Okay, welcome back. So the first question we asked at the end of the last segment was, can you ever have a system that has unsteady mass flow and steady energy transfer? And the answer is no, never. Think about it mathematically just in terms of the governing equations. If you have an unsteady mass flow term, what you're saying is you need to include this term explicitly, right? This would be the term from the conversation of mass, and now, the second part of that statement said they're steady energy transfer. So that's saying that the change in the energy associated with control volume 0. You can't possibly have these two assumptions occur at the same time. If there's mass in the control volume, it has energy associated with it. And if that mass transfer is unsteady, then the energy transfer has to be unsteady. And the second question is very similar, when can you have flow only flow into a system and no flow out? The only way you can have mass transfer in one direction is if you retain the unsteady flow terms, that there's mass accumulating in the system. The mirror image to this is what system can actually have only flow out? Well, that would be essentially depleting a tank or something, you know, that sort of example. That only happens if you deplete the mass in the control volume. So again, let's, my example here is if you have only flow in, it's just like when you inflate a balloon. Your conservation of, of mass expression would look like this. Okay, so you would, that's perfectly allowed, but you have to have this term be nonzero. Okay, and we will cover some unsteady equations or unsteady examples. They're pretty tough and for reasons that we can hint at right here. So imagine if you wanted to, to solve this equation. Imagine if you wanted to do some analysis here. Well, you would want to integrate this expression, on both sides. This is the easiest integral we could ever do, this could be easy if we make some very strong bold assumptions, in particular, what happens to the mass flow rate as a function of time. So if you think about our balloon example, it may be a little bit harder to conceptualize blowing up a balloon, but let's imagine deflating a balloon. The flow rate is not steady as a function of time. In particular, as the pressure in the balloon decreases as the balloon deflates, the flow rate decreases as a function of time. So, this term here, we would need to understand how does that term vary as a function of time in order for us to evaluate that integral. So these can be tricky systems if we can't invoke bold assumptions like some sort of time dependence for that mass flow rate. Okay. So we're starting to see the tip of the iceberg on how these system, how are systems that we consider very simple things, inflating a balloon, can be really complicated in terms of the actual analysis. So we're going to do some examples. It's the best way for you to become skilled at using these state relations and the conversation relations. Okay, so let's start with water compressed, this is an example of compressing water, and we're going to start from an initial state where we have saturated water vapor. So, immediately, you should be thinking aha, that has a quality of one, and I have a pressure, so I have quality and pressure, my system is fully defined in terms of being able to find the thermodynamic properties. I'm going to compress it, so we know the pressure should increase, which indeed it does, to a high pressure of 1 megapascal and a temperature of 250 degrees. The water is actually cooled during this compression process, and we are given that cooling, the specific energy of the cooling, not a heat transfer rate. We're given the specific heat transfer. So again, the units and the language here are hopefully cuing you to some very specific interpretations. For example this would tell me that I have a lowercase q. I have the heat transfer normalized for heat per unit mass. Okay. So we're asked to analyze this process, we are trying to understand, does it require work or does it produce work? We don't know. And, evaluate the specific work of the process. So again, specific work, I'll try and make that a lowercase, case w, so you could see it here, is the total work divided by the mass in the system, and then, we're going to sketch the process on a p-v diagram. When you do any analysis, you should use these steps, these procedures. The very first thing you need to do is define a system, fine. We're going to go through and define the system. We know we are compressing by some mechanism, we don't necessarily know how. water. Okay. So, we're going to place that water into some arbitrary volume here, my little piston here. We know that the water is cooled during this process, so we know that we have some heat transfer out of the system. To be more specific, we know that we have a specific heat transfer out of the system. We're going to guess that the work is into the system because we need to make an assumption so we are going to guess that there's work in, and we know that work in is going to have a negative sign associated with it. So, if we get an answer that is negative, then we know our guess was correct, okay? we are going to neglect kinetic and potential energy effects, because in our little conceptual system here, it's a piston, you know, we're not going to have a whole lot of changes in kinetic and potential energy. Okay. so we define the system and we look at the system and again, our tools are going to be the conservation of mass and the conservation of energy as shown here, and the state relations. But we can invoke state relations only on an as-needed basis. Okay, so this is a closed system. We don't see any mass crossing the system boundary, and so, we know that the conservation of mass for this system is really simplified. The mass is a constant, and I'll draw redraw my little figure here of my piston and my control surface. Okay and we're going to look at the conservation of energy, and again, start with the conservation of energy in its most general form, which says that all changes in energy within the controlled mass have to be balanced by the heat transfer and the work transfer that occur on that mass. And again we're given this information, we're going to neglect kinetic and potential energy effect so the left-hand side of this expression becomes this. and what we want