 An ideal regenerative Rankine cycle with a closed feedwater heater uses water as the working fluid. The turbine inlet is operated at 500 psi and 600 degrees Fahrenheit, and the condenser is maintained at 5 psi. Steam is supplied to the feedwater heater at 40 psi before being expanded into the condenser. Determine and complete the following. First, why the proportion of mass flow rate that leaves the turbine early to feed the closed feedwater heater, the specific work-in relative to the cycle as a whole, and the thermal efficiency. So note that in this problem, we still have the medium pressure occurring at 7 and 3 for the closed feedwater heater, but instead of compressing it up to the high pressure before mixing, we are expanding it down to the low pressure before mixing. That means that one pump is accomplishing all of the work and that one, two, five, and six all have the mass flow rates of the cycle. Let's start by identifying our independent intensive properties for the state points so that we can then determine the specific work-in, the specific Q-in, the specific work-out, and the specific Q-out, and then the thermal efficiency. I will start this process by pointing out that we still have three pressures, 500 psi, 5 psi, and 40 psi. Which state points have the high pressure? Two, five, and six. Which state points have the low pressure? Eight, four, and one. That leaves three and seven with the intermediate pressure. So I will populate one independent intensive property for all eight of our state points, using pressure. Next I will consider the pump and the turbine. I was given no indication as to an operational efficiency, so I will assume that they are both 100% efficient. That means that the process from one to two is assumed to be isentropic because an isentropic process represents the ideal work for a turbine and a pump. And then I assume that S7 and S8 are all equal to S6. That leaves me with state point one, three, four, five, and six. Left hanging. State six is easy. We have a temperature. So state six has a temperature of 600 degrees Fahrenheit. Next I'll assume that the condenser just condenses, and in this case is also a mixing chamber, but the condenser condenses, and as soon as the water has condensed, it leaves the condenser. It's just a mechanical device where water condenses and then it falls to the bottom and leaves. There's nothing to refrigerate it, to sub-cool it, to hyper-compress it. There is just a condensing process and then the water leaves, so we assume it leaves as a saturated liquid. Then three, four, and five. Well, I can make the same assumption about state three that I did in the previous problem where I had assumed based on the same concept of using latent energy for the heat exchange that the water that leaves at three has condensed and leaves. And then because I'm assuming that our closed feed water heater is operating ideally because I was given no indication otherwise, I'm going to use the same relationship for maximum conditions, which is a temperature at five being equal to a temperature at three. That leaves me with state point four. And to answer that question, let me pose a question. What do we know about expansion valves? Well, if we were to set up an energy balance on the expansion valve, I have no opportunities for heat transfer nor work. Therefore, this open system undergoing a steady process would have its energy balance simplified down to the sum in November theta is equal to the sum out of m dot theta, theta contains enthalpy, kinetic energy and potential energy. So if I neglect changes in kinetic and potential energy, I'm left with m dot three h three is equal to m dot four h four. And then if the mass flow rate at three has to equal state four, because it's a steady device with one inlet and one outlet, then h three must equal h four. Therefore, we treat expansion valves as being isenthalpic, that is, of a constant enthalpy. So I use x one and p one to look up s one, then I use s two and p two to look up h two. I guess I also look up h one using x one and p one. And I use x three and p three to look up h three and use h four being equal to h three to announce that h four is known. And then I use t five and p five to look up h five. So I have to go back to state three look up the saturation temperature corresponding to our 40 psi. Then I use t six and p six to look up h six and s six. And I use s six to look up h seven and h eight using the pressure at the medium pressure and the low pressure respectively. So I have everything I need to perform all eight enthalpy lookups. So let's just assume that we had them and move on. The next process would be to calculate the specific work in the specific queue in the specific workout and the specific queue out. The specific work in is going to be occurring in the one and only pump. Let me back up a second. We're defining y as the mass flow rate that leaves the turbine early. So I'm calling y m dot seven over m dot six. And then the remainder one minus y is m dot eight over m dot six. And I recognize that one two five and six are all the same. And they are all the mass flow rate through the cycle. Then eight stands alone. And seven three and four are all the same. So specific work into the cycle would be the total power input divided by m dot cycle, which would be m dot one times h two minus h one divided by m dot cycle because m dot one is equal to m dot cycle. That means my specific work in is just h two minus h one per queue in I'm looking at the boiler boiler also has the mass flow rate through the cycle flowing through it, which means that I'm just going to be left with h six minus h five. For our specific workout, we have the same exact energy balance that we have had for the previous two examples. So I'm going to use h six minus y h seven minus the quantity one minus y h eight. And if you want to see why unintended, you can go back and watch those videos again, h six minus y h seven minus the quantity one minus y h eight. And then and then specific you out is going to be the complicated one this time because I have multiple inlets and outlets. So let's just walk through that energy balance. I have entering mass flow rate at eight and four. Let me double check those state point numbers. And because I have a steady device that is an open system, my energy balance is going to simplify quite a bit. One to skip a few I can jump to writing q, as well as the sum in of m dot theta is equal to q dot out plus the sum out of m dot theta because there's no opportunities for heat transfer in nor work. And it's an open system operating steadily. And then theta contains enthalpy plus specific kinetic