1. Dimensional Analysis
In engineering the application of fluid mechanics in designs make much of the
use of empirical results from a lot of experiments. This data is often difficult
to present in a readable form. Even from graphs it may be difficult to
interpret. Dimensional analysis provides a strategy for choosing relevant data
and how it should be presented.
This is a useful technique in all experimentally based areas of engineering.
If it is possible to identify the factors involved in a physical situation,
dimensional analysis can form a relationship between them.
The resulting expressions may not at first sight appear rigorous but these
qualitative results converted to quantitative forms can be used to obtain any
unknown factors from experimental analysis.
2. Dimensions and units
Any physical situation can be described by certain familiar properties e.g.
length, velocity, area, volume, acceleration etc. These are all known as
dimensions.
Of course dimensions are of no use without a magnitude being attached. We
must know more than that something has a length. It must also have a
standardised unit  such as a meter, a foot, a yard etc.
Dimensions are properties which can be measured. Units are the standard
elements we use to quantify these dimensions.
In dimensional analysis we are only concerned with the nature of the
dimension i.e. its quality not its quantity. The following common
abbreviation are used:
length = L
mass = M
time = T
force = F
temperature = Q
In this module we are only concerned with L, M, T and F (not
Q). We can represent all the physical properties we
are interested in with L, T and one of M or F (F can be represented by a
combination of LTM). These notes will always use the LTM combination.
The following table (taken from earlier in the course) lists dimensions of
some common physical quantities:
Quantity 
SI Unit 
. 
Dimension 
velocity 
m/s 
ms^{1} 
LT^{1} 
acceleration 
m/s^{2} 
ms^{2} 
LT^{2} 
force 
N
kg m/s^{2} 
kg ms^{2} 
M LT^{2} 
energy (or work) 
Joule J
N m,
kg m^{2}/s^{2} 
kg m^{2}s^{2} 
ML^{2}T^{2} 
power 
Watt W
N m/s
kg m^{2}/s^{3} 
Nms^{1}
kg m^{2}s^{3} 
ML^{2}T^{3} 
pressure ( or stress) 
Pascal P,
N/m^{2},
kg/m/s^{2} 
Nm^{2}
kg m^{1}s^{2} 
ML^{1}T^{2} 
density 
kg/m^{3} 
kg m^{3} 
ML^{3} 
specific weight 
N/m^{3}
kg/m^{2}/s^{2} 
kg m^{2}s^{2} 
ML^{2}T^{2} 
relative density 
a ratio
no units 
. 
1
no dimension 
viscosity 
N s/m^{2}
kg/m s 
N sm^{2}
kg m^{1}s^{1} 
M L^{1}T^{1} 
surface tension 
N/m
kg /s^{2} 
Nm^{1}
kg s^{2} 
MT^{2} 
 3. Dimensional Homogeneity
Any equation describing a physical situation will only be true if both sides
have the same dimensions. That is it must be dimensionally homogenous.
For example the equation which gives for over a rectangular weir (derived
earlier in this module) is,
The SI units of the left hand side are m^{3}s^{1}.
The units of the right hand side must be the same. Writing the equation with
only the SI units gives
i.e. the units are consistent.
To be more strict, it is the dimensions which must be consistent (any set of
units can be used and simply converted using a constant). Writing the equation
again in terms of dimensions,
Notice how the powers of the individual dimensions are equal, (for L they are
both 3, for T both 1).
This property of dimensional homogeneity can be useful for:
 Checking units of equations;
 Converting between two sets of units;
 Defining dimensionless relationships (see below).
4. Results of dimensional analysis
The result of performing dimensional analysis on a physical problem is a
single equation. This equation relates all of the physical factors involved to
one another. This is probably best seen in an example.
If we want to find the force on a propeller blade we must first decide what
might influence this force.
It would be reasonable to assume that the force, F, depends on the
following physical properties:
diameter, d
forward velocity of the propeller (velocity of the plane), u
fluid density, r
revolutions per second, N
fluid viscosity, m
Before we do any analysis we can write this equation:
F = f ( d, u, r, N,
m )
or
0 = f_{1} ( F, d, u,
r, N, m )
where f and f_{1
}are unknown functions.
