Okay. Whenever we're trying to define a state, we need to take information about the process, both before and after that state, when we're considering a cycle. So, for this case, remember we were looking for the enthalpy at the exit stage of the first turbines. So, if this is state one, we wanted the information at state two. Now, from one to two, we know that that's an isentropic process, and then from two to three, we know that that's an isoberic process. So to define state two, we need to realize that P2 is given. We know the pressure is 25 bar and that, that's equal to P3. In this case I told you the reheater has an inlet pressure of 25 bar and that is exactly P2. Conversely, I could have told you. The reheater has an exit pressure of 25 bar and you should have mapped that successfully to the fact that that's P3. so that gives us the information from 2 to 3. We then need to look upstream to determine the rest of the information to fully define the state and that's the isentropic nature of the process. And so, we use the entropy that we found from our online steam calculator. And between those two pieces of information, pressure and entropy are always independent. As we see in our figure, and as we show here, state two is in the super heat region. And again, our hint here is that we don't like steam turbines to experience phase change. And so we're going to consider those generally to always be super heat region. So, we would go look up in our steam tables. So if we had more extensive online calculators, steam tables online, we would find that for a pressure of 25 bar and an entropy that was given at state one of 6.9, so S1 equals 6.904 kilojoules per kilogram kelvin. Which equals S2, we come up with an enthalpy of 3149 kilojoules per kilogram. Okay, at this point we now have enthalpy values for essentially all stated information at every single state condition in this process. So we can determine whatever information we want. Now, our original question that we were asked is to calculate the cycle efficiency of the power plant. And we're going to do that, and we're going to pick up some different addition information along the way that's going to be very informative about telling us scale, magnitude, and how we can interpret the information from this type of a cycle. Okay, so the next step is let's find the heat transfer into the cycle. Now, remember from our process diagram let's go ahead and draw that again. We have our steam generator. And here's where we add heat. And remember because there's a re-heater, there's two points in this cycle where we add heat. So this is QN1. We had our first stage of the turbine, and then we had our re-heater. And that's where we added heat a second time. So, this is state one. This is state two. Here's state three, and again to complete the diagram, here's my second stage of the turbine, my condenser, and then here is my pump. And we would finish our labeling, it looks like that. Okay, so we want to find the heat transfer into the cycle. We're going to invoke all the assumptions we typically do for these types of turbo machinery. So, we're going to assume that we have steady state. Steady flow. So all those time derivatives are going to go to zero. We're going to consider that the kinetic, the changes in kinetic and potential energy are negligible. And so, we don't have to worry about those and that leaves us for each of these component. Recall, all we're going to have to consider is the heat transfer, the work transfer and the enthalpy. Okay. For the turbines, we're going to consider those adiabatic. So, they'll have work out of the turbine but all turbines are considered adiabatic. And for our steam generators, of course, they, the heat exchangers, so any heat exchanger, there's no work transfer. Okay, so I'm going to simplify this conservation of energy analysis to be the very simplified form that we had determined before. And I'm just going to cut straight to the chase just so we can get to some numbers faster here. So if we consider the heat transfer in, into the steam generator, that's going to be given by a balance between the enthalpy at the exit and an enthalpy at the entrance. So, H1, minus H6. That's heat transfer into the system, so we expect this number to be greater than 0, because that's our sign convention. And if we go ahead and plug in the numbers that you've been collecting in the past units, we're going to have an enthalpy at state one of 3625.8 minus the enthalpy at state 6 which is 426.5. And again, these are normalized by the mass flow rates so this is a kilo joules per kilogram number and if we wanted to be precise here we could label this as a q, lower case q. Qn1 and I like to keep the, I want to keep it normalized by mass flow rate just for a little while. Because we're going to actually determine this mass flow rate in just a couple of moments, well a few more than a couple of moments. And if you go ahead math we get a number of 3199.3 kilojoules per kilogram. Are added in the steam generator. So, this the amount of energy on a per mass basis, added to the water in the steam generator. We go through that exact same process for the reheater to determine the second portion of heat added to our system. So, for the reheater, we have heat addition in 2. That's this