A.Kourmatzis AMME2261 - FLUID MECHANICS 1 2018
Module 4: Applied Fluid Dynamics 1 - Dimensional Analysis
Module 4: Applied Fluid Dynamics 1-
Dimensional Analysis
These lectures closely follow:
Introduction to Fluid Mechanics (8th Ed.) by Philip J. Pritchard, Wiley, 2010
and have been modified from previous offerings by Dr. Matthew Cleary, AMME
Module 4: Applied Fluid Dynamics 1-
Dimensional Analysis
Topics:
– Experimental fluid mechanics
– The Buckingham Pi Theorem
– Determining non-dimensional groups
– Significant non-dimensional groups
– Flow similarity and model experiments
A.Kourmatzis AMME2261 - FLUID MECHANICS 1 2018
Module 4: Applied Fluid Dynamics 1 - Dimensional Analysis
Introduction
Analytical solution of the fundamental equations of fluid dynamics is very
difficult.
Substantial simplification (e.g. assuming the flow is inviscid) improves the
chances of finding an analytical solution but the application is limited.
Numerical solution of the full, un-simplified equations is possible but limited
by computational power. (4th year elective on CFD).
Experimental fluid mechanics therefore remains as the only viable method
for many flows of practical importance.
But what experiments do we conduct? What properties do we vary?
Pritchard, Chapter 7
A.Kourmatzis AMME2261 - FLUID MECHANICS 1 2018
Module 4: Applied Fluid Dynamics 1 - Dimensional Analysis
Introduction
As an example, how would we determine drag on a sphere?
With basic knowledge of fluids we can guess that drag force depends on:
- the size of the sphere
- the fluid speed
- the fluid viscosity
- the fluid density .
Symbolically or more formally
Naively, we start by varying D. About 10 different diameters should give a
curve of FD vs D (for constant V, m and r )
Then we repeat each those 10 experiments for 9 other values of V (already
100 experiments!), and so on …
The whole exercise requires 104 experiments. Then there is the analysis!
Pritchard, Chapter 7
 rm,,,VDfFD    .0,,,, rmVDFg D
A.Kourmatzis AMME2261 - FLUID MECHANICS 1 2018
Module 4: Applied Fluid Dynamics 1 - Dimensional Analysis
Introduction
Fortunately, there is a much better way.
We can reduce the five variables FD , D, V, m and r to only two non-
dimensional parameters
Conduct experiments varying the independent non-dimensional parameter
on the RHS of equation. 10 experiments should suffice.
Advantages:
1. Fewer experiments. Less data to analyse.
2. Greater convenience. There is no need to find 100 different fluids with
different densities and viscosities.
3. Generalised data set. Not limited to specific spheres and specific fluids.
Pritchard, Chapter 7
.
22 






