Understanding Rankine Cycle and Efficiency

All right, well, good morning everybody. I hope you had a nice weekend. Isn't it nice that the world didn't blow up over the weekend? Especially with what's going on in the Middle East these days, the rhetoric gets hot. Hopefully the bombs don't fall. Anyway, yeah, you know, I grew up in the era where we had to hide under our desks in grammar school. You know, as a drill they had every day. Well, not every day, every week. Ah, we'll get into that some other time. So, again, good morning everybody. I have the new homework set here on the board. Five problems from chapter 10. Again, I'll do next Wednesday. As a reminder, last week's homework will be due this coming Wednesday. And please make special note of problem 119 that I modified it a little bit on you, so make sure you're doing the right version of 9-119. Additionally, I just want to remind you that today is the last day to show proof of having met the prerequisites. And I got that for all but one student, Mr. Mumby. So if you don't mind that one student, just make sure you show me before the end of the day. that you've passed the prerequisites for this class. So, with that, let's just get right back to the Rankine cycle. Now, the Rankine cycle is the cycle that we use to model a steam power plant. Hopefully, this was made clear last time. We're no longer dealing with gas problems, right? We're not talking about air standard cycles. These are not gas power cycles. We're dealing with steam now, H2O if you will, that's moving through the cycle. Now last time we looked at the basic schematic. We have a pump that is going to move liquid from a low to a high. pressure. We're then going to take that liquid, which of course is just water, and we're going to add heat to it in a device that we call a boiler, and that turns the water into the vapor phase, so steam. Although it's possible it could be saturated steam, it could be a high-quality two-phase mixture, but certainly the intention is to turn most of the liquid into steam, hopefully all of it into steam. This steam now goes through a turbine. where work is done. And as we expand through the turbine, the pressure drops. And then we go into a heat exchanger, which we call a condenser. And here the heat is removed. Whatever steam we had coming out of the turbine or it could be, again, could be saturated, could be high quality two phase. But nonetheless, we're going to condense the vapor back into a liquid and we go right back into the pump again. So this is our basic ranking cycle. Why don't we just call it a simple ranking cycle? Okay, we've got work input from the pump. We've got work production or work output by the turbine. We have heat input into the boiler. And then we have heat rejection through the condenser. Please keep in mind that unlike the Brayton cycle where the heat. rejection takes place essentially into the atmosphere. Here it generally doesn't go into the atmosphere, at least not directly. The heat would actually be rejected into a separate water supply. We'd actually call it a sink because this is the heat sink, right? The heat rejected from the cycle. goes into that heat sink. So what I show here is kind of a curly line with a couple lines coming off of it. You know, that's supposed to represent some other flow path. This could be water from the local river or the local ocean. More likely it's going to be a clean water supply from a device called a cooling tower. And we'll talk about cooling towers later in this course. But nonetheless, we're going to reject our heat into water typically. A completely separate stream of water. We're not going to worry about it right now. We'll just simply recognize that that's where the heat output is. is going. But this is our simple ranking cycle, okay? And as I mentioned last time, in many ways it's similar to the Brayton cycle, at least in that the heat transfer takes place at constant pressure. And that would apply both to the heat input as well as heat rejection. These are constant pressure processes, so that's similar to the Brayton cycle. And then also work is done isentropically. At least... For the ideal ranking cycle, okay, eventually, actually very shortly, we're going to talk about the non-ideal ranking cycle where the work is not isentropic anymore, right? The work would be first assumed to be isentropic, which will allow us to find the ideal exit conditions from the turbine and the pump. But then we're going to have to apply the isentropic efficiency just like we did for the Brayton cycle to determine the actual outlet conditions from the turbine and the pump and thus to find the actual work. So, this is just review from last time. Please note that I also showed this on a TS diagram relative to the saturation lines. And generally, we would show the minimum and maximum pressure lines. You know, these are lines of constant pressure. So, the maximum and the minimum would be like such. Now, please note that state point one is always going to be a saturated liquid. So, one is always a sat liquid. This is just based on the design and operations of the condenser. The condenser condenses to a saturated liquid, but it doesn't remove more heat than that. Just to the saturated liquid. We pump up to point two to the high pressure line. This is a vertical line. I know the diagram I put on the board last time, some of these lines look a little skewed, but this is supposed to be a vertical line. After all, it's isentropic. We then add heat until... We come out of the boiler into our turbine and then we do work. And again, this is done in isentropic way, at least in the ideal cycle. So, again, a vertical line. Then we reject our heat through the condenser. Now, please note that this diagram I had on the board last time and again from today doesn't represent every possible situation. Granted, points one and two are always going to be as is shown here because point one is always going to be a saturated liquid. But point three doesn't have to be superheated as shown here. Point three could be exactly on the saturated vapor line. Point three could be a two-phase mixture? We really don't know. It just depends on the problem at hand. It depends how much heat we're adding to the water in that boiler. If we add enough heat, then certainly it should be enough to change the phase of all the liquid water into a vapor. If we don't provide enough heat, then we may end up with a high-quality two-phase mixture. And of course, we can't have all these different point threes. It just depends on the problem, right? You'll have either there or there or there. And then, of course, as we expand isentropically, well, again, we could end up... at state point four anywhere. I mean, we could have a state point four right on the saturated vapor line if we had the appropriate state point three. So any of these are options. And quite frankly, it doesn't even matter, right? The only thing that matters is where we're going to get our thermodynamic property data. If it's superheat, we go into table A6. If it's two-phase, then we have to use, or saturated, then we have to use data from A4 or A5. But the basic methods of solution don't change. This just is intended to represent all the possible possibilities as we are over here on the right-hand side of the diagram. Again, though, you're not going to have all those cases in one problem. You'll just have one vertical line. You'll just have to make sure you're really clear when you're solving that problem where the actual thermodynamic state points are. So are there any questions just in general about the nature of the Rankine cycle? Okay. If you want to take a look at a typical Rankine cycle power plant, I'll put a I mean, they're all over the place. Probably the closest big one is Etiwanda, which is where the 10 and the 15 freeways cross, a little bit north of there. You can't miss it as you're driving up the 15. Just look eastbound and you'll see all these big smokestacks and, well, that's a ranking cycle. That's a steam power plant. There's others that are closer, but again, we'll not get into that right now. Anyway. So, let's start looking now at the equations that would be appropriate for this type of analysis. Again, we're specifically looking at the ideal case. And still, though, the simple ranking cycle. In this particular case, let's note again that for all situations, we're assuming there's no change in kinetic potential energy. As usual, right, there's no significant velocity or velocity changes through any of these devices. There's no significant height changes. So those are the are going to be neglected. And of course, we're also going to note that during the work processes, remember in the ideal case, work is done isentropically, which means adiabatic reversibly, which means there's no heat transfer, right? So there's no heat transfer during work. And also during heat transfer, Well, there's no work, right? I mean, heat transfer is done in two very passive heat exchangers. They're just big shells with a bunch of tubes in them. There's no miscellaneous work, right? There's no shaft work. There's no mixing work. There's no work at all. So, there's no work terms. So therefore, let's look at the first law. And of course, the first law, well, regardless of which of the four processes we're dealing with, the first law basically leaves us with nothing but enthalpy change, right? Heat transfer transfer will equal the enthalpy change. Work will equal the enthalpy change. So for the first law, let's now go to each device and we'll start with the pump. We know that the pump is work input. And again, this is done without any heat transfer, with no kinetic and potential energy change. So the first law just tells us this is the enthalpy change. So this is just H2 minus H1. Let's also note, however, that something that you, again, should have covered, which is discussed in, I guess, is chapter 7. I think somewhere near the end of chapter 7 in your textbook. It also shows that when we have a liquid, and certainly this is a liquid, right, as we go from point 1 to point 2, when we have a liquid for the ideal case, not only is the pump work equal to the enthalpy change, but it's also equal to the specific volume at the inlet to the pump times the pressure change. So you can actually use either of these processes, but again, for the ideal case only. So that's how we're going to calculate the pump work using either one of these two equations. When we deal with the turbine, again, the turbine work is our work output. And again, this is just the enthalpy change as we go from 3 to 4. So just H3 minus H4. Keep in mind again, all these terms are absolute values, right? We know the pump work, the magnitude is positive, but it is work input. Technically, it's negative work, but. We'll just let those work terms have their sign included within the various equations like the thermal efficiency equation. Again, a reminder of what we already know. So here's our turbine. This is our turbine work equation. And there's really nothing more we can add to this. this. With regards to the heat transfer, when we look at the boiler, the heat input in the boiler is again just the enthalpy change, so H3 minus H2. And within the condenser, the magnitude of the heat output, the heat rejection, as we go from 4 to 1, it's just H4 minus H1. So these equations are really quite straightforward. And ultimately, of course, we're interested in finding the thermodynamic efficiency. That certainly doesn't change. These are still heat engine cycles. We're still trying to find the relationship between the amount of work we have as our benefit, right, the network output, compared to the amount of heat that we have to pay for, our heat input. So this doesn't change. This is still the network over the heat input. And of course, I could just plug in all the numbers from above if I want to. Maybe I'll just leave it in a more general form. Turbine work minus compressor work over heat input is our thermal efficiency. And of course, all those terms are just based on enthalpy data. And then, And then lastly, I might note that we're not always just interested in work per unit mass or heat transfer per unit mass. We're also often interested in the rate that work is done. In other words, how much power is being produced by this facility or for that matter, what is the rate of heat transfer in the boiler or in the condenser? So in general, we would also just show that the power is the mass flow rate times the work per unit mass and the rate of heat transfer. is the mass flow rate times the heat transfer per unit mass. And again, this would apply both for work in or work out, as well as heat in or heat out. And again, this is just a reminder of information that hopefully you already know. So, perhaps the best way to illustrate how to utilize all these in a problem is to actually go