 An ideal regenerative Rankine cycle with an open feedwater heater uses water as the working fluid. The turbine inlet is operated at 500 psi and 600 degrees Fahrenheit and the condenser is operated at 5 psi. Steam is supplied to the feedwater heater at 40 psi before being pumped and mixed with the feedwater. Determine and complete the following. First, why the proportion of mass flow rate through the turbine that exits early to serve the closed feedwater heater. Then the specific work in that's relative to the cycle itself. Then lastly the thermal efficiency. Like our previous example I'm going to try to identify two independent intensive properties about all nine of our stay points. The presence of the closed feedwater heater means that I either have to pump my fluid back up to the high pressure before mixing or use a trap or expansion valve to drop it back down to the low pressure before mixing. In either case I'm going to have additional steps in the process. I have to have a separate place for the water to join back together and that is essentially an open feedwater heater in and of itself. It's just that the purpose isn't to actually accomplish any sort of heat transfer. The goal is just to allow the streams to mix together so I have additional devices. That's why I have nine stay points instead of seven. First I recognize that I still have three pressures and let's think through those together. Which stay points have the low pressure? That's right eight because it's the outlet of the turbine once it has expanded all the way and one because the condenser operates isobarically and then the high pressure would be five and six but the high pressure also includes four, nine and two. Does that make sense? Because the mixing chamber has to occur isobarically and the heat exchange process within the closed feedwater heater also has to be isobaric. So that means the separate streams in the closed feedwater heater that's seven to three and two to nine are both going to be isobaric. So two is equal to nine which is equal to four which is equal to five. Therefore the high pressure includes six and five and four and nine and two. That leaves us with seven and three at the medium pressure. So pump one is pumping from the low pressure all the way up to the high pressure, two is just going from medium to high. So nine, four, five, two and six are high, two, four, five, six, nine. Then eight and one are low. That leaves us with seven and three as the medium pressure. Those pressures give me one half of my independent intensive properties required to get to all the rest of the lookups. Next I will consider the operation of the pumps and turbine because I was given no indication as to the operating efficiency of the turbine or the pumps. I will assume that they are occurring isentropically meaning that they're 100% efficient. So S2 is going to equal S1, S4 is going to equal S3 and S6 is equal to S7 and S8. That leaves me with four stay points unaccounted for. One of those is fulfilled by the temperature I was given at the inlet to the turbine that would be state point six. So I will say T6 is 600 degrees Fahrenheit, leaving me with three unaccounted for. The next one is easy. The condenser is assumed to only condense. So I assume that the outlet of the condenser is a saturated liquid. Therefore, I'm assuming X1 is zero. And similarly, I can assume that the closed feedwater heater is allowing the stream at seven to condense. And that's where it's getting its energy to push into the stream from two to nine. So the condensation of the steam is actually what's accomplishing the heating process. And I'm assuming that once it condenses, it leaves because there's not enough temperature difference to really do a whole lot. So T3, no. So X3 is also assumed to be zero. And again, I will indicate that that's an assumption that I'm making with an asterisk. So I have five and nine unaccounted for. Five will come from an energy balance on the mixing chamber. Once I know everything about four and nine, knowing five is just a matter of combining the two. If I set up an energy balance on the mixing chamber, I recognize that it's assumed to be adiabatic. And there's no opportunities for work to occur. And if I neglect any changes in kinetic and potential energy, I will end up with the sum in of m dot h is equal to the sum out of m dot h, meaning m dot nine h nine plus m dot four h four is equal to m dot five h five. If I know the proportion of mass flow rates at nine and four, I can calculate h five. So I will just write h five comes from energy balance on mixing chamber. That leaves me with state nine. So the key to state nine is similar to the assumption that we made about the operation of the regenerator back in the Brayton cycle. If we didn't have enough information to deduce otherwise, we assume that regenerator operated with 100% effectiveness, meaning as much heat as could be transferred was transferred. And back in that analysis, we had a heat exchanger where the flows were flowing in opposite directions. So a high temperature here was driven to a lower temperature here at a high temperature here was driven to a lower temperature here. And if I were to plot out the position, let's call this hot in hot out hot out hi ho cold in cold out co seco. And then I am plotting temperatures relative to x position. And if I had the worst heat exchanger in the world, hot input would be here. Bold input would be here. And absolutely nothing would happen in between the two. So the temperature of see out would be the same as see in. And the temperature of hot out would be the same as the temperature of hot in. And if they were working just a little bit, then the change in enthalpy, which for air is directly correlated to a change in