 Consider a reheated Rankine cycle using steam as the working fluid. Steam leaves the boiler and enters the turbine at 4 MPa and 400°C and is expanded in the low pressure turbine stages to 10 kPa. An engineer is considering adding a 400 kPa reheat stage to this facility which would bring the steam back up to 400°C before expanding again. I want us to determine the thermal efficiency of this facility both with the reheat stage added and without the reheat stage. And essentially what we're doing here is an analysis of two different Rankine cycles. In one of them we have a simple Rankine cycle wherein we are compressing first in the pump from 10 kPa to the high pressure of 4 MPa then we enter the boiler bringing the temperature up to 400°C again. So note states 1, 2 and 3 are going to be the same across the two. The difference is now we enter a high pressure turbine expand from 4 MPa down to 400 kPa before going into the boiler again increasing the temperature back up to 400°C before expanding from 400 kPa down to 10 kPa. That means that we have a couple of additional state points and the state point at the entrance to the condenser is going to be different. Let's set up two different system diagrams and identify the four state points associated with the simple version that is without the reheat stage and the six state points that are relevant to the reheat variation of this problem. So in the simple version of this problem, the non-reheat version of our Rankine cycle we only have two pressures, the high pressure which is 4 MPa I believe, yep 4 MPa and the low pressure which is 10 kPa. I know that states 1 and 4 are at the low pressure and states 2 and 3 are at the high pressure. In the reheat version of this problem I have a low pressure of 10 kPa at states 6 and 1 a high pressure of 4 MPa at states 2 and 3 and then an intermediate pressure or medium pressure at states 4 and 5 and that's the reheat pressure which is 400 kPa. So those pressures give me half of my required independent intensive properties. For the rest of them I have to consider what I assume about the phases and what I assume about the operation of the devices. I assume that the pump and turbines operate isentropically because I wasn't given enough information to deduce otherwise which means that I can say S2 in the non-reheat version is equal to S1 and S4 is equal to S3 and then in the reheat version I can say S2 is equal to S1 S4 is equal to S3 and S6 is equal to S5. Then I assume that the entrance to my pump is a saturated liquid. I assume that because the condenser is only condensing. The condenser is not super cooling, it's not compressing the substance, it's just a chamber with a hole in the bottom where water condenses and then leaves. So X1 in both cases is assumed to be 0. And I will add that that is an assumption we make unless we are given enough information to deduce otherwise. If you were told something about the operation of your pump like it is super cool by 20 degrees Celsius or it is in the compressed liquid region by this amount that would indicate your condition at state 1. Since we don't know anything like that we assume that it's just a saturated liquid. Then our final independent intensive property comes from the fact that we were told the entrance to the turbine is 400 degrees Celsius meaning T3 and T3 are both 400 degrees Celsius. And then I was told that the reheat stage brings the temperature back up to 400 degrees Celsius which means T5 is also 400 degrees Celsius. So I have enough independent intensive properties identified so as to be able to look up whatever I need. At state 1 I'm going to want H1 and S1 using S1 to look up H2. Then using T3 and P3 I can look up H3 and I can use S3 to look up H4. Also note because I have a pump I can make the assumption that I have an incompressible fluid across the pump and you look up V1 and then use the shortcut to get to H2 instead of using entropy. On the reheat iteration I have the same values as earlier for states 1, 2 and 3 meaning that I can actually just pull those over once I've looked them up for part A. At state 4 we begin to diverge. We have a different value because we have a different pressure. State 5 we can look up H5 and S5 and then use that entropy for H6. For my property lookups I'm going to be using the steam tables in the back of the Wiley textbook. I would encourage you to consider pausing the video now and trying these lookups on your own as practice and then unpausing and watching me go through it to verify that you got the same results. So now is your chance to pause. 