## Tuesday, November 22, 2011

### MrsDrPoe: Dimensional Analysis, Part II

Happy Thesis Tuesday!  Today we'll resume our discussion on dimensional analysis by further investigating the Buckingham Pi Theorem.

This theorem is applied to a problem through several simple steps:

Step 1: List all the variables that are involved in the problem.  This step is crucial- without properly identifying all the necessary variables, we will not be able to obtain the correct dimensionless groups.  Each of the properties should be independent.  For example, if density and specific weight are important, we could use rho and gamma, rho and g, or gamma and g, but to use all three would be redundant (since gamma = rho*g).

Step 2: Express each of the variables in terms of basic dimensions.  We will be looking at these quantities in terms of length (L), time (T), mass (M), and temperature (theta), since these are the basic dimensions for SI and BG unit systems.

Step 3: Determine the required number of pi terms.  This number is equal to k-r where k is the number of variables in the problem (from Step 1), and r is the number of reference dimensions required to describe these variables (from Step 2).  Note: if k-r = 1, skip to Step 5.

Step 4: Select a number of repeating variables, where the number required is equal to the number of reference dimensions.  Here we should choose r number of variables from our list of independent variables; there are our "repeating variables."  It is best to choose ones with the simplest representation in basic dimensions (for instance: choose a distance, L, instead of a velocity, L/T).

Step 5: Form a pi term by multiplying one of the non-repeating variables by the product of the repeating variables, each raised to an exponent that will make the combination dimensionless.  For example: Pii = ui*u1^(ai)*u2^(bi)*u3^(ci), where ui is a non-repeating variable, u1, u2, and u3 are repeating variables, and a, b, and c are exponents.  Note: it is acceptable for a, b, or c to equal zero as long as at least one is non-zero.

Step 6: Repeat Step 5 for each of the remaining non-repeating variables.

Step 7: Check all resulting pi terms to make sure they are dimensionless.  It's easy to make mistakes through this process; check your work by writing each of the variables in terms of their basic dimensions (or even units) to see if all the basic dimensions cancel out.

Step 8: Express the final form as a relationship among the pi terms, and think about what it means.  For example: Pi1 = f(Pi2, Pi3, ..., Pik-r), where Pi1 is the pi term containing the dependent variable.

While at this point you may not believe what I said about this process being simple, next week, I'll prove it to you with an example problem.  We'll also look at some common pi terms/dimensionless number in fluid mechanics.  Happy studying!