through an example problem. So, we really have everything we need already. Maybe I just want to remind you before I move forward that when we deal with an isentropic work process, whether it's compression from one to two or expansion from three to four, we certainly treat this differently than we did when we were dealing with our gas power cycles. I mean, we're not going to use relative pressure. anymore, relative specific volumes. We're not going to use ideal gas equations that relate pressure and temperature and volume ratios. Those only apply to your ideal gas type problems, right? The ones where we have air standard cycles to model our gas power cycles. Here we have to actually use the thermodynamic property data. We have to remember that it's an isentropic process from one to two, right? We have to look up the value of the entropy at point one and then say, well, S1 equals S2 and then add our known entropy at two and some other property which typically is going to be the peak pressure. Then we go back into our tables and we look up the enthalpy, OK? Same thing from three to four. We know state point three, that's typically given. It's isentropic from three to four, so we look up S3, that equals S4. And at S4 and at the minimum pressure, which would be P4, well, we just look up the enthalpy at point four. So we're going to use our. steam tables. We could also just call them water tables rather than any kind of ideal gas table. So please make sure you're very careful on this. So let's begin. Let's look at problem 10-12 in your version of the textbook. So let's give this a second to pop up. So, as the projector is warming up, the screen is coming down. OK. Hopefully, you've written these equations into your notes already. In fact, maybe I'll stop the screen halfway down. Uh-oh, I pushed the wrong button. So, we'll be able to say many of the equations anyway. Nonetheless, this is problem 1012. So, let me just read it to you briefly. We have a steam power plant that's operating on the simple ideal Rankine cycle between the pressure limits of 3 megapascals and 50 kilopascals. scale. So clearly we're given the maximum and minimum pressures in the cycle. We're given the temperature of the steam at the inlet to the turbine. So that's 0.3 of 300 degrees Celsius. And we know the mass flow rate of the steam through the cycle as 35 kilograms per second. second. Show this on a TS diagram and determine the thermal efficiency and the net power output. So this is a very typical problem of this particular type. I don't think there's any point in me, I take that back. There is a point in me showing this on a TS diagram because we're going to have to figure out which of those four possibilities shown in the previous diagram are accurate. So let's just show this again. Let's show our constant pressure lines, one, two. Over here, we don't know exactly where three and four are yet. We're going to have to figure that out. So let's just proceed with the problem before we finish the diagram. So we have our pressure limits. So, we're given P2 and P3. That's our maximum pressure of 3 megapascal. And we have our minimum pressure, which is P1 and P4, which is 50 kilopascal. We're given the turbine inlet temperature of 300 degrees Celsius. And we're given the mass flow rate for the steam moving through the cycle of 35 kilograms per second. OK. So we want to show it on the TS diagram, find the thermal efficiency, and also find the net power output. OK. So, let's just go through this exactly as I explained just a moment ago. We'll start at state point one. So, we know the pressure at point one. Now, you may say, well, don't we need a second thermodynamic property in order to do determine the state. But we do have that second property, right? We know that for every condenser, it's a saturated liquid coming out of the condenser. So, point one has to be a saturated liquid. OK. even if it doesn't say so, it is, it always is. I mean, it really just has to do with the design of the condenser. The bottom of the condenser is where the liquid resides, right? It's got a higher density. The steam, the vapor in the condenser resides in the top portion of the condenser. So you just draw the water off the bottom of the condenser. It's a two-phase mixture in that condenser, but you're only pulling off the saturated liquid portion. So it has to be a saturated liquid. I mean, it is. It always is. So we know that data. So we're going to have to go into the appropriate table, which is table 8. 5, this is our pressure table. We simply go to 50 kilopascals and we look up the value of HF. So H1 is HF and that's 340.54 kilojoules per kilogram. But we're also going to need the entropy at this point. Well, either we're going to need the entropy at this point or we're going to need the specific volume. Let me explain. We know that point 2 is going to be a compressed liquid, right? However, tables in your book, table A7, are pitiful at best. There's only like six pressure entries and the first one is at some whopping pressure. like, I don't know, 50 megapascals, something like that. It's pretty much useless to us for all reasonable problems. So rather than even try to use table A7, which again we could theoretically use, we might have to do some extrapolation as well as interpolation. And we always want to avoid that if we can. But we don't really have to, right? We could just say that the work associated with the pump is V1 times P2 minus P1. We don't actually have to go through the process of saying S1 is equal to P2 minus P2. We can just go equals S2 and then going into our table A7 at S2 and P2 and looking up the value of enthalpy and finding the difference between that enthalpy and the enthalpy of 0.1, this is the easier way to do it. So let's do that. Let's just look up V1, which is also the same as VF. And that's 0.001030 meters cubed per kilogram. OK. So that's the data that we need. And now we can find the pump work. So 0.001030 meters cubed per kilogram. below 103 meters cubed per kilogram. P2 is 3 megapascals. So, let's convert that to kilopascals so that the units work out properly. I would note that a kilopascal is a kilonewton per square meters. And a kilonewton meter is a kilojoule. And the units all cancel out appropriately. And we're just left with, well, work with the units of kilojoules per kilogram. kilogram. So we go through the mathematics and we get 3.04 kilojoules per kilogram. By the way, the conversion factors are all identities, the ones I just mentioned. But if you're doing the same problem in the British system of units, you definitely have to be careful, right? There's no identities there. You're going to have cubic feet per pound mass and then pressure units of pounds force per square inch. So you'll definitely have to use the conversion factors that are in the inside back cover of your textbook or the electronic version of the inside. back cover your textbook, and apply those conversion factors appropriately in order to get your work in the proper units. And now again, I like to talk about units. It's usually the easiest way to determine whether you did a problem correctly or not. If the units come out fine, then you probably did it right. If the units don't come out fine, you probably made a mistake. So anyway, we have our pump work, which we'll note is also equal to h2 minus h1. And therefore, h2 is just going to be the pump work plus h1. So that's that's just going to be 343.58 kilojoules per kilogram. All right. So there's our enthalpy at one and two. And now let's just move on to state point three. So now we do have two properties. We have both the pressure and temperature at 3. And we should be able to see pretty clearly that at this particular given pressure, that is at 3 megapascals. given temperature is greater than the saturation temperature. Clearly, this is superheated. So we go into table A6, which is our superheat table. So now we just look up the data. Again, it's pretty straightforward. We just look up the enthalpy. As usual, there's a series of sub tables in table A6, right? You go to the sub table that's labeled 3 megapascals or perhaps it says 3000 kilopascals. You go down to the appropriate temperature line on that sub table. And sometimes by interpolation, sometimes not. We just read off the value of enthalpy. So we get the enthalpy now of 2994.3 kilojoules per kilogram. But we're definitely going to need the entropy as well. So the entropy at 0.3. is 6.5412 kilojoules per kilogram per Kelvin. OK. So that's all we need at 0.3. Now we go from 3 to 4. So we know two properties really at 0.4. We know the pressure at 4. And we know the entropy at 4, which equals the entropy at 0.3. So before we really know which table to go into, we have to ask ourselves, well, is it superheated, saturated, or too two-phase mixture. So again, that's pretty easy to do. I mean, if we just go into table A5 initially and just compare at this pressure the value of the entropy of the saturated vapor to the given entropy, what we'll actually find here is that The saturated vapor is going to have a greater entropy than the entropy of 0.4. So if 0.4 is less than that of a saturated vapor, well, clearly we're in between SF and SG. I mean, this is a two-phase mixture. Maybe another way to write this would be to just note that SF is less than S4, which is less than SG. perhaps that would be a little more understandable. Let's just leave it in that form. Well, of course, we know therefore that it's a two-phase mixture. And it turns out that table A5 is indeed the correct table for us to be in. It's a two-phase or saturated mixture. So, great. We stick with table A5. Now what we need to do is just remember that we have to use the given thermodynamic property of entropy to find the quality at point four. And then we use that quality to find the enthalpy. So we remember that for a two-phase mixture, S4 equals SF plus a quality at 4 times SFG. And again, we're still at P4, right? P4 is the same as P1. That's 50 kilopascals. So at 50 kilopascals, again, we've already noted that it's a two-phase mixture. So now let's just look up the numbers and find the quality. So let's see. S4 is the same as S3, so 6.5412. And then we just have to look to look up SF and SG, I'm sorry, SF and SFG. So, SF is 1.0912 kilojoules per kilogram. The quality at four is my unknown. And SFG, which is the difference in entropy between SG and SF, right, between the saturated vapor and saturated liquid. So, this is listed as 6.5019. And this allows me to find the quality at state 0.4. and it's.8382. And now that we have the quality at 4, let's use the same equation that we did above but now with enthalpy. H4 equals HF plus the quality times HFG. So again, let's just plug in the numbers. So at 50 kilopascals, HF is 340.54. The quality is.8382. And HFG is. 2304.7. All these are in kilojoules per kilogram. So this gives me 2272.3 kilojoules per kilogram. And now we have the enthalpy at all of our state points. So now we can just finish up the problem. So before I erase the diagram, in fact I'm not going to erase it, let's at least complete it. We know that point four is a two-phase mixture. We know that point three is superheated. So that actually finishes the TS diagram for us. Now we can find the other terms, heat input is H3 minus H2. So I'll just let you subtract one from the other. We get 2650.7 kilojoules per kilogram. Now we find the network. which is just the difference between the turbine and the pump work. So we've actually already calculated the pump work. So this is H3 minus H4. That's the turbine work minus the pump work. Well, again, I'll let you just plug in the numbers on your own. We have H3, we have H4. The net work is 718.9. kilojoules per kilogram. So, we can find the thermal efficiency, the network over the heat input. We get.7, I'm sorry,.271 or 27.1% if you prefer. And then, let's also find the net power. So, it's 35 kilojoules per second times 718.9 kilojoules per kilogram. I'm sorry, mass flow rate is not kilojoules per second, it's kilograms per second. So, kilograms per second times kilojoules per kilogram gives you kilojoules per second, which is the same thing as a kilowatt. And we get 25.2 times. 