3, 2, 1. Hey you're back! How did it go? How did that triple interpolation go? Did you enjoy it? How about that pentuple interpolation? How about that one well? Did you struggle at all with the interpolation values for the S3 and S5 values that you had in part B? I hope not. I hope it all went well. Just to double check that we're all on the same page let's walk through this together shall we? At state 1 we are looking for H1 and S1 that's a saturated liquid at the low pressure which is 0.1 bar, 10 kPa which is 0.1 bar. For that I'm going to jump into our saturated liquid tables by pressure I'm going to find 0.1 bar and I'm going to read off HF and SF those values are 191.83 and 0.6493 respectively. So 191.83 is our H1 and that was kilojoules per kilogram and S1 was 0.6493. Stay one done! Next up is state 2 wherein I had an entropy of 0.6493 and a pressure that is the high pressure which is 4 MPa which is 40 bar. The first thing to do at state 2 is to fix the phase so that would be done by going into our saturation tables by pressure finding 40 bar and then comparing our S value to SF and SG. I should have a compressed liquid because I'm going up from the saturated liquid line but just to double check here I see 40 bar has an SF and SG value of 2.7964 and 6.0701 respectively. My S2 is less than SF which means that I have a compressed liquid. So jumping over to my compressed liquid tables I can scan down until I find 40 bar and I see, oh no, I don't happen to have a 40 bar sub-table. So what that means is I'm going to be interpolating between the 25 bar and 50 bar sub-tables and since my entropy value doesn't happen to perfectly correspond to either of these two rows see here I can highlight those to make it easier to follow perhaps. Because my interpolation doesn't lie directly on one of the two rows I have to interpolate between two rows and between two tables at the same time. So I'm doing a triple interpolation for both entropy and enthalpy and it's the last interpolation that unites the two so I do two intermediate stepping stone lookups for enthalpy two intermediate stepping stone lookups for entropy and then the final entropy that I have gets us to what we actually want for enthalpy. So it's more of a pent-uple lookup a pent-uple interpolation if you will and to help visualize this I can draw a pressure sub-table that's representing what we're doing we are building a table of temperature, enthalpy, and entropy values and I'm going to make this a better table here let's just go overboard here with this temporary table shall we straight line boom they weren't even long enough great 30 times the charm okay temperature column done enthalpy column done enthalpy column done so I want to look up an enthalpy value 40 degrees Celsius and 40 bar and an entropy value at 40 degrees Celsius and 40 bar and then 80 degrees Celsius and 40 bar again for both enthalpy and entropy so I'm essentially doing this lookup then this lookup then this lookup and then this lookup and then I'm using my entropy to get us to this value so four intermediate lookups before the actual interpolation that gets me to H2 for this I'm going to need our calculator and I will start with 40 degrees Celsius and 40 bar for enthalpy and then 80 degrees Celsius and 40 bar for enthalpy and then 40 degrees Celsius and 40 bar for entropy and then 80 degrees Celsius and 40 bar for entropy so wake up the calculator jump back into our compressed liquid tables what I'm doing first is interpolating for 40 degrees Celsius and 40 bar for enthalpy so 40 minus 25 divided by 50 minus 25 and that is equal to the value that I want minus the value at 40 degrees Celsius and 25 bar which is 169.77 divided by the value at 50 bar and 40 degrees Celsius which is 171.97 minus 169.77 and I'm looking for X furthermore since I'm doing this like a million times in a row I'm going to plug these in symbolically to help me iterate a little bit faster so first look up is at an A value of 169.77 and a B value of let's say B there we go and a B value of 171.97 giving me an enthalpy value of 171.09 and then next I'm interpolating for 80 degrees Celsius and 40 bar so 338.85 and 336.86 and then next entropy value at 40 degrees Celsius which is going to be 171.97 and 169.77 and then entropy at 80 degrees Celsius 338.85 and no well what was I doing I just did enthalpy again let's try that again and then entropy value at 40 degrees Celsius is 0.5705 and 0.5715 and then entropy value at 80 degrees Celsius which is 1.0720 and 1.0737 so my four stepping stone values are respectively 171.09, 338.054 0.5709 0.07268 so again just to show what I'm doing visually here can actually populate those values so I had 171.09, 171.09 and then 338.054 and then 0.5709 and 1.07268 and then I'm using my entropy value which is 0.6493 to get to our enthalpy value here so our final interpolation this one for all the potatoes is 0.6493 minus 0.5709 divided by 1.07268 minus 0.5709 and that is equal to x minus 171.09 divided by 338.054 minus 171.09 and I get a value of 197.177 which I don't even need to write on the table because I'm writing it up here 197.177 