Government Engineering College - Bhuj Subject:- Fluid mechanics

TOPIC:- BUCKINGHAM’S PI THEOREM EXAMPLE

Example No:- 1Show 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

NdfdVF ,22

Dimension of the variable involved :

Number of variables n = 6 Number of fundamental dimension m = 3Number 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

2MLT

1LT

11 TML3ML

1T

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:

3333

2222

1111

cba

cba

cba

Vd

VVd

FVd

20:

130:

10:

)()()(

1

111

1

213111000

1111

bT

bcaL

cM

MLTMLLTLTLM

FVdcba

cba

1

2

2

1

1

1

1

b

a

c

• SO, Substituting the value of a1, b1, c1 in equation,

• For term:

• For term:

1

22122

11 Vd

FFVd

2

10:

30:

10:

)()()(

2

212

2

223212000

2222

bT

bcaL

cM

MLTMLLTLTLM

VVdcba

cba

1

1

0

2

2

2

b

a

c

V

NdVVd 011

2

3

)()()( 233313000

3333

MLTMLLTLTLM

Vdcba

cba

• 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 1113

1 23

0,,221

dVV

Nd

Vd

Ff

dVV

Ndf

Vd

F,

22

3 .33 /1

dVV

NdfdVF ,22

Example No:- 2Show 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

Dimension of the variable involved :

Number of variables n = 6 Number of fundamental dimension m = 3Number of dimensionless л term (n – m) = 3

Sr. No.

Variable Symbol

Dimension

1. Lift 2. Mass

densityρ

3. Velocity V4. Depth d L5. Angle of

incidence6. Viscosity μ

LF

2MLT3ML1LT

11 TML

00TML

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:

3333

2222

1111

cba

cba

cba

Vd

Vd

FVd L

1

20:

130:

10:

)()()(

1

111

1

213111000

1111

bT

bcaL

cM

MLTMLLTLTLM

FVdcba

cba

L

2

2

1

1

1

1

b

a

c

Substituting the value of a1, b1, c1 in equation,

For term:

For term:

1

22122

1 Vd

FFVd LL

2

10:

30:

0:

)()()(

2

222

2

00023212000

bT

bcaL

cM

TLMMLLTLTLMcba

0002 Vd

3

10:

130:

10:

)()()(

3

333

3

1133313000

bT

bcaL

cM

TMLMLLTLTLMcba

0

0

0

2

2

2

b

a

c

dV

Vd 1113

1

1

1

3

3

3

b

a

c

Substituting the value of , , in equation(1)

…..Ans.

1 23

dVVdF

dVVd

F

dVVd

Ff

L

L

L

,

,

0,,

22

22

22

Example No:- 3Fluid 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

fVo2

o

Sr. No.

Variable Symbol Dimension

1. Shear stress

2. Viscosity μ

3. Density ρ

o11 TML

21 TML

3ML

Number of variables n = 5 Number of fundamental dimension m = 3Number 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

1111

cba

cba

Vd

Vd o

1

o

cba

Vd 1111

• For term:

20:

130:

10:

)()()(

1

111

1

2113111000

bT

bcaL

cM

TMLMLLTLTLMcba

2

0

1

1

1

1

b

a

c

2

12011 VVd o

o

2

10:

130:

10:

)()()(

2

222

2

1123212000

bT

bcaL

cM

TMLMLLTLTLMcba

1

1

1

2

2

2

b

a

c

dVdV

Vd 11112

Substituting the value of , in equation(1),

….Ans.

21

dVfV

dVf

V

dV

Vf

o

o

o

12

12

21 ,

Example No:- 4This 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

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:

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

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

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.

Example No:- 5We 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

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:

(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:

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:

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

Thank you…