m
r
r
VD
f
DV
FD
A.Kourmatzis AMME2261 - FLUID MECHANICS 1 2018
Module 4: Applied Fluid Dynamics 1 - Dimensional Analysis
Introduction
The following figure shows the functional relationship (data combined from
various experiments)
Wide range of applicability. e.g. drag on golf ball, hot air balloon, red blood
cell, settling volcanic dust particle, …
Pritchard, Chapter 7
A.Kourmatzis AMME2261 - FLUID MECHANICS 1 2018
Module 4: Applied Fluid Dynamics 1 - Dimensional Analysis
Buckingham Pi Theorem
How do we determine the non-dimensional parameters?
The Buckingham Pi Theorem states that the functional relationship between
n dimensional parameters
can be transformed into a corresponding relationship between n - m
independent, non-dimensional parameters
where m is (usually) the minimum number of dimensions (e.g. M, L, t, T, q)
required to define all the parameters
Buckingham-Pi does not predict the functional form of G; that is what the
experiments are for!
Pritchard, Chapter 7
  0,,, 21 nqqqg 
  0,,, 21  mnG 
.,,, 21 nqqq 
A.Kourmatzis AMME2261 - FLUID MECHANICS 1 2018
Module 4: Applied Fluid Dynamics 1 - Dimensional Analysis
Buckingham Pi Theorem
Pritchard, Chapter 7
• If there are n physical variables describing a physical process which contain m basic
dimensions where n>m, then this process can be described with n-m non-dimensional
quantities.
• We call these “π” groups, so there would be π1 , π2 , πn-m non-dimensional quantities
• For instance if there are five physical variables u1, u2, u3, u4, u5 which contain basic
dimensions of mass (M), length (L) and time (T) then there are 5-3=2 non-dimensional
quantities to describe this situation
• Assuming no ratios between u1, u2 and u3 can be made non-dimensional then the
following two equations can be written in order to derive the two dimensionless
numbers or “π” groups.
• u1, u2 and u3 are called the “repeated variables” and u4 and u5 the independent
variables.
• π1 =(u1)
a(u2)
b(u3)
c(u4)
• π2 =(u1)
d(u2)
e(u3)
f(u5)
A.Kourmatzis AMME2261 - FLUID MECHANICS 1 2018
Module 4: Applied Fluid Dynamics 1 - Dimensional Analysis
Buckingham Pi Theorem
1. List the set of fluid dynamic variables involved. Set size = n.
This requires experience. If an important parameter is omitted an incomplete
picture will emerge. Inclusion of an unimportant parameter results in additional
Pi groups and perhaps the need for additional experiments but is otherwise not a
problem. IF YOU THINK A PARAMETER IS IMPORTANT INCLUDE IT!
2. List the set of primary dimensions (e.g. M, L, t). Set size = m.
3. Work out the dimensions of all the variables from Step 1.
4. Select m variables that between them cannot form a non-dimensional group.
These are called the repeating variables.
5. In turn, combine each of the remaining n – m variables with repeating variables in
such a way that non-dimensional groups are formed.
Pritchard, Chapter 7
A recipe for determining the  Groups
A.Kourmatzis AMME2261 - FLUID MECHANICS 1 2018
Module 4: Applied Fluid Dynamics 1 - Dimensional Analysis
Example 4.1: Drag force on a smooth sphere
Obtain a set of non-dimensional parameters.
Pritchard, Chapter 7
Buckingham Pi Theorem
A.Kourmatzis AMME2261 - FLUID MECHANICS 1 2018
Module 4: Applied Fluid Dynamics 1 - Dimensional Analysis
Example 4.1: (continued)
Pritchard, Chapter 7
Buckingham Pi Theorem
A.Kourmatzis AMME2261 - FLUID MECHANICS 1 2018
Module 4: Applied Fluid Dynamics 1 - Dimensional Analysis
Example 4.1: (continued)
Pritchard, Chapter 7
Buckingham Pi Theorem
A.Kourmatzis AMME2261 - FLUID MECHANICS 1 2018
Module 4: Applied Fluid Dynamics 1 - Dimensional Analysis
Example 4.2 (Extra Practice): A slightly more complex example, but using
exactly the same steps: Determine the non-dimensional numbers of a
heated pipe-flow. The flow has some heat input (q) leading to some
temperature change, the pipe has a diameter (D), length (L), and air
flowing at a velocity (u). Relevant thermal properties of the air are the
specific heat capacity (cp) and the thermal conductivity (k). You should
also take physical properties of density and viscosity to be relevant.
Buckingham Pi Theorem
A.Kourmatzis AMME2261 - FLUID MECHANICS 1 2018
Module 4: Applied Fluid Dynamics 1 - Dimensional Analysis
What if we chose different repeating variables?
In example 4.1 we chose r, V, D as the repeating variables to get
Could we have selected different repeating variables? Yes.
If we had chosen r, m, D as the repeating variables we’d get
Pritchard, Chapter 7







VD
f
DV
FD
r
m
r 22







m
r
m
r VD
f
FD
22
This is OK but since r and m appear in
both  groups we cannot separate the
relative effects of the them.
Buckingham Pi Theorem
A.Kourmatzis AMME2261 - FLUID MECHANICS 1 2018
Module 4: Applied Fluid Dynamics 1 - Dimensional Analysis
Buckingham Pi Theorem
It is possible to select different repeating variables and determine different 
groups. We could then conduct experiments and get different but equivalent
functional forms.
However, experience indicates that the most useful information is obtained
by following the following rules.
1. Never select the dependent variable (e.g. drag force) as a repeating
variable.
2. If possible avoid choosing viscosity as a repeating variable.
3. Usually it works out best to use density, velocity and length as repeating
variables (from experience, this usually results in suitable parameters for
correlations, and also they are easy to measure!).
Pritchard, Chapter 7
A.Kourmatzis AMME2261 - FLUID MECHANICS 1 2018
Module 4: Applied Fluid Dynamics 1 - Dimensional Analysis
Significant  Groups
Important forces in fluid dynamics.
Inertial force: which is proportional to
Viscous force: ``
Pressure force: ``
Gravity force: ``
Surface tension:
Pritchard, Chapter 7
dt
du
Vma r
2233 LV
VL
V
L
t
V
L rrr 
A
dy
du
A m  VLL
L
V
mm 2
ΔP A 2ΔPL
3gLrgVmg r
L
A.Kourmatzis AMME2261 - FLUID MECHANICS 1 2018
Module 4: Applied Fluid Dynamics 1 - Dimensional Analysis
Significant  Groups
Often the inertial force dominates the flow, and we consider the ratio of the
other forces to it.
viscous / inertia :
pressure / inertia:
gravity / inertia:
surface tension / inertia:
Pritchard, Chapter 7
VLr
m
2V
P
r