kilojoules per kilogram and thank you intermediate table you have done your duty now I'll be gone because I need the space that's our enthalpy two done again you could have used the shortcut you could have looked up v1 and then used the incompressible pump work equation which is the specific work of the pump which is equal to h2 minus h1 is approximately equal to v1 times p2 minus p1 could have used that equation to approximate h2 but since we're here to do property lookups let's just do them as accurately as possible now I have T3 400 degrees Celsius and the high pressure which was again 4 megapascals which is 40 bar first question here is what is the phase at state 3 for that I will go into our saturation tables by pressure I will find 40 bar and I will look at excuse me the saturation temperature which is 250.4 I see that my temperature which is 400 degrees Celsius is higher than the saturation temperature at 40 bar which means that I must have a superheated vapor so I go into my superheated water vapor tables and I scan until I find the 40 bar table and look at that we actually have a 40 bar sub table I almost forgot that could happen that's convenient therefore my h3 and s3 values are going to be 3213.6 and 6.769 respectively 3213.6 3213.6 3213.6 kilojoules per kilogram and s3 which is equal to 6.769 6.769 kilojoules per kilogram Kelvin and then for h4 we're going to use our entropy which is again 6.769 and we're going to compare that to sf and sg at our low pressure to determine if we have a compressed liquid a superheated vapor or something in between so I'm going to go back to my pressure sub tables and find 0.1 bar zoom in entirely too much and then I see that my sf at 0.1 bar is 0.6493 kilojoules per kilogram Kelvin and my sg value is 8.1502 my entropy value at state 4 is 6.769 which is between those two values which means that I have a saturated liquid vapor mixture therefore my interpolation is going to look like s4 minus sf divided by sg minus sf which by the way is among our definitions for the quality at state 4 is equal to h4 minus hf divided by hg minus hf and I could look up the quality here as well if I wanted to but I don't need it to be able to continue so I won't I'm just going to jump from entropy all the way to enthalpy so my interpolation is going to go come on calculator that entropy at state 4 which is equal to state 3 is 6.769 hold on establish the frame of our interpolation 6.769 minus sf at 0.1 which was 0.6493 divided by sg which is 8.1502 minus sf which is 0.6493 is equal to I am going to scroll over to the left I can figure out how to get my hand pulled back I'll have to comment that's fun go away comment so h which is what I need jump back to the calculator minus 191.83 divided by 2584.7 minus 191.83 and that gives me a h4 which is 2144.08 2144.08 the only thing that my quality at state 4 would be useful for here would be I don't know if I asked us to graph this on a ts diagram but when have I ever asked us to do that I mean that's totally unlikely right anyway now that we have all four enthalpies we can proceed to calculate the work in the queue in the work out and the queue out and I will point out here that I could just jump all the way to thermal efficiency if I wanted to plugging in everything symbolically but doing that would assume that you know how to build these equations and I just want a little bit more practice at that so character building in the meantime and then once we have those quantities I can calculate the thermal efficiency so thermal efficiency is equal to net work out divided by queue in and that's what we actually wanted so I have a simple ranking cycle go away calculator so work in is going to be h2 minus h1 because it's the specific work across the pump queue in is going to be h3 minus h2 work out is going to be h3 minus h4 and queue out is going to be h4 minus h1 and then I'm going to have basically the same framework here for part b so while these are still blank I'm just going to copy that over it's too much over send it a little bit like radio protocol there this is too much over anyway I'm going to plug in my h1, h2, h3 and h4 values to get my work in, my queue in, my work out and my queue out those are 197.177 minus 191.83 giving me a work in of 5.347 and then 3213.6 minus 197.177 which gives me a queue in of 3,016.42 my work out would be 3213.6 minus 2144.08 and my queue out would be 2144.08 minus 191.83 so I have work in queue in work out and queue out 5.347 kilojoules per kilogram and then 3,016.42 kilojoules per kilogram and then 1069.52 kilojoules per kilogram and 195.25 kilojoules per kilogram then my work out is going to be 1069.52 minus 5.347 and I get 1064.17 and then just to double check that I built those equations correctly I will take queue in minus queue out and are you ready for it to be the same? here we go