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Module 4: Applied Fluid Dynamics
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)