 In the previous section what we did is we came up with the process diagram as well as the process schematic for the Rankin cycle with Regen and Reheat. What we're going to do now is we're going to go through and we're going to determine the values at each of the different states, so the property values. And I'll refer back to the process diagram and the process schematic. Whenever you're doing this, the best place to start the process is where you know the information and usually what we'll be doing is we'll be starting right at the saturated liquid line when you're going into the pump. So this would be location one, which on our schematic diagram would be there. So let's go ahead and start the process of determining the properties and the state information and then we'll work our way through and eventually get to the first law. So to begin with, state one, we can go into the steam tables and there we're dealing with a saturated liquid. So what I would do is I would pull out the enthalpy value as well as the specific volume at state one. And with that what we'll want to do is we'll want to be able to determine the enthalpy at state two and in order to do that we'll be using the steady flow equation for work into a pump. And recall that is one that we can evaluate knowing the specific volume at one and then the pressure change between states two and state one. And so that is the work and that work is going to be equal to the change in enthalpy between states two and state one. So looking back on our TS diagram, what we're looking at here is the work going from state two to state one and that's what we've just evaluated. And so with that what we can then do is determine the value of H2. So that gives us state one and state two. The next thing we want to do is we want to determine the value at state three. So let's look at our TS diagram. So the next state that we're interested in is right here, state three. Again that is on the saturated liquid line. So we're just going to use the same process that we did before. And the pressure for this one is going to be the pressure at which let's look back on our TS diagram. That is the pressure at which we're pulling the steam off as part of the bleed stream. So that was a 0.8 megapascals. So that is the pressure that you need to look at when you go into your steam tables. Then we'll pull off the specific volume because we're going to use the work equation again in order to determine information at state four. Except here we have the pressure difference between states four and three. So with that we know H3 and so we can directly determine H4, H4 turns out to be, okay. So if we look at our TS diagram we now have state one, state two, state three, state four. The next state that we need to determine is state five. And state five we are told the temperature, we know the pressure is 10 megapascals and the temperature is 550 degrees C. So with that you can look in your steam tables but you'll find that that is at a superheat condition or superheated steam. And so what I'm going to pull off is the enthalpy value there. And another thing that it's useful to pull off at this point is going to be oops sorry there's the problem statement. One other thing that will be important to pull off, so we're trying to find information at state five and notice five to six, seven that is an isentropic process. So at this point it is beneficial to pull out the entropy at state five. So that is another value that we're going to pull from the tables. So we have the entropy at state five. Now in determining state six and if we look on the TS diagram state six, state seven are the same. Well we know the entropy is going to be the same between five and six and seven. But we need to determine the enthalpy at that point. We know the pressure is 0.8 megapascals. And so we will use the entropy enabling us to determine the value of the enthalpy. And so for this you're probably going to have to do a little bit of interpolation in the tables. But what we have is S6 is equal to S5. And so we know the pressure here it is 0.8 megapascals. And so what we do is we look at our tables and we find that we are still superheated. You know we can estimate the temperature. And the other thing that we get is the enthalpy of six which is going to be the enthalpy of seven as well. Okay, so let's look at our TS diagram and see how much we've been able to extract. So we have state five, we have state six and seven. Now what we still need to determine is states eight and nine. And in order to get state eight we know the pressure is 0.8 megapascals and it goes up to 500 degrees C. So we can go into the superheated tables and get that. And then to get property information at state nine we're going to use the same thing that we did between five and six and that is we're going to get the entropy at state eight which will enable us to determine values at state nine. So let's create a new page and we will do that on this new page. So we're after state eight. So this is the conditions after you go through the reheat boiler. So that is the value of enthalpy and entropy at state eight would be that. That'll be kilojoules per kilogram Kelvin. And finally state nine what we do is we use the fact that expansion from eight to nine is an isentropic process. Now we know if we go back the pressure in our condenser is 10 kPa. So what we can do is we can look in the steam tables at the value of entropy at 10 kPa. And when we do that the the value