 Alright, so in this segment what we're going to do is solve an example problem applying Buckingham Pi. We've looked at the techniques, that's what we did in the last segment, the step-by-step. There are six steps for doing Buckingham Pi analysis and what this does is it provides a mechanism by which we can take the parameters or variables that we might have in an experiment and we can collapse them into non-dimensional groups and and then you can use those to be able to collapse your data and plan your experiments. So we're going to return to the very first thing that we looked at at the beginning of this lecture which was the problem of a drag or of computing or measuring I should say drag on a sphere. And if you recall that was the one where we said that we would have to do 10 to the 4 or 10,000 experiments in order to determine what the relationship of the drag on a sphere is and that makes no sense whatsoever. And so what we are going to do is come up with the Pi variables or the non-dimensional groups that characterize drag on a sphere. And so we talked about the main variables or parameters. We had force on the sphere. We had density of the gas flowing over it or of the liquid, velocity, diameter, and then our dynamic viscosity. And what we want to do in this problem, we are asked to find a set of dimensionless groups. So we will use the Buckingham Pi technique. Let's go step by step. So those are the parameters we see n is equal to 5 and that's the number of parameters for the problem that we're investigating. The primary dimensions for this problem, we don't have temperatures. We don't have to worry about theta for temperature. All we have if you look at all the dimensions of each of the terms with theoretically I should write out what I'm not going to. Actually I'll do that in the next step. But we have mass, we have length, and we have time. So with that we have three primary dimensions. The next step here, looking at our primary dimension, so force, we can rewrite that as mass, length, time squared, velocity is length per unit time, diameter is a length, density is mass per length cubed, and viscosity that is mass per length time. So there we can see we have r equals three primary dimensions because all we see in here is mass, length, and time. So with that step four is to come up with repeating parameters and we should have m equals r equals three repeating parameters. And these are parameters that will reappear in our different non-dimensional groups that we come up with. And four fluid mechanics, quite often good ones to use are the density, the velocity, and the diameter. And people have done this over and over and over and then they conducted the experiments. So that's how they're able to know that these are good ones. The same sort of thing in turbo machinery. You go through Buckingham pie analysis and there are certain ones that you would pick. But that comes through practice. So don't get too worried about how we picked row V and D. It just turns out that they're good ones for fluid mechanics. And then step five is the process of coming up with our dimensionless groups. So we have n minus m is equal to five minus three equals two. So that tells us that we should have two dimensionless groups resulting through this process. And what we have, we said that each of those is going to consist of row V D. And so we combine it with the remaining variables or parameters. So one of them is going to have force, row V D. And the other one is going to have viscosity, row V D. So those are the basis for our two dimensionless groups. And we'll cluster those together and come up with a dimensionless number. Currently they're dimensional, but we have to come up with exponents for that. So let's look at the process of making this dimensionless. So what I should do is correct that. I said dimensional, but at that stage they're really dimensional. Because if I were to combine those in that form they would have dimensions. And we want to make them dimensionless. That's what we're doing now. So pi one, this will be our first pi parameter. And we take our force. And the technique that we do, I leave your first variable or parameter as is. And then raise the other ones to different exponents. So density will be raised to the A, velocity B, diameter to the C. And this is then going to result in. And I'm going to pull in the dimensions for each of the terms that we have in this group. So we have that. And we've raised it to powers A, B, and C. We want to equate this to mass. We want to have zero dimensions. Length. We want to have zero dimensions. And time. We want to have zero dimensions. So we're going to get three equations, three unknowns. And this is the fun part of Buckingham pi analysis, is solving these equations. So we take our mass. And when we look at the mass term. And what you do is you look at all the contributions to mass that are on the left and you equate it to zero on the right. We get a one plus an A is equal to zero. And that tells us right off the bat that A is equal to minus one. And then we go to length. And for length, we have one minus a three A for the second one. And then we have a B. And then we have a C. So that one's going to remain a little inconclusive because we still have a little bit of work to do. So let's park that for now. And let's look at time, which is the third. So on the