Problem 46031. Construct dimensionless parameters

Summary
Write a function to construct dimensionless parameters from a list of variables, a matrix indicating their dimensions, and a vector identifying the variables to use to make the other variables dimensionless. More details are given in "Problem statement" below.
Introduction
Dimensional analysis and the Buckingham π theorem
The Buckingham π theorem is a key result exploited in dimensional analysis in engineering, physics, and applied mathematics. It states that if a problem involves m variables and n physical dimensions (e.g., length, time, mass), then the problem can be described with m-n dimensionless groups, called π groups. In other words, the "physics does not depend on a specific unit system".
To find the dimensionless groups, we follow these steps:
  1. Choose n variables that cannot form a dimensionless group by themselves
  2. Form the product of one of the remaining m-n variables with the variables in step 1, each raised to different exponents
  3. Determine the exponents that would make the product (or group) dimensionless
  4. Repeat steps 2 and 3 for the remaining variables
Example
Suppose we are interested in the drag F (i.e., a force) on a cylinder of length L and diameter d immersed in a fluid of density ρ and dynamic viscosity μ flowing at velocity V. Because the problem involves 6 variables and 3 dimensions (length, time, and mass), it can be described with 3 dimensionless groups. For step 1 above, choose V, d, and ρ. We could choose other lists of 3, but notice that we could not choose V, d, and L because d and L form a dimensionless group themselves (and we have no way of removing mass dimensions.)
For step 2, start with the drag F. Following the step, we get
pi_1 = F V^{a1} d^{a2} rho^{a3}
where a1, a2, and a3 are exponents. Because pi1 must be dimensionless and because the dimensionless of F, V, d, and ρ are ML/T^2, L/T, L, and M/L^3, respectively, we can solve a system of three equations to find the exponents and write
pi1 = F V^{-2} d^{-2} rho^{-1}
which is related to a quantity in fluid mechanics called the drag coefficient. We then repeat steps 2 and 3 with the two remaining variables and find
pi2 = L/d
and
pi3 = mu/(V d rho)
or a geometric ratio and the inverse of the Reynolds number.
Problem statement
Write a function that constructs dimensionless parameters from a cell array vars with the variable names, a matrix dims with the exponents for the dimensions in rows (length, time, mass) for each variable in the columns, and a vector indx indicating the variables chosen in step 1 above. The function should return a cell array Pi with the dimensionless groups in strings and a vector a of size m x m-n with the exponents of the normalizing variables. For the example above, the input could be specified as
vars = {'F', 'V', 'd', 'L', 'rho', 'mu'};
indx = [2 3 5];
dims = [ 1 1 1 1 -3 -1;
-2 -1 0 0 0 -1;
1 0 0 0 1 1];
and the output would be
Pi = {'F V^{-2} d^{-2} rho^{-1}', 'L d^{-1}', 'mu V^{-1} d^{-1} rho^{-1}'}
a = [-2 0 -1;
-2 -1 -1;
-1 0 -1]
Further specifications:
  • If the variables chosen in indx will not work (i.e., the number is incorrect or some form a dimensionless group themselves), return a = NaN and Pi = {'Error in choosing normalizing variables'}.
  • If the exponent of a variable is zero, omit the variable from the string.
  • If the exponent of a variable is one, keep the variable but do not show the exponent.
  • If the exponent of a variable is a positive integer, indicate it with a caret (^) and no braces.
  • If the exponent is anything else, indicate it with a caret and braces.
  • If the exponent is not an integer, express it as a fraction. Hint: MATLAB has a function that will help.
Solve

Solution Stats

62.5% Correct | 37.5% Incorrect
Last Solution submitted on Oct 14, 2024

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