4 Buckingham Pi theorem

As suggested in the last section, if there are more than 4 variables in the problem, and only 3 dimensional quantities (M, L, T), then we cannot find a unique relation between the variables. The best we can hope for is to find dimensionless groups of variables, usually just referred to as dimensionless groups, on which the problem depends. In fact, this is quite a good approach, for reasons that will be discussed in the next section.

Buckingham Pi is a procedure for determining dimensionless groups from the variables in the problem.

4.1 The Rules

Let us assume that there are n = 3 dimensional quantities to consider - mass, length and time. The problem involves m = 6 variables, denoted A . . . F. In general we can derive m - n dimensionless groups, often denoted ₁, ₂..., using the following procedure

Write down the dimensions for all variables A . . . F
Select n of the variables - say A, B, C. These are called the repeating variables, and will appear in all the terms. Note that there are certain restrictions on our choice :
- None of the repeating variables can be dimensionless
- No two repeating variables can have the same overall dimension. For instance, D, the pipe diameter, and r, the roughness height, both have dimension of L, and so cannot both be used as repeating variables.
Select one other variable - say D. Some combination of A, B, C, D is dimensionless, and forms the first term or dimensionless group. We can find the combination by dimensional analysis, by writing the group in the form $a b c d a b c d TT1 = A B C D , so [ ] = [A] [B] [C] [D]$ Equating coefficients gives 3 equations for 4 unknowns, so we can express all the coefficients in terms of just one.
Repeat this procedure with the repeating variables and the next variable, so use A, B, C, E. Continue until no variables are left.
Having worked out all the dimensionless groups, the relationship between the variables can be expressed as a relationship between the various groups. Typically we can write this as one group (for example ₁ as a function of the others $TT1 = f(TT2,TT3,...)$

4.2 Worked example

This is probably best illustrated by a worked example. The head loss in a horizontal pipe in turbulent flow is related to the pressure drop p, and is a measure of the resistance to flow in the pipe. It depends on the diameter of the pipe D, the viscosity and density , the length of the pipe l, the velocity of the flow v and the surface roughness . We start by listing the dimensions of these parameters

D	[L]
v	[LT^-1]
	[ML^-3]
p	[ML^-1T^-2]
	[ML^-1T^-1]
l	[L]
	[ ]

We will choose D, v and as repeating variables. Our first dimensionless group involves p, in the form

a b c d a - 1b -3c -1 -2 d D v r Dp , so [ ] = [L] [LT ] [M L ][M L T ]

This gives the relations

0	= c + d
0	= -b - 2d
0	= a + b - 3c - d

equating coefficients of M, T and L respectively. Writing them all in terms of d,

c	= -d
b	= -2d
0	= a - 2d + 3d - d a = 0

Resubstituting

( )d v-2dr-dDpd = Dp- is dimensionless, so TT1 = Dp rv2 rv2

We can repeat the process with to get the second dimensionless group :

Davbrcmd, so [] = [L]a[LT-1]b[M L- 3]c[M L -1T-1]d

Equating coefficients

0	= c + d
0	= -b - d
0	= a + b - 3c - d

Solving again in terms of d gives c = -d, b = -d and a = -d, thus

( m )d m D- dv-dr-dmd = rDv- is dimensionless, so TT2 = rDv-

In fact, we recognise this group as the Reynolds number, written upside down, so

TT2 = rDv- m

The next dimensionless group will involve l with dimension L. However one of the repeating variables is the diameter D, and so the ratio of the two is already dimensionless. So the next dimensionless group is

TT3 = l- D

Finally,

is already dimensionless anyway, being the ratio of the height of the roughness to the pipe diameter D. so

TT4 = j

From this analysis we have successfully determined that the turbulent flow in a roughened pipe depends on a head loss parameter

TT1 = Dp2 rv

together with the Reynolds number, the ratio of the pipe length to diameter, and the surface roughness coefficient. We can write this relation as a relation between the groups

TT1 = f(TT2,TT3,TT4)

or in full

Dp2 = f(rDv-,-l,j) rv m D

This is as far as we can go using dimensional analysis. Experiment however shows that the pressure drop depends linearly on the length of the pipe, so we can make this relation explicit :

Dp- -l rv2 = D f1(Re,j)

We can rearange this a little further, rewriting it as

( )( ) Dp- v2 l- rg = 2g D f2(Re,j)

which is of course the Darcy equation for head loss in a pipe.

4.3 Helpful hints

There are a number of helpful short cuts that can simplify matters :

If a quantity is dimensionless, it is a term already
If any two variables have the same dimensions, their ratio will be a term
In the final expression, the dimensionless groups can appear in any functional form. In particular, the expression may depend on ^-1 as easilly as on , so any dimensionless group can be inverted if that is more convenient.
Any term may be expressed in terms of the others.

All of these short cuts were used in the previous example.

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