temperature is the same in both the hot stream and the cold stream, meaning that I end up with lines that move up and down by the same amount. And if I were to extrapolate this to ideal circumstances, I end up with a line directly connecting the two. So for a heat exchanger with flows in opposite directions like this one, ideal operation is marked by the temperature of the cold outlet being the same as the temperature of the hot inlet and the temperature of the cold inlet being the same as the temperature of the hot outlet. But our situation is different. In our situation, we have flows that are going more towards each other. So if I draw this as a box, I have the cold stream entering and then undergoing a whole bunch of surface area before leaving again. Let's stick with our naming convention here. That was see in and see out. And then hot in is steam. So we are spraying that all over these coils. It is allowed to condense and then leave. So if we're assuming that as much heat as can be transferred is transferred, then heat transfer will continue until the temperatures are the same. But here I'm not referring to the temperature of the inlet of one stream and the outlet of the other. No, I'm referring to the temperature of the two outlet streams. So in conclusion, ideal operation of a closed feed water heater like this one is marked by the temperature of the outlet streams being the same. So the assumption I make about state point nine is that it is the same as T3. And with that I have two independent intensive properties which theoretically would define all nine state points and looking up all my properties is just a matter of putting in the time. So in an effort to continue our analysis of the problem before we get bogged up in the property lookups, let's assume that we had done that, okay? Poof, we have all nine enthalpies or rather we have eight of them. What do we do with those enthalpies? Well, I want us to determine the specific work in, the specific Q in, the specific work out, and the specific Q out. Starting with work in, why don't you try that on your own first? What do you get for an equation for the specific work in to the cycle? Did you get one minus Y times H2 minus H1 plus Y times H4 minus H3? Excellent. Here we are defining seven or rather the mass flow rate that leaves as seven relative to six as Y and whatever remains one minus Y. So the proportion of the mass flow rate that leaves early, m dot seven over m dot six is defined as Y and whatever is left over is one minus Y. So if 25% of the stream at six leaves at seven, the remaining 75% must leave at eight. Then I recognize that the mass flow rate at six is the same as five and those two state points are what I'm calling m dot cycle, the overall mass flow rate through my cycle and that stream is split, some of it goes into eight, one, two, and nine and some of it goes into seven, three, and four. So in my workout equation, excuse me, in my work in equation, I'm taking the total power input and dividing by m dot cycle, the total power input is going to be the power input to both pumps. So the power input to pump one plus the power input to pump two, and the power input to pump one is going to simplify down to the power input is equal to the mass flow rate through pump one, which is either m dot one or m dot two times the quantity H2 minus H1. Therefore I could write the power of pump one as m dot one times the quantity H2 minus H1. And then for pump two, the power of pump two could be written as m dot three or four times the quantity H4 minus H3. Therefore w dot in is equal to m dot one times H2 minus H1 plus m dot three times H4 minus H3. And then I'm dividing that entire quantity by m dot cycle. And because m dot one divided by m dot cycle is the same as m dot eight divided by m dot six, that means I'm writing it as one minus y. And then because m dot three divided by m dot cycle is the same as m dot seven divided by m dot six, I'm writing that as y. So my equation should be one minus y times the quantity H2 minus H1 plus y times the quantity H4 minus H3. Then Q in occurs in the boiler and the boiler has the mass flow rate of the cycle. So Q dot in divided by m dot cycle is going to simplify to m dot five times H6 minus H5 divided by m dot cycle, which is just one times the quantity H6 minus H5, which I can write as H6 minus H5. And then for our workout, if we set up an energy balance on the turbine, we're going to end up with the sum in of m dot H is equal to the power output plus the sum out of m dot H. Therefore m dot six H6 is equal to w dot out plus m dot seven H7 plus m dot eight H8. And when I divide that entire quantity by m dot cycle, I'm left with H6 is equal to the specific workout plus y times H7 plus one minus y times H8. When I rearrange that equation to write specific workout, I should get the entrance H6 minus y times the first outlet minus one minus y times the second outlet. And then for Q out, I'm only analyzing the condenser because remember the mixing chamber in the closed feed water heater are assumed to be well insulated. Therefore Q dot out would be m dot eight times the quantity H8 minus H1 dividing that quantity by m dot cycle would yield m dot H over m dot cycle times H8 minus H1, which simplifies down to one minus y times H8 minus H1. So once again, I'm left with a relationship that is only a function of our enthalpies and why. So in order to be able to finish the problem, I have to determine why. And to do that, I'm going to need to perform an energy balance on a device about which I know everything. Because I don't know the workout that rules out the turbine because I don't know Q out I don't I can't analyze the condenser because I don't know the work in I can't analyze either of the pumps because I