Buckingham Pi Theorem and Steps for obtaining Pi terms

Okay, so the next question is, how many pi terms am I gonna need for this set of parameters that is given to me? Okay, in order to accomplish that, there's something called Buckingham Pi theorem. Okay, here's what the Buckingham Pi theorem states. If a flow phenomenon depends on k physical parameters that involve R basic dimensions, then there exists only k minus R independent pi terms that can be formed from these physical parameters. So let's explain this for a minute. Okay, what is the k, number of parameters? Okay, so if I go up here, in an example that I've formed at the beginning of this session, so I said that there is an FD, l, velocity, viscosity, and density. So in this particular case, it is five. You can see that I have five parameters in this particular case. Okay, if I go down here, and that involve R basic dimension, dimensions. As we discussed is the general dimensions section, we have M, L, T, or F, L, T. So the R is equal to three in this particular case. Okay, so then there exists five minus three is equal to two independent pi terms that can be formed from these physical parameters. Okay, before solving an example, I want to establish clear steps for you to follow for the success in this particular section. Okay, let me go ahead and write the steps. I'll be right back. Okay, so here are the steps. The first step is identifying what are the parameters that affect my fluid flow. Okay, in this particular course, you will be given this information. However, in real-life applications, if I'm interested in a drag force of an airplane wing, then I have to go ahead and find out what the parameters are that will impact my drag force. Obviously, this can be many, many number of parameters in real life. Okay, step number two, I'm expressing these parameters, these K parameters, each of them in terms of this M, L, T, or F, L, T system. Okay, and we went over this process in the previous session. Okay, after going through step 2, then I go to step 3. And looking at the parameters that have been identified, I do see whether I used all of the M, L, T, or F, L, T systems. Okay, one point I'd like to highlight is, you cannot simply go ahead and pick which is convenient for this particular sets of parameters. As an example, you cannot call force is equal to force and then go ahead and call mass is equal to mass if there are two separate parameters that needs to be analyzed. You have to be consistent, either M, L, T only, or F, L, T only. Okay? Okay, in step number four, you're going to select R value that you identified in number three. Most of the cases, it will be number three, will be the R. Okay, it will be M, L, T or F, L, T. But in real life, most of the parameters will be functions of M, L, T, okay? Or F, L, T. So you're going to select, let's say, three out of these, let's say, five parameters. Okay, and this is the step that I see most of the issues. Okay, so I put here two caveats. Number one, the combined, as a combination of these three parameters, as an example, must contain all the basic dimensions. If your, your case, depends on M, L, and T, the combination of these three parameters must represent M, L, and T. If they do not have mass, for instance, that's not a good selection of this step number four. And the second section is, each parameter, each every single parameter, must be dimensionally independent. So for instance, if I have a velocity one and velocity two, as two parameters that impact my drag force, then I cannot select both the velocities as repeating variables. Okay, so you must be very careful in this particular step. Step number five, let's say I have three parameters, then I'm gonna select one more. With these four parameters, I will go ahead and find my pi terms. And I will show you how to do this with an example. Okay, and then the last step is, I'm gonna repeat the step number five, which is to having these four parameters, as an example, and I'm gonna repeat this k minus R minus 1 times. Let's say that k is 5, R is 3, then I will repeat this only one time, because 5 minus 3 minus 1 is 1.