10 to the third kilowatts. We can also show this as 25.2 megawatts if we prefer. But this then is the power output, the net power output. And that finishes this problem. So any questions on this? You've seen problems like this before. but perhaps not all in one problem, right? You've already looked at pump problems. You've already looked at turbine problems. You've already looked at heat exchanger problems. So now it's just a matter of bringing them all together into one complete thermodynamic cycle. So what would be the next thing we need to do? At this point, what we need to do is look at the non-ideal cycle. So certainly this cycle has some merit, but it's not particularly realistic. We know that there's going to be friction losses in the real world anyway. We know there's going to be heat transfer losses in the real world anyway within the turbine or within the pump. So we really have to identify those losses. And the way we do it is through isentropic efficiencies, just like we've done for our other cycles, our Brayton cycles specifically. So let's now look then at the non-ideal but still simple Rankine cycle. Because it's still a simple ranking cycle, there's no need to redraw the schematic diagram. But certainly there will be a slight change in the TS diagram. Let's still show the constant pressure lines. Well, still the same state point. One is going to still represent the inlet to the pump. Three is going to still represent the inlet to the turbine. But now we're going to have to use the same notation 2s and 2a that we did with. the Brayton cycle to represent the ideal and actual compression process in the pump. We're going to have to use the 4S and the 4A to represent the ideal and then actual expansion process in the turbine. So as we go from 1 to 2S, 2S would be directly above point 1. And of course, 2A, the actual discharge from the pump, is a little bit to the right of that on this maximum pressure line. And then here, 4s is going to be directly below.3. But in the real case,.4a is actually going to be to the right of.4s along that constant pressure line. So this would be a general diagram that illustrates this process. Please note that, again, we don't know exactly where point three is. And therefore, we don't know exactly where point four S and four A are, right? If point three is closer to the saturation line or even into the two-phase region, then it's very possible that both four S and four A are both going to be two-phase. phase mixtures. It's also possible if three were a little bit further up, higher temperature, then perhaps 4s and 4a would both be superheated vapors. We just don't know for sure. Again, from an analytical point of view, it doesn't matter at all. All that affects is where we get our data. So this is a cycle on this particular type of diagram. So the question then is how does this affect any of our equations? Well, the equations just don't change by that much. The heat input is still going to be the heat input from point 2 to point 3, but now we just want to specify 2a to point 3. Whoops. So this is h3 minus h2a. So again, the only difference is that we're replacing our 2 with a 2a in the previous equations. The pump work, our work input is going to be h2a minus h1. This is not going to be equal to h2a. to the specific volume times the pressure difference anymore because this is no longer a ideal process, right? It's not ideal. We have to consider the losses in the pump and in the turbine. So you can't use a relationship that only applied for the ideal case. completely analogous to when we talked about compressors, right, in the Brayton cycle. If it were ideal, we'd have used relative pressures. But in the non-ideal case, we don't, right? There's an actual enthalpy at the discharge from, well, what would have been the compressor in that cycle. So let's just keep this in mind. Furthermore, we would note that as we deal with the pump, we're going to have to deal with the pump's isentropic efficiency, which is, of course, the ideal over the actual work. And, of course, the ideal work. is V1 times P2 minus P1. I mean, we could also show this as H2S minus H1, but again, it's usually easier for us and we usually only have that data available anyway for us to assume that the pump work for the ideal case is just V delta P rather than using compressed liquid tables, which frankly we don't have. And then, of course, on the denominator, we would just have H2A minus H1. All right, so there's the pump work and the isentropic efficiency equation. The turbine work is our work output. And this is going to be H3 minus H4A. And then, of course, we're going to have to use the turbine's isentropic efficiency, which is defined as the actual over the ideal work. If you would like to, so that you don't confuse the ideal and actual pump work with the ideal and actual turbine work, you can put little subscripts p and t. You know, you can put a little p by the pump and put a little t by the turbine. Is it necessary? No, but it might help you as you're learning this material. Anyway, the actual work is h3 minus h4a. And then, of course, the ideal work would be h3 minus h4s. So we're still going to have to go through the process of assuming. that it's an ideal work process from 1 to 2s or from 3 to 4s. But then we're going to have to apply the isentropic efficiency, which will generally be given in order for us to calculate the actual work terms. OK. And then, of course, we'll have our thermodynamic efficiency equation, which is still just going to be the network over the heat input. And the network is still the turbine minus the pump work over the heat input, which is still Qn. equation. So we could just use these equations from directly above in order to find the thermodynamic efficiency. So again, that hasn't changed at all. The only thing that's changed is the way that we're treating the pump and the turbine. OK. Well, if that's the case, then why don't I go through an example problem of this too? There's really nothing more that needs to be discussed. And in fact, why don't we just redo the same problem? Why don't we just do problem 10-12 but this time instead of assuming that it's isentropic, it's not isentropic anymore in the turbine and the pump. So let's do the same problem but given pump and turbine isentropic efficiency. So why don't we say the pump's isentropic efficiency is 80% and the turbine's is 90%. And we're still going to do it everything else. else exactly the same as we did before. Now, one reason I've done this is simply because it's going to save me a lot of time because most of the problem is identical, right? The only thing that isn't identical is our use of the isentropic efficiencies. So, H1 and V1 are the same as before. So, those don't change at all, right? H3 and S3 are identical. exactly the same as before. So those don't change at all. What we called point two previously is now called 2s. So. The previous 0.2 is now 2S. And for that matter, the previous 0.4 is now 0.4S. So nothing really changes much in this particular problem. Again, the big change is just the way that we treat the pump and the turbine. OK. So. if what we previously called point 2 is now point 2s, let's continue then and consider that pump. And the pump has an isentropic efficiency associated with it. And we know that it's going to be the ideal over the actual enthalpy change. And of course, ideal enthalpy change is the same as V delta p. p. And this we call the pump work, but it would be the ideal pump work, right? I mean, we're looking at 2s now. So this would be the ideal pump work. And it would be over h2a minus h1. So, So we can use this now to calculate H2. So the pump efficiency is.8. The ideal pump work, this V delta P term, we had already calculated previously. That was just 3.04 kilojoules per kilogram. So we do it exactly the same way. No need to redo it. H2A is my unknown. H1 is the same term that we had before, which is what? 340.54. And this allows me to find the value of H2. So I can do that. find the actual enthalpy at.2 and it's 343.62 kilojoules per kilogram. OK. So that's all we really have to do here. Next, we move on to the turbine. Now again, we're just using the isentropic efficiency of the turbine, which in this problem is 0.9. And this is going to be H3 minus H4 actual over H3 minus H4F. And again, we consider that what we called H4 in the previous problem is now H4S and H3 is exactly the same. So, there's only one unknown in this particular equation. So, we have 2994.3. minus H4A, or 2994.3 minus what was previously just H4, it's now H4S, so 2272.3. All the units are the same in kilojoules per kilogram. So, we can find the actual enthalpy as 2344.5 kilojoules per kilogram. All right, so. have all the thermodynamic property data we need. Let's just finish up the problem. So the heat input is H3 minus H2A. If we go through the mathematics here, we end up with 2650. 0.68 kilojoules per kilogram. And the net work, again, the turbine minus the compressor work, or H3 minus H4A minus H2A. I'm sorry, I said compressor work, turbine work minus the pump work. So H2 minus H3A minus H2A minus H1. Again, we have all this enthalpy data, H1 and H3 are identical to the first time we did this problem. So the network then becomes 646.72 kilojoules per kilogram. And the thermal efficiency then is just the ratio of the above. So we get.244 or 24.4%. And the net power is just the mass flow rate times the work per unit mass net. So again, I'll let you plug in the numbers. And we get 22.6. 10 to the third kilojoules per second or 22.6 megawatts. Of course, a kilojoule per second is a kilowatt, so we can show those units if we prefer. All right, so there is that particular problem. Any questions on how we solve this one? Please note that the numbers are significantly different. Well, maybe I shouldn't say significantly, but noticeably different. In this particular case, our net work is, you know, what, about 10% or so less. The heat in. going to be a little bit less because we end up having a somewhat higher enthalpy coming out of the pump than we did if we assumed that it was not ideal, I'm sorry, than if we assumed that it was ideal. But still, the net effect is a reduction in the thermodynamic efficiency, which is what we'd expect. I mean, in the first case, we didn't consider any of the losses, right? Now, we're specifically considering the losses within the turbine and the pump. Again, heat transfer losses, friction losses, fluid flow losses. So, these losses are all built into the efficiency term that you're given. And that would be that. Okay. So, any questions then on this problem? Great. So, let us move on. So, one thing that I should note is that There are ways that we can improve the thermodynamic efficiency of a Rankine cycle. And some of these ways are similar to what we have used in the Brayton cycle to improve its overall thermodynamic efficiency. One thing that could be done is to use the Brayton cycle to improve its overall thermodynamic is regeneration. But regeneration is a little more complicated with regards to a steam cycle than it is for a gas power cycle. In fact, before I even get into that anymore, I just want to say that I'm going to talk about the I just realized I didn't put the TS diagram on the board for this last example problem that I was doing. I think that it's not going to be a problem. If you just look at the previous diagram, it looks essentially the same. The only thing we don't know for sure is where state point four is. We're not completely sure where might point four go. We're not completely sure if four A is going to be a two-phase mixture or superheated. We don't know for sure. I mean, we'd have to look into the data and figure that out. Now, is it that important? Eh, perhaps not. But let's just finish up the problem anyway. So, I need more erasers. Okay, so there's one, here's 2a, here's 3. here's 4s and 4a, we just don't quite know yet. OK. So all we really have to do is go into our property tables at 50 kilopascals and compare our value of the enthalpy at 4a to the value of hg. Is 4a less than, equal to, or greater than the value of hg at that particular time? particular pressure. So that's what we really should have done as we finished up this problem. And now I have to hunt for the answer. Let's see. OK. So, HG is about 2645 and my value of H4 is, H4A is 2344. So, it would look like we're still in the two-phase region. For less than HG, I mean, assuming that I looked up the numbers properly, which I don't always do. So therefore, it's two-phase. So it turns out for this particular problem. going to be right around there. OK. Now again, that assumes that I looked up the numbers properly. I'm just kind of doing this on the fly because I didn't actually see the results in here. So hopefully everything I've done is accurate. Anyway, one more time, any questions on this particular problem now that it's finished finally? All right. Back to what we were talking about before. Okay, so what could we possibly do to improve the thermodynamic efficiency of a Rankine cycle? One thing that we can do is regeneration, but I mentioned previously that we have a little bit of a problem with regeneration in the same context at least as regeneration. generation in the Brayton cycle. Keep in mind that in the Brayton cycle, the discharge from our turbine was always at a relatively high temperature, okay, several hundred degrees, certainly well above the ambient air. air temperature and even well above the temperature of the ambient air that's compressed from point one to point, from