don't know Q and I can't analyze the boiler. That leaves me with the mixing chamber and the closed feed water heater. But remember at this point in the analysis, we haven't yet figured out H5. We can do everything else, but we can't do H5. Therefore, the only device about which I know everything is going to be the closed feed water heater itself. So an energy balance on the closed feed water heater will yield y. And then I can use an energy balance on the mixing chamber to determine H5. And then I can determine the specific work in the specific Q in the specific workout the specific Q out, then I can determine the thermal efficiency at which point I'm done with the question. So energy balance on our closed feed water heater. And for that, I will draw a big vertical line. So that ended up being an awfully busy drawing. But the point is, we have a cool stream at two to nine that is being heated up by the condensation of steam, which comes in at seven. And the result of that condensation leaves at state eight. If I set up an energy balance on this control volume, I have steady state operation of an open system. So I'm going to skip the first couple of steps and write E dot in is equal to E dot out. And then because it is an open system, the energy could cross the boundary as heat transfer or work. And because it is assumed to be adiabatic, I can neglect the heat transfers. I have no opportunity for work to occur. So I neglect the works. And remember that data contains enthalpy specific kinetic energy and specific potential energy. And I'm neglecting changes in kinetic and potential energy. Therefore, I'm left with the sum in of m dot h is equal to the sum out of m dot h. And then I have entering mass flow rates in the form of states two and seven. So I can write that as m dot two times h two plus m dot seven times h seven. And then I have exiting mass flow rates in the form of the other two states nine and eight. Is that eight? No, it's three. And that's confusing. Let's correct that diagram I just drew. Okay, two comes in, seven comes in, nine goes out, three goes out. We got it. Okay, then our sum out would be m dot nine h nine plus m dot three h three. And you're expecting me to now divide by m dot cycle and you're right that would work. But before I do that, I can make my life a little bit easier by recognizing that m dot two and m dot nine are the same. And m dot seven and m dot three are the same. So if I bring them together, say by taking m dot seven times h seven minus h three is equal to m dot two times h nine minus h two. Now I have two fewer mass flow rates to have to deal with. So I'm not seven times h seven minus h three is equal to m dot two times h nine minus h two. So I'm saying the energy absorbed by the cool stream is equal to the energy exiting the warm stream multiplied by mass to get rates. Cool. That all makes sense. Now I'm going to divide everything by m dot cycle. And then I'll have m dot seven divided by m dot cycle, which is the same as m dot seven divided by m dot six. So that simplifies to because I want to know for that joke to work, you had to say it alone. I just operate under the assumption that you are all participating audibly in those rhetorical questions that I ask you. And then m dot two divided by m dot cycle is equal to m dot eight divided by m dot six, which is one minus y. And then you know, it's time for algebra to solve for y. So I'm going to foil the right hand side first. I don't really know why I'm narrating. You guys can probably figure out what I'm doing. But here I am minus first outside as h two inside minus y times h nine last plus h two times y. That's equal to y times h seven minus y times h three. Now I want to get all the y's together. So I'm going to say y times the quantity h seven minus h three plus h nine minus h two y times the quantity h seven minus h three plus h nine minus h two. I'm still narrating my algebra. If I hadn't spoken, I could have time lapsed this. But you know what? It's fun to do algebra together. Then solving for y yields h nine minus h two divided by the quantity h seven minus h three plus h nine minus h two and nine minus two full stream divided by seven minus three plus nine minus two. Yep, I'm cool with it. So at this point in our analysis, if we had looked up all of the enthalpies except for state five, I would have enough information to calculate y and then armed with our new y. I could calculate h five by performing an energy balance on the mixing chamber. I think I can fit that in here. I have entering mass at state. Let's double check this time as opposed to resorting to memory nine and four nine and four. And it's leaving at five and the mixing chamber itself is adiabatic and has no opportunity for work. So if I were to set up the worst control volume ever, perform an energy balance on that control volume because it's steady and it's open and I'm neglecting changes in kinetic and potential energy and there are no opportunities for heat transfer or work. I'm going to do one to skip a few all the way down to the sum in of m dot h is equal to the sum out of m dot h. I have one exiting mass two entering masses. So m dot nine h nine plus m dot four h four is equal to m dot five h five therefore I'm going to get rid of this page count. Therefore h five is equal to that's not what I wanted read for some reason h five is equal to m dot four h four divided by a m dot five plus it's a little bit too much like a nine so I'm going to correct that. That's a much better four yeah good fours. That is a good four absolutely nothing say it again and h five okay and if you expected me to have divided by m dot cycle I mean what I did was solve for h five by dividing both sides by m dot five which is also equal to m dot cycle so it's