Today I want to talk about the last topic in fluid mechanics, topic five, similitude and dimensional analysis. And, in this segment we will be looking at basic dimensions and units, the Buckingham Pi Theorem and similitude. And, now we'll first discuss basic dimensions and units. So, firstly the issues and topics that will be discussed here are similitude and dimensional analysis enables us to analyze fluid flow problems by consideration purely of the dimensions of the variables involved. And the questions that we have are, how can we reduce the number of variables in complex fluid flow problems? And also, how do we design experimental studies of experiments on complex fluid flow, to present the results in the most complex, compact way? And finally, if we make a small-scale model of a fluid flow fluorometer in the laboratory, how do we scale up the results we obtain in the model to full size, and how do we ensure that the small-scale model correctly reproduces the full-scale conditions? So out of these the first three topics falls under the general heading of dimensional analysis, which we'll talk about now. And the last topic comes under the heading of similitude and modeling. Now, firstly, I want to talk about some basic dimensions and units in physical problems. So, we have what are called sometimes the primitives or primary dimensions, are space, which is just the length, time, mass, and force. And all of the variables are generally expressible in terms of these primary dimensions. For example, speed. Speed is just distance over time, or distance per unit time. So in terms of fundamental dimensions, the dimensions are L divided by T, which we can write as LT to the minus 1. And we usually put this into, in these square brackets here, to denote that we're talking about dimensions. So those are primitives or fundamental dimensions, which are differentiated from units. The units are the particular system of units that we use to describe them. For example, speed or velocity would have units of meters per second, feet per second, or miles per hour. Whatever system of units we're using. Now, in the list there, I had four primitives. Space, time, mass, and force. But is force really a fundamental dimension? Well, no it isn't, because it's related to mass and acceleration by Newton's second law of motion. So in other words the dimensions of force are the same as the dimensions of mass times acceleration. So, usually we only have three fundamental dimensions. Either mass, length, and time, or force, length, and time. Now, the other thing which is really important is that any equation must be dimensionally homogeneous. And what that means is that all additive terms in an equation must have the same dimensions. An example is the Bernoulli equation, P over gamma plus v squared over 2g plus z is equal to a constant. And in this case, obviously, we see that we have z here, which is an elevation. In other words, that has dimensions of length, and therefore, all the other terms in this equation must also have dimensions of length. Which indeed they do, you can check that for yourself. And so also must the constant on the other side have dimensions of length. So all additive terms at any physically real equation must have the same dimensions. If they don't, you know there's something wrong with that equation. So all variables that we're interested in can, generally speaking, be expressed in terms of those three primary dimensions, either F L and T or M L and T depending. For example, acceleration is LT to the minus 2. Density is mass per unit volume ML to the minus 3, etcetera. All are expressible in terms of the three fundamental dimensions. Now, the next very important principle in dimensional analysis is the Buckingham Pi Theorem, which was first formulated in 1914. And here is the section from the reference handbook. And the Buckingham Pi Theorem states this. That, let's suppose we have some variable, which we'll denote by U1, is some function of other variables U2, U3, etcetera, up to Un. In other words, U1 is some function of these other variables or it depends on these variables. The theorem states that if R is the number of primary dimensions, for example, mass, length, and time, needed to describe the end variables, then we can write capital pi 1 is some function of pi 2, pi 3, etcetera, up to pi M, where each capital pi here is a dimensionless group; in other words, it has no dimensions, no units. And the number of dimensionless groups m is equal to the number of variables minus the number of dimensions. And this is the basic theorem of dimensional analysis due to Buckingham. Now usually, problems probably won't involve actual dimensional analysis. Usually what we'll do is just invoke a number of very common dimensionless groups that occur in fluid mechanics all the time, which I'll talk about as we go along. And this is a listing of the major dimensionless groups, for example, the Reynolds number is Vl over nu, a Froude number, v over square root of gl. And these are the forces or situations where these dimensionless groups are particularly important, which I'll discuss more when we discuss similitude in a moment. So, most often in problems involving dimensional analysis, We will look for familiar groups from this listing here, rather than try to derive them from first principles. This is the corresponding section from the reference handbook where they list these different dimensionless groups, for example, the ratio of inertia to pressure force is here. It's given by this ratio. And again, I'll discuss this more as we move onto similitude later on. Let's do an example. So the question is the drag force F on a sphere in a fluid flow. So let's suppose we have a sphere here with a fluid flowing past here at some velocity V. The diameter of the sphere is d, and the resulting drag force on this object is F. So we're given that the drag force is a function of diameter, d, velocity, V, and the fluid properties, the density row and viscosity, V, mu. The number of independent dimensionless groups can be formed to describe the results of experiments is what, 1, 2, 3, or 4? So here are the steps to approach a problem like this. The solution is, first we list the variables involved. So F here, and F is called the dependent variable. In other words, it depends on the other variables and the other variables. And the other variables are called the independent variables. So in this case we have the drag force, in some function of diameter, velocity, density, and viscosity. The next step is to list the dimensions of each quantity. So F is a force which is mass times acceleration. So in MLT units the dimensions of force are MLT to the minus 2. D is a diameter, is a linear length. So it's dimensions are length, L. V, velocity, is LT to the minus one. Rho is density, or mass per unit volume, ML to the minus 3. Viscosity, a little bit more tricky, but I'll leave you to convince yourselves of this. The dimensions of viscosity are ML to the minus 1, T to the minus 1. So here is our basic equation for the number of independent dimensionless groups, M is equal to N minus R. In this case M is equal to 5. We have one, two, three, four, five variables. I'm sorry, M is equal to 5, and the number of dimensions involved is 3, mass, length, and time. So therefore, the answer is 2. So, the closest answer is B, two dimensionless groups. And, just to continue with that typical dimensionless groups for this problem would be like this, F over rho d squared V squared is some function of rho VD over mu, would be two acceptable dimensionless groups in this case. Now remember that the Buckingham Pi theorem gives you the minimum number of dimensionless groups which can be formed too. The maximum number is actually greater than that. And these are not the only dimensionless groups that can be formed. But they are two correct ones, for this problem. Normally, we change the definition slightly. And instead of the F over rho d squared V squared, we write it like this. We put in effect to half here, which you might recognize comes from the Bernoulli equation, and instead of D squared, we put an area of, like a frontal area, which is proportional to diameter squared, and that ratio is called CD, the drag coefficient. And the ratio of the right hand side, rho VD over mu, is just the Reynolds number. Which is important in any situation where viscous forces are important. So this case, the problem would reduce to drag coefficient is some function of Reynolds number only. Next question, which of the following is a non dimensional grouping where F is a force, rho is density, A is area, and U is velocity? So we're given four different alternatives here and only one of these is a true non dimensional group. Now in this case none of these groups are exactly the standard form that we had from the list of tables earlier. So just look for one which is similar to one of these groups. And in this case, the one that's most, you would expect, is the drag coefficient F over one half rho AV squared, from which you recognize the closest one that looks like that is B. So, that's a way without actually going through each one to determine its dimensions and make sure it's dimension is, just look for an already familiar group to get your answer. And that will solve many dimensional analysis problems.