hey look guys it's the same ray 1064.17 kilojoules per kilogram and I will remind you here that these numbers being the same doesn't mean that your enthalpy values are correct it just means that you built these four equations correctly as we get to more complicated variations on the Rankine cycle the complexity of those equations will increase and as a result the probability of something going wrong while you're building them will also increase so network out 1064.17 divided by queue in which was 3016.42 gives us a thermal efficiency of 35.3% and with that I can call part A done and move on to part B part B begins with the same three state points because the independent intensive properties which define states 1, 2, and 3 are the same the h1, h2, h3 values will also be the same so I will just copy those over but note that because the pressure is different at state 4 our state 4's are different so don't get too heavy handed here then at state 4 we have an entropy of 6.769 and a pressure of 400 kPa which would be 4 bar so the first decision like with state 4 earlier is what phase do I use and to fix our phase we go back into our saturation tables we find 4 bar and then I grab my highlighter tool and then forget that I have my highlighter tool when I go to move the page around later on as is tradition and then I'm going to compare my entropy value which is 6.769 to SF and SG I see that my S value is between the two therefore I have a saturated liquid vapor mixture so I'm going to use the same interpolation as earlier I'm going to say h4 minus hf divided by hg minus hf is equal to S4 minus SF divided by Sg minus SF and because I don't actually care about the quality I don't need the quality as a stepping stone so I will just do the interpolation in one fell swoop so I'm going to say let's just grab hf and hg while we're here with the highlighter so that we don't accidentally misgrab that number switch back to the hand tool scroll up just to make sure that that's hf and hg and indeed it is so our interpolation our interpolation will look like 6.769 6.769 6.769 minus 1.7766 divided by 6.8959 minus 1.7766 and that's equal to x minus 604 0.74 divided by 2738.6 minus 604.74 and that gives me an enthalpy of 2685.7 and then at state 5 I have a temperature of 400 degrees celsius and a pressure of 400 kPa which is again 4 bar so I'm going to go into my saturation tables at 4 bar I'm going to switch back to the hand tool I'm going to see that the saturation temperature is 143.6 degrees celsius my temperature is greater than the saturation temperature at 4 bar which means that I still have a superheated vapor so I'm going to jump over to my superheated vapor tables and I'm going to scan until I find 4 bar and unfortunately I do not have a pressure sub table corresponding to 4 bar which means that I have to interpolate and worse yet I have to interpolate between 3 bar on one page and 5 bar on the other so prepare yourselves for lots of scrolling so my interpolation is going to go 4 minus 3 and I press the wrong button divided by 5 minus 3 and that is equal to the thing that I'm looking for minus the value at 4 bar divided by the value at 5 bar which is 3271.9 minus the value at 4 bar solving for x so previous page going down 3 bar 400 degrees celsius and it will be 3275 so 3275 and 3275 let's just verify 3275 for 400 degrees celsius and 3 bar and 400 degrees celsius and 5 bar is 3271.9 cool so my enthalpy at state 5 is going to be 3273.45 3273.45 same process for entropy therefore my entropy will be interpolated between something and 7.7938 then jumping back to 3 bar down 400 degrees celsius 3 bar is 8.0330 right entropy yep 400 degrees celsius 3 bar 8.0330 8.0330 ready here we go again 8.0330 and i get 7.9134 kilojoules per kilogram kelvin and then h6 is going to have the same entropy and the low pressure which was 10 kilopascals which is 0.1 bar so as you are probably getting the hang of we have to first look at sf and sg to determine the phase at a pressure of 0.1 bar on the saturation tables by pressure i can see that sf is 0.6493 and sg is 8.1502 my entropy lies between the two which means that i have a saturated liquid vapor mixture so i'm going to take s6 minus sf divided by sg minus sf is equal to h6 minus hf divided by hg minus hf so i am going to take my entropy i'm going to scooch this down so i can read everything all at once my entropy which was 7.9134 minus 0.6493 and i'm going to divide by 8.1502 minus 0.6493 and that is equal to x minus 191.83 divided by 2584.7 minus 191.83 that gives me an enthalpy value of 2509.16 and just to make a sanity check my entropy value was much closer to sg than it was to sf which means that our enthalpy value should be much closer to hg than it is to hf so these are all strong indications that i probably did that at least mostly correct my biggest fault when