ME 424 – Fluid Machinery
Placement:
4th Year Mechanical Engineering Student
Subject: Additional Example for Dimensional Analysis
1. It is known that the variation of pressure, ∆p, within a static fluid is dependent upon the specific
weight of the fluid and the elevation difference, ∆z . Using dimensional analysis, find the form of the
hydrostatic equation for pressure variation.
Given:
pressure, ∆p,
specific weight of the fluid
the elevation difference, ∆z
Required:
form of the hydrostatic equation for pressure variation
Solution:
The variables with units and dimensions are listed below.
S. No.
Variable
Unit
Dimension
1
N/m2
F/L2
∆p
2
N/m3
F/L3
γ
3
m
L
∆z
There are three parameters and two dimensions. So one π term can be identified.
π 1 = ∆pγ a ∆z b
F Fa b
⋅ ⋅L
L2 L3a
Equating the indices of F and L,
(1) 1 + a =0
(2) -2 – 3a + b = 0
Then a = -1, and -2 – 3(-1) + b = 0; b = - 1.
F 0L0 =
Therefore;
π1 = ∆pγ −1∆z −1
π1 =
∆p
γ∆z
∆p = f (γ∆z ) (answer)
2. Water sloshes back and forth in a tank as shown in Fig. P7.13. The frequency of sloshing, ω, is
assumed to be a function of the acceleration of gravity, g, the average depth of the water, h, and
the length of the tank, L. Develop a suitable set of dimensionless parameters for this problem using
g and L as repeating variables.
Given:
frequency of sloshing, ω
1
ME 424 – Fluid Machinery
Placement:
4th Year Mechanical Engineering Student
Subject: Additional Example for Dimensional Analysis
acceleration of gravity, g
average depth of the water, h
length of the tank, L
Required:
set of dimensionless parameters
Solution:
The variables with units and dimensions are listed below.
S. No.
Variable
1
ω
2
g
3
h
4
L
Unit
1/s
m/s2
m
M
Dimension
1/T
L/T2
L
L
There are four parameters and two dimensions. So two π term can be identified. Repeating variable
g and L.
π 1 = ωg aLb
1 La b
⋅
⋅L
T T 2a
Equating the indices of L and T,
(3) a + b = 0
(4) -1 – 2a = 0
Then a = -1/2, and -1/2 + b = 0; b = 1/2.
L0T 0 =
Therefore;
π1 = ωg −1 2L1 2
π1 =
ωL1 2
g1 2
Also
π 2 = hgaLb
La b
⋅L
T 2a
Equating the indices of L and T,
(5) 1+a + b = 0
(6) – 2a = 0
Then a = 0, and 1 + 0 + b = 0; b = -1.
L0T 0 = L ⋅
Therefore;
π 2 = hg 0L−1
π2 =
h
L
2
ME 424 – Fluid Machinery
Placement:
4th Year Mechanical Engineering Student
Subject: Additional Example for Dimensional Analysis
(π1 ,π 2 )
 ωL1 2 h 
 12 , 
g
L 

Or
ωL1 2
h
= f 
12
g
L
3. Assume that the flowrate, Q, of a gas from a smokestack is a function of the density of the ambient
air, ρa, the density of the gas, ρg, within the stack, the acceleration of gravity, g, and the height and
diameter of the stack, h and d, respectively. Use ρa, d, and g as repeating variables to develop a set
of pi terms that could be used to describe this problem.
Given:
flowrate, Q
density of the ambient air, ρa
density of the gas, ρg
acceleration of gravity, g
height and diameter of the stack, h and d
Required:
set of pi terms
Solution:
The variables with units and dimensions are listed below.
S. No.
Variable
Unit
Dimension
3
1
Q
m /s
L3/T
3
2
kg/m
M/L3
ρa
3
kg/m3
M/L3
ρg
2
4
g
m/s
L/T2
5
h
m
L
6
d
m
L
There are six parameters and three dimensions. So three π term can be identified. Let ρa, d and g
the repeating variables.
π 1 = Qρaad bg c
a
c
L3  M 
 L 
⋅  3  ⋅ (L )b ⋅  2 
T L 
T 
Equating the indices of M, L and T,
(1) 3 – 3a + b + c = 0
(2) a = 0
(3) -1 – 2c = 0
Therefore, c = - ½ ; 3 – 3(0) + b – ½ = 0, b = - 5/2
M 0L0T 0 =
π 1 = Qρa0d −5 2g −1 2
π1 =
Q
d g
52 12
3
ME 424 – Fluid Machinery
Placement:
4th Year Mechanical Engineering Student
Subject: Additional Example for Dimensional Analysis
Then
π 1 = ρg ρaad bg c
a
c
M M
 L 
⋅
⋅ (L )b ⋅  2 
3  3 
L L 
T 
Equating the indices of M, L and T,
(1) 1 + a = 0
(2) – 3 – 3a + b + c = 0
(3) – 2c = 0
Therefore, c = 0, a = - 1, - 3 – 3(-1) + b + 0 = 0, b = 0
M 0L0T 0 =
π 2 = ρg ρa−1d 0g 0
π2 =
ρg
ρa
And
Then
π 3 = hρaad bg c
a
c
M
 L 
M 0L0T 0 = L ⋅  3  ⋅ (L )b ⋅  2 
L
 
T 
Equating the indices of M, L and T,
(1) a = 0
(2) 1 – 3a + b + c = 0
(3) – 2c = 0
Therefore, c = 0, 1 – 3(0) + b + 0 = 0, b = -1
π 3 = hρa0d −1g 0
π3 =
h
d
π 1 = f (π 2 ,π 3 )
Answer:
ρ h
Q
= f  g , 
52 12
d g
 ρa d 
4