Buckingham Pi Theorem: Choosing Common Variables

In summary, the Buckingham Pi Theorem states that in a system with n variables and k units, we can form p dimensionless groups where p = n-k. These groups can be used to relate variables with different units. In order to choose the correct variables for the groups, we want them to cover all the units in an independent way. In the example given, Da, ρ, and d were chosen as the common variables because they represent the units of length squared, mass, and distance, respectively. However, different combinations of variables could have been chosen depending on the context of the problem and the relationship between the variables.
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Homework Statement


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, let's 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
π2=Dadρedfv
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?
 
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  • #2
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.
 

Related to Buckingham Pi Theorem: Choosing Common Variables

1. What is Buckingham Pi Theorem and why is it important in science?

The Buckingham Pi Theorem, also known as the π theorem, is a mathematical concept used in dimensional analysis to determine the relationships between physical variables in a problem. It is important in science because it allows for the simplification and generalization of complex systems, making them easier to analyze and understand.

2. How does Buckingham Pi Theorem work?

The theorem states that if there are n variables involved in a problem and they can be expressed in terms of k independent basic dimensions, then there will be n-k dimensionless products (or π terms) that can be formed from these variables. These π terms will have the same value for all physically similar systems, allowing for the reduction of variables in the problem.

3. What are common variables and why are they important in Buckingham Pi Theorem?

Common variables are physical variables that are shared between different systems or problems. They are important in Buckingham Pi Theorem because they allow for the formation of dimensionless π terms, which can then be used to represent the relationships between the variables in a problem.

4. How do you choose common variables in Buckingham Pi Theorem?

To choose common variables, you must first identify the basic dimensions involved in the problem. Then, look for variables that have the same dimensions and can be combined to form dimensionless π terms. These variables should be chosen in a way that eliminates as many dimensions as possible, making the problem easier to solve.

5. What are the limitations of Buckingham Pi Theorem?

Buckingham Pi Theorem is limited to problems that involve physical quantities that can be expressed in terms of a set of basic dimensions. It also assumes that the relationships between variables are linear. Additionally, it may not work for problems that involve non-physical quantities, such as social or economic factors.

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