# Buckingham pi theorem

1. Jun 8, 2015

### 582153236

1. The problem statement, all variables and given/known data
In buckingham pi theorem, you have p=n-k dimensionless groups (π1, π2,...)
where n=number of total variables and k=number of total units among the variables

For example, lets say we want to relate:
ρ~m/L3 (density)
μ~m/L*t (viscosity)
v~L/t (velocity)
d~L (distance)
Da~L2/t (diffusivity)
k~L/t (mass transfer coefficient)

In this case, n=6, k=3 (m, L, t) so we have 3 non dimensional groups.

In the solution to this problem, 3 common variables are chosen, Da, ρ, d, such that:
π1=Daaρbdck
π3=Dagρhdiμ

So in this case, Da, ρ and d are chosen as the common variables among all groups. I am wondering why these three variables were chosen specifically. Can a different combination of variables be chosen to achieve the correct answer? Also, were three variables chosen because there are three units?

2. Jun 9, 2015

### fzero

There are 3 units so we must choose 3 variables that cover all the units in an independent way. By independent, we mean that if we had a 7th variable that was an area $A$ with units of length squared, then we wouldn't want to choose $d$ and $A$, since their units are directly related.

If we let $[a]$ denote the units of the quantity $a$, then we can see that $L = [d]$, $m = [ \rho d^3]$ and $t = [d^2/D_a]$, so we can express each unit in terms of these variables. The powers that appear here will be reflected in the exponents of the $\pi_i$ that you wrote down.

We also see that we could have chosen different variables if we wanted. For example, $[D_a] = L^2/t$ and $[v]=L/t$, so we could have chosen $(d,\rho,v)$ since we can solve $t = [d/v]$ if we wanted. A good choice of variables might be suggested by the context of the problem. For example, we might expect some of the variables to be dependent on the other ones. Then we might choose the independent variables to set the units and use the $\pi_i$ to express the dependent variables.