point one to point two. So it makes a lot of sense to use regeneration in a Brayton cycle because we have a higher temperature exhaust than our inlet air to the combustion chamber. We can exchange some heat and heating up the inlet air into the combustion chamber means using less heat in the combustion chamber. Great. But that's not the case here, right? I mean in this particular case. going through the compression process in the pump, we're actually at a temperature that's considerably higher than the temperature that leaves that turbine, right? How can you use regeneration in that case? I mean, certainly there'd be no way to recover heat at a low temperature and somehow be able to transfer it to a higher temperature. I mean, heat transfer doesn't work that way. So we're not going to be able to use regeneration in the same sense as we use regeneration in or Brayton cycle. So just keep that in mind. Yes, there is still something called regeneration, but it's not going to be done quite the same way. The other thing that we could do theoretically is by going through the compression process in the pump in multiple stages, but that doesn't really help us out very much. I mean, look at the previous example problem and look at the total amount of pump work. There's only about three kilojoules per kilogram of pump work and we can compare that. that to the amount of net work of well over 600 kilojoules per kilogram. So the pump work is not a significant contributor to the efficiency or perhaps I should say inefficiency of the cycle. We're using what? Less than a half a percent of the turbine's outlet work in order to provide pumping. Now there's no such thing as back work ratio for a Rankine cycle. But the back work ratio is an illustration of that, right? The back work ratio is the ratio of how much work is required to be done. by the compression device versus how much work is produced by the turbine. We saw that for the gas turbine cycles, that could easily be 50% or even higher. For here, it's a fraction of a percent. So trying to minimize the pump work is never going to be cost effective, never, ever, ever. You know, some of the pumps that are used are absolutely massive in steam power plants. They cost for a single pump millions of dollars. To save maybe a small fraction of a fraction of a percent in pump work. work and having to spend millions of dollars for that extra pump, not to mention some sort of intercooler in between, that just makes absolutely no sense at all. So we're not even going to consider that. However, we can increase the work output if we do, pardon me, if we do work output in stages similar to what we saw for the Brayden cycle. Will allow us to go through an expansion process down to some intermediate pressure and then we're going to reheat back into the engine. back up to some elevated temperature and then allow the steam to pass through a second turbine in order to produce more work. And it turns out that if we compare the extra amount of work that's produced by working in stages to the extra amount of heat that's going to be required during the reheat process, this should actually improve the thermodynamic efficiency. In other words, the net work attributable to this process of reheat is actually going to be more significant than the actual work that's being done. than the amount of heat transfer. And if we look at just the efficiency of that individual process, again, that's going to be a little bit higher. We're going to get a little more work per unit of heat transfer than we did for the original part of this process. The efficiency is going to go up. So that's what we're going to talk about first is reheat. OK. So no longer can we call this the simple ranking cycle. We're going to have to just simply call it the ranking cycle. This time with reheat. So first, Let's keep in mind that there's actually a second reason why we may want to use reheat. Certainly one reason is to improve the thermodynamic efficiency. Another reason has to do with temperature limits. And I believe we talked about this as well with regards to the Brayton cycle. If you add heat all at once, you end up with a really high exit temperature from, well, what would be the boiler in this case. Sometimes those temperatures are so high that you're not really able to find the. the appropriate materials for use in your steam turbine and you have very serious, well, let's say materials concerns in that design. So rather than increase to some really elevated temperature, we could just increase to some modest temperature, do some work, and then reheat back up to that modest temperature. We'll still have a high thermodynamic efficiency, but we're not going to have to worry about the materials problems associated with this high temperature steam, which is moving through the turbine. So what would the cycle look like? We understand its need. Now let's look at it. So here we have our pump. And as usual, we'll just enter the pump at state point one. Big P for pump. Here's my work input from the pump. We're now going to go through our boiler. So here's the boiler where my heat input takes place. We're going to leave the boiler and go to the first turbine. So let's put some state points. I see two entering the boiler, three into the turbine. OK. Now we're going to expand through this first turbine until we get to state point four. And now we're going to go back into our furnace. Keep in mind the boiler is just one section of the furnace. You've got this massive, massive furnace. It could be 10 stories high. and there'll be different heat exchanges within it. One heat exchange is certainly the biggest one. It's going to be what we call the boiler. But within the furnace there's going to be other heat exchangers. The one we're talking about now we simply call the reheater. So often we would just place the reheater at least on our schematic diagram immediately adjacent to the boiler because in the real world they are adjacent to each other. Anyway, they're different sections. They're different heat exchangers but they're still next to each other. So now we come out. of the turbine at.4 and we go through the tubes of this heat exchanger that we call the reheater. And we come out of this reheater at.5. And now we have to go back into a second turbine. So here's the second turbine. And work is going to be done from both of these turbines. Heat input, by the way, is also going to be provided not only to the boiler but also to the reheater. So both of these heat input terms as well as both of these work terms have to be considered. And then coming out of the turbine, finally we go back to our condenser. So just big C for condenser. And we leave in, let's see, six. has to come out of this turbine. Seven would come out of the condenser and then flow back into the pump. So that's what this cycle would look like on a schematic diagram. Now one thing I will note here, unlike the Brayton cycle, there's no requirement to have the pressure ratios the same across each turbine. turbine stage, OK? And there's also no need to have the same temperature going into each turbine or the same temperature going out of each turbine, OK? It doesn't have to be that way for the Rankine cycle. So you could be given any of number of different pressures that the reheater operates at. I will note, however, that the basic rule of heat input or heat exchange at constant pressure still applies to this part of the ranking cycle, OK? It's still a ranking cycle. The heat transfer is still done at constant pressure. And not only does this include the boiler as well as the condenser, but also the reheater. OK. So that part of the process hasn't changed. Yes? Just a quick question. So point one and point seven, shouldn't they be exactly the same? Point, oh, yes. No, actually I'm sorry. I, no, no. Yeah, you're right. I've missed, I don't need a.7, so that's a good point. Here's my heat rejection in the condenser. And I really should pay more attention to my own notes because I don't have any.7 in my notes. So, but that's a good observation. Yeah. Once we reject the heat in our condenser, we're really going all the way down to the conditions of.1. So, yes, thank you for pointing that out. Okay. Now, what would this look like on the TS diagram? And again, we'd have to consider whether we're talking about the ideal case or the non-ideal case. So I suppose we should just begin by looking at the ideal case with reheat. Well, we'd have something like this. So again, we'll show this on a TS diagram relative to the saturation lines. Let's show a minimum and maximum pressure. But let's also note that the reheater is going to operate at some intermediate pressure, right? We're leaving the first turbine, coming out of that turbine at an intermediate pressure before we reheat at that constant pressure. And then we expand through the next turbine all the way down to, well, the pressure of the condenser. So there will be some intermediate pressure in this problem. So we might as well show it as well. And this would be simply the pressure associated with the reheater. So let's just go through the process. So we start at point one. We're going to compress not up to the pressure of the reheater, but all the way up to the pressure of the boiler. So, point two should be directly above point one at the maximum pressure. We're now going to add heat until we get to point three. We're now going to expand until we get to the pressure of the reheater. So we just move vertically downwards until we get to.4. Again, with the understanding that.4 might be two-phase. It might be saturated vapor. It might be superheat. We don't know. It just depends on the problem. Nonetheless, we reheat up to whatever the temperature of.5 is. By the way, I have shown 5 and 3 as being the same, but it doesn't have to be. It is very common to have the same temperature. coming out of the boiler as well as coming out of the reheater, but it's not necessarily going to be true. So read the problems carefully. But nonetheless, we come out of.5 and then we expand through the second turbine down to.6 and ultimately reject heat to.1. OK. So let's just again make sure that we include all the right terms, right? We have heat input in the boiler and the reheater. We have turbine work out. output from each of the two turbines. And I suppose it doesn't hurt to show the pump work input and the heat rejection within the condenser. So this is just an illustration then, both schematic as well as on a thermodynamic property diagram of what this particular Rankine Cycle III heat would look like. Again, let me note, no need. to have the pressure ratio the same across each turbine. No need to have the same temperature at inlet or outlet from each turbine. Okay, so ever so slightly different, but the same basic process, right? We're still talking about constant pressure heat input. Still talking about work being done in two stages instead of one. Again, out there in the real world, it is certainly possible that you could have multiple reheaters, but I I've honestly never seen that. So theoretically possible, but typically it's not done. And I've never seen any situation with multiple pumps. Sometimes you'll have multiple pumps and turbines that are in parallel with each other because sometimes it's good for maintenance reasons instead of having one device that works at 100% capacity. You'll have two devices that work at a fractional capacity so that you don't have to shut the system down every time you need to go through maintenance. You can run one pump with one turbine and do maintenance on the other devices. In fact, what I've generally seen is you typically have three work devices that run in parallel to one another, particularly for the pumps. They're all at 50% capacity, so you can run two pumps at 100% while doing maintenance on the third. third pump and then, you know, cycle that third pump back in and keep running the system without having to shut it down. Turns out that pumps are really the biggest maintenance issue. You know, rotating equipment in general tends to be the biggest maintenance issue in a power plant. So, you know, having multiple pumps that operate at fractional capacities is good from a maintenance point of view. Yeah, it costs you more money, but it would eliminate downtime and therefore you're still able to make more money than the extra money it costs you to buy the extra pump. Again, things to think about in the wonderful world of thermodynamics. And with that, we're done for the day. So we'll continue this discussion next time. We'll look at more example problems. And that's that. So again, the one student that hadn't shown me proof of prerequisite, please don't forget to do so. And again, you've got homework due next time. I still have some old homework from last time that's up here. So pick up your work if you would. See you all on Wednesday.

