 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. Determine and complete the following. First, why the proportion of mass flow rate that is brought out of the turbine early to feed the open feedwater heater, then the specific work-in relative to the cycle as a whole, and then the thermal efficiency. I will begin by trying to identify two independent intensive properties about all seven state points. Once I have those identified, looking up whatever properties I need is just a matter of putting in the time. I will start that process by posing the question, how many different pressures do I have? You're right, I have three. They were given to us 500, 5, and 40. Now, which state points have which pressure? Well, 4 and 5 have the high pressure, and 1 and 7 have the low pressure. Remember that all heat exchange processes in the Rankine cycle are assumed to happen at a constant pressure, meaning that 6, 2, and 3 are all going to be maintained at a constant pressure. That open feedwater heater still counts as a heat exchange process, even though no heat actually crosses the boundary. Furthermore, since the open feedwater heater is just a box with three holes, there's not much opportunity for any pressure changes to be meaningful relative to 5, 40, and 500. So we have a high pressure at 4 and 5, we have a low pressure at 7 and 1, and then we have the medium pressure at 2, 3, and 6. Remember that within Imperial Unit Notation Schemes, PSIA indicates an absolute pressure as opposed to PSIG, which indicates a gauge pressure. I will assign those pressures to our seven state points. Next, I recognize that I was given a temperature of 600 degrees Fahrenheit at the turbine inlet, which would be state point 5. And then I recognize that because I was given no indication as to the ice and traffic efficiency of the pumps nor the turbine, I'm going to assume that all three are 100% efficient, meaning that S2 is equal to S1, S4 is equal to S3, and S7 and S6 are both equal to S5. So at state 5, I have enough information to fix our entropy, which is then used to look up properties at states 6 and 7. At state 1, I need one more property to look up S1. And at state 3, I need one more property to look up S3. My next observation will be that the condenser is assumed to only condense. It doesn't subcool or supercool the substance. It just allows the water vapor to condense, and then once it has condensed, it leaves. So the output of the condenser, state point 1, is assumed to be a freshly condensed, saturated liquid. That assumption is pretty cut and dry. The condenser operates in a very reliable and predictable manner. But we're also going to make the same assumption about the open feed water heater. We're going to assume that 6 and 7 are positioned in such a way so as to allow state 3 to be a saturated liquid. I think that'll make a little bit more sense once we look at our TS diagram. But for now, just assume that state 3 is also a saturated liquid. We want our pumps to only handle liquid, not a mixed stream of liquid and vapor. I will put an asterisk here to know that that is an assumption that we are making. Now that I have two independent intensive properties defined for all 7 state points, looking up what we need is just a matter of putting in the time. And what we're going to want is enthalpies. Let's assume for the moment that we had done that. Let's not spend any more example problem time on that. Let's say that we had looked up H1, H2, H3, H4, H5, H6, and H7. What would we do now? Well, traditionally we would calculate the specific work in, the specific Q in, the specific work out, and the specific Q out. Again, those might not be exactly what I need to be able to answer the individual questions in order, but it's a good character building nonetheless. So let's try that here. Let's determine a specific work in, a specific Q in, a specific work out, and a specific Q out. I will begin with work in. What is the specific work in to this cycle? I'll give you a hint. It is not H4 minus H3 plus H2 minus H1. Why is it not? Because we are now starting to deal with multiple streams of mass flow rate. We define our specific work in relative to the cycle by describing it as the power input to the cycle, divided by what we call the mass flow rate through the cycle, which is the biggest mass flow rate appearing. It is the overall mass flow rate. So when the streams split and are recombined, we are describing the combined stream. The power input would be the power input to the two individual pumps. And then we recognize that our power input to both pumps is just going to be mass flow rate times change in enthalpy. So I could describe the power input to pump one as m dot one or two times the quantity H2 minus H1. And then the power input to pump two would be m dot three or m dot four times the quantity H4 minus H3. Then we are defining this entire thing by m dot cycle. And at this point, we have a problem and that problem is we don't actually know the mass flow rates. We have no indication as to how big this cycle is. Since we can't calculate the mass flow rates, we can't calculate the specific work in the specific you in the specific workout in