6.1. Variables to be Considered

So, step 1 is, you have to list all the variables that are involved in the problem. In our case, we know that listing the variables was, one is pressure per unit length, something that needs to be find out. Then there is a diameter D, there is the density ρ, then there is viscosity µ and the velocity V.

In the first step of dimensional analysis, it is crucial to identify and list all relevant variables that affect the problem at hand. In the example of pipe flow, we look for the variables that might influence the flow characteristics. Here we have five key variables: pressure per unit length (which we need to determine), diameter of the pipe (D), density of the fluid (ρ), viscosity (µ), and the flow velocity (V). Each of these variables plays a specific role in defining the flow and must be accounted for in our analysis.

Think of a recipe where each ingredient is essential to prepare a dish successfully. If you forget to list one ingredient, the final product may not turn out as expected. Similarly, in fluid dynamics, omitting a variable can lead to incorrect analysis and results.

Second step is, we have to express each of these variables in terms of basic dimensions, very simple to do. So, we have to write down the dimensions of all these 5 variables. So, velocity is LT- 1 here, µ is FL - 2T, so, delta pl is FL – 3. These dimensions, if you recall, we had written some slides ago when we were trying to explain it through experimental procedure and the D is L. Whatever remaining it is ρ, it is FL – 4 T 2.

Once we have listed all the relevant variables, the next step involves expressing each variable in terms of its basic dimensional units. This means identifying the fundamental quantities represented by each variable. For example, velocity (V) has the dimensions of length divided by time (LT^-1), viscosity (µ) is expressed as force times length squared over time (FL^-2T), pressure per unit length (∆pl) is given in force per unit volume (FL^-3), diameter (D) simply has the dimension of length (L), and density (ρ) is mass per unit volume (ML^-3). This helps us form a standardized basis for comparison among different variables.

Imagine trying to compare fruits, but one person describes them by weight, another by size, and another by taste. To have a meaningful discussion, everyone needs to agree on a common metric, like weight or volume. Similarly, converting all variables into basic dimensions allows us to analyze and understand them collectively.

After that, we have to determine the unique number of pi term. So, what we have seen? The unique number of pi terms. How? So, going to tell you now, first I will tell you and then do all the. So, first we have to know what is k, k is total number of variables. So, in our case, 1, 2, 3, 4, 5, or we can count here, as well, 1, 2, 3, 4, 5. So, k was 5. And what our miniµm numbers of reference dimensions that are there? So, that you can see by looking at the dimensions of this variable. So, we have length L also is there, T is also there, F is there. Is there any other term? Every all this 5 of these terms has basic dimensions length T and F, F, L, T. So, that means, suppose if there was no F in any of the 5 variables, then reference dimensions could have been 2. Here, L is also there, T is also their, F is also there, so, r will be 3. So, number of Pi terms is going to be k – r, according to the Buckingham Pi theorem.

In step 3, we calculate the number of dimensionless groups, known as Pi terms, that we can form from the listed variables. We start by determining 'k', which is the total number of variables (5 in our case). Next, we identify 'r', the number of reference dimensions. Given that we have three basic dimensions: length (L), time (T), and force (F), we find that r = 3. The Buckingham Pi theorem informs us that the number of Pi terms we can form is given by the expression k - r. Therefore, in our scenario, we have 5 - 3 = 2 Pi terms, indicating we can formulate two dimensionless groups to effectively analyze our system.

Think of baking a cake. If you have a recipe with 5 ingredients but only have 3 measuring cups to make a standardized measurement, you will need to combine some ingredients (like flour and sugar) into groups to simplify your cooking process. Here, the Pi terms are like your simplified measurements aimed at making the baking process easier and more coherent.

Now, step 4, we have to select a number of repeating variables, where the number required is equal to the number of reference dimension. So, repeating variables will be equal to the number of reference dimension. In this case, we have how many references dimension? 3. So, number of repeating variables that will be there in all the dimensionless Pi terms will be 3, I mean, repeating variables. So, 3 repeating variables and this will depend upon, 3 repeating variables for this case, not always, for this case of pipe flow. And why? Because the repeating variable should be equal to the number of reference dimensions, in our case it is 3.

In step 4, we choose the repeating variables, which are necessary to formulate the Pi terms. The number of repeating variables must correspond to the number of reference dimensions identified earlier, which in this case is three (F, L, T). This selection is critical as each of these repeating variables should be dimensionally independent of one another. The repeating variables help in establishing the relationship among the dimensionless Pi terms in a cohesive manner that reflects the problem dynamics accurately.

Imagine a group of friends working on a team project. Each friend brings a unique skills set, and for the project to be successful, there should be three distinct roles: a leader, a researcher, and a presenter. Each role represents a repeating variable. If any friend attempts to take on multiple roles, the team dynamic suffers, just as we cannot have repeating variables that can be derived from one another in dimensional analysis.

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