Okay, welcome back. So, I asked you to write the simplified conservation of mass equation for a hose being used to water the garden. So you would hopefully had drawn again, we'd talked about this before, get into the practice of drawing your system diagram. So we'll just use some arbitary loop here that shows our water coming in. Okay, okay. So here's my hose connected to a faucet on one end or hydriant on one end. And then we have the water coming out on the other end. So my system boundaries are defined within the hose, up to the exit point of the hose. And we're going to see that system boundary is going to be very important we only want to define boundaries when we know properties, we know states and things like that. That's going to be really important for us to develop again good habits. So we have mass that leaves the system into the garden and we have mass of water that enters the system and that's from the spigot. And since we assume the hose is being used to water the garden, we're not just starting or just finishing. We're going to assume the system is steady or unvarying as a function of time. So, again, to develop good practices always start with the most general form of the equation. Don't simplify it on the fly, that's where you make mistakes. So always start with your expression for control volume which says the change in the control volume, the change of the mass within the control volume is balanced by all the mass in and all the mass out. Then you can simplify, and then you can even keep track of your approximations and simplifications. You can say, hey, that's a steady system. So there is no time variance, and that term is equal to 0. And you can say, hey, each of these summations can be simplified to one term entering and one term leaving. And what we see is that for this simple system there is only one flow rate in the system because mn is the mass flow rate in is equal to mass flow rate out and so we could even drop the subscript notation. OK? So again we're going to gentle our way in or increase the complexity in these systems as we go on. Okay, so for our system, we, for our systems that we're going to consider in this class. We're going to consider only 1D flow. We're going to assume all flow is normal to the boundaries at the inlets and exits and that all intensive properties are uniform across the inlet and exit areas. So you have to ask yourself when you do your analysis, if any of these approximations are invalid you need to really back up and look at maybe a more complex way to interpret a problem, interpreting a system. So you could imagine if you treated all the water entering the Hoover Dam as being at one temperature that, that probably not be a great approximation. There's likely to be quite a bit of variability in the temperature from the top to the bottom. Or from the top of the lake to the bottom of the lake. Things like that. So that's a pretty extreme example. But you do need to understand that we have invoked some pretty severe approximations in this class. Just so we can kind of move forward. Get through this information quickly. Okay, what we do want to talk about the conservation of mass. and a little bit of a multidimensional form again. And this will help you understand when the approximations break down. So I'm going to draw a let me switch back to the red, which might be a little easier for you to see. So assume we have some arbitrary surface. And we have some differential area, some very small portion of that area where we know mass goes across the system boundary. And in this little example, we're going to assume that in fact the mass goes at an, at an angle that's not normal to the surface. So we're going to say that that has a velocity, the mass as it travels through this little piece of the differential area here, actually has an angle associated with it. And then we can set up, you know, the classic geometric form if we set up a normal angle to the surface, that we have a component of that velocity that's normal. And then we have a component of that velocity that's tangential. So that's v tangent, velocity component that's tangential, there. Okay. so again, and this is the velocity associated with the mass that's crossing the system boundary. So, we know that for mass crossing that little differential control volume for some time period Delta T. And we're going to take the limit, you can see I hope where this is headed, where we're going to take the limit where delta T is very, very small. That that amount of mass crossing the system boundary is simply given by the volume of that slant cylinder, so if we project this area here. let me be more, let me be a little bit more clear here. And do this and take these off. And I'm going to simplify this sketch so we can see it perhaps a little better. okay. So if this is the velocity of the mass. This is the velocity vector of the mass exiting that system. We can take this area here and know that it's going to be projected in this direction and that's the slant cylinder. So this is a slant cylinder and it's essentially the cylinder of space or the volume that's swept out by the mass, moving through this little differential control volume. So, we know that the mass, sorry, there we go. The mass crossing that little differential area of the time period delta v is simply the volume of the slant cylinder times the density of the, of the fluid, the cross, so it's product, mass is simply volume times density. So, we can write that volume of the slant cylinder as simply being the product of ooh, sorry. I made an error there, let's fix that too. That's not a volume, that's simply the velocity vector times that delta T. In the normal, so the normal component of the velocity vector times delta t times the differential area, that's the volume of the slant cyllinder. Okay. So if we take both sides of this expression here, so the mass crossing the differential control area, dA, for a time period delta t. Let's take the delta t from each side, so divide by delta t and take the limit. And what we get is now a mass per unit time unit on the left hand side. So we end up with a m dot on the left hand side and on the right hand side, we have an integral over the area. Over the total area of the density times the normal velocity component times dA. Okay, this is really important, this is a fundamental definition of the mass flow rate. Now again, in our class, we're going to assume 1d flow where all flow is normal to the boundaries at the inlets and exits and all the intensive properties are uniform across the system exit. So that allows us to take this expression and simplify it to density. We're going to assume that density is constant across the area but we can drop all of the, normal notation here, like this. So ultimately what we're able to do... Sorry. Let me squeeze my interval in there. Let's do this. And since we're going to further invoke that the velocity is going to be a constant across those differential areas. And even further, assuming constant. [SOUND] V. Constant velocities. We get the more powerful [SOUND] expression where the mass flow rate is simply the product of the density with the velocity of the fluid and the area of the inlet or outlet. Okay. So this is the expression. That's very, very powerful for us. Okay? So, now I want to get into, go back to, you know, let's use all these tools that we have, so we're going to take all this information that we have on state relations. We just revisited the conservation of mass, but then looked at how it's affected by being a control volume. So we're going to work through this problem a bit at a time, so what, our example is going to consider is a steam turbine. So, this is turbines ,the big turbines you find in nuclear power plants, things like that, coal fired power plants. We're given information about the pressure and temperature and the velocity at the entrance to the turbine. We're told it's operating at steady state conditions and that steam leaves a turbine as a dry saturated vapor at a pressure of ten bar. The inlet diameter of the turbine is 0.45 meters and the outlet diameter of the turbine is 0.36 meters. So I'm also going to start introducing you to the way that we we denote some of these, specific types of equipment. So, turbines we typically show as this, trapezoid. Sideways trapezoid [LAUGH]. I had to remember what shape that was. And we typically say mass enters on one side, and we have mass that leaves on the other side. And you can see they typically start with small diameters and end with large diameters. So that's the orientation for the system. And again, we would draw our control surface here. Our control surface around the system. So, I've given you kind of a general layout, a general schematic. the, I'll give you again a kind of a foreshadow. Ultimately, turbines of course are supposed to generate power, so when we start talking about the conservation of energy, we're going to add this energy transfer. we're going to add this notation to show the energy transfer in the system by work transfer. You also should realize what type of work transfer that is. So I want you to think about that too. For right now what I want you to just consider is do you have enough information given what we have in the problem statement for you to determine the mass flow rate of steam through the turbine. If you do I want you to write that expression in variable form. Write down how you would determine the mass flow rate. If you don't, tell me what you need, what additional information do you need in order for you to define that flow rate. And we'll talk about that next time.