the specific you out, right? We can. We can because we don't actually care about the mass flow rates themselves. All that really matters for this calculation is the relative proportion of the mass flow rates. So if we were to write this as a proportion of mass flow rates, I'm essentially splitting the denominator here. So I'm writing m dot one over m dot cycle times the quantity H2 minus H1 plus m dot three divided by m dot cycle times H4 minus H3. It's not the mass flow rates that matter. It is the proportion of mass flow rates that matters. But what is mass flow rate of the cycle anyway? Well, since we are splitting the stream in the turbine and then recombining in the open feed water heater, three, four and five will all have the overall mass flow rate. And then some percentage of that will go into the stream at six and the leftover percentage will go into seven one and two. If you think of it like if 25% of the mass flow rate at five goes into state six, then whatever's left, the remaining 75% must go into the stream at seven. And those proportions are what we care about in these calculations. That proportion of mass flow rate at five that leaves as six, we could indicate with a variable instead of having to keep track of it. We call that variable Y. Therefore, Y is the proportion of mass flow rate five, which leaves at six, and then whatever's left over will leave at seven. So if Y was 0.25, 25% of the steam from five is leaving at six and the remaining 75% is leaving at seven. So now we can begin to write our specific quantities in terms of either Y, one minus Y or one. And remember m dot one is equal to m dot two, which is equal to m dot seven, m dot six stands alone, and then m dot three, m dot four and m dot five are also all equal to each other. And this quantity m dot three or m dot four or m dot five is what we're calling m dot cycle. So in this calculation, does m dot one over m dot cycle become one Y or one minus Y? You're right. It becomes one minus Y because m dot one is equal to m dot seven and m dot five is equal to m dot cycle. Therefore, m dot one over m dot cycle is equal to m dot seven over m dot five, which is one minus Y. Then, does m dot three over m dot cycle become one Y or one minus Y? You're right. It's one because m dot three and m dot cycle are equivalent. Therefore, I write my specific work in as one minus Y times h two minus h one plus h four minus h three. If you had just written h two minus h one plus h four minus h three, you are neglecting to account for the fact that different mass flow rates flow through those two pumps. The mass flow rate through pump one is not the same as the mass flow rate through pump two. So I can't just add the specific quantities together. Cool. One quantity down. Three more to go. What is the specific Q in to this cycle? You're right. It's h five minus h four. The reason it's just h five minus h four is because the mass flow rate going through the boiler is the mass flow rate of the cycle. So if I were to write Q dot in divided by m dot cycle and then substitute in m dot four times the quantity h five minus h four divided by m dot cycle, that would simplify down to one times the quantity h five minus h four. Now the specific workout is a little bit more complicated than just taking a difference in enthalpy. The reason it's a little bit more complicated this time is because I have more than one inlet and outlet. Both six and seven are exiting mass flow rates. So in order to yield the correct equation, I should really step back and do an energy balance. I recognize that I have steady state operation. So it's going to be most convenient for me to divine all three terms by dt. At which point I have d e d t is equal to e dot in minus e dot out. d e d t is zero because the turbine is assumed to operate steadily. Therefore e dot in will equal e dot out. For an open system energy could cross the boundary as heat transfer work or the sum in of m dot theta. Then I neglect heat transfers because I have an isentropic turbine isentropic implies adiabatic have a turbine. So I'm neglecting the work in and remember that theta contains enthalpy plus specific kinetic energy plus specific potential energy. And then I have one inlet and two outlets. So I'll write this as m dot five h five is equal to what am I doing? This should be an outlet. Got all confident my copy and paste excuse me that's equal to the power output on the right plus the sum of the two m dot h terms. So that would be m dot six h six plus m dot seven h seven. Therefore my power output would be m dot five h five minus m dot six h six minus m dot seven h seven. And then I'm dividing this entire quantity by and that cycle. So first question what is m dot five divided by m dot cycle? Is it one y or one minus y? You're right. It's one. So my specific work in would begin with one times h five minus then is m dot six over m dot cycle y one minus y or one. You're right. It's why y times h six minus is m dot seven over m dot cycle y one minus y or one. It's one minus y. Therefore my workout my specific workout for the cycle would be h five minus y h six minus one minus y h seven. And I will point out here that