3.2. Step 2: Express each variable in terms of basic dimensions

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-4T2. So, this is step 2.

In this chunk, we are focusing on the second step of dimensional analysis, which is to express variables in terms of their basic dimensions. Basic dimensions typically refer to mass (M), length (L), and time (T). Each variable in hydraulic engineering can be expressed as a combination of these dimensions.

Here are the variables mentioned:

1. Velocity (V) - This is expressed as LT^-1, meaning it has dimensions of length per time.
2. Viscosity (µ) - This is expressed as FL^-2T, which signifies how the fluid resists flow and is dependent on both mass and time.
3. Pressure per unit length (∆pl) - This is represented as FL^-3, indicating force per length over area.
4. Diameter (D) - This is simply represented as L, as diameter is a length measurement.
5. Density (ρ) - This variable is expressed as FL^-4T², combining mass and volume dimensions with time. The importance of this step lies in converting real-world measurements into a mathematical framework that allows for comparisons and calculations in fluid dynamics.

Imagine you are cooking. Each ingredient has a specific measurement (like cups, grams, and teaspoons). In the same way, in fluid dynamics, each variable has its 'measurement' in terms of basic dimensions. Just as in cooking where you need the right amounts of ingredients to make a good dish, in fluid dynamics, you need to understand the basic dimensions of physical quantities to effectively analyze fluid behavior.

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. So, 2 Pi terms, this is just to explain you.

This chunk discusses the next steps needed for dimensional analysis, particularly focusing on determining the number of Pi terms using Buckingham's Pi theorem. Here's a breakdown of the process:

• Total Number of Variables (k): We start with identifying how many variables are involved in our analysis. In our case, we have five variables (pressure, diameter, density, viscosity, and velocity), so k = 5.
• Reference Dimensions (r): Next, we look at the basic dimensions present in these variables. We identify these as length (L), time (T), and force (F). Since every variable contributes to these dimensions, we determine that r = 3.
• Number of Pi Terms: The Buckingham Pi theorem states that the number of dimensionless Pi terms (π) is equal to the number of variables minus the number of reference dimensions (k - r). Therefore, we have 5 variables minus 3 reference dimensions, resulting in 2 unique Pi terms. This is crucial because these dimensionless groups simplify our analysis significantly.

Think of it like organizing a party. You have a certain number of guests (variables) and specific seating arrangements (dimensions). If the venue can only accommodate three different seating styles (reference dimensions), then the way you organize your guests into those styles (Pi terms) helps to streamline and simplify the party planning. Just as fewer seating styles can make it easier to plan the seating arrangement, having a smaller number of dimensionless groups makes fluid analysis easier.

Now, step 4, we have to select a number of repeating variables, where the number required is equal to the number of reference dimensions. 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.

The focus of this chunk is on selecting repeating variables essential for forming dimensionless Pi terms. The steps are as follows:

• Repeating Variables: According to the Buckingham Pi theorem, the number of repeating variables must equal the number of reference dimensions. Since we identified three reference dimensions (length, time, and force), we must select three repeating variables.
• Importance of Independence: These repeating variables must be dimensionally independent, meaning you can't derive one from the others. This independence is necessary for accurate dimensional analysis and helps create valid dimensionless groups that do not contain redundancy.

Imagine a recipe where you need three core ingredients to make a dish: flour, sugar, and eggs. You can’t substitute one for another because they each serve their unique functions in the recipe. In a similar way, the repeating variables in a dimensional analysis provide unique contributions and must be independent to ensure the recipe for analysis works perfectly.

Now, step 5 is, we have to form a Pi term. Now, how is that is formed? We have to form a Pi term by µltiplying one of the non repeating variables by the product of the repeating variables each raised to an exponent that will make the combination dimensionless. So, this means now, for in our particular case, where we had 5 variables, pressure drop per unit length, µ, D, ρ and V. We chose 3 as repeating variables, which means we were left with 2.

In this step, we learn how to form dimensionless Pi terms using the chosen repeating and non-repeating variables. Here’s how it works:

• Product Formation: You select one of the non-repeating variables (e.g., pressure drop, ∆pl) and multiply it by the product of the repeating variables (e.g., density, diameter, velocity), each raised to a power that needs to be solved.
• Results in Pi Terms: This product must be dimensionless, meaning all dimensions must cancel out. The aim is to create a dimensionless group that conveys a relationship between the variables under consideration. In essence, this step allows us to condense complex relationships into simpler, more manageable formats that can be analyzed further.

Consider making a smoothie. You have a recipe that requires bananas (a non-repeating ingredient) and other ingredients like milk and yogurt (repeating ingredients). To get the perfect smoothie consistency (dimensionless term), you need to find the right ratios (exponents) for each ingredient. Like ensuring the banana balances the other ingredients, forming Pi terms balances physical properties in fluid dynamics for analysis.

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