essentially the same thing anyway which is fun. So the four divided by m dot five is equal to m dot seven divided by m dot six which is y so this is y times h four and then m dot nine is equal to m dot eight and then m dot nine divided by m dot five would be equal to m dot eight divided by the cycle which is one minus y and that gives me everything I need to calculate h five. So let's recap our outline for the moment. We are going to look up h one, h two, h three, h four, h six, h seven, h eight, h nine and then we're going to use our energy balance on our closed feedwater heater to determine why and then we are going to use our energy balance in the mixing chamber to determine our h five and then once we have y and all nine h's we are going to calculate the specific work in the specific q in the specific workout and the specific q out and again because this is a thermo two example problem and I'm assuming that you guys are good at property table lookups I'm not going to waste any more example problem time looking up the properties on camera instead we are just going to jump to having all of the enthalpies except for state five are you ready here we go three two one and with that we have eight of our nine enthalpies again I included my work in full detail here on the attached pages so if you download the pdf below this video you can follow along with the lookups if you want to try them on your own and I would encourage you that you do and then our next step is going to be to calculate y from knowing h nine, h two, h seven and h three so I'm going to take nine minus two which is going to require our calculator friend nine minus two two three seven point three five minus one three three point two seven and then we are dividing that quantity by another quantity seven minus three that's one zero five five no one zero eight five point zero two minus three was two three six point one six plus nine minus two plus two three seven point three five minus one three three point two seven that gives us a y value of zero point one zero nine two two and armed with our new y we now have everything we need to calculate h five with the exception of a little bit more space so I'm going to take that y value times h four oh no I moved the white that I had used to block out the page number that's interesting anyway so y times h four which was two thirty nine point two and I'm adding to that one minus y it's not minus calculator come on you're better than this times h five which was wait times h five what what what four and nine what am I doing I remember I was making a joke about the song war sounding like the song for so multiplying by h nine not h five and that h nine value was two three seven point three five and that would be two three seven point five five two it's useful at this point to sanity check that number we are mixing together the streams at four and nine so our state five should be somewhere between states four state four's enthalpy and state nine's enthalpy so it should be somewhere between two thirty nine point two and through thirty seven point three five since our y value is so small I should expect a number that is so much closer to h nine than it is to h four because I'm taking a bigger quantity times h nine plus a smaller quantity times h four that's why our value of two thirty seven point five five two being so close to two thirty seven point three five compared to two thirty nine point two is such a good sign and now I have all nine of my enthalpies so I have everything I need to be able to continue I will calculate the work in the q in the workout in the q out which I will do first by scooching some stuff around and I do what how did I break okay let's try that again which I will do by scooching some stuff around hopefully correctly this time and that is probably enough space to finish the problem so one minus y times h two minus h one so one minus y times h two minus h one one thirty three point two seven minus one thirty point one seven plus y times h four minus h three h four minus h three was two thirty nine point two minus two thirty six point one six and I get a work in of approximately three next is q in which is easy that's just six minus five so I take twelve ninety eight point three minus two thirty seven point five five two I get one thousand sixty point seven five just occurs to me now how many decimal points I used on work in but you know what we're into it already so let's just continue then I'm taking one two nine eight point three minus y times h seven and h seven was one thousand eighty five point oh two minus one minus y times h eight which is nine five four point one eight giving me three hundred and twenty nine point eight three and then after such complicated calculations such as work in and work out key out should be a breeze I'm taking one minus y times h eight minus h one so nine five four point one eight minus one thirty point one seven which gives me seven hundred and thirty four point zero one two and with those quantities I have enough to calculate the network out the net he transfer in and then the thermal efficiency so first up I have workout minus work in which is three hundred and twenty nine point eight three minus three point zero nine three four five according to my work that is three hundred twenty six point seven three six and then I take one thousand and sixty and I divide by seven hundred and thirty four and I get three hundred and twenty six point seven three six as well again those values being the same implies that I built my equations for work in queue and work out in queue out correctly which is a little bit more reassuring as they get more complicated hundred and twenty six point seven three six that leaves us with thermal efficiency which is three