i'm working quickly like this is i make typos when i'm typing so fingers crossed that didn't happen on any of those property lookups anyway i have my six enthalpies so i can calculate our work in q in work out in q out work in is h2 minus h1 because it occurs in the pump q in is h3 minus h2 plus h5 minus h4 because q in is going into the process from 2 to 3 and also the process from 4 to 5 so i have to account for both of those similarly my work out will be h3 minus h4 plus h5 minus h6 because i have to add together the work of the two turbines so 3 minus 4 plus 5 minus 6 and then my q out is just a boring h6 minus h1 because the q out is occurring in the condenser and with those equations established i can calculate a work in a q in a work out in a q out my work in is going to be the same because it's the same driving enthalpies 5.347 and then we are taking h3 minus h2 plus h5 minus h6 i don't know why i pulled up the solve function so 3 minus 2 which is 3213.6 minus 197.177 plus 5 minus 4 which is 3273.45 minus 2685.7 2685.7 3273.45 that gives me 3604.17 kilojoules per kilogram then 3 minus 4 plus 5 minus 6 3213.6 minus 2685.7 plus 3273.45 minus 2509.16 giving me 1292.19 and lastly i have 6 minus 1 which is 2509.16 minus 191.83 giving me 2317.33 and then with those quantities established i can take 1292.19 minus 5.347 and i get 1286.84 and then i compare that 23604. 36004.17 minus 2317.33 and they should be the same and hooray they are the same and i get 1286.84 and 1286.84 those quantities are still in kilojoules per kilogram and when i take that number divided by qin i get 36004.17 i get a thermal efficiency of 35.7 so with that i have finished what the problem asked me for but you guys probably know what i want to do right i want to plot this on a ts diagram you know just for fun right because that's how you keep your ts diagram skills sharp so i'm going to jump to a new page i'm going to establish some axes like i'm going to be able to fit two on those there we go and then this is the entropy axis i want to note that with an s and this is the temperature axis i'm going to be with a t and i'm going to want to plot my state points relative to the saturation lines so i will draw some saturation lines do that better and then really long in the x axis that's kind of terrible it should be a little bit more curvy but that'll work for now and then i'm going to draw some lines of constant pressure some isobars i'm going to do one at the low pressure and then one at the high pressure which i will draw as being here and remember that in the superheated vapor region these get more swoopy the further to the right you get so i'm going to draw my state point 1 is being right here so saturated liquid on the low pressure and then state point 2 is being directly above it which is right here and then state 3 which is at the high pressure and 400 degrees celsius 400 is a little bit higher than the critical point so i will draw that like here and then i will split between hold on this is 4 mega pascals this is 10 kilopascals and then i will split my diagram so i draw a separate one for a and b i will copy and paste this over hopefully it fits down here it doesn't cool so i guess i will open up another new sheet of paper and i will paste that diagram and we will jump back to this one so my state 4 is directly below state 3 which puts it right about here therefore my process from 1 to 2 is a vertical line and then 2 to 3 follows this line a constant pressure and then 3 to 4 is a vertical line again and 4 back to 1 is a horizontal line because it's an isophar is it a process that occurs at a constant pressure under the saturation dome so this region enclosed is my network out and remember that that's because it's q in minus q out which is also equal to 1 minus 4k and the area under the curve from 2 to 3 represents my q in so you can get a visual representation for the thermal efficiency by imagining this proportion divided by the entire area under 2 to 3 now let's compare and contrast that to what it would look like with the reheat cycle for that i'm going to need another line of constant pressure and i will draw that right about here this is 400 kilopascals so now state 4 is still directly below state 3 we were really close to the saturated vapor lines but not quite there still under the dome and then we follow back up to the same temperature 5 and then down again to state 6 which is also close to the saturated vapor line but not quite there so my process from 1 to 2 is a vertical line and i have a constant pressure process from 2 to 3 and then a vertical line from 3 to 4 constant pressure process from 4 back up to 5 and then a vertical line from 5 to 6 and constant pressure process from 6 back to 1 so my network out is still the region under this curve we've just added this region here does that make sense