Hopefully the bombs don't fall. Anyway, yeah, you know, I grew up in the era where we had to hide under our desks in grammar school. You know, as a drill they had every day. Well, not every day, every week.

Ah, we'll get into that some other time. So, again, good morning everybody. I have the new homework set here on the board. Five problems from chapter 10. Again, I'll do next Wednesday.

As a reminder, last week's homework will be due this coming Wednesday. And please make special note of problem 119 that I modified it a little bit on you, so make sure you're doing the right version of 9-119. Additionally, I just want to remind you that today is the last day to show proof of having met the prerequisites. And I got that for all but one student, Mr. Mumby. So if you don't mind that one student, just make sure you show me before the end of the day.

that you've passed the prerequisites for this class. So, with that, let's just get right back to the Rankine cycle. Now, the Rankine cycle is the cycle that we use to model a steam power plant.

Hopefully, this was made clear last time. We're no longer dealing with gas problems, right? We're not talking about air standard cycles. These are not gas power cycles. We're dealing with steam now, H2O if you will, that's moving through the cycle.

Now last time we looked at the basic schematic. We have a pump that is going to move liquid from a low to a high. pressure.

We're then going to take that liquid, which of course is just water, and we're going to add heat to it in a device that we call a boiler, and that turns the water into the vapor phase, so steam. Although it's possible it could be saturated steam, it could be a high-quality two-phase mixture, but certainly the intention is to turn most of the liquid into steam, hopefully all of it into steam. This steam now goes through a turbine. where work is done. And as we expand through the turbine, the pressure drops.

And then we go into a heat exchanger, which we call a condenser. And here the heat is removed. Whatever steam we had coming out of the turbine or it could be, again, could be saturated, could be high quality two phase. But nonetheless, we're going to condense the vapor back into a liquid and we go right back into the pump again. So this is our basic ranking cycle.

Why don't we just call it a simple ranking cycle? Okay, we've got work input from the pump. We've got work production or work output by the turbine.

We have heat input into the boiler. And then we have heat rejection through the condenser. Please keep in mind that unlike the Brayton cycle where the heat. rejection takes place essentially into the atmosphere. Here it generally doesn't go into the atmosphere, at least not directly.

The heat would actually be rejected into a separate water supply. We'd actually call it a sink because this is the heat sink, right? The heat rejected from the cycle. goes into that heat sink. So what I show here is kind of a curly line with a couple lines coming off of it.

You know, that's supposed to represent some other flow path. This could be water from the local river or the local ocean. More likely it's going to be a clean water supply from a device called a cooling tower.

And we'll talk about cooling towers later in this course. But nonetheless, we're going to reject our heat into water typically. A completely separate stream of water.

We're not going to worry about it right now. We'll just simply recognize that that's where the heat output is. is going. But this is our simple ranking cycle, okay? And as I mentioned last time, in many ways it's similar to the Brayton cycle, at least in that the heat transfer takes place at constant pressure. And that would apply both to the heat input as well as heat rejection.

These are constant pressure processes, so that's similar to the Brayton cycle. And then also work is done isentropically. At least...

For the ideal ranking cycle, okay, eventually, actually very shortly, we're going to talk about the non-ideal ranking cycle where the work is not isentropic anymore, right? The work would be first assumed to be isentropic, which will allow us to find the ideal exit conditions from the turbine and the pump. But then we're going to have to apply the isentropic efficiency just like we did for the Brayton cycle to determine the actual outlet conditions from the turbine and the pump and thus to find the actual work. So, this is just review from last time. Please note that I also showed this on a TS diagram relative to the saturation lines.

And generally, we would show the minimum and maximum pressure lines. You know, these are lines of constant pressure. So, the maximum and the minimum would be like such. Now, please note that state point one is always going to be a saturated liquid.

So, one is always a sat liquid. This is just based on the design and operations of the condenser. The condenser condenses to a saturated liquid, but it doesn't remove more heat than that. Just to the saturated liquid. We pump up to point two to the high pressure line.

This is a vertical line. I know the diagram I put on the board last time, some of these lines look a little skewed, but this is supposed to be a vertical line. After all, it's isentropic.

We then add heat until... We come out of the boiler into our turbine and then we do work. And again, this is done in isentropic way, at least in the ideal cycle. So, again, a vertical line. Then we reject our heat through the condenser.

Now, please note that this diagram I had on the board last time and again from today doesn't represent every possible situation. Granted, points one and two are always going to be as is shown here because point one is always going to be a saturated liquid. But point three doesn't have to be superheated as shown here.

Point three could be exactly on the saturated vapor line. Point three could be a two-phase mixture? We really don't know.

It just depends on the problem at hand. It depends how much heat we're adding to the water in that boiler. If we add enough heat, then certainly it should be enough to change the phase of all the liquid water into a vapor. If we don't provide enough heat, then we may end up with a high-quality two-phase mixture. And of course, we can't have all these different point threes.

It just depends on the problem, right? You'll have either there or there or there. And then, of course, as we expand isentropically, well, again, we could end up...

at state point four anywhere. I mean, we could have a state point four right on the saturated vapor line if we had the appropriate state point three. So any of these are options.

And quite frankly, it doesn't even matter, right? The only thing that matters is where we're going to get our thermodynamic property data. If it's superheat, we go into table A6. If it's two-phase, then we have to use, or saturated, then we have to use data from A4 or A5.

But the basic methods of solution don't change. This just is intended to represent all the possible possibilities as we are over here on the right-hand side of the diagram. Again, though, you're not going to have all those cases in one problem.

You'll just have one vertical line. You'll just have to make sure you're really clear when you're solving that problem where the actual thermodynamic state points are. So are there any questions just in general about the nature of the Rankine cycle?

Okay. If you want to take a look at a typical Rankine cycle power plant, I'll put a I mean, they're all over the place. Probably the closest big one is Etiwanda, which is where the 10 and the 15 freeways cross, a little bit north of there. You can't miss it as you're driving up the 15. Just look eastbound and you'll see all these big smokestacks and, well, that's a ranking cycle. That's a steam power plant.

There's others that are closer, but again, we'll not get into that right now. Anyway. So, let's start looking now at the equations that would be appropriate for this type of analysis. Again, we're specifically looking at the ideal case.

And still, though, the simple ranking cycle. In this particular case, let's note again that for all situations, we're assuming there's no change in kinetic potential energy. As usual, right, there's no significant velocity or velocity changes through any of these devices.

There's no significant height changes. So those are the are going to be neglected. And of course, we're also going to note that during the work processes, remember in the ideal case, work is done isentropically, which means adiabatic reversibly, which means there's no heat transfer, right? So there's no heat transfer during work. And also during heat transfer, Well, there's no work, right?

I mean, heat transfer is done in two very passive heat exchangers. They're just big shells with a bunch of tubes in them. There's no miscellaneous work, right?

There's no shaft work. There's no mixing work. There's no work at all. So, there's no work terms. So therefore, let's look at the first law.

And of course, the first law, well, regardless of which of the four processes we're dealing with, the first law basically leaves us with nothing but enthalpy change, right? Heat transfer transfer will equal the enthalpy change. Work will equal the enthalpy change.

So for the first law, let's now go to each device and we'll start with the pump. We know that the pump is work input. And again, this is done without any heat transfer, with no kinetic and potential energy change.

So the first law just tells us this is the enthalpy change. So this is just H2 minus H1. Let's also note, however, that something that you, again, should have covered, which is discussed in, I guess, is chapter 7. I think somewhere near the end of chapter 7 in your textbook.

It also shows that when we have a liquid, and certainly this is a liquid, right, as we go from point 1 to point 2, when we have a liquid for the ideal case, not only is the pump work equal to the enthalpy change, but it's also equal to the specific volume at the inlet to the pump times the pressure change. So you can actually use either of these processes, but again, for the ideal case only. So that's how we're going to calculate the pump work using either one of these two equations.

When we deal with the turbine, again, the turbine work is our work output. And again, this is just the enthalpy change as we go from 3 to 4. So just H3 minus H4. Keep in mind again, all these terms are absolute values, right? We know the pump work, the magnitude is positive, but it is work input. Technically, it's negative work, but.

We'll just let those work terms have their sign included within the various equations like the thermal efficiency equation. Again, a reminder of what we already know. So here's our turbine.

This is our turbine work equation. And there's really nothing more we can add to this. this. With regards to the heat transfer, when we look at the boiler, the heat input in the boiler is again just the enthalpy change, so H3 minus H2. And within the condenser, the magnitude of the heat output, the heat rejection, as we go from 4 to 1, it's just H4 minus H1.

So these equations are really quite straightforward. And ultimately, of course, we're interested in finding the thermodynamic efficiency. That certainly doesn't change.

These are still heat engine cycles. We're still trying to find the relationship between the amount of work we have as our benefit, right, the network output, compared to the amount of heat that we have to pay for, our heat input. So this doesn't change. This is still the network over the heat input. And of course, I could just plug in all the numbers from above if I want to.

Maybe I'll just leave it in a more general form. Turbine work minus compressor work over heat input is our thermal efficiency. And of course, all those terms are just based on enthalpy data. And then, And then lastly, I might note that we're not always just interested in work per unit mass or heat transfer per unit mass.

We're also often interested in the rate that work is done. In other words, how much power is being produced by this facility or for that matter, what is the rate of heat transfer in the boiler or in the condenser? So in general, we would also just show that the power is the mass flow rate times the work per unit mass and the rate of heat transfer.

is the mass flow rate times the heat transfer per unit mass. And again, this would apply both for work in or work out, as well as heat in or heat out. And again, this is just a reminder of information that hopefully you already know.

So, perhaps the best way to illustrate how to utilize all these in a problem is to actually go through an example problem. So, we really have everything we need already. Maybe I just want to remind you before I move forward that when we deal with an isentropic work process, whether it's compression from one to two or expansion from three to four, we certainly treat this differently than we did when we were dealing with our gas power cycles. I mean, we're not going to use relative pressure.

anymore, relative specific volumes. We're not going to use ideal gas equations that relate pressure and temperature and volume ratios. Those only apply to your ideal gas type problems, right?

The ones where we have air standard cycles to model our gas power cycles. Here we have to actually use the thermodynamic property data. We have to remember that it's an isentropic process from one to two, right?

