Buckingham Pi Theorem

The Buckingham Pi theorem is used in dimensional analysis to obtain a set of dimensionless numbers that represent a certain physical model.

The Buckingham Pi theorem states that the total number of relevant parameters (n) can be grouped into (n-m) independent dimensionless groups. The number m is usually equal to the minimum independent dimensions required to specify the dimensions of all the relevant parameters.

Example:

Note: my excuses if this doesn't display right in your browser; I had to use greek symbols in some way.

Determine the dimensionless parameters describing the pressure drop ΔP in a pipe with diameter D, and length L.

    <-- L -->
________________________

D
________________________

First we have to specify the important parameters for flow through a pipe; ΔP: pressure drop, ρ: density, V: velocity, μ viscosity, L: pipe length, D: pipe diameter. The pressure drop ΔP will be a function of five variables: ΔP= f(ρ, V, μ, L, D). If we had chosen more than five variables, they will cancel each other in the dimensional analysis. If we had chosen less than five variables, the system can not be fully described without adding more parameters.

We write down the chosen parameters as functions of their primary dimensions; Mass {M}, Length {L}, Time {t}, and Temperature {T}:
{ρ}={M//L³}
{ΔP}={ML/t²/L²}={M/(Lt²)}
{μ}={M/(Lt)}
{V}={L/t}
{L}={L} {D}={L}

There are three primary dimensions involved: M, L, and t. Therefore, we can reduce the total number of dimensional parameters to (6-3)=3. We can select any set of 3 dimensional parameters that include all the primary dimensions involved in this problem. For instance, we select ρ, V and D: We set up dimensionless Π groups, by combining the parameters selected previously, combined with the other parameters (such as ΔP, μ and L). The first group: Π₁=ρ^aV^bD^cΔP. In this group a, b, and c exponents are needed to non-dimensionalize the group:
{M/L³}^a{L/t}^b{L}^c{ML/t²}=M⁰L⁰t⁰
Therefore: a+1=0, -3a+b+c+1=0, -b-2=0 -> a=-1, b=-2, c=0.

This solves the exponents of the first Π group. Similarly, we can solve the remaining Π groups:
Π₁=ΔP/(ρV²)
Π₂=μ/(ρVD)
Π₃=L/D

The functional relationship can be written as Π₁=f₂(Π₂, Π₃) or:
ΔP/(ρV²)=f₂(μ/(ρVD), L/D)

The second Π term represents a Dimensionless Number called the Reynolds Number, Re. Therefore we can write that the pressure drop in a pipe is dependent on the Reynolds Number, and the ratio of the length and diameter of the pipe:
ΔP/(ρV²)=f₂(Re, L/D).