energy plus specific potential energy. So if we assume that changes in kinetic and potential energy are negligibly small, that means I'm left with the sum in of m dot h. There are two inlets so that would be m dot eight h eight plus m dot four h four is equal to q dot out plus m dot one h one, the one and only outlet. Therefore q dot out is going to be m dot eight h eight plus m dot four h four minus m dot one h one. And then if I divide everything by m dot cycle, I'm going to be left with m dot eight divided by m dot cycle which is one minus y plus m dot four divided by m dot cycle which is y minus y times h four minus m dot one divided by m dot cycle which is just one h one. So with these four equations, I can calculate my specific works and heat transfers. From that I can determine my network and that heat transfer and then thermal efficiency. All I need to do that are my eight enthalpies which you'll remember I totally know by now because I totally looked them up and y. So in order to be able to complete the problem, we are going to have to determine y. And remember in order to compute y, we have to perform an energy balance on a device about which we know everything. We can't analyze the boiler, the turbine, the condenser nor the pump because I don't know Q in, work out, Q out nor work in. The expansion valve is boring. There's not much to do there. Therefore I'm left with the energy balance on the closed feed water heater. So that energy balance is something that we've done before so I will go through it relatively quickly. So I'm going to have the sum in of m dot h is equal to the sum out of m dot h because there's no opportunities for heat transfer nor work and changes in kinetic and potential energy are negligibly small. So I'm going to write that as m dot two h two plus m dot seven h seven. That is equal to the sum out of m dot theta, which is going to be m dot five h five plus seven dot three h three. And then I'm going to group together my mass flow rates to limit how many y's I have to keep track of. So two and five will be together and that's increasing in energy. So I'm going to want to write that as m dot two times h five minus h two. That would be the right hand side of my equation. So I'm left with m dot seven times h seven minus h three five minus two seven minus three energy gained energy lost. That makes sense. Next I will divide everything by m dot cycle m dot two divided by m dot cycle is one because m dot two is equal to m dot six and m dot cycle and m dot six are the same. So note here that if you had just copied over the energy balance on the closed feed motor heater complete with all of the algebra below it from the previous problem, you would have an incorrect answer because the mass flow rates are different. Seven and three are both y here. So m dot seven divided by m dot cycle is just going to be y times h seven minus h three. Therefore, y is equal to h five minus h two divided by h seven minus h three. So at this point, all that's left to do to finish the question is look up our eight enthalpies, plug them into this relationship down here to determine why plug our enthalpies and our y values into these relationships to determine the work in the queue in the workout in the queue out and then compute a network out and a thermal efficiency. So again, because the property lookups are supposed to be something that you master in thermal one, it's a little bit outside the scope of thermal two for me to spend a whole bunch of time in this example problem video looking those properties up. So I'm just going to look them up and cut that out of this video. If you want to see where I got the numbers, I will include the work with this PDF so that you can follow it. That will be linked in the description below the video. But in the meantime, here we go, you ready? All eight enthalpies three, two, one. Now that we have all eight enthalpies calculating wise, just a matter of computing some numbers together. For that, we will use our calculator, wake up calculator. And I am going to take h five minus h two, which is 237.35 minus 133.27 divided by seven minus three, which is 1,085.02 minus 236.16. That yields 0.1226. Now we can take that number and plug it into our relationships for work in queue and work out in queue out. So I'm beginning with h two minus h one, which is 133.27 minus 130.17, which gives us 3.1. And then h six minus h five would be 1298.3 minus 237.35, giving me 1060.95. And then h six minus y times h seven. So 1298.3 minus y times h seven minus the quantity one minus y times h eight yields 328.078. And then one minus y times h eight, which is this quantity I already have, probably would have been faster just to start over. But here we are, one minus y times h eight plus y times h four, which was 236.16 minus h one, which is 130.17. And I get 735.972. And then from those quantities, I can determine a network out a net heat transfer in and a thermal efficiency. So 328 and change minus three and change yields 324.978. And then Q in minus Q out is the same number, which indicates that we built those equations correctly. And then we take that number, and we divide by Q in to yield thermal efficiency of 30.6%. So part a asked for y, part b asked for specific work in, and part c asked for thermal efficiency, which means that we are done with our calculations. What I would like to do next is probably unsurprising, like to plot this on a TS diagram. So state one is going to be here, low pressure, assumed to be a saturated liquid, then state two is directly above it, all the way on the high pressure side. You can draw that a little bit better. Also, let's switch to black. So it's easier to see. And two, and then from state two, we are gaining a little bit of heat before we get to state five. And state five was a compressed liquid. So I'm going to leave that in our compressed liquid region. I'm going to call that, oh, I don't know right about here. And then it goes all the way up to state six, which is our maximum temperature and pressure that we'll call right here. And then that expands down to seven and eight, which are directly below it. I believe seven and eight were both saturated liquid vapor mixtures and they are quality of 0.9 and 0.82. So my graph isn't perfectly to scale here, but hopefully it gets the point across. And then seven gives up some of its heat going to three, which is a saturated liquid. And then we have a line of constant enthalpy from three to four. Remember, those look like this. So encountering the low pressure line. And then that is meeting the process from eight to one. So we end up with process line like this. And that's our diagram.