These can be expanded into an infinite series which can itself be reduced to
F = K d^{m} u^{p}
r^{q} N^{r} m^{s}
where K is some constant and m, p, q, r, s are unknown constant
powers.
From dimensional analysis we
 obtain these powers
 form the variables into several dimensionless groups
The value of K or the functions f and
f_{1} must be determined from experiment. The
knowledge of the dimensionless groups often helps in deciding what experimental
measurements should be taken.
5. Buckingham's
p theorems
Although there are other methods of performing dimensional analysis, (notably
the indicial method) the method based on the Buckingham
p theorems gives a good generalised strategy for
obtaining a solution. This will be outlined below.
There are two theorems accredited to Buckingham, and know as his
p theorems.
1^{st} p theorem:
A relationship between m variables (physical properties such as
velocity, density etc.) can be expressed as a relationship between mn
nondimensional groups of variables (called p
groups), where n is the number of fundamental dimensions (such as mass,
length and time) required to express the variables.
So if a physical problem can be expressed:
f ( Q_{1} , Q_{2} , Q_{3}
,���, Q_{m} ) = 0
then, according to the above theorem, this can also be expressed
f ( p_{1} ,
p_{2} , p_{3}
,���, Q_{mn} ) = 0
In fluids, we can normally take n = 3 (corresponding to M, L, T).
2^{nd} p theorem
Each p group is a function of n
governing or repeating variables plus one of the remaining variables.
6. Choice of repeating variables
Repeating variables are those which we think will appear in all or most of
the p groups, and are a influence in the problem.
Before commencing analysis of a problem one must choose the repeating
variables. There is considerable freedom allowed in the choice.
Some rules which should be followed are
 From the 2^{nd} theorem there can be n ( = 3) repeating
variables.
 When combined, these repeating variables variable must contain all of
dimensions (M, L, T)
(That is not to say that each must contain M,L and T).
 A combination of the repeating variables must not form a
dimensionless group.
 The repeating variables do not have to appear in all
p groups.
 The repeating variables should be chosen to be measurable in an
experimental investigation. They should be of major interest to the
designer. For example, pipe diameter (dimension L) is more useful and
measurable than roughness height (also dimension L).
In fluids it is usually possible to take r, u
and d as the threee repeating variables.
This freedom of choice results in there being many different
p groups which can be formed  and all are valid.
There is not really a wrong choice.
7. An example
Taking the example discussed above of force F induced on a propeller
blade, we have the equation
0 = f ( F, d, u, r,
N, m )
n = 3 and m = 6
There are m  n = 3 p groups, so
f ( p_{1} ,
p_{2} , p_{3}
) = 0
The choice of r, u, d as the repeating
variables satisfies the criteria above. They are measurable, good design
parameters and, in combination, contain all the dimension M,L and T. We can now
form the three groups according to the 2nd theorem,
As the p groups are all dimensionless i.e. they
have dimensions M^{0}L^{0}T^{0} we can use the principle
of dimensional homogeneity to equate the dimensions for each
p group.
For the first p group,
In terms of SI units
And in terms of dimensions
For each dimension (M, L or T) the powers must be equal on both sides of the
equation, so
for M: 0 = a_{1} + 1
a_{1} = 1
for L: 0 = 3a_{1} + b_{1} + c_{1} + 1
0 = 4 + b_{1} + c_{1}
for T: 0 = b_{1}  2
b_{1} = 2
c_{1} = 4  b_{1} = 2
Giving p_{1} as
And a similar procedure is followed for the other p
groups. Group
For each dimension (M, L or T) the powers must be equal on both sides of the
equation, so
for M: 0 = a_{2}
for L: 0 = 3a_{2} + b_{2} + c_{2}
0 = b_{2} + c_{2}
for T: 0 = b_{2}  1
b_{2} = 1
c_{2} = 1
Giving p_{2} as
And for the third,
For each dimension (M, L or T) the powers must be equal on both sides of the
equation, so
for M: 0 = a_{3} + 1
a_{3} = 1
for L: 0 = 3a_{3} + b_{3} + c_{3} 1
b_{3} + c_{3} = 2
for T: 0 = b_{3}  1
b_{3} = 1
c_{3} = 1
Giving p_{3} as
Thus the problem may be described by the following function of the three
nondimensional p groups,
f ( p_{1} ,
p_{2} , p_{3}
) = 0
This may also be written:
