How can I develop a dimensionless quantity using several dimensional values?

I need to develop a dimensionless number using gas flow rate, diameter, rotational speed, power and...

Answer 1

[Mukul Rao]

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 :

http://www.mathworks.com/matlabcentral/fileexchange/40890-buckinghams-pi-theorem-in-matlab

[Nasim Hashemi]

Hi Rao Thank you so much for your response :) I will use this code.

[Mukul Rao]

No problem, glad you found the information acceptable .