 In this lecture what we'll be doing is we'll be solving a couple of example problems involving the rank and cycle and we'll begin by looking at a simple rank and cycle just so that we get some practice at solving these types of problems. So there's our problem statement what we're dealing with is a simple ideal rank and cycle and we're told that it's operating between the pressure limits the boiler pressure 3 MPA and down to the condenser pressure of 50 kPa. We're told that the steam at the inlet to the turbine is at 400 degrees C so that would correspond to the 3 MPA boiler pressure. Now the mass flow rate is given at 25 kilograms per second and we're asked to do a couple of things. One is to show the cycle on a TS diagram which is always a good idea to be able to figure out what's going on with respect to the saturation lines. The second thing that we're after is the thermal efficiency of the cycle and finally the net power output so that would be a power output of megawatts. So that's the problem that we have before us and whenever you're doing these types of problems it's always a good idea to begin by writing down the information you know and we will also write out the TS diagram so let's get started by doing that. So that's the information that we have before us now. Now looking at the TS diagram for this cycle it's not too complex so we write our two-phase line for the cycle and remember we always start at the compressed liquid region over on the left that is state one we go through the pump that takes us up to state two which then goes into the boiler we follow the lines and we go into a superheated state that takes us to state three and then we come through our turbine and we expand to some point right now we don't know where we are expanding to it could be superheated it could be right on the saturated vapor line or it could be in the two-phase region and so that will be state four and that takes us back into the condenser. So what we'll be doing here whenever you're solving one of these problems it's a bit of a puzzle that you're solving and you usually start like any puzzle you start with the conditions that you know so consequently what we'll do we'll start at state one and it will go into the steam tables for that and it's what we'll do you go into the steam table and here you would be looking at saturated liquid or saturated water we know this here is a 50 kPa and this here is at three mega Pascal so you go into the saturated or the steam table saturated liquid 50 kPa and you pull off the enthalpy and I'm going to skip the units but that would be kilojoules per kilogram and another thing to pull off at this point let's get the specific volume and after you do a number of these problems you'll know which things to pull out of the steam table at which point but you'll see why we want the specific volume in a moment so that's a specific volume at state one as well as the enthalpy now we have a pump which takes us from one to two at the initial state here we have no idea how to estimate what state two is but what we'll be doing is we'll be taking advantage of our reversible steady flow work equation and that enables us to say the work in the pump is the specific volume at one multiplied by the change in pressure between state two and state one and that is equal to the change in enthalpy so by knowing the enthalpy at state one we can use this equation in order to determine the enthalpy at state two because if we look here we pulled this off the steam table and we know p2 that's the boiler pressure p1 is the condenser pressure so we know everything in that equation we also know each one and that enables us to get h2 so when we calculate this what we find is the value is 3.0385 and again that would be kilojoules per kilogram and so we can sub it back in and solve then for h2 and when we do that we find that h2 is that so the next thing to do we have state one and we figured that out we figured out state two the next place we want to go is to state three so in order to do that what we have to do first of all is determine are we superheated and so you go and you look at three megapascals and 400 degrees C are you in a saturated region are you superheated and for this case it'll be superheated so that takes us to state three so you pull out the value of the enthalpy there and let's pull out the entropy we're going to need that and I'll show you in a moment why we're doing that and that would be kilojoules per kilogram kelvin for entropy it's different than enthalpy now state three we just pulled off the steam table it's superheated now we're going to use the value of enthalpy there in order to figure out where state four is right now we don't know where it is but we know the entropy and we can say they're assuming that this is going to be an isentropic expansion process in the turbine we're told no different that enables us then to get a direct link between three and four because we can rate entropy at three is equal to entropy at four annoying the pressure we can then determine state four so that is the next step that we will be doing here so we said s3 was equal to s4 and then we know that this is occurring at 50 kPa and it turns out for that value we are in the two phase region and so those are the values off the steam table for SF and then that would be SFG and from that we can then solve for the quality at state four eight nine six six so this is really a cycle that's not that great because ideally you don't want to have expansion into the two phase region or you get the drop what's in the drop in efficiency but knowing that we can then determine the enthalpy at state four so with that we have all of the states specified because we just determined state four so we have one two three and four and so now what we can do basically go ahead and start applying the first law between components in order to determine the thermal efficiency if we go back to the problem statement they want us to get the thermal efficiency and the net power output so we're going to work on the thermal efficiency and we can write that the thermal efficiency is the network and divided by the heat and and the net work is equal to the work out of the turbine minus the work that we put into the pump now applying the first law for a steady flow device with no heat transferred at the turbine and no kinetic or potential energy that turns out to be just h3 minus h4 and then similarly for the pump that is h2 minus h1 we know those values so we can now substitute them so we get that for the network and the heat and the heat and we can evaluate if we go back to our TS diagram that's going to be what's going on in the boiler so this is where we have Q in right here so we're going from state two up to state three as we go through the boiler knowing the values of enthalpy so there we have work net and Q in and with that we can go back and take each of these and plug them into this equation up here so let's go about doing that and then that will enable us to determine the thermal efficiency of this cycle and just like we would expect it's not the greatest thermal efficiency it's 28.4 percent that's because we're expanding quite far into the two phase region and you lose efficiency as a result of that now the that was the first part of the problem let's see here so we have the thermal efficiency the next thing and we wrote out the TS diagram the next thing they want us to do is calculate the net power output so let's tackle that so from before we wrote work net was 820.45 kilojoules per kilogram the network output is just going to be the mass flow rate multiplied by the work per unit mass that we just calculated and what we get is that this plant has a power output of 20.5 11 megawatts so that concludes this example problem you can see it's kind of straightforward and simple but whenever you're solving these problems write out the TS diagram to begin with you don't always know all the states and so sometimes you kind of got to jump around a little bit had we drawn this a little differently here with expansion down into the superheated region I would have had to have redrawn the diagram but I made a guess that it went into the two phase region which in in the end it did as we could tell by the quality but anyway so that that's the TS diagram it came out correct in the first place so that concludes that example problem the next segment we'll take a look at a more challenging example problem again dealing with the ranking