2. Reason for Dimensional
Analysis…
Usually we can’t determine all
essential facts based on theory
alone
Experiments!
Erosion Experiment
We can greatly reduce the
number of tests needed by using
dimensional analysis
We can derive easier set-ups
using similarity laws!
3. Similarity Laws
Allow us to use small-scale models and
convenient fluids…
Predict the performance of a PROTOTYPE (Full-size
device) from tests on a MODEL
Geometric Similarity
Kinematic Similarity
Dynamic Similarity
4. Geometric Similarity
Model (m) and its prototype (p) have identical
shapes but differ only in size
GOAL: Flow patterns must be geometrically similar
8. Dynamic Similarity
IN ADDITION TO KINEMATIC SIMILARITY,
corresponding forces are in the same ratio in
both prototype and model
9. Dynamic Similarity
m
p
r
F
F
F
Force Scale Ratio:
Forces acting on fluid element:
2
2
2
4
2
3
2
2
2
3
:
:
:
:
L
V
T
L
T
L
L
ma
F
Inertia
L
F
Tension
Surface
L
E
A
E
F
Elasticity
VL
L
L
V
A
dy
du
F
Viscosity:
pL
pA
F
Pressure:
g
L
mg
F
Gravity
I
T
v
v
E
V
P
G
11. Dimensionless Numbers
These Ratios give rise to dimensionless
numbers:
1. Reynold’s Number – Re or R
- Good for flow through completely filled conduits,
submarine or other object moving through water
deep enough that it does not produce surface
waves, airplane traveling at speeds below that at
which compression occurs
LV
LV
LV
V
L
F
F
V
I
2
2
Re
12. Dimensionless Numbers
- To be dynamically equivalent in system where
viscous and inertia forces dominate:
p
m
p
m
LV
LV
Re
Re
13. Dimensionless Numbers
2. Froude Number – Fr or F
- Good for flow in open channels, wave action set
up by a ship, forces of a stream or bridge pier, flow
over a spillway, the flow of jet from an orifice
gL
V
gL
V
gL
V
L
F
F
Fr
G
I
2
3
2
2
14. Dimensionless Numbers
Note that in some cases all three forces are
dominate: gravity, friction, and inertia
To achieve dynamic similarity, we must satisfy both Re and
Fr
Only way is to use fluids of different kinematic viscosity
2
/
3
p
m
p
m
L
L
15. Dimensional Analysis
Usually we can’t determine all
essential facts based on theory
alone
Experiments!
Erosion Experiment
We can greatly reduce the
number of tests needed by using
dimensional analysis
16. Dimensional Analysis
Technique called Buckingham Pi Theorem
Arranges parameters into lesser number of
dimensionless groups of variables
Based on Mass-Length-Time System (MLT)
Let X1, X2, X3, … , Xn be n dimensional variables
We can write a dimensionally homogeneous
equation relating these variables as…
0
,...,
,
,
, 4
3
2
1
n
X
X
X
X
X
f
17. Dimensional Analysis
k
n
k
n
,...,
0
,...,
,
2
1
2
1
Technique called Buckingham Pi Theorem
We can rearrange this equation into the following
where is another function and each is an
independent dimensionless product of some of the
X’s
18. Dimensional Analysis
Steps in Buckingham Pi Theorem:
Let us focus on an example as we work through the
steps: Drag force (FD) on submerged sphere as it
moves through a stationary, viscous fluid
STEP 1: Identify all variables and count the number
of variables (n)
n = 5 0
,
,
,
,
V
D
F
f D
19. Dimensional Analysis
Steps in Buckingham Pi Theorem:
STEP 2: List the dimensions of each variable in the
MLT system and find the number of fundamental
dimensions (m)
m = 3 (M, L, T)
LT
M
L
M
T
L
V
L
D
T
ML
FD
3
2
20. Dimensional Analysis
Steps in Buckingham Pi Theorem:
STEP 3: Find the reduction number, k
k = Usually equal to m (cannot exceed m,
rarely less than m)
k = try to find m dimensional variables that
cannot be formed into a
dimensionless group
If m are found, then k = m; if not reduce k
by one and retry
21. Dimensional Analysis
Steps in Buckingham Pi Theorem:
STEP 3:
m = 3 (M, L, T)
So k = m = 3!
LT
M
T
L
L
L
M
DV
3
22. Dimensional Analysis
Steps in Buckingham Pi Theorem:
STEP 4: Determine n-k (This is the number of
dimensionless groups needed!)
n-k = 5-3 = 2
Step 5: Select k variables to be primary (repeating)
variables that contain all m (M, L, T) dimensions
V
D
23. Steps in Buckingham Pi Theorem:
Step 5: Select k variables to be primary (repeating)
variables that contain all m (M, L, T) dimensions
Step 6: Equate exponents of each dimension on both
sides of Pi term (M0L0T0):
Dimensional Analysis
D
c
b
a
c
b
a
F
V
D
V
D
2
2
2
2
1
1
1
1
0
0
0
1
1
1
3
1
1
1
1
1
T
L
M
LT
M
T
L
L
L
M
V
D
c
b
a
c
b
a
24. Steps in Buckingham Pi Theorem:
Step 6: Working with the 1st Pi Term:
Dimensional Analysis
1
1
1
1
1
1
1
1
1
1
1
1
1
0
0
0
1
1
1
3
1
1
1
1
1
Re
1
,
1
,
1
0
1
:
0
1
3
:
0
1
:
DV
V
D
c
b
a
c
T
c
b
a
L
a
M
T
L
M
LT
M
T
L
L
L
M
V
D
c
b
a
c
b
a
25. Steps in Buckingham Pi Theorem:
Step 6: Working with 2nd Pi Term:
Dimensional Analysis
2
2
2
2
1
2
1
1
1
1
1
1
1
1
0
0
0
2
1
1
1
3
2
2
2
2
2
2
,
1
,
2
0
2
:
0
1
3
:
0
1
:
V
D
F
F
V
D
b
a
c
c
T
c
b
a
L
a
M
T
L
M
T
ML
T
L
L
L
M
F
V
D
D
D
c
b
a
D
c
b
a
26. Steps in Buckingham Pi Theorem:
Step 7: Rearrange the Pi groups as desired:
Dimensional Analysis
Re
Re
2
2
2
2
1
1
2
2
1
V
D
F
V
D
F
D
D
27. Use dimensional analysis to arrange the following
groups into dimensionless parameters:
(a)
(b)
Example
V
L
V
