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 m³s^-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:

1. Checking units of equations;
2. Converting between two sets of units;
3. 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 = f₁ ( F, d, u, r, N, m )

where f and f₁ 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

1. obtain these powers
2. form the variables into several dimensionless groups

The value of K or the functions f and f₁ must be determined from experiment. The knowledge of the dimensionless groups often helps in deciding what experimental measurements should be taken.

5. Buckingham's

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:

So if a physical problem can be expressed:

f ( Q₁ , Q₂ , Q₃ ,………, Q_m ) = 0

then, according to the above theorem, this can also be expressed

f ( p₁ , p₂ , p₃ ,………, Q_m-n ) = 0

In fluids, we can normally take n = 3 (corresponding to M, L, T).

2^nd p theorem

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

1. From the 2^nd theorem there can be n ( = 3) repeating variables.
2. 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).
3. A combination of the repeating variables must not form a dimensionless group.
4. The repeating variables do not have to appear in all p groups.
5. 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 ( p₁ , p₂ , p₃ ) = 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 M⁰L⁰T⁰ 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 = a₁ + 1

a₁ = -1

for L: 0 = -3a₁ + b₁ + c₁ + 1

0 = 4 + b₁ + c₁

for T: 0 = -b₁ - 2

b₁ = -2

c₁ = -4 - b₁ = -2

Giving p₁ 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 = a₂

for L: 0 = -3a₂ + b₂ + c₂

0 = b₂ + c₂

for T: 0 = -b₂ - 1

b₂ = -1

c₂ = 1

Giving p₂ 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 = a₃ + 1

a₃ = -1

for L: 0 = -3a₃ + b₃ + c₃ -1

b₃ + c₃ = -2

for T: 0 = -b₃ - 1

b₃ = -1

c₃ = -1

Giving p₃ as

Thus the problem may be described by the following function of the three non-dimensional p groups,

f ( p₁ , p₂ , p₃ ) = 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

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 ( p₁ , p₂ , p₃ ……… p_m-n ) = 0

Then the following manipulations are permitted:

1. Any number of groups can be combined by multiplication or division to form a new group which replaces one of the existing. E.g. p₁ and p₂ may be combined to form p_1a = p₁ / p₂ so the defining equation becomes
   f ( p_1a , p₂ , p₃ ……… p_m-n ) = 0
2. The reciprocal of any dimensionless group is valid. So f ( p₁ ,1/ p₂ , p₃ ……… 1/p_m-n ) = 0 is valid.
3. Any dimensionless group may be raised to any power. So f ( (p₁ )², (p₂ )^1/2, (p₃ )³……… p_m-n ) = 0 is valid.
4. Any dimensionless group may be multiplied by a constant.
5. Any group may be expressed as a function of the other groups, e.g.
   p₂ = f ( p₁ , p₃ ……… p_m-n )

In general the defining equation could look like

f ( p₁ , 1/p₂ ,( p₃ )ⁱ……… 0.5p_m-n ) = 0

10. Common

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 non-dimensional 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^-1T^-1

p:(force/area) ML^-1T^-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 m-n = 5 - 3 = 2 non-dimensional groups.

For the first group, p₁:

M] 0 = c₁ + 1

c₁ = -1

L] 0 = 3a₁ + b₁ - 3c₁ - 1

-2 = 3a₁ + b₁

T] 0 = -a₁ - 1

a₁ = -1

b₁ = 1

And the second group p₂ :

(note p is a pressure (force/area) with dimensions ML^-1T^-2)

M] 0 = c₂ + 1

c₂ = -1

L] 0 = 3a₂ + b₂ - 3c₂ - 1

-2 = 3a₂ + b₂

T] 0 = -a₂ - 2

a₂ = - 2

b₂ = 4

So the physical situation is described by this function of non-dimensional numbers,

The question wants us to show :

Take the reciprocal of square root of p₂: ,

Convert p₁ by multiplying by this new group, p_2a

then we can say