bcaL
cM
TMLMLLTLTLM
cba
1
1
1
2
2
2



b
a
c




 dV
dV
Vd   1111
2

15.  Substituting the value of , in equation(1),
….Ans.
21































dV
fV
dV
f
V
dV
V
f
o
o
o
1
2
12
21 ,

16. Example No:- 4
This example is elementary, but demonstrates the general
procedure: Suppose a car is driving at 100 km/hour; how long does
it take it to go 200 km? This question has two fundamental
physical units: time t and length , and three dimensional
variables: distance D, time taken T, and velocity V. Thus there are
3 − 2 = 1 dimensionless quantity. The units of the dimensional
quantities are:
The dimensional matrix is:
The rows correspond to the dimensions , and t, and the columns
to the dimensional variables D, T, V. For instance, the 3rd column,
(1, −1), states that the V (velocity) variable has units of

17.  For a dimensionless constant we are
looking for a vector such that the matrix
product of M on a yields the zero vector [0,0]. In linear
algebra, this vector is known as the kernel of the
dimensional matrix, and it spans the nullspace of the
dimensional matrix, which in this particular case is one-
dimensional. The dimensional matrix as written above is
in reduced row echelon form, so one can read off a kernel
vector within a multiplicative constant:

18. If the dimensional matrix were not already reduced, one
could perform Gauss–Jordan elimination on the
dimensional matrix to more easily determine the kernel. It
follows that the dimensionless constant may be written:
or, in dimensional terms

19.  Since the kernel is only defined to within a multiplicative
constant, if the above dimensionless constant is raised to
any arbitrary power, it will yield another equivalent
dimensionless constant.
 Dimensional analysis has thus provided a general
equation relating the three physical variables
which may be written

20.  where C is one of a set of constants, such that .
The actual relationship between the three variables is
simply so that the actual dimensionless equation
( ) is written:
• In other words, there is only one value of C and it is
unity. The fact that there is only a single value of C and
that it is equal to unity is a level of detail not provided by
the technique of dimensional analysis.

21. Example No:- 5
 We wish to determine the period T of small oscillations in
a simple pendulum. It will be assumed that it is a
function of the length L , the mass M , and the
acceleration due to gravity on the surface of the Earth g,
which has dimensions of length divided by time squared.
The model is of the form
 (Note that it is written as a relation, not as a
function: T isn't written here as a function of M, L,
and g.)
 There are 3 fundamental physical dimensions in this
equation: time t, mass m, and length l, and 4
dimensional variables, T, M, L, and g. Thus we need only
4 − 3 = 1 dimensionless parameter, denoted π, and the
model can be re-expressed as

22. where π is given by
for some values of a1, ..., a4.
The dimensions of the dimensional quantities are:
for some values of a1, ..., a4.
The dimensions of the dimensional quantities are:
The dimensional matrix is:

23.  (The rows correspond to the dimensions t, m, and l, and
the columns to the dimensional variables T, M, L and g.
For instance, the 4th column, (−2, 0, 1), states that
the gvariable has dimensions of .
 We are looking for a kernel vector a = [a1, a2, a3, a4] such
that the matrix product of M on a yields the zero vector
[0,0,0]. The dimensional matrix as written above is in
reduced row echelon form, so one can read off a kernel
vector within a multiplicative constant:

24.  Were it not already reduced, one could
perform Gauss–Jordan elimination on the
dimensional matrix to more easily
determine the kernel. It follows that the
dimensionless constant may be written:
•In fundamental terms:

25.  which is dimensionless. Since the kernel is only defined to
within a multiplicative constant, if the above dimensionless
constant is raised to any arbitrary power, it will yield another
equivalent dimensionless constant
 This example is easy because three of the dimensional
quantities are fundamental units, so the last (g) is a
combination of the previous. Note that if a2 were non-zero there
would be no way to cancel the M value—therefore a2 must be
zero. Dimensional analysis has allowed us to conclude that the
period of the pendulum is not a function of its mass. (In the 3D
space of powers of mass, time, and distance, we can say that
the vector for mass is linearly independent from the vectors for
the three other variables. Up to a scaling factor, is the
only nontrivial way to construct a vector of a dimensionless
parameter.)
 The model can now be expressed as

26. Thank you…