this entire region is still network out the region that i added is proportionally adding more network out than it is q in which means that i'm going to improve my thermal efficiency another way to think of that is to consider how this would look if i had 3 reheating stages let's say green for that if i had split this across 3 stages i would end up with something that looks like this and if i split it across 4 stages and go down up down and if you were to extrapolate that all the way out to infinity if you had an infinite number of reheating stages you would essentially have a horizontal line all the way to the low pressure which means that you have essentially an isothermal workout process which means as you add reheating stages you are getting closer to a car no workout if that logic makes sense so the more reheat stages you add the higher your thermal efficiency but that butts up against the reality of having to build this in reality the addition of a reheating stage is going to add some mechanical complexity you have to pay money for that and at a certain point you cross the line of diminishing returns so adding to your thermal efficiency by incurring a cost doesn't make up the cost if the increase in thermal efficiency is too small to actually get a return on that investment so in reality you see the point of diminishing returns appearing at about 2 or 3 stages so you might see 2 frequently you might see 3 infrequently but you hardly ever see 4 if that logic makes sense the next question i want to pose is why did the engineer choose 400 kPa i mean that could be chosen arbitrarily but at what point is it better or worse to use a higher reheating pressure or a lower reheating pressure to answer that question we really should consider this calculation over and over again and for that let's open it up in matlab if you set this up in matlab you get something that looks like this note that i'm using x-steam to perform my property lookups and x-steam is expecting pressures in bar which means that i'm defining my low medium and high pressures in bar i also have a high temperature established but the rest of the calculations are all the same we look up all of our properties using the x-steam function we take our difference in enthalpies to determine our work in, queue in, work out, and queue out and we determine a thermal efficiency and when i run this calculation in matlab i get a thermal efficiency of 38.87% so that's going to be slightly more accurate than our hand calculations because x-steam is doing a better job of interpolating than we are we're using linear interpolation for everything it's using for the actual curve fit of the data to get better values it also has the benefit of not losing any accuracy to rounding but a thermal efficiency of 35.87% is pretty close to what we had and the advantage of this is that we can make changes like what if i were to increase the reheating pressure from 4 bar to i don't know 8 bar is that going to increase or decrease my thermal efficiency well we can see we're now at 36.3 and if i were to increase that from 8 say to 16 we dropped it a little bit so the function of how the reheat pressure affects thermal efficiency is a little bit more complicated than just it always increases it by increasing the pressure or it always decreases it by decreasing the pressure so to get a better idea of what the ideal reheat pressure is that would optimize the thermal efficiency we should consider all of the possible reheat pressures between the lower bound which would be equal to the low pressure and the upper bound which would be equal to the higher pressure matlab can do that all relatively quickly so i'm just setting this all inside of a loop i'm having it iterate through all of the possible reheat pressures starting at the low pressure and ending at the high pressure and what we get is a relationship that looks like this so we can see that increasing the reheat pressure improves the thermal efficiency up until around 12 bar maybe and then at that point increasing the reheat pressure decreases the thermal efficiency so generally speaking a higher reheat pressure is better overall but we can optimize the thermal efficiency by shooting for a reheat pressure of maybe 11 or 12 and matlab can actually spit out that maximum value the maximum thermal efficiency and the reheat pressure corresponding to it if we wanted it to that's the advantage of having the calculations done in a computer as opposed to completing them by hand but don't get too comfortable with completing them electronically because i still expect you to be able to handle the calculations by hand because when you take the Fe you're going to be expected to look up the properties and compute some calculations with a hand calculator and a set of tables and with that i think we can consider this problem done