depending on which source you're looking at some textbooks prefer to write this in terms of y a little bit more easily. So they will take h five minus y times h six minus h seven plus y times h seven and then they will bring together the y terms so they can factor it out and write this as h five minus h seven plus y times the quantity h seven minus h six. And the reason that they do that is I don't know presumably to write it in terms of only one y so that they can solve for y a little bit more conveniently. This is also an interesting way of writing out the specific works of the individual sides of the turbine. I don't I'm of the opinion that writing it this way is no more convenient and running it this way. So I'm not going to do that but just a heads up. You might see it appear like that and it's not that it's a different equation. It's just rewritten based on what someone else deems is convenient. So we're left here with our specific you out and for our specific you out. We need to look at our condenser. So again specific you out would be the total rate of Q out divided by m dot cycle. The total rate of Q dot out would be m dot seven or m dot one times the quantity h seven minus h one and dividing that by m dot cycle would yield one minus y times h seven minus h one and just to be consistent here. I will write both of these as their rate term divided by m dot cycle. So if someone only looks at the sheet and doesn't watch the video. I can hopefully follow what's going on at least a little bit more easily. Okay. It is convenient for us to write out the specific works and heat transfers in terms of just enthalpies and the y value because I have enough information to get all the h's all I need now to finish the problem is why. Now how do I determine why the trick to determining why is I need to perform an energy balance on something about which I know everything. So I can't perform it on the boiler because I don't know Q in can perform it on the pumps because I don't know working I can't perform it on the turbine because I don't know workout and I can't perform it on the condenser because I don't know Q out that leaves me with the open feed water heater. I know the open feed water heater itself is antibiotic because just like the regenerator in the Brayton cycle there is no meaningful opportunity for heat to cross the boundary. We assume all of the heat transfer occurs internally and there's no opportunities for work. Which means if I were to write out an energy balance on the open feed water heater just like with the turbine we are going to end up with E dot in is equal to E dot out and just like the turbine E dot in could be Q dot in plus work that in plus the sum of m dot theta and E dot out could be Q dot out plus work that out plus the sum out of m dot theta. I'm neglecting works and heat transfers and inside of the theta term I'm neglecting changes in specific kinetic energy and specific potential energy which leaves me with the sum in of m dot H is equal to the sum out of m dot H. I have two and let's states two and six have one outlet state three. So I can write this as m dot two H two plus m dot six H six is equal to m dot three H three. Now what? Well now just like in the turbine. It behooves me to divide everything by m dot cycle that would allow me to write this in terms of y and one minus y and one instead of mass flow rates. So is m dot two divided by m dot cycle y one minus y or one you're right it's one minus y is m dot six divided by m dot cycle y one minus y or one you're right it's y and is m dot three divided by m dot cycle y one minus y or one. You're right it's one. So now I want to solve this equation for y because remember we have enough information to determine all of our enthalpies. So I'm going to jump into determining why knowing these enthalpies to do that and we'll do a little bit of algebra here. So all it would take to finish our problem is looking up our seven enthalpies. Once we have them we can calculate y once we have y we can calculate the specific work in the specific key in the specific work out and the specific key out and once we have those we can determine the thermal efficiency and in that process we would have answered part a b and c since this is thermal two and property lookups are assumed to be mastered in thermal one I'm not going to spend any more example problem time on the property lookups themselves. So now I'm going to cut to the end of the property lookup process so that we can spend more time on the thermal two side of things. I will leave my work attached to this document so if you want to download the PDF of this work from the box below the video you are welcome to try to follow along with what I'm doing but I'm not going to dedicate more video time to it so you ready here we go three two one and with that we have all seven enthalpies and the entropy is it took to get us there with those enthalpies my next step would be to calculate y so y is going to be h3 minus h2 divided by h6 minus h2 and for that we need to go back up to these numbers h3 minus h2 is going to be 236.16 minus 130.28 divided by h6 minus h2 1,085.02 minus 130.28 and I get a y value of 0.111 and now