to do is we have information it's on per mass basis, we know the mass is constant. So let's go ahead and simplify this expression by dividing it in the left-hand side and right-hand side by the mass in the system. So what we get is the u at the final state. The internal, specific internal energy at the final state minus the specific internal energy at the initial state is simply given by this expression here. Okay. So, we need to know the state. So what do we know? What don't we know? Okay. We're given the heat transfer. We're asked to find the work transfer. We have we need to have information from the initial state and the final state. Do we have enough information in order for us to fully define the initial state and the final state. So that's step three, find any missing or required state information. Well, we go and we look at what we know about state one. At the initial state, at one, we know that the faulty is equal to one and we know that the pressure is equal to 200 kPa. So that system is completely defined. That state is completely defined. At state two, we're told that we have a temperature of 250 degrees C, and we have a pressure of 1 megapascal. we don't know where we are in, with the perspective of the saturation region with this information yet. That's actually pretty advanced skill, to be able to take your state information, specifically pressure and temperature, because those are the only independent parameters. Those are the only variables that we're concerned about whether or not they're independent or dependent, based on state. So, it's only those two that we know are dependent when we're in the saturation region. So we look at these and you, you'd have to go to your tables or your your online your, your worksheets and things like that, and you'd have to be able to identify what state am I in? Am I in the subcooled liquid, am I in the saturation region, or am I in the super heat region? For this problem, I want to simplify the process a little. these state conditions are actually outside the range of what our online calculator will provide, without buying the official full version. So these conditions, sure, they're, they're well-known, with, there are lots of steam table datas, data sets out there that you could look up the information for these two states. But our online calculator can't do it. so what I want to do is tell you they're fully defined. This is in fact in the super, the second state, the final state is in fact in the super heat region. And I'm going to give you the internal energy for each of these states. Knowing that, again, part of your learning process is you're going to have to become familiar with steam tables. Whether they're, you know, tabulated versions, or whether they're online calculators, and understand how to use them and interpret the information. So we will come back to looking at tabulated information in this class, but for right now, I just want to work this example, just saying, hey, look, you can identify that they're two independent intensive properties. For these two states, so the states are fully defined. So you would take this information and look up the internal energy, the specific internal energy, and you would get a value of 2,587.9 kilojoules per kilogram. For state one, and for state two, you would get a value of 2709.9 kilojoules per kilogram. If you have access to a more complete set of steam tables, or if you already have another thermodynamic textbook that has the state steam tables listed tabulated form. You should go ahead and practice looking up this information. If you don't just assume we'll build that a little bit later, how to use the tables in this class. so again right now practice with that if you can, if not don't worry. We'll come back to it. Okay, so now we have the internal energy, for the initial, the final state and the initial state. We can come back to the conservation of energy and we can solve for the variable we were interested in which is the work transfer, and so, I'll just pop that back on the screen here. So again, here's our governing equation if we solve for the work transfer on this screen. We can see that that we have is essentially the work transfer comes under this side, and we end up with mu plus u2. So, this is the equation we're trying to solve. So we go back to take that under the next screen, or under our next slide. And what we have is, and if you plug in the numerical values, we have 93 kilojoules per kilogram. remember that's heat transfer in so it has to have a negative sign associated with it, because that's our sine convention, heat transfer out. So the water is cool during the process. So as we substitute in the numerical values from the state parameters, and from what the given information was, we need to remember that the water was cool during this process. So as we take that value for the heat transfer, we have to include that negative sign, because remember, that's our sign convention. Heat transfer out of the system is negative so we have minus 93 kilojoules per kilogram here. plus 2587.9 minus 2709.9 Both in units of kilojoules, but kilogram and everything has the same units. So everything looks nice and consistent. And we punch that in our calculator and what we get is a value of minus. We can see already, we know it's going to be negative. This term is negative, and this term, the, negative the internal energy at the final state is greater than. The internal energy at the initial state, so we know we're going to have a negative term on the work. What does that mean? That means work in is the correct assumption here, because recall back up here, we assumed, so we assume work in, okay? And so, so we've learned that to compress of, this was a saturated vapor at the initial state. So compress, to compress a saturated vapor to 100, of water to 100 megapascals requires that we put work into the system, even if we cool the vapor during that compression process. Okay, so, what I want you to think about now, we've gone through this analysis. Do you have enough information to determine the total work for the process? We evaluated the specific work, the work that kilojoules per kilogram, the work per unit mass in the system. Do we have enough information to find the total work? If we do, do the calculation. If not, tell me what information you need to find the total work in the system.