Today, we're going to start on a new knowledge area. Number 11 on the FE outline. Hydraulics and Hydrologic systems. Here is the outline as presented in the, FE review manual. And the topics are Basic hydrology, Hydraulics, Pumping systems, Water distribution systems, Reservoirs, Groundwater and Storm sewer collection systems. However, again, I will follow a slightly different outline, that makes somewhat more logical sense and deals more with the essential areas. And firstly, and today, we'll start looking at the Momentum Theorem. And then later on we'll look at Flow in Pipes, Analysis of Pumps and Turbines, Flow in Open Channels, Hydrology, and Groundwater. So here again is the outline, and this segment today we will, first complete a basic, principle of fluid mechanics, which is the Momentum Theorem. And here, we will look at the concept of a control volume, the most general form of the Momentum Theorem and then, limited to one-dimensional systems, and finally, do some examples to illustrate it. Now, basically, the Momentum Theorem is Newton's second law. Which was previously applied to, to systems, which we know from solid mechanics is in this form, force is equal to mass times acceleration, as applied to a particular system like a block of wood, a block of metal, or something. We can rewrite that in this form, that acceleration of course is the rate of change of velocity with respect to time, dV by dt. Which in turn, because the mass is constant, we can take the mass inside the parentheses, and rate the, that equation as force, is d by dt of mV. But mV of course, is the momentum of the system. So we can think of Newton's second law as stating that the force is, is equal to the rate of change of momentum of the system. In other words, it's an equation for conservation of momentum. So that is the familiar form of the Momentum Theorem. But, in fluid mechanics, we're not concerned generally speaking with the system. We are concerned with the region, or a volume in the flow. Suppose for example, we had a beam with a flow of fluid around it like air or water if it's a piling in the ocean. Then we're interested in the force exerted by this flowing fluid on the beam. So, individual particles of fluid follow some trajectory and then travel off downstream. And, of course, once they are passed the beam and traveled off into the distance, we are no longer concerned with them anymore. So, what we are concerned with is this region around the beam here. The region, and what occurs inside this region. So, this, in turn, leads us to the concept of a control volume. So, for example, in this case, this region might be defined by a control volume which I'll denote by cv around the beam. And we are concerned with what happens inside this control volume. So the shape of the control volume depends on the particular problem that you're analyzing, and a good choice of the control volume allows a much simpler solution to the problem. So in the first case here, it could be as simple as the flow in a pipe, in which case the control volume is just a cylindrical shape. Or, if it's more complex like a jet engine here, the control volume might completely encompass the jet engine. So, those are fixed control volumes. But the control volume could also be moving or it could be deforming. For example, in the case of the balloon here which is collapsing, the control volume might follow the interior surface of the balloon, so as the balloon contracts, the control volume also contracts. In other words, it deforms. The particular, peculiar properties if you like of a control volume is that generally speaking for us it will be fixed in space but it has the peculiar properties that it allows fluid and matter and mass to pass freely through its boundaries. The most generally statement of the Momentum Theorem in fluid mechanics as applied to a controlled volume is this equation right here, which I'll describe that more in a little bit more detail. This equation here, the Momentum Theorem, we can cons, think of as a fluid mechanical version of Newton's second law. Here is the equation again, the Generalized Momentum Theorem, and in this equation the term on the left hand side is the sum of all the forces. The sum of the forces acting on the fluid in the control volume. The first term on the right hand side, the d by dt term, is the rate of change with time of momentum of the system, or the fluid. Which we won't be concerned with, for reasons that I'll show in a minute. And the final term here. The integral over the control surface is the net flux of momentum, the net flux of momentum out of the control volume. Out of the CV. So the fluid is flowing, it's carrying momentum with it, and that term represents the net flux of momentum. So the next step is that most of the instances that we'll be concerned with here are steady flows. So that simplifies the equation right away. That this term, the first term on the right hand side goes out and leaves us with this equation here. Now, again, the equation in this form is rather complex. It involves a surface integral over on the right hand side. And most of the cases that we'll be interested in, will be applied to one dimensional systems. For example, this strange looking, pipe here. Where we have flows coming in here at V1. Flow going out here. Another flow coming in here, etc. So we have one-dimensional flows, by which I mean we can average the properties over the inlet and outlet stations. And, in that case, the integral term here becomes replaced by simple products and summations. So, for one dimensional systems, our integral term on the right hand side becomes this, where the first term here. The summation over all the outlets of V rho V.A minus the summation over all of the inlets, V rho V.A.. So now you see we no longer have the integrals. All of the integrals have been replaced by simple products. Like velocity times area. Now, that simplifies even further because this term in parentheses here, V.A, the dot product of the area and the velocity, is just the volume flow rate Q. So, our equation becomes what we so, show here. But that again can be simplified because rho Q, density times volume flow rate, is mass flow rate, m.. So, we get to this equation here. And finally, if we have a situation where we have just one inlet. Which I'll label one. And one outlet, which I'll label two. For example, here, I have a pipe which is diverging. I have flow coming in here at velocity V1, exiting at station two at velocity V2. In that case, mass conservation says that the mass flow rate at one is equal to the mass flow rate at two, is equal to m., the mass flow rate through the system. So then, our equation reduces to this, summation f is equal to m., V2 minus V1. Which, again, I can write as rho Q V2 minus V1, where Q is the volume flow rate and the mass flow rate is equal to rho times Q. So this is the final form of the Momentum Theorem, probably the most common way that we'll be using it in the FE exam. And it's important to note, of course, that this is a vector equation, the velocities and the forces are vectors, so generally speaking we'll have to apply this equation multiple times in different directions to find velocities or force components in the different coordinate directions. Here is the corresponding equation from the FE reference manual. And now let's take a look at some examples in the next video.