Okay, welcome back. So the question we wanted to answer was, do you have enough information to calculate the total work for the process? The answer is, no, you don't. And you might have thought, well I have the state condition, so I can get the specific volume, which is part of what you need. But, remember, what we calculated, was the work transfer. Everything was on a per unit, mass basis. So, you do have information about the specific volume, at the initial and the final state. But, in order for you to extract the mass from the system, which is what you need to calculate the total work, you need to be able to take from either the initial or final state, you need to know the volume of the system. Okay. So, unless you have the volume, you can't calculate the mass, from the thermodynamic properties. If you're given the volume, you could calculate the mass. And once, you have the mass, you could calculate the total work. So total work means this, there. So, once you have the total volume, and the specific volume, you could calculate the mass. And then, you could use that for the total work term. But with the information you have, and the given problem statement, you don't have enough information to calculate any total quantities. Okay, so now we're gonna go back to that earlier turbine example, we worked on. Where we really looked at only the conservation of mass, at our previous consideration of this problem, so I'm going to remind you what the problem statement was. We were given the entrance conditions to the turbine, we knew the pressure and the temperature, and the velocity of the fluid. We were told it was a steady state system. And we were told the steam, left the turbine at a dry saturated vapor condition, so we knew it had aquality of 1, and a pressure of 10 bar. Using that information, we recognized, that, in fact, we had a fully defined state at the entrance, and a fully defined state at the exit of the turbine. We were able to find the specific volumes. Based on the specific volumes, we were able to find the mass flow rate. And we also needed the diameter of the turbine, inlet and outlet. So that information was also provided for us. So again, we went through, we used this expression here, to find the mass flow rate. And then, we also used that expression in order to find the velocity at the exit. So now what we want to do is, we have all this information, all this state information, mass flow rate information. What's the power generated by the turbine at these conditions? So we want power, so we know it's a work transfer term, on a rate basis. Okay, so we wanna know what's the power generated by the turbine on a rate basis? So the work transfer on a rate basis. So we go back and we look at the conservation of energy, just like we always do. And again, we're gonna start with the most general form, and I'm gonna squeeze it in here. And again, I hope that the more you use this expression, the more instinctive it will be for you to just write this down. And again, the familiarity will really breed good habits here. So again, just like we did for the conservation of mass, if it was steady system there, it has to be a steady state system here in the conservation of energy. We're going to assume it's adiabatic, We are not going to neglect the kinetic energy effects. We wanna keep that for now, because that's actually something that I wanted to explore in this question. So we're gonna neglect only the potential energy at this time. So we're gonna neglect potential energy effects. Okay, so we look at the simplified form of that expression, and it looks just like the one we had developed in the previous segment. We were looking at generic forms, for these different turbo machinery devices. So, in other words, we had the rate of work transfer is governed by the sum of the enthalpy plus the kinetic energy at the inlet, minus those parallel terms At the outlet. Okay. The velocities are all known at both the entrance and the exit. We actually determined that in our previous visitation of this question, when we first looked at it. So we have to go back to our steam tables, and get the enthalpy information, so we could plug it into this expression. So we know that the enthalpy, once we go to that online calculator, does have these conditions, so you can go check that. We find the enthalpy at the inlet, specific enthalpy at the inlet, and the specific enthalpy at the exit. And again, at the entrance state, we know pressure and temperature, it's in the super heat, that's a fully defined state. At the exit, we have the quality and the pressure, that's a fully defined state, so we can find these enthalpies. So now, we go and take these values, and we plug them back into this equation here. And I'm gonna keep the pieces separate for just a moment. And I'm gonna group them, so that, we have the enthalpy terms first. So the 3504 minus the 2777 kilojoules per kilogram, minus the velocity terms, which are going to be 64 meters per second, squared, divided by 2. No. [LAUGH] Sorry, I keep being careful here. Minus 3.75 meters per second, squared, divided by 2, end bracket. So what we have here is the energy associated with the mass, the internal energy of the flow, and the flow work. Right? So this is the energy of flow work, plus internal energy, and this is the energy transfer associated with the kinetic energy, the changes in kinetic energy. So if we go ahead and crank through the math, what we find is this first term is equal to 727 kilojoules per kilogram. And the second term here, is equal to 2.04 kilojoules per kilogram. Okay. We're gonna remember these numbers. We're now gonna plug in the mass flow rate, which we had found earlier. And we go through the math, and at all times, you're being very careful about your units, and making sure everything's consistent. And what we find is that the work transfer for this system is 143,256 kilowatts or 143 megawatts. That is a respectable turbine. And it's also a realistic turbine. It's a realistic power for a turbine. So these turbines, at this point in time, we've developed what they call microturbines, which aren't really micro. They're about the size of a typical bathroom. They're huge, okay? Micro is just relative to their big brothers and big sisters, that are used in the large scale stationary power generation. But those turbines can generate a whole lot of power. And megawatt turbines are pretty common, their pretty well within the comfort zone of what we can design and develop. So that's a pretty big amount of energy that's being delivered out of that turbine. Now, what I wanna do in the next part of this question is shown here. How much of the power, that was generated by that turbine, is attributable to kinetic energy, and how much was attributable to the flow work and the internal energy? So, what we can do is, let's call this the work transfer associated with the kinetic energy, compared to the total work energy generated, or the total power that was generated out of the turbine. So we take that 2.04 and divided it by our total, which is 729 kilojoules per kilogram. And what you find is that's equal to 0.28%. So when we did this calculation, let's go back to the previous slide here. We were looking at velocities, that we might've thought were really respectable, 64 meters per second of steam flow, isa pretty high, fast speed fluid. That's pretty moving, pretty quickly. But, relative to the internal energy, and the flow work in that system, the kinetic energy is negligible, it's trivial. So what we've done, in this process, we've said wow, if only less than 1% of the power out of the turbine is attributable to kinetic energy. In other words, 99.7% of the power generated by that turbine, was due to flow work, and the internal energy of the fluid. So what we've done is justify that whole assumption of, when we consider turbines, we don't care about the kinetic energy. We don't care about the potential energy. It's nothing compared to the 2000 pound gorilla in the room, which is the internal energy, and the flow work. And that's why we also, typically, consider them adiabatic, cuz while there is heat transfer, there's not a lot, relative to the energy transfer of that fluid motion, the fluid transfer, in the system. Okay. You should feel a whole lot of confidence about understanding how we make these assumptions, and feel justified, and how you might justify assumptions, that you invoke in your analysis. So, you can go through, and do the calculations to determine what level of resolution, in all of those perimeters, do you need. And turbines don't need kinetic energy. They're, again, dominated by the fluid transfer facts. So, now what I wanna ask you is, how would heat losses affect the power generated by that turbine? And again, you can think of this just conceptually, like you can imagine the turbine in your mind's eye, and think about what that affect would have on the state parameters. Or, go ahead and look at the conservation of energy equations, and see what would happen mathematically. Any mechanism that works for you is fine, but that's our question to start with next time.
