Here is an example to determine the Reynolds Number given the dynamic viscosity (kg.m-1.s-1) , density rho (Kg/m^3), some length scale D (m) and velocity V (ms-1) . Please find the explanation in the code comments:
%Elementary Dimensions of mu are M,T,L
%mu = M / (LT)
%Elementary Dimensions of density rho are M,L
%rho = M / L^3
%Elementary Dimensions of length scale D is L
%D = L
%Elementary Dimensions of Velocity is L,T
%V = L/T
%There are 4 physcial variables mu,rho,D and V and 3 physical dimensions
% L,T,M
%By Buckingham's pi theorem, there is 1 dimensionless variable
%pi1 = (rho)^a1 * (V) ^b1 * (D) ^ c1 * mu
% Considering dimensions only, 1 = (M / L^3)^a1 * (L/T) ^b1 * (L) ^ c1 * M / (LT)
% or (LT) /M = M^(a1) * L^(-3a1 + b1 +c1) * T^(-b1)
%Hence comparing both sides, we are solving
% a1 = -1 ; -3a1 + b1 +c1 = 1; -b1 = 1;
Coeff = [1 0 0;-3 1 1;0 -1 0];
rhs = [-1 ; 1 ;1];
solution = Coeff\rhs;
fprintf('a1 = %f\tb1=%f\t,c1=%f\t\n',solution)
%The final result is the Reynolds number Re = 1/pi1 = rho*V*D/ mu
Note that it is certainly possible to create some function that accepts the powers of the physical dimensions as the input and auto-generates the required powers to create the dimensionless variables. Here is an example from file exchange that I believe does this :
