6.3. Resulting Pi Terms

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 dimensions. 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. So, we choose 3 independent variables as repeating variables, so, this is also important. So, those 3 repeating variables should be independent, one should not, one cannot be obtained from the other.

In this chunk, we are discussing the selection of repeating variables for dimensional analysis using Buckingham Pi theorem. The repeating variables must equal the number of reference dimensions; here, we have 3. This means we'll select 3 repeating variables that are independent of one another, meaning no variable should be derivable from the others. For instance, if you consider density, dynamic viscosity, and a characteristic length, choosing two of these as repeating variables might render the third redundant.

Think of selecting team members for a project. You need to ensure that each selected member brings a unique skillset that contributes to the overall team functionality. If one member's skill overlaps significantly with another's, their inclusion may not add value, akin to how repeating variables should not be interdependent.

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 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.

In step 5, we form our first Pi term. This involves selecting a non-repeating variable (for example, pressure drop per unit length) and multiplying it by the product of the selected repeating variables raised to some undetermined exponents. The goal of this multiplication is to form a dimensionless group, meaning that when we analyze the dimensions, they should cancel out, resulting in a unitless term.

Consider mixing ingredients for a smoothie. You choose one primary fruit (the non-repeating variable) and a mix of spices or additives (the repeating variables). The proportions of fruit and additives must be balanced so that the result is a delicious smoothie with consistent flavor, similar to how dimensional analysis requires balancing variables to achieve a dimensionless term.

So, step 5 still continuing because we are forming the, so, you see, I will just take. So, F appears here, so, F to the power 1 and F to the power c. So, this will be, there is only 2 Fs, so it will become 1 + C is equal to 0. For L, it is going to be minus 3 + a + b – 4c is equal to 0. And similarly, for T, minus b + 2c is equal to 0.

After forming a Pi term, we need to determine the exponents by equating the dimensions to zero. For example, if we have 'F' appearing in our term with a power, say 1 for the non-repeating variable and some 'c' for a repeating variable, we set up an equation by summing the powers of 'F' to zero, which helps us solve for unknown exponents a, b, and c. This process is similar for the dimensions of length (L) and time (T).

Think of solving a puzzle where you have to find the right piece (exponent) that completes the picture (dimensionless group). Each piece must fit perfectly with others, just as each exponent must work with the others to balance the dimensional equation.

Now, the 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 confirm that our derived Pi terms are dimensionless. This is important because any term used in analysis must not carry units. We do this by checking that when we sum the powers of mass (F), length (L), and time (T) in the Pi terms, they equate to zero. If all resulting equations are valid and dimensionless, we can use them confidently in our analysis.

Imagine double-checking a recipe to ensure you didn't leave out any ingredient. Each measurement must be precise and correct for the final dish to turn out as expected. This dimension-checking step ensures your final groups are valid and usable, just like ensuring the accuracy of your recipe.

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