Dimensional Analysis and General Dimensions

Today I'm gonna start a new section titled Dimensional Analysis, Similitude, and Modeling. Okay. So first off, this dimensional analysis is actually a branch of mathematics. Okay. And the whole point is, I can organize the variables that define my flow into sets of dimensionless groups. I'll show you how to do it. Okay, but before I go ahead with the mathematical approach, I'm gonna talk a minute about why we're doing this. Okay. So far, if you watched all the videos, you have noticed that I obtained some algebraic solutions to the problems that I posed, right? But in some cases, this is not quite possible. Okay. The flow may be too complicated, I may have too many parameters that changes the behavior of my flow, so I may not be able to fully understand my flow with an algebraic equation. Okay, so in this case, what I do is typically I go to the experimental route. So I do some experiments, try to analyze the data, fit some curves to the data, and get some equations based on my experimental data. Now, the problem is, experiments are expensive. Okay. Regardless of what you're dealing with, experiments will take significant amount of time, money, and resources. Okay. This is applicable to academia as well as industry as well. What I would like to do in this analysis is to reduce the amount of time, effort, and money spent on the experiments. Okay. And the first approach is called dimensional analysis. So I'll start by that. So let's start. First of all, this is a simple method of predicting physical phenomena. For instance, if I'm interested in the drag force of an object, okay, F D typically represents the drag force. Okay. This will be a function of multiple parameters. How we in this particular approach, I want to minimize from the real life much more manageable. Okay. Let's say that I have some kind of a length, okay, I have a velocity, and typically fluid effects are represented by the viscosity and density. Okay. So let's say that this is the case. And again, in real life that's much more complicated than this. Okay, the goal here is to collect these one, two, three, four, five parameters into a smaller number. I will quantify soon what I mean by a smaller number non-dimensional terms, so that I can conduct my experiments faster and cheaper. And as you will see soon, these smaller number of non-dimensional terms will be a function of the parameter. In this particular case, it will be two. Okay. So I can reduce from five to two. These dimensionless groups are called the Pi terms. If I have N number of Pi terms, then we'll write this relationship. This Pi N can be represented as a function of Pi 1, Pi 2, Pi 3, all the way to Pi N minus 1. Next step is to find this N value. I mentioned that if we have five parameters, we are gonna end up with two non-dimensional terms, right? However, before I go ahead and find my N value, I will have to introduce a concept to you. Okay. And this concept is called general dimensions. So what I mean by general dimensions is this is related to the units. Every physical parameter has a unit associated with it. Okay. And in fluid mechanics, M (mass), length, and time, or force, length, and time is sufficient to represent flow properties. I would like you to note that there's also temperature. Sometimes we may need to include temperature as well, but I will not be going in that direction today. Okay. So basically there's two units: MLT and FLT. Okay, let me give you an example. For instance, let's look at the density, right? Density is obtained by mass divided by volume, right? The mass, the unit of mass is, in terms of if I’m using MLT system, it will be mass divided by volume, will be length cube, right? So from here I will obtain M L minus 3 as my density unit. Let me do the same analysis for the FLT system. So for that, I will use the same formula, obviously, right? But now, this time around I cannot use mass because I'm using FLT, right? So now what I need to do is, I need to represent mass in terms of a force. So you can see from F is equal to ma; is force will be equal to mass times acceleration, which is L T minus 2, right? So from here, if I go ahead and find my M, I will get myself F L minus 1 T square, right? So then I’ll simply go ahead and insert it there. So it's gonna be F L minus 1 T square. The volume is still L cube, right? And if I go ahead and find it, you will see that I will get obtain F L minus 4 and T square. So that is my density in the FLT system. I would like you to note that I was able to accomplish this because I used something called dimensionally homogeneous. What it means is that the units on the left-hand side must be equal to the units on the right-hand side. This is applicable for all the courses or all the physical phenomena as well.