8. Wrong choice of physical properties.
If, when defining the problem, extra  unimportant  variables are introduced
then extra p groups will be formed. They will play
very little role influencing the physical behaviour of the problem concerned and
should be identified during experimental work. If an important / influential
variable was missed then a p group would be missing.
Experimental analysis based on these results may miss significant behavioural
changes. It is therefore, very important that the initial choice of variables is
carried out with great care.
9. Manipulation of the
p groups
Once identified manipulation of the p groups is
permitted. These manipulations do not change the number of groups involved, but
may change their appearance drastically.
Taking the defining equation as: f (
p_{1} , p_{2}
, p_{3} ��� p_{mn}
) = 0
Then the following manipulations are permitted:
 Any number of groups can be combined by multiplication or division to
form a new group which replaces one of the existing. E.g.
p_{1} and p_{2}
may be combined to form p_{1a} =
p_{1} / p_{2}
so the defining equation becomes
f ( p_{1a}
, p_{2} , p_{3}
��� p_{mn} ) = 0
 The reciprocal of any dimensionless group is valid. So
f ( p_{1} ,1/
p_{2} , p_{3}
��� 1/p_{mn} ) = 0 is valid.
 Any dimensionless group may be raised to any power. So
f ( (p_{1} )^{2},
(p_{2} )^{1/2}, (p_{3}
)^{3}��� p_{mn} ) = 0 is
valid.
 Any dimensionless group may be multiplied by a constant.
 Any group may be expressed as a function of the other groups, e.g.
p_{2 }= f (
p_{1} , p_{3}
��� p_{mn} )
In general the defining equation could look like
f ( p_{1} ,
1/p_{2} ,( p_{3}
)^{i}��� 0.5p_{mn} ) = 0
10. Common p
groups
During dimensional analysis several groups will appear again and again for
different problems. These often have names. You will recognise the Reynolds
number rud/m. Some
common nondimensional numbers (groups) are listed below.
Reynolds number
inertial, viscous force ratio
Euler number
pressure, inertial force ratio
Froude number
inertial, gravitational force ratio
Weber number
inertial, surface tension force ratio
Mach number
Local
velocity, local velocity of sound ratio
11. Examples
The discharge Q through an orifice is a function of the diameter d,
the pressure difference p, the density r,
and the viscosity m, show that
,
where f is some unknown function.
Write out the dimensions of the variables
r: ML^{3} u: LT^{1}
d: L m: ML^{1}T^{1}
p:(force/area) ML^{1}T^{2}
We are told from the question that there are 5 variables involved in the
problem: d, p, r, m
and Q.
Choose the three recurring (governing) variables; Q, d,
r.
From Buckingham's p theorem we have mn = 5  3 =
2 nondimensional groups.
For the first group, p_{1}:
M] 0 = c_{1}^{ }+ 1
c_{1} = 1
L] 0 = 3a_{1} + b_{1}  3c_{1}  1
2 = 3a_{1} + b_{1}
T] 0 = a_{1}  1
a_{1} = 1
b_{1} = 1
And the second group p_{2} :
(note p is a pressure (force/area) with dimensions ML^{1}T^{2})
M] 0 = c_{2} + 1
c_{2} = 1
L] 0 = 3a_{2} + b_{2}  3c_{2}  1
2 = 3a_{2} + b_{2}
T] 0 = a_{2}  2
a_{2} =  2
b_{2} = 4
So the physical situation is described by this function of nondimensional
numbers,
The question wants us to show :
Take the reciprocal of square root of p_{2}:
,
Convert p_{1 }by multiplying by this new
group, p_{2a}
then we can say