We have to look up the value of the entropy at point one and then say, well, S1 equals S2 and then add our known entropy at two and some other property which typically is going to be the peak pressure. Then we go back into our tables and we look up the enthalpy, OK? Same thing from three to four. We know state point three, that's typically given.

It's isentropic from three to four, so we look up S3, that equals S4. And at S4 and at the minimum pressure, which would be P4, well, we just look up the enthalpy at point four. So we're going to use our.

steam tables. We could also just call them water tables rather than any kind of ideal gas table. So please make sure you're very careful on this. So let's begin. Let's look at problem 10-12 in your version of the textbook.

So let's give this a second to pop up. So, as the projector is warming up, the screen is coming down. OK.

Hopefully, you've written these equations into your notes already. In fact, maybe I'll stop the screen halfway down. Uh-oh, I pushed the wrong button.

So, we'll be able to say many of the equations anyway. Nonetheless, this is problem 1012. So, let me just read it to you briefly. We have a steam power plant that's operating on the simple ideal Rankine cycle between the pressure limits of 3 megapascals and 50 kilopascals.

scale. So clearly we're given the maximum and minimum pressures in the cycle. We're given the temperature of the steam at the inlet to the turbine.

So that's 0.3 of 300 degrees Celsius. And we know the mass flow rate of the steam through the cycle as 35 kilograms per second. second.

Show this on a TS diagram and determine the thermal efficiency and the net power output. So this is a very typical problem of this particular type. I don't think there's any point in me, I take that back. There is a point in me showing this on a TS diagram because we're going to have to figure out which of those four possibilities shown in the previous diagram are accurate. So let's just show this again.

Let's show our constant pressure lines, one, two. Over here, we don't know exactly where three and four are yet. We're going to have to figure that out. So let's just proceed with the problem before we finish the diagram.

So we have our pressure limits. So, we're given P2 and P3. That's our maximum pressure of 3 megapascal. And we have our minimum pressure, which is P1 and P4, which is 50 kilopascal. We're given the turbine inlet temperature of 300 degrees Celsius.

And we're given the mass flow rate for the steam moving through the cycle of 35 kilograms per second. OK. So we want to show it on the TS diagram, find the thermal efficiency, and also find the net power output. OK. So, let's just go through this exactly as I explained just a moment ago.

We'll start at state point one. So, we know the pressure at point one. Now, you may say, well, don't we need a second thermodynamic property in order to do determine the state.

But we do have that second property, right? We know that for every condenser, it's a saturated liquid coming out of the condenser. So, point one has to be a saturated liquid.

OK. even if it doesn't say so, it is, it always is. I mean, it really just has to do with the design of the condenser. The bottom of the condenser is where the liquid resides, right? It's got a higher density. The steam, the vapor in the condenser resides in the top portion of the condenser.

So you just draw the water off the bottom of the condenser. It's a two-phase mixture in that condenser, but you're only pulling off the saturated liquid portion. So it has to be a saturated liquid. I mean, it is. It always is.

So we know that data. So we're going to have to go into the appropriate table, which is table 8. 5, this is our pressure table. We simply go to 50 kilopascals and we look up the value of HF. So H1 is HF and that's 340.54 kilojoules per kilogram. But we're also going to need the entropy at this point.

Well, either we're going to need the entropy at this point or we're going to need the specific volume. Let me explain. We know that point 2 is going to be a compressed liquid, right?

However, tables in your book, table A7, are pitiful at best. There's only like six pressure entries and the first one is at some whopping pressure. like, I don't know, 50 megapascals, something like that.

It's pretty much useless to us for all reasonable problems. So rather than even try to use table A7, which again we could theoretically use, we might have to do some extrapolation as well as interpolation. And we always want to avoid that if we can.

But we don't really have to, right? We could just say that the work associated with the pump is V1 times P2 minus P1. We don't actually have to go through the process of saying S1 is equal to P2 minus P2. We can just go equals S2 and then going into our table A7 at S2 and P2 and looking up the value of enthalpy and finding the difference between that enthalpy and the enthalpy of 0.1, this is the easier way to do it.

So let's do that. Let's just look up V1, which is also the same as VF. And that's 0.001030 meters cubed per kilogram. OK. So that's the data that we need.

And now we can find the pump work. So 0.001030 meters cubed per kilogram. below 103 meters cubed per kilogram. P2 is 3 megapascals. So, let's convert that to kilopascals so that the units work out properly.

I would note that a kilopascal is a kilonewton per square meters. And a kilonewton meter is a kilojoule. And the units all cancel out appropriately.

And we're just left with, well, work with the units of kilojoules per kilogram. kilogram. So we go through the mathematics and we get 3.04 kilojoules per kilogram.

By the way, the conversion factors are all identities, the ones I just mentioned. But if you're doing the same problem in the British system of units, you definitely have to be careful, right? There's no identities there. You're going to have cubic feet per pound mass and then pressure units of pounds force per square inch. So you'll definitely have to use the conversion factors that are in the inside back cover of your textbook or the electronic version of the inside.

back cover your textbook, and apply those conversion factors appropriately in order to get your work in the proper units. And now again, I like to talk about units. It's usually the easiest way to determine whether you did a problem correctly or not. If the units come out fine, then you probably did it right.

If the units don't come out fine, you probably made a mistake. So anyway, we have our pump work, which we'll note is also equal to h2 minus h1. And therefore, h2 is just going to be the pump work plus h1.

So that's that's just going to be 343.58 kilojoules per kilogram. All right. So there's our enthalpy at one and two. And now let's just move on to state point three.

So now we do have two properties. We have both the pressure and temperature at 3. And we should be able to see pretty clearly that at this particular given pressure, that is at 3 megapascals. given temperature is greater than the saturation temperature. Clearly, this is superheated. So we go into table A6, which is our superheat table.

So now we just look up the data. Again, it's pretty straightforward. We just look up the enthalpy. As usual, there's a series of sub tables in table A6, right? You go to the sub table that's labeled 3 megapascals or perhaps it says 3000 kilopascals.

You go down to the appropriate temperature line on that sub table. And sometimes by interpolation, sometimes not. We just read off the value of enthalpy.

So we get the enthalpy now of 2994.3 kilojoules per kilogram. But we're definitely going to need the entropy as well. So the entropy at 0.3.

is 6.5412 kilojoules per kilogram per Kelvin. OK. So that's all we need at 0.3.

Now we go from 3 to 4. So we know two properties really at 0.4. We know the pressure at 4. And we know the entropy at 4, which equals the entropy at 0.3. So before we really know which table to go into, we have to ask ourselves, well, is it superheated, saturated, or too two-phase mixture.

So again, that's pretty easy to do. I mean, if we just go into table A5 initially and just compare at this pressure the value of the entropy of the saturated vapor to the given entropy, what we'll actually find here is that The saturated vapor is going to have a greater entropy than the entropy of 0.4. So if 0.4 is less than that of a saturated vapor, well, clearly we're in between SF and SG. I mean, this is a two-phase mixture. Maybe another way to write this would be to just note that SF is less than S4, which is less than SG.

perhaps that would be a little more understandable. Let's just leave it in that form. Well, of course, we know therefore that it's a two-phase mixture.

And it turns out that table A5 is indeed the correct table for us to be in. It's a two-phase or saturated mixture. So, great. We stick with table A5.

Now what we need to do is just remember that we have to use the given thermodynamic property of entropy to find the quality at point four. And then we use that quality to find the enthalpy. So we remember that for a two-phase mixture, S4 equals SF plus a quality at 4 times SFG. And again, we're still at P4, right? P4 is the same as P1.

That's 50 kilopascals. So at 50 kilopascals, again, we've already noted that it's a two-phase mixture. So now let's just look up the numbers and find the quality.

So let's see. S4 is the same as S3, so 6.5412. And then we just have to look to look up SF and SG, I'm sorry, SF and SFG.

So, SF is 1.0912 kilojoules per kilogram. The quality at four is my unknown. And SFG, which is the difference in entropy between SG and SF, right, between the saturated vapor and saturated liquid.

So, this is listed as 6.5019. And this allows me to find the quality at state 0.4. and it's.8382. And now that we have the quality at 4, let's use the same equation that we did above but now with enthalpy.

H4 equals HF plus the quality times HFG. So again, let's just plug in the numbers. So at 50 kilopascals, HF is 340.54.

The quality is.8382. And HFG is. 2304.7. All these are in kilojoules per kilogram. So this gives me 2272.3 kilojoules per kilogram. And now we have the enthalpy at all of our state points.

So now we can just finish up the problem. So before I erase the diagram, in fact I'm not going to erase it, let's at least complete it. We know that point four is a two-phase mixture.

We know that point three is superheated. So that actually finishes the TS diagram for us. Now we can find the other terms, heat input is H3 minus H2. So I'll just let you subtract one from the other. We get 2650.7 kilojoules per kilogram.

Now we find the network. which is just the difference between the turbine and the pump work. So we've actually already calculated the pump work. So this is H3 minus H4.

That's the turbine work minus the pump work. Well, again, I'll let you just plug in the numbers on your own. We have H3, we have H4. The net work is 718.9. kilojoules per kilogram.

So, we can find the thermal efficiency, the network over the heat input. We get.7, I'm sorry,.271 or 27.1% if you prefer. And then, let's also find the net power. So, it's 35 kilojoules per second times 718.9 kilojoules per kilogram. I'm sorry, mass flow rate is not kilojoules per second, it's kilograms per second.

So, kilograms per second times kilojoules per kilogram gives you kilojoules per second, which is the same thing as a kilowatt. And we get 25.2 times. 10 to the third kilowatts.

We can also show this as 25.2 megawatts if we prefer. But this then is the power output, the net power output. And that finishes this problem.

So any questions on this? You've seen problems like this before. but perhaps not all in one problem, right? You've already looked at pump problems. You've already looked at turbine problems.

You've already looked at heat exchanger problems. So now it's just a matter of bringing them all together into one complete thermodynamic cycle. So what would be the next thing we need to do?

At this point, what we need to do is look at the non-ideal cycle. So certainly this cycle has some merit, but it's not particularly realistic. We know that there's going to be friction losses in the real world anyway.