I can proceed to calculating the specific work in the specific Q in the specific workout and the specific Q out and for that I am going to open up a new page and copy these equations over and then give us some space to work so first up I had one minus y times h2 minus h1 plus h4 minus h3 so that's going to be one minus this quantity here times h2 minus h1 130.28 minus 130.17 plus h4 minus h3 and that was 239.17 minus 236.16 don't really need the parentheses but I'm in 3.1078 BTUs per pound mass then Q in is just h5 minus h4 so 128.3 minus 239.17 that's 1059.13 and then I want my specific workout which is h5 minus yh6 minus 1 minus yh7 so that's 128.3 minus 0.11899 times 1085.02 minus the quantity 1 minus y times 954.18 negative 329.61 and then Q out was 1 minus y times h7 minus h1 so minus oh sorry you can't see my calculator 1 minus y times h7 minus h1 954.18 minus 130.17 giving me 732.629 the next I want the network out and the net he transfer in so workout minus work in is going to be this quantity here minus this quantity here I get 326.502 and then Q in minus Q out is going to be this quantity here minus this quantity here I got the same number which is a sign that I built those equations correctly it's not quite as easy to take for granted once you start involving y into things and then the thermal efficiency which is the network out divided by the heat transfer in and I get 30.8% so A was down here B was here and C is down here and that's all I needed for the problem but you guys know what I want to do next I want to draw a TS diagram wouldn't really be an example problem we didn't draw TS diagram right and my TS diagram is going to be drawn relative to the saturation lines which I will draw the best of my ability that's quite a squiggle there at the end that will do for now shouldn't quite curl up at the end but you know TS diagram is going to TS diagram now I want three lines of constant pressure corresponding to 5, 40, and 500 and I will draw those somewhat arbitrarily I don't actually want this to be to scale all I want is that relative positioning of state points these correspond to 5 psi, 40 psi, and 500 psi and for states 1 and 3 I have a quality of 0 so those are going to be the easiest to start with the quality of 0 on the low pressure and medium pressure line so 1 is whoops the 1 is here and 3 is here and then 2 and 4 are directly above then state point 5 was a superheated vapor and at a temperature of 467.13 degrees Fahrenheit so I'm going to draw that here then state 6 and 7 are directly below it I should have a saturated liquid vapor mixture for both of them which I do the quality of state 6 was about 0.91 and the quality for state 7 was about 0.82 which is pretty close to my drawing again this isn't intended to be perfect in terms of scaling just to show the relative positioning of our state points and our cycle diagram goes like this here let me switch these to black then we have a straight line from 1 to 2 a straight line from 3 to 4 4 should be out in the compressed liquid region I don't know what I was doing there 3 to 4 also a straight vertical line that I'm exaggerating a little bit so that it's actually visible on the TS diagram and then 4 to 5 is a line of constant pressure 5 to 6 goes straight down 5 to 7 also goes straight down so 5 to 6 are on top of each other and then 7 back to 1 and 3 is going to be the meeting point 2 and 6 so earlier when I said I thought 3 would make a little bit more sense when we talk about our TS diagram what I meant by that was we are primarily using state 2 to condense state 6 so we are bringing some of the steam out it's a mixture and we are using the condensation of the steam to heat up our stream at 2 because remember that 2 and 6 are not actually that different in terms of temperature the real potential for energy to be moved around is in the latent energy of the steam so we are condensing state 6 in order to heat up state 2 and we assume that that box is built in such a way that we mix the streams together and we allow the result, the liquid at the end of the process to leave as a saturated liquid you could design it to try to accomplish some other purpose but since the operation of the pump relies on the fluid being a liquid we really want to shoot for a liquid so it's possible that we could end up slightly in the compressed liquid region at state 3 as well but we are assuming ideal operation of everything including but not limited to the regenerator itself so that's our TS diagram and the work, the network out is still the region enclosed by the diagram but the magnitude itself is a little bit more complicated to try to visualize because it's not just one of these areas if that makes sense we have the area enclosed by the mass flow rate that goes from 1 to 2 to 3 to 4 to 5 to 7 to 1 and we also have the region that's enclosed by 3 to 4 to 5 to 6 so it's a little bit of a doubling up of the area under the curve but since the goal is to try to increase our network out for a given amount of Q in the process is still just 4 to 5 we can get more network out by stacking the TS diagram on top of itself if that logic makes sense and that concludes this example