hundred and twenty six point seven three six divided by q in which was one thousand and change and I get thirty point eight percent so if I were to pose the question how does this efficiency compare to the previous problems efficiency well it seems like they're pretty close to the same right thirty point eight three compared to thirty point eight zero but our thermal efficiency in the closed feed water heater is reflective of the differences between the two types of regenerative ranking cycles closed feed water heaters are always going to be less efficient than open feed water heaters that might cause you to ask yourself why then would anyone ever use a closed feed water heater they're more mechanically complicated they're less effective why would I want one well when we are talking about different ways of implementing a regenerative ranking cycle the advantage of the closed feed water heater is that you can then accomplish the overall cycle with fewer pumps so if I were to expand the stream from seven to three back down to the low pressure and mix it together with the stream at one minus y and have them compressed in a single pump I would save money on pumps and the cost savings might be enough to offset the slight drop in thermal efficiency with a closed feed water heater especially when we start to encounter effectiveness of the feed water heater because it's real easy to build an effective open feed water heater it's not particularly easy to build an effective closed feed water heater to the same level but that savings might offset the increased cost of complexity and the slight loss of thermal efficiency the last thing I want us to do is to draw a TS diagram of this cycle for that I'm going to start with the same axes and saturation lines as the previous example nope third time's a charm and on this expertly drawn TS diagram I am going to indicate three lines of constant pressure one low pressure one medium pressure one high pressure and of course this is hugely exaggerated because the goal of this diagram is just to indicate rough shape what I'm looking for is the positioning of our state points relative to one another and I begin by recognizing that states one and three both had assumed a saturated liquid so one and three are going to be here and here that was 8.3 right yep and then two and four are going to be directly above them and I'm going to draw this in black you hopefully draw a little bit more contrast and then again I have six seven and eight all in a nice little column here so then the remainder of our state points and then for the remainder of our state points we're just going to be looking at this region up here and I got a little bit overconfident when I was drawing my state points here two is not on the medium pressure two is on the high pressure because again one to two is going all the way up to the high pressure so I'm left with five and nine nine and five are both going to be on the high pressure line they are just going to be between four and two I will draw that like this so two gains a little bit of energy to nine four and nine mixed together to yield five five while it was a constant pressure process up to six seven comes back to three before going up to four it comes back to one and there's our TS diagram for this cycle if we were to analyze this problem in MATLAB instead of performing the calculations by hand we end up with something that looks like this so MATLAB is running through the calculations that we had done by hand using x-team to perform all the property lookups and then determining why using our values for enthalpy calculating a specific work in specific Q in a specific workout and a specific Q out and then calculating a thermal efficiency MATLAB also has the benefit of being able to plot the TS diagram accurately so on this TS diagram everything is actually to scale I will point out that in my drawing we had exaggerated everything in the compressed liquid region so as to actually indicate how the state points lie relative to one another but in the compressed liquid region the lines of constant pressure appear so close together that they are essentially inseparable when we're looking at an actual graph so when we look at the MATLAB code it's impossible to differentiate those isobars in the compressed liquid region and those state points over here referring to two nine five and four appear all on top of one another another advantage of performing this calculation in MATLAB is that it allows us to perform changes to our given information and see how that affects the results I could say consider the regenerator pressure and ask the question what happens if the regenerator pressure is changed from 40 to 50 we could change it to 50 and watch how this changes our thermal efficiency went from 31 on the nose to 31.07 we saw that it increased and if I wanted to I could wrap this code in a loop and have it perform the calculation oh I don't know maybe a thousand times and graph the results this takes a second because it has to do a lot of calculations here let me make that figure a little bit bigger so hopefully it's easier to read bring it over so on this figure I'm showing thermal efficiency as a function of regenerator pressure in the center we can see that the optimum thermal efficiency is going to occur at a regenerator pressure of I don't know maybe 80 or 90 psi and then on the left graph I'm showing how y changes as a function of regenerator pressure so we can't just have a y value of 0.8 it is a function of the pressure at which we are extracting the steam and then on the right graph I'm just plotting thermal efficiency as a function of that y value so the y value corresponding to a regenerator pressure that yields the highest thermal efficiency is about 0.15