We know there's going to be heat transfer losses in the real world anyway within the turbine or within the pump. So we really have to identify those losses. And the way we do it is through isentropic efficiencies, just like we've done for our other cycles, our Brayton cycles specifically. So let's now look then at the non-ideal but still simple Rankine cycle. Because it's still a simple ranking cycle, there's no need to redraw the schematic diagram.

But certainly there will be a slight change in the TS diagram. Let's still show the constant pressure lines. Well, still the same state point. One is going to still represent the inlet to the pump. Three is going to still represent the inlet to the turbine.

But now we're going to have to use the same notation 2s and 2a that we did with. the Brayton cycle to represent the ideal and actual compression process in the pump. We're going to have to use the 4S and the 4A to represent the ideal and then actual expansion process in the turbine.

So as we go from 1 to 2S, 2S would be directly above point 1. And of course, 2A, the actual discharge from the pump, is a little bit to the right of that on this maximum pressure line. And then here, 4s is going to be directly below.3. But in the real case,.4a is actually going to be to the right of.4s along that constant pressure line. So this would be a general diagram that illustrates this process.

Please note that, again, we don't know exactly where point three is. And therefore, we don't know exactly where point four S and four A are, right? If point three is closer to the saturation line or even into the two-phase region, then it's very possible that both four S and four A are both going to be two-phase. phase mixtures. It's also possible if three were a little bit further up, higher temperature, then perhaps 4s and 4a would both be superheated vapors.

We just don't know for sure. Again, from an analytical point of view, it doesn't matter at all. All that affects is where we get our data. So this is a cycle on this particular type of diagram.

So the question then is how does this affect any of our equations? Well, the equations just don't change by that much. The heat input is still going to be the heat input from point 2 to point 3, but now we just want to specify 2a to point 3. Whoops. So this is h3 minus h2a. So again, the only difference is that we're replacing our 2 with a 2a in the previous equations.

The pump work, our work input is going to be h2a minus h1. This is not going to be equal to h2a. to the specific volume times the pressure difference anymore because this is no longer a ideal process, right? It's not ideal. We have to consider the losses in the pump and in the turbine.

So you can't use a relationship that only applied for the ideal case. completely analogous to when we talked about compressors, right, in the Brayton cycle. If it were ideal, we'd have used relative pressures.

But in the non-ideal case, we don't, right? There's an actual enthalpy at the discharge from, well, what would have been the compressor in that cycle. So let's just keep this in mind. Furthermore, we would note that as we deal with the pump, we're going to have to deal with the pump's isentropic efficiency, which is, of course, the ideal over the actual work.

And, of course, the ideal work. is V1 times P2 minus P1. I mean, we could also show this as H2S minus H1, but again, it's usually easier for us and we usually only have that data available anyway for us to assume that the pump work for the ideal case is just V delta P rather than using compressed liquid tables, which frankly we don't have. And then, of course, on the denominator, we would just have H2A minus H1. All right, so there's the pump work and the isentropic efficiency equation.

The turbine work is our work output. And this is going to be H3 minus H4A. And then, of course, we're going to have to use the turbine's isentropic efficiency, which is defined as the actual over the ideal work. If you would like to, so that you don't confuse the ideal and actual pump work with the ideal and actual turbine work, you can put little subscripts p and t.

You know, you can put a little p by the pump and put a little t by the turbine. Is it necessary? No, but it might help you as you're learning this material.

Anyway, the actual work is h3 minus h4a. And then, of course, the ideal work would be h3 minus h4s. So we're still going to have to go through the process of assuming. that it's an ideal work process from 1 to 2s or from 3 to 4s. But then we're going to have to apply the isentropic efficiency, which will generally be given in order for us to calculate the actual work terms.

OK. And then, of course, we'll have our thermodynamic efficiency equation, which is still just going to be the network over the heat input. And the network is still the turbine minus the pump work over the heat input, which is still Qn.

equation. So we could just use these equations from directly above in order to find the thermodynamic efficiency. So again, that hasn't changed at all.

The only thing that's changed is the way that we're treating the pump and the turbine. OK. Well, if that's the case, then why don't I go through an example problem of this too? There's really nothing more that needs to be discussed. And in fact, why don't we just redo the same problem?

Why don't we just do problem 10-12 but this time instead of assuming that it's isentropic, it's not isentropic anymore in the turbine and the pump. So let's do the same problem but given pump and turbine isentropic efficiency. So why don't we say the pump's isentropic efficiency is 80% and the turbine's is 90%. And we're still going to do it everything else.

else exactly the same as we did before. Now, one reason I've done this is simply because it's going to save me a lot of time because most of the problem is identical, right? The only thing that isn't identical is our use of the isentropic efficiencies.

So, H1 and V1 are the same as before. So, those don't change at all, right? H3 and S3 are identical. exactly the same as before.

So those don't change at all. What we called point two previously is now called 2s. So.

The previous 0.2 is now 2S. And for that matter, the previous 0.4 is now 0.4S. So nothing really changes much in this particular problem. Again, the big change is just the way that we treat the pump and the turbine. OK.

So. if what we previously called point 2 is now point 2s, let's continue then and consider that pump. And the pump has an isentropic efficiency associated with it. And we know that it's going to be the ideal over the actual enthalpy change.

And of course, ideal enthalpy change is the same as V delta p. p. And this we call the pump work, but it would be the ideal pump work, right? I mean, we're looking at 2s now.

So this would be the ideal pump work. And it would be over h2a minus h1. So, So we can use this now to calculate H2.

So the pump efficiency is.8. The ideal pump work, this V delta P term, we had already calculated previously. That was just 3.04 kilojoules per kilogram.

So we do it exactly the same way. No need to redo it. H2A is my unknown.

H1 is the same term that we had before, which is what? 340.54. And this allows me to find the value of H2. So I can do that. find the actual enthalpy at.2 and it's 343.62 kilojoules per kilogram.

OK. So that's all we really have to do here. Next, we move on to the turbine.

Now again, we're just using the isentropic efficiency of the turbine, which in this problem is 0.9. And this is going to be H3 minus H4 actual over H3 minus H4F. And again, we consider that what we called H4 in the previous problem is now H4S and H3 is exactly the same.

So, there's only one unknown in this particular equation. So, we have 2994.3. minus H4A, or 2994.3 minus what was previously just H4, it's now H4S, so 2272.3. All the units are the same in kilojoules per kilogram.

So, we can find the actual enthalpy as 2344.5 kilojoules per kilogram. All right, so. have all the thermodynamic property data we need. Let's just finish up the problem. So the heat input is H3 minus H2A.

If we go through the mathematics here, we end up with 2650. 0.68 kilojoules per kilogram. And the net work, again, the turbine minus the compressor work, or H3 minus H4A minus H2A. I'm sorry, I said compressor work, turbine work minus the pump work. So H2 minus H3A minus H2A minus H1. Again, we have all this enthalpy data, H1 and H3 are identical to the first time we did this problem.

So the network then becomes 646.72 kilojoules per kilogram. And the thermal efficiency then is just the ratio of the above. So we get.244 or 24.4%. And the net power is just the mass flow rate times the work per unit mass net. So again, I'll let you plug in the numbers.

And we get 22.6. 10 to the third kilojoules per second or 22.6 megawatts. Of course, a kilojoule per second is a kilowatt, so we can show those units if we prefer.

All right, so there is that particular problem. Any questions on how we solve this one? Please note that the numbers are significantly different.

Well, maybe I shouldn't say significantly, but noticeably different. In this particular case, our net work is, you know, what, about 10% or so less. The heat in. going to be a little bit less because we end up having a somewhat higher enthalpy coming out of the pump than we did if we assumed that it was not ideal, I'm sorry, than if we assumed that it was ideal. But still, the net effect is a reduction in the thermodynamic efficiency, which is what we'd expect.

I mean, in the first case, we didn't consider any of the losses, right? Now, we're specifically considering the losses within the turbine and the pump. Again, heat transfer losses, friction losses, fluid flow losses. So, these losses are all built into the efficiency term that you're given. And that would be that.

Okay. So, any questions then on this problem? Great. So, let us move on. So, one thing that I should note is that There are ways that we can improve the thermodynamic efficiency of a Rankine cycle.

And some of these ways are similar to what we have used in the Brayton cycle to improve its overall thermodynamic efficiency. One thing that could be done is to use the Brayton cycle to improve its overall thermodynamic is regeneration. But regeneration is a little more complicated with regards to a steam cycle than it is for a gas power cycle. In fact, before I even get into that anymore, I just want to say that I'm going to talk about the I just realized I didn't put the TS diagram on the board for this last example problem that I was doing. I think that it's not going to be a problem.

If you just look at the previous diagram, it looks essentially the same. The only thing we don't know for sure is where state point four is. We're not completely sure where might point four go. We're not completely sure if four A is going to be a two-phase mixture or superheated. We don't know for sure.

I mean, we'd have to look into the data and figure that out. Now, is it that important? Eh, perhaps not.

But let's just finish up the problem anyway. So, I need more erasers. Okay, so there's one, here's 2a, here's 3. here's 4s and 4a, we just don't quite know yet.

OK. So all we really have to do is go into our property tables at 50 kilopascals and compare our value of the enthalpy at 4a to the value of hg. Is 4a less than, equal to, or greater than the value of hg at that particular time? particular pressure. So that's what we really should have done as we finished up this problem.

And now I have to hunt for the answer. Let's see. OK. So, HG is about 2645 and my value of H4 is, H4A is 2344. So, it would look like we're still in the two-phase region. For less than HG, I mean, assuming that I looked up the numbers properly, which I don't always do.

So therefore, it's two-phase. So it turns out for this particular problem. going to be right around there. OK.

Now again, that assumes that I looked up the numbers properly. I'm just kind of doing this on the fly because I didn't actually see the results in here. So hopefully everything I've done is accurate. Anyway, one more time, any questions on this particular problem now that it's finished finally? All right.

Back to what we were talking about before. Okay, so what could we possibly do to improve the thermodynamic efficiency of a Rankine cycle? One thing that we can do is regeneration, but I mentioned previously that we have a little bit of a problem with regeneration in the same context at least as regeneration.

generation in the Brayton cycle. Keep in mind that in the Brayton cycle, the discharge from our turbine was always at a relatively high temperature, okay, several hundred degrees, certainly well above the ambient air. air temperature and even well above the temperature of the ambient air that's compressed from point one to point, from point one to point two.