left hand side, we have a minus two. And then the next place of time appears is in this term. So we have a minus B. And then that's on the right hand side equal to zero. So that tells us that B is equal to minus two. So that's a little bit more information. Now what we can do is we can take the A and the B and plug it into our L equation. So let's go ahead and do that. And from that, we can solve for C. And we get C is equal to minus two. So the pi group, pi one, the one with the force, turns out to be the following. We have force and then density is to the power minus one. So we have density in the denominator. Velocity is to the minus two. So we have velocity squared in the denominator. And then the diameter is raised to the power minus two. And so we have D squared. So that's what we get for our pi one group. And we'll check that, but that should be dimensionless. So let's continue on to pi two. So what we're going to do, we're going to continue using the same base that we had here, but we're going to interchange F. We're going to put in the dynamic viscosity in order to get pi two. So looking at pi two, again I'm going to go through the process of raising these to some unknown power at this point. And our process is determining what A, B, and C is. So let's expand that out. Okay, and we equate that to all of those raised to the zero on the right hand side. Again, going through each of the primary dimensions, we start with mass. And what we have on the left hand side is a one plus an A. And on the right hand side, that's a zero. So from that, we get A is minus one. Looking at the length. Again, we can't conclude anything. Let's move on to time. So from time, we get B equals minus one. Plug A and B back into the length equation. We get C is minus one. So coming back to our pi parameter up here, we can rewrite that. Pi two is we have viscosity and everything else is going to be in the denominator. And so they're all raised to the power one. So we have rho Vd in the denominator. And that is our second non-dimensional pi group through this process for a sphere. So looking at Buckingham pi, the next step is step six, where we have to check the dimensions. So let's do that. So we find those are both dimensionless, which is good. That means that we did our work right. And what we can write is that pi one is going to be equal to some function of pi two. And writing and substituting in pi one and pi two. So we get that for a relationship. And this functional form F, we don't know that yet. This is determined via experiment. Another thing that we can say here, looking at mu over rho Vd, that is one over the Reynolds number. So that's a very common non-dimensional number that we use in fluid mechanics, where the Reynolds number equals rho Vd over mu. So that's good. We get something popping out of there that we recognize. Another thing, rho V squared, that one, and then d squared, this is an area, or it's proportional to area. And this is related to dynamic pressure, which we saw when we looked at Bernoulli's. And so that is related to one half rho V squared. And sometimes we give it the symbol Q infinity for dynamic pressure. So what we're seeing on the left-hand side is something with our drag force divided by Q infinity times some area. And that then is eventually moving in the direction of what we would call a drag coefficient. And that is something, so we could say Cd is proportional to this. There's actually a factor, the factor of one half that is missing from this form of the equation. But just by using dimensional analysis alone, a lot of information is coming out of this process. And so you can see the efficiency and the technique and really the efficiency. I'll write it out in the next paragraph here and then we'll discuss that. So what we have here is that by going through Buckingham Pi, we've been able to show that Cd, our drag coefficient, which we had as being proportional or related to force, rho V squared over d squared, or some area, was a function of 1 over Reynolds number. And so what that means is that this is the functional relationship that we're after. And in order to do the experiments then, all you need to do is you need to vary Reynolds number. And if we do 10 tests, you could have 10 velocities. And with that, you would get some sort of functional relationship with 10 points. Now, I don't know if it's linear or non-linear. So it might be flat. It probably is flat for a drag coefficient. But anyways, let me erase that just in case I goofed up here. The drag coefficient, unless you're going through some sort of transition, usually doesn't change that much. But you may find that you get some sort of relationship like that if you have Cd and then Reynolds number down here. So anyways, what that is showing is that in order to do this experiment, you don't have to do 10,000 experiments like we were originally seeing. You can actually just do 10. And that's going through a little bit of analysis. It goes long ways. And so that is really the value of the Buckingham pie approach. All of a sudden, by going through this approach, we find the important non-dimensional parameters that are governing the experiment that we're looking at. And so that is Buckingham pie. We'll take a look at another example in the next segment and then we'll look at other aspects of dimensional analysis and fluid mechanics.