Today, I want to discuss topic three of our outline of knowledge area ten, fluid mechanics, the continuity and energy equations. So this is topic three, continuity and energy equations and in this segment, we will first discuss the equations for continuity and conservation of mass and show some applications, and then the energy equation, also known as the Bernoulli Equation. So today, we'll be looking at continuity and mass conservation. This is the relevant section from the reference handbook and it discusses the conservation of mass and continuity in one-dimensional systems. One-dimensional meaning that the properties are essentially constant over the cross section, for example, a flow in a pipe or can be averaged over the cross section. So I consider one-dimensional flow in a pipe. The one dimension being the horizontal distance X, measure down the pipe. So the first thing that's important is that the mass flow rate, which is crossing any cross section in this pipe. For example, if I take a cross section across here, then there's a mass flow rate crossing here and that mass flow rate we usually denote by m dot. And the equation for mass flow rate m dot is rho AV, where rho is fluid density, A the correctional area and V the velocity. So this is a mass flow rate. Therefore, generally speaking, it has dimensions of mass per time, for example, kilograms per second or pounds mass per second. A related quantity is the volumetric flow rate, which we usually denote by Q and the volumetric flow rate is just the cross-sectional area times velocity. And volumetric flow rate has dimensions of volume or length cube per time, for example, cubic meters per second or cubic feet per second, which we often abbreviate as csf or in typical American Engineering Units, gallons per minute. So the relationship between mass flow rate and volume metric flow rate is m dot is equal to rho times Q. For steady flow, in other words, nothing is changing with time. For steady flow, the mass is conserved. So the mass flow rate, which is crossing section one upstream must be exactly the same as the rate at which mass is passing downstream. So therefore, m dot 1 = m dot 2 or rho 1 A1V1 is rho 2 A2V2. And for an incompressible flow, in other words, the density is constant, which is the kind of fluid that we're most concerned with in this course, then this reduced to A1V1=A2V2 or Q1=Q2. In other words, the volumetric flow rate is constant or conserved in the pipe. If the pipe is converging, like we've drawn in the diagram up at the top here. In other words, A2 is less than A1, then V2 is greater than V1, so the flow is accelerating. Conversely, if the flow is diverging, in other words, the cross-sectional area is increasing, then the flow is decelerating. The velocity is decreasing. If we have multiple inlets and outlets, for example, merging or bifurcating flows, such as in this Y shape branch here, then the appropriate equation is that the summation of the mass flow out is the equal to the summation of all of the mass flow rates in or summation in rho Q is equal summation out rho Q. And again, for an incompressible flow, the density is constant and cancels out. Leaves us with the sum of the volume flow rates in is equal to the sum of the volume flow rates out. So in this example, what's entering in here m dot 1 = m dot 2 + m dot 3. Or for an incompressible fluid, Q1=Q2+Q3. Now let's do some simple examples and we're given that the water is flowing through the converging pipe as shown. If the water is incompressible, the velocity is decelerating, accelerating, constant or cannot tell? Well, that's fairly straightforward. In this case, we see that A1, the upstream area is greater than A2. Therefore, from continuity, V1 is less than V2. In other words, the flow must be accelerating, because the area is decreasing. Now a numerical example related to that. We're given that water is flying through the pipe and the mass flow rate upstream is 20 pounds, mass per second, density of water is given and the diameter 2 is 6 inches. What is the velocity V2? It's most nearly, which of these answers? So our solution is first of all, the mass conservation equation, m dot = rho 1A1V1 is rho 2A2V2. So rearranging the downstream velocity V2 is the mass flow rate divided by rho 2A2. Plugging in, we've got that is equal to m dot over rho 2 and the area is pi by 4 d squared. Substituting in the values given, we have 1.63 fit per second, so the closed stanza is B. To continue with that, we can also compute the volume metric flow rate. The volumetric flow rate is the mass flow rate divided by the fluid density, which in this case is 20 divided by 62.4, which is 0.321 cubic feet per second or converting to common engineering units about 144 gallons per minute is the flow rate through the pipe. The next problem, we have water flowing through the bifurcating pipe as shown and we're given that the conditions at station one is 4 meters per second and the area is 0.1 meters squared per second. We're also given that the velocity at V2 is 10.8 meters per second. If the water is incompressible, the velocity at station three, this station is most nearly, which of these? So for the solution, we have an incompressible fluid. Therefore, we can say that the volume flow rate is conserved. In other words, what's flowing in is equal to the sum of what's flowing out. Q1 is Q2+Q3 or Q3, which is what we are trying to find is Q1- Q2. Substituting the definitions of volume flow rate, velocity times area, we get this equation. And finally, substituting for rearranging for V3, the ancient philosophy, which is what we're trying to find is equal to this expression. And plugging in the numbers, we get that, which gives us velocity V3 is 2.63 meters per second. So the best answer is D.