So it makes a lot of sense to use regeneration in a Brayton cycle because we have a higher temperature exhaust than our inlet air to the combustion chamber. We can exchange some heat and heating up the inlet air into the combustion chamber means using less heat in the combustion chamber. Great. But that's not the case here, right?

I mean in this particular case. going through the compression process in the pump, we're actually at a temperature that's considerably higher than the temperature that leaves that turbine, right? How can you use regeneration in that case?

I mean, certainly there'd be no way to recover heat at a low temperature and somehow be able to transfer it to a higher temperature. I mean, heat transfer doesn't work that way. So we're not going to be able to use regeneration in the same sense as we use regeneration in or Brayton cycle. So just keep that in mind. Yes, there is still something called regeneration, but it's not going to be done quite the same way.

The other thing that we could do theoretically is by going through the compression process in the pump in multiple stages, but that doesn't really help us out very much. I mean, look at the previous example problem and look at the total amount of pump work. There's only about three kilojoules per kilogram of pump work and we can compare that. that to the amount of net work of well over 600 kilojoules per kilogram.

So the pump work is not a significant contributor to the efficiency or perhaps I should say inefficiency of the cycle. We're using what? Less than a half a percent of the turbine's outlet work in order to provide pumping.

Now there's no such thing as back work ratio for a Rankine cycle. But the back work ratio is an illustration of that, right? The back work ratio is the ratio of how much work is required to be done.

by the compression device versus how much work is produced by the turbine. We saw that for the gas turbine cycles, that could easily be 50% or even higher. For here, it's a fraction of a percent. So trying to minimize the pump work is never going to be cost effective, never, ever, ever. You know, some of the pumps that are used are absolutely massive in steam power plants.

They cost for a single pump millions of dollars. To save maybe a small fraction of a fraction of a percent in pump work. work and having to spend millions of dollars for that extra pump, not to mention some sort of intercooler in between, that just makes absolutely no sense at all. So we're not even going to consider that. However, we can increase the work output if we do, pardon me, if we do work output in stages similar to what we saw for the Brayden cycle.

Will allow us to go through an expansion process down to some intermediate pressure and then we're going to reheat back into the engine. back up to some elevated temperature and then allow the steam to pass through a second turbine in order to produce more work. And it turns out that if we compare the extra amount of work that's produced by working in stages to the extra amount of heat that's going to be required during the reheat process, this should actually improve the thermodynamic efficiency.

In other words, the net work attributable to this process of reheat is actually going to be more significant than the actual work that's being done. than the amount of heat transfer. And if we look at just the efficiency of that individual process, again, that's going to be a little bit higher. We're going to get a little more work per unit of heat transfer than we did for the original part of this process.

The efficiency is going to go up. So that's what we're going to talk about first is reheat. OK.

So no longer can we call this the simple ranking cycle. We're going to have to just simply call it the ranking cycle. This time with reheat.

So first, Let's keep in mind that there's actually a second reason why we may want to use reheat. Certainly one reason is to improve the thermodynamic efficiency. Another reason has to do with temperature limits.

And I believe we talked about this as well with regards to the Brayton cycle. If you add heat all at once, you end up with a really high exit temperature from, well, what would be the boiler in this case. Sometimes those temperatures are so high that you're not really able to find the.

the appropriate materials for use in your steam turbine and you have very serious, well, let's say materials concerns in that design. So rather than increase to some really elevated temperature, we could just increase to some modest temperature, do some work, and then reheat back up to that modest temperature. We'll still have a high thermodynamic efficiency, but we're not going to have to worry about the materials problems associated with this high temperature steam, which is moving through the turbine.

So what would the cycle look like? We understand its need. Now let's look at it. So here we have our pump.

And as usual, we'll just enter the pump at state point one. Big P for pump. Here's my work input from the pump.

We're now going to go through our boiler. So here's the boiler where my heat input takes place. We're going to leave the boiler and go to the first turbine. So let's put some state points.

I see two entering the boiler, three into the turbine. OK. Now we're going to expand through this first turbine until we get to state point four.

And now we're going to go back into our furnace. Keep in mind the boiler is just one section of the furnace. You've got this massive, massive furnace.

It could be 10 stories high. and there'll be different heat exchanges within it. One heat exchange is certainly the biggest one.

It's going to be what we call the boiler. But within the furnace there's going to be other heat exchangers. The one we're talking about now we simply call the reheater. So often we would just place the reheater at least on our schematic diagram immediately adjacent to the boiler because in the real world they are adjacent to each other.

Anyway, they're different sections. They're different heat exchangers but they're still next to each other. So now we come out.

of the turbine at.4 and we go through the tubes of this heat exchanger that we call the reheater. And we come out of this reheater at.5. And now we have to go back into a second turbine. So here's the second turbine.

And work is going to be done from both of these turbines. Heat input, by the way, is also going to be provided not only to the boiler but also to the reheater. So both of these heat input terms as well as both of these work terms have to be considered.

And then coming out of the turbine, finally we go back to our condenser. So just big C for condenser. And we leave in, let's see, six. has to come out of this turbine.

Seven would come out of the condenser and then flow back into the pump. So that's what this cycle would look like on a schematic diagram. Now one thing I will note here, unlike the Brayton cycle, there's no requirement to have the pressure ratios the same across each turbine. turbine stage, OK? And there's also no need to have the same temperature going into each turbine or the same temperature going out of each turbine, OK?

It doesn't have to be that way for the Rankine cycle. So you could be given any of number of different pressures that the reheater operates at. I will note, however, that the basic rule of heat input or heat exchange at constant pressure still applies to this part of the ranking cycle, OK?

It's still a ranking cycle. The heat transfer is still done at constant pressure. And not only does this include the boiler as well as the condenser, but also the reheater. OK. So that part of the process hasn't changed.

Yes? Just a quick question. So point one and point seven, shouldn't they be exactly the same?

Point, oh, yes. No, actually I'm sorry. I, no, no.

Yeah, you're right. I've missed, I don't need a.7, so that's a good point. Here's my heat rejection in the condenser. And I really should pay more attention to my own notes because I don't have any.7 in my notes.

So, but that's a good observation. Yeah. Once we reject the heat in our condenser, we're really going all the way down to the conditions of.1. So, yes, thank you for pointing that out. Okay.

Now, what would this look like on the TS diagram? And again, we'd have to consider whether we're talking about the ideal case or the non-ideal case. So I suppose we should just begin by looking at the ideal case with reheat. Well, we'd have something like this. So again, we'll show this on a TS diagram relative to the saturation lines.

Let's show a minimum and maximum pressure. But let's also note that the reheater is going to operate at some intermediate pressure, right? We're leaving the first turbine, coming out of that turbine at an intermediate pressure before we reheat at that constant pressure. And then we expand through the next turbine all the way down to, well, the pressure of the condenser. So there will be some intermediate pressure in this problem.

So we might as well show it as well. And this would be simply the pressure associated with the reheater. So let's just go through the process. So we start at point one. We're going to compress not up to the pressure of the reheater, but all the way up to the pressure of the boiler.

So, point two should be directly above point one at the maximum pressure. We're now going to add heat until we get to point three. We're now going to expand until we get to the pressure of the reheater.

So we just move vertically downwards until we get to.4. Again, with the understanding that.4 might be two-phase. It might be saturated vapor. It might be superheat.

We don't know. It just depends on the problem. Nonetheless, we reheat up to whatever the temperature of.5 is. By the way, I have shown 5 and 3 as being the same, but it doesn't have to be. It is very common to have the same temperature.

coming out of the boiler as well as coming out of the reheater, but it's not necessarily going to be true. So read the problems carefully. But nonetheless, we come out of.5 and then we expand through the second turbine down to.6 and ultimately reject heat to.1. OK.

So let's just again make sure that we include all the right terms, right? We have heat input in the boiler and the reheater. We have turbine work out.

output from each of the two turbines. And I suppose it doesn't hurt to show the pump work input and the heat rejection within the condenser. So this is just an illustration then, both schematic as well as on a thermodynamic property diagram of what this particular Rankine Cycle III heat would look like.

Again, let me note, no need. to have the pressure ratio the same across each turbine. No need to have the same temperature at inlet or outlet from each turbine.

Okay, so ever so slightly different, but the same basic process, right? We're still talking about constant pressure heat input. Still talking about work being done in two stages instead of one. Again, out there in the real world, it is certainly possible that you could have multiple reheaters, but I I've honestly never seen that.

So theoretically possible, but typically it's not done. And I've never seen any situation with multiple pumps. Sometimes you'll have multiple pumps and turbines that are in parallel with each other because sometimes it's good for maintenance reasons instead of having one device that works at 100% capacity. You'll have two devices that work at a fractional capacity so that you don't have to shut the system down every time you need to go through maintenance.

You can run one pump with one turbine and do maintenance on the other devices. In fact, what I've generally seen is you typically have three work devices that run in parallel to one another, particularly for the pumps. They're all at 50% capacity, so you can run two pumps at 100% while doing maintenance on the third.

third pump and then, you know, cycle that third pump back in and keep running the system without having to shut it down. Turns out that pumps are really the biggest maintenance issue. You know, rotating equipment in general tends to be the biggest maintenance issue in a power plant.

So, you know, having multiple pumps that operate at fractional capacities is good from a maintenance point of view. Yeah, it costs you more money, but it would eliminate downtime and therefore you're still able to make more money than the extra money it costs you to buy the extra pump. Again, things to think about in the wonderful world of thermodynamics.

And with that, we're done for the day. So we'll continue this discussion next time. We'll look at more example problems. And that's that. So again, the one student that hadn't shown me proof of prerequisite, please don't forget to do so.

And again, you've got homework due next time. I still have some old homework from last time that's up here. So pick up your work if you would.

See you all on Wednesday.

Transcript for:Understanding Rankine Cycle and Efficiency

Transcript for:
Understanding Rankine Cycle and Efficiency