of the entropy there is going to be. And what we find is that we're in the two phase region at 10 kPa. So we use the equation that enables us to determine enthalpy when we're in the two phase region. And we want to solve for x nine. So you see 96 percent is in the vapor phase. So we're quite close to the saturated vapor line. But we have 4 percent droplets or water in the mix. Which means we're going to lose some efficiency there. And finally H nine we can get the enthalpy value from there as well. So let's go back to our T.S. diagram. We've now finally determined state information at eight as well as at nine. So with that we've specified all of the state information. There still is some information that is missing before we can really solve the problem. And one main piece of information that still remains to be determined is the mass fraction. What percentage of mass are we stripping off and sending into the open feedwater heater as well as sending through the reheat. So in order to determine the mass fraction what we're going to do is we're going to apply the first lot to the open feedwater heater. Sometimes when you're solving these problems it's kind of like solving a big puzzle. You don't always know exactly which direction you need to go as you're going through it. And so you solve one piece of information that gives you another that gives you another. But in this case we will apply the first law to open feedwater heater. So let's take a look at where that gets us. Now in here when we're dealing with a open feedwater heater we will assume that the open feedwater heater is well insulated. There are no moving boundaries and consequently there is no work being done. Kinetic energy and potential energy we will neglect. And with that the equation we get is basically just looking at mass flux times the enthalpy coming in and that will be equal to mass flux exiting enthalpy exiting. So with that equation we can look back at our schematic and what we see coming into the open feedwater heater we have information at state 7, information at state 2 and what is leaving is information at state 3. So the two inlets are 7 and 2 and the one exit is 3. And so those are the things that we put into our equation. So what is coming in? So we get that equation there. Now what we're going to do and we want to be able to reduce this a little further ideally we want to get it to enable us to determine the mass fraction y. A couple of other things that we know through the continuity equation remember that is mass conservation we can write m dot 3 equals m dot 5. So let's take a look back at our diagram see what that is saying. We're saying 3 is equal to 5. That's basically the mass flow rate going through the boiler is the same as the mass flow rate at 3 and it's the same as mass flow rate at 5. So that's a pretty straightforward one. Another thing that we can say is mass flow rate at 2 is mass flow rate at 5 which is the boiler mass flow rate minus m dot 7. So let's look back and see what that is saying m dot 7 is what we're stripping off and sending through the feedwater heater. So with that comment we were saying m dot 2 mass flow rate at 2 which is basically what goes through the low pressure turbine is equal to the boiler feed minus what we're stripping off and putting into the open feedwater heater. So we can do that with mass conservation and we can then make substitutions into our first law applied to the open feedwater heater. Specifically what we can do is we can take this m dot 2 and we can substitute it in here and the other thing we can do is m dot 3 equals m dot 5 that is this equation here. We can substitute that there and what we're going to do after we substitute these we're going to divide by the mass flow rate coming through the boiler m dot 5. If you recall back when we defined the ratio of the mass fraction it was expressed as being m dot 7 what we're stripping off divided by the mass flow rate going through the boiler and so that's why we want to divide by m dot 5. So let's go ahead and do that in the next slide. So there we have our equation after making the substitutions now what I'm going to do is divide through by m dot 5 so the mass flow rate going through the boiler and when we do that we get 1 minus y notice that that's the first thing because the second term here we have m dot 7 divided by m dot 5 that's our definition of mass fraction y multiplied by enthalpy at 2 plus y times enthalpy at 7 is equal to 1 times enthalpy at 3. And if we look back at our schematic we said 1 was times h3 well there we have 100% mass flow coming out so that makes sense that it would be 100% and we have 1 minus y times h2 1 minus y times h2 and the other thing we have is y mass fraction enthalpy stream 7 and we get y times enthalpy stream 7 so it kind of relates to the mass fraction that we have now in this equation here we know all of the values of enthalpy because we've determined all of our state information so we can now go ahead and directly solve for y and when you do that you get the mass fraction as being 2 0 1 8 so we're stripping off approximately 20% of the stream to go the steam to go into the open feed water heater. So that will I'll stop at this point and then in the next segment what we're going to do we're going to apply the first law to a lot of the different components that are within the system and the final step to the solution will be then going ahead and trying to determine the thermal efficiency of the cycle.