2V
gL
LV 2r

A.Kourmatzis AMME2261 - FLUID MECHANICS 1 2018
Module 4: Applied Fluid Dynamics 1 - Dimensional Analysis
Significant  Groups
Reynolds number: ratio of inertia force to viscous force
Euler number: ratio of pressure force to inertial force
Froude number: ratio of inertial force to gravitational force
Weber number: ratio of inertial force to surface tension
Pritchard, Chapter 7
μ
ρVL
Re
2
2
1
Eu
ρV
ΔP

gL
V
Fr

r LV 2
We 
A.Kourmatzis AMME2261 - FLUID MECHANICS 1 2018
Module 4: Applied Fluid Dynamics 1 - Dimensional Analysis
Flow Similarity
Many practical devices are too large and expensive to test at full size.
Perform testing on scale models.
How do we predict force on prototype from force measured on scale model?
What conditions must be met to ensure similarity of model and prototype?
1. Geometric similarity : same shape, proportional linear dimensions
2. Kinematic similarity : velocities in same direction and differ by a
constant scale factor
3. Dynamic similarity : forces are parallel and differ by a constant scale
factor
Dynamic and kinematic similarity imply geometric similarity.
Pritchard, Chapter 7
A.Kourmatzis AMME2261 - FLUID MECHANICS 1 2018
Module 4: Applied Fluid Dynamics 1 - Dimensional Analysis
Flow Similarity
From previous example; dimensional analysis revealed that
Dynamic similarity occurs if
Pritchard, Chapter 7
 Re
22
f
VD
f
DV
FD 






m
r
r
prototype
22
model
22
prototypemodel
prototypemodel
Then
ReRe i.e.





























DV
F
DV
F
VDVD
DD
rr
m
r
m
r
A.Kourmatzis AMME2261 - FLUID MECHANICS 1 2018
Module 4: Applied Fluid Dynamics 1 - Dimensional Analysis
Flow Similarity
Example 4.3: (Pritchard)
Pritchard, Chapter 7
A.Kourmatzis AMME2261 - FLUID MECHANICS 1 2018
Module 4: Applied Fluid Dynamics 1 - Dimensional Analysis
Flow Similarity
Example 4.3: (continued)
Pritchard, Chapter 7
A.Kourmatzis AMME2261 - FLUID MECHANICS 1 2018
Module 4: Applied Fluid Dynamics 1 - Dimensional Analysis
Flow Similarity
Example 4.3: (continued)
Pritchard, Chapter 7
A.Kourmatzis AMME2261 - FLUID MECHANICS 1 2018
Module 4: Applied Fluid Dynamics 1 - Dimensional Analysis
Incomplete similarity
It is relatively simple to achieve dynamic similarity when only a single non-
dimensional number is involved.
What about when there are multiple non-dimensional numbers to match?
Consider this example.
Since a ship operates at an interface of two fluids of vastly different
densities, velocity waves (a gravitational phenomena) are produced and this
results in wave drag. Additionally, viscous friction results in viscous drag.
Total drag is the sum of the wave and viscous drag forces.
Pritchard, Chapter 7
A.Kourmatzis AMME2261 - FLUID MECHANICS 1 2018
Module 4: Applied Fluid Dynamics 1 - Dimensional Analysis
Incomplete similarity
Determine  groups using Buckingham Pi:
1. List primary variables: FD, V, l, m, r, g
2. List dimensions: M, L, t
3. List dimensions of primary variables: FD, V, l, m, r, g
4. Select repeating variables: r , V, l
5. Create non-dimensional groups: (exercise to try at home!)
Pritchard, Chapter 7
FD V l m r g
ML / t2 L / t L M / Lt M / L3 L / t2
Fr
gl
VVlVl
lV
FD  22221 Re, , m
r
r
A.Kourmatzis AMME2261 - FLUID MECHANICS 1 2018
Module 4: Applied Fluid Dynamics 1 - Dimensional Analysis
Incomplete similarity
Full dynamic similarity requires Frm = Frp and Rem = Rep
From the Fr equality we get
From the Re equality we get
Pritchard, Chapter 7
p
m
p
m
p
p
m
m
pm
l
l
V
V
gl
V
gl
V
FrFr



p
m
p
m
p
m
p
pp
m
mm
pm
l
l
V
V
lVlV






ReRe
A.Kourmatzis AMME2261 - FLUID MECHANICS 1 2018
Module 4: Applied Fluid Dynamics 1 - Dimensional Analysis
Incomplete similarity
If we now substitute the velocity ratio obtained by Fr number equality we get
A typical model to prototype length scale ratio is 1/100 which would require
that
But mercury is the only liquid with  less than water and even then only by
one order of magnitude. (Also impractical to fill a towing tank with mercury).
So we are unable to achieve full dynamic similarity! Novel approaches
needed, or an excuse to attach lasers to a ship! :
http://www.scienceinpublic.com.au/media-releases/indonesiaferry
Pritchard, Chapter 7
2/3









p
m
p
m
p
m
p
m
l
l
l
l
l
l


1000
1

p
m


A.Kourmatzis AMME2261 - FLUID MECHANICS 1 2018
Module 4: Applied Fluid Dynamics 1 - Dimensional Analysis
Incomplete similarity
In the case of drag on a ship we are able to estimate viscous drag
analytically (3rd year). Therefore model tests are conducted ensuring
similarity of the Froude number while corrections are made between the
model and prototype viscous drag.
Pritchard, Chapter 7
model prototype
A.Kourmatzis AMME2261 - FLUID MECHANICS 1 2018
Module 4: Applied Fluid Dynamics 1 - Dimensional Analysis