one here. Again, we're going to normalize everything by the mass flow rate. And we're going to have when we plug in the values here, this is going to be for H3 minus H2, enthalpy at state 3 minus the enthalpy at state 2. And that's going to give us values of 3686.8 minus 3149.0 on the kilojoule per kilogram basis which gives us a total of 537.8 kilojoules per kilogram. So, to be very precise in our language, this is the net heat that we're interested in here. The net heat into the cycle, which is the sum of these two contributions. The contribution from the steam generator and the contribution from the reheater. So Q in total. Again, all normalized on a mass flow rate, is equal to Q in 1 plus, which is a sum of the 3,199 and the 538 kilojoules per kilogram. Giving us a net value of 3,737.1 kilojoules per kilogram are added to the cycle, between those two heat exchangers. Okay, we're going to take that information. And remember, for us to determine the cycle efficiency, we need to have, remember cycle efficiency. We'll just do a reminder of that. Is the work transfer for the cycle, divided by the heat transfer in. And we remember the work transfer for the cycle, the net work transfer is identically equal to the heat transfer for the cycle, divided by the heat transfer in. Okay. So, we've just found the denominator that we need for this calculation. We still need the numerator. We can determine either net work transfer or net heat transfer, one of the two, but we don't need to do both. Since we've already started with our heat transfer calculations, I think we should continue to work with the heat transfers. So, that's what we'll do. So, we're going to find the net heat transfer for the cycle. Remember, we need to know all the heat transfer in, and all the heat transfer out. Now again, we've already found all the heat transfer in. And again, we can make these on a rate basis. Now we need to find the heat transfer out. There's only one heat exchanger where we reject heat, that's the condenser, and so if we want to find the heat transfer out. Again, normalized on mass basis, that's simply the enthalpy difference across the condenser. So, that's going to be h5 minus h4. And if we go ahead and substitute in the values that we determined for those two enthalpies, we get 417.46 minus 2756.4 kilojoules per kilogram. And we get a number when we punch that into our calculator, which is minus 2,338.9. It's a negative number, which is, as we would expect, because remember, this is heat transfer out of the system. So, to find that cycle efficiency, just like we had on the previous slide, we need to know the net heat transfer for the cycle divided by the net heat transfer in. Go ahead and put these on a rate basis. Everything normalized by the one mass flow rate that's consistent for all the components in this example. In this cycle, because there's only one loop. So, we have one mass transfer in this system. Again, the net heat transfer for the cycle is simply the sum of the heat transfer in and out. So if we go ahead and do that calculation, we get 537.8 plus 3,199.3 minus 2,338.9 all divided by net heat transfer in, which was 3737.1. Which is a value in the numerator here and that heat transfer is 1398.2 kilojoules per kilogram. And then the denominator, we have 3737.1 kilojoules per kilogram, so our cycle efficiency is dimensional, as we would expect it to be. And we get a cycle efficiency of 37.4%. Okay, next. I want to give you some food for thought. We're still going to keep working on this problem. We're still going to keep looking at the numbers. But before we do this, I want you to think about some of the power issues that we face in the United States and abroad. What you're looking at here is a diagram of where the electricity production is by coal within the United States. And we're looking at terawatt hours here in this figure. But more importantly, what you see is that power production is typically by coal where there are coal reserves. So if you're not aware of it, Illinois has significant coal reserves, so does the Midwest, in general, and Texas. And so, what you see is power production by coal is, of course, where generally located where the coal reserves are. The turbo US annual power capacity is about 340 gigawatts of power generated typical capacity within the United States. That's about 50% of the power generation in the United States is by coal. 90% of these power plants are over 25 years old. At least 25% or 50 gigawatts of the coal fired capacity is expected to be retired within the next ten years. They're beyond any terms of re-licensing of those facilities. Nuclear energy within the United States is expected to retire another 40 gigawatts or more of power in the same time period. So that's about 90 gigawatts of power is coming offline in the next ten years. What's going to be the most likely energy carrier for the new, or next generation, of stationary power plants? And this really is a, just sit back and think about it question. We really don't have the tools, yet, for us to identify which one of the energy carriers is most likely to be replacing or coming online in the next ten years. But we'll discuss that when we get started next time, and we'll continue to look at that example of a steam power plant. Thank you.
