2. Lecture-24

So, the 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. So, first step we have done. We have listed all the variables that are involved in the problem.

In the first step of dimensional analysis, we need to identify and list all the variables relevant to the flow problem we are studying. In this case, the variables include:

1. Pressure per unit length (where we need to find out its value),
2. Diameter of the pipe (D),
3. Density of the fluid (ρ),
4. Viscosity (µ),
5. Velocity of the fluid flow (V). Understanding these variables is critical because they will form the basis for the next steps in our analysis.

Think of this step like preparing a shopping list before going grocery shopping. If you want to make a meal, you first need to know which ingredients you'll need, just like we need to know which variables are relevant for our analysis.

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. So, this is step 2.

In this step, we convert each variable into its base dimensions, which helps us analyze the relationships between them. The dimensions are as follows:

1. Velocity (V) has dimensions of Length over Time (LT⁻¹).
2. Viscosity (µ) has dimensions of Force over Length squared and Time (FLT⁻²).
3. Pressure per unit length (∆pl) is a force per unit length which results in dimensions of Force per Length (FL⁻²).
4. Density (ρ) has dimensions of Mass per Volume (ML⁻³), which can be rewritten in dimensional terms as FL⁻³T².
5. Diameter (D) is simply a length (L). Understanding these dimensions is crucial for the next steps in dimensional analysis.

Imagine you are constructing a new piece of furniture. Just like you would decide on the material dimensions (like feet, inches, and pounds), in dimensional analysis, we are breaking down complex measurements into their simpler, base types for clarity.

Now, step 3 says, determine the required number of pi terms. So, in our, in this particular case, the basic dimensions are F, L, T. So, that means basic dimensions are 3 total, same thing what I have written before. Then the number of Pi terms are the number of variables, 5 minus the number of basic dimensions, 3. So, there should be 2 Pi terms for this case. We have already written it and seen it. So, 3 steps we have done. First step was, determining the, writing the all the variables, second writing their dimensions and third number determining the number of Pi terms. This forms very crucial first 3 steps.

In Step 3, we calculate how many dimensionless parameters, called Pi terms, we can derive from our variables. To do this, we use the Buckingham Pi theorem, which states that the number of Pi terms (π) equals the number of variable dimensions (k) minus the number of fundamental dimensions (r). Here, we have:

• Total variables (k) = 5
• Basic dimensions (r) = 3 (Force, Length, Time) So π = k - r = 5 - 3 = 2. This indicates we will form 2 dimensionless groups to describe the system.

Think of this step as narrowing down the most critical aspects of what makes a great recipe. Just as you focus on key ingredients instead of every possible herb or spice, we focus on key dimensionless groups that describe fluid flow instead of all variables.

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 this step, we choose the repeating variables, which must equal the number of fundamental dimensions. Here, we have 3 reference dimensions (length, force, and time), so we need to select 3 repeating variables from our previously identified variables. These should be dimensionally independent; that is, none should be able to be derived from the others. This is crucial for forming dimensionless groups that properly characterize the system under study.

Selecting repeating variables can be compared to selecting primary ingredients when baking a cake. Just like you need flour, sugar, and eggs as your core ingredients, we select core variables that will help us build our dimensionless relationships.

So, now, the step 5. So, step 4 was, we have to select the number of repeating variables, we have chosen that. 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 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. 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.

Step 5 involves creating dimensionless Pi terms from our selected variables. You multiply one of the non-repeating variables (like pressure drop) by the product of the repeating variables to the appropriate powers. For example, if we denote the non-repeating variable as ∆p and repeating variables as D, V, and ρ, we can express the first Pi term as:

• Pi1 = ∆p * D^a * V^b * ρ^c where a, b, and c are unknown exponents to be determined. This term's goal is to be dimensionless, leading us to find a specific functional relationship amongst the variables.

Think of making a cocktail where you need to combine various ingredients in the right proportions to achieve the perfect taste. Here, you want to mix your repeating variables (like base spirits) and non-repeating (flavor elements) to create balanced Pi terms that make sense dimensionally.

Now, the step 6 of a general procedure is, we have to repeat this step 5, if there are more number of repeating variables that are left. So, we have to repeat step 5 for each of the remaining repeating variables. There will be cases where you will not get only 2 dimensionless group, but let us say, 3, 4, 5 depends how many. In another way, we could also have chosen DV and µ, as another repeating group.

In Step 6, if there are still non-repeating variables available after forming the initial group of Pi terms, we repeat Step 5 for those variables. This process allows us to generate additional dimensionless groups, which may represent different physical phenomena or conditions in our analysis.Additionally, it is essential to maintain the distinction between different repeating variables to avoid implicit equations, which complicate the analysis.

Think about preparing a comprehensive report where you've gathered various data points for your analysis. Just as you delve deeper by analyzing each category for more insights, you continue forming Pi terms to achieve a complete understanding of the situation.

Now, step 7 is, you have to check all the resulting pi terms to make sure that they are dimensionless. That is an important step. So, here we check again, we actually have done the same steps, I mean, and same checking of this term in this, I mean, last lecture. So, we can see, Pi 1 is F to the power 0 L to the power 0 T to the power 0 and similarly, this is also 0.

In Step 7, we must confirm that our previously calculated Pi terms are, in fact, dimensionless. We do this by confirming that when we compile the terms into their dimensions, they yield zero for each base dimension (Force, Length, Time). This process ensures that we have correctly followed the methodology of dimensional analysis, allowing us to use Pi terms effectively for physical relationships.

It’s like proofreading a final essay for any grammatical errors. Just as you double-check your work to ensure clarity and correctness, we ensure our Pi terms are dimensionless, confirming we’ve followed proper procedures.

Now, in the end, express the final form as a relationship among Pi terms and think about what it means. So, the final step is Pi 1 is a function of Pi 2, Pi 3. Here, we have only 2 so we can simply write; Pi 1 is a function of Pi 2. So, Pi 1 is a function of Pi 2, or we can simply write in a reverse format as well, D delta L ρ V square is equal to, if this is dimension, so 1 by this is also dimensions so we have written this and this is as I told you, is the Reynolds number.

In this final step, we express the relationship between our dimensionless groups. For example, we can state Pi1 is a function of Pi2, creating an equation that correlates the two Pi terms. This relationship reveals critical insights about the flow characteristics, such as showing that the pressure drop relates to the Reynolds number, which is fundamental in fluid dynamics and helps us understand how different conditions affect flow.

Imagine summarizing a complex story into its key themes or plots. Just like you would distill the essential elements to convey the core message of a story, we summarize the relationships between our Pi terms to understand the fundamental behaviors of fluid dynamics.

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