 Steam is the working fluid for an ideal Rankine cycle. Saturated vapor enters the turbine at eight megapascals, and saturated liquid exits the condenser at 0.0075 megapascals. The net power output of the cycle is 100 megawatts. Determine the thermal efficiency, the backwork ratio, the massful rate of steam in kilograms per hour, the rate of heat addition, the rate of heat rejection, and then, if the condenser was cooled by a stream of cooling water entering the condenser at 15 degrees Celsius, determine the minimum stream mass flow rate in kilograms per hour, which would prevent the cooling water from exceeding 35 degrees Celsius. I will begin with a system diagram. Next, I will begin to parse out the properties that I know about my four state points. Remember that it takes two independent, intensive properties to fully define a state point, so my goal is going to be to try to identify two independent, intensive properties for all four state points. Having those would mean that I could look up whatever it was that I needed. First, let me pose the question, how many pressures do we have? There are two. Because remember, like the Brayton cycle, all heat addition and rejection processes in the Rankine cycle are assumed to be isobaric. So the condenser occurs isobarically, and the boiler occurs isobarically. Therefore, P2 is equal to P3, which is the high pressure, and P4 is equal to P1, which is the low pressure. So I will identify those pressures over here. P high is eight megapascals. P low is 0.0075 megapascals. And then state one and four are at the low pressure, and two and three are at the high pressure. Next, I recognize that I was told saturated vapor enters the turbine, and that would be state three. And I will show that in my properties as a quality. X3 is equal to one because it's a saturated vapor, and the saturated liquid exits the condenser, that would be state one. So I will write that as X1 is equal to zero. States one and three are now fully defined, from which I can look up whatever I want. What I will want to look up is eventually going to be enthalpy for my energy balances. But in the meantime, I will also want S1. Why? Because our compression and expansion processes are assumed to be isentropic unless we're given enough information to deduce otherwise. We have not been given that information. Therefore, we assume S2 is equal to S1, and S4 is equal to S3. So at state one, I want to look up H1 and S1. With that S1, I can look up H2. At state three, I will want to look up H3 and S3. And at state four, I will use that S3 to look up H4. And with that, I have enough information to get started on my property lookups. I want the four enthalpies, and I need S1 and S3 to be able to get there. I also am going to add specific volume one into the mix for reasons that will make sense later. For now, let's just add it to the lookups that I'm requiring us to do. Our working fluid is water, which means that our tables for these properties are going to be tables A2 through A5. And remember, the first part of the lookup process is going to be to fix the phase at all four state points. Once I know the phase, I can go into the correct table and evaluate the properties that I need. If you want a review of the property lookup process, I have some videos under Thermo One's chapter three videos. That I think might be helpful. But for now, let's start with state one, assuming that you guys totally remember how to do all of these property lookups. So at state one, I have a pressure of 0.075 bar and a quality of zero, meaning I'm going to grab the saturated liquid property from my saturation tables by pressure. Remember that the saturation tables A2 and A3 contain the same information. One is just listed in even increments of pressure. One is listed in even increments of temperature. So if I jump into table A3 and find 0.075, I see that I don't happen to have a row corresponding to 0.075 bar, which means I'm going to have to interpolate between 0.06 and 0.08. I will write that out as 0.075 minus 0.06 divided by 0.08 minus 0.06. That will represent the proportion of the way I am between 0.06 and 0.08. And I will apply that same proportion to specific volume, specific enthalpy, and specific entropy. Those are the VF, HF, and SF values at 0.075 bar. And to make this a little bit easier to follow, as I'm scrolling all over the place, I will highlight these two rows so that we can just grab the values as we see fit. And when I'm typing this on my calculator, I will type it using the solve function so that hopefully, it's a little bit easier for you guys to reverse engineer what I'm doing. When you calculate it yourself, of course, you would probably do the algebra and actually compute a number as opposed to trying to leave the calculator to do the algebra. Anyway, 0.075 minus 0.06 divided by 0.08 minus 0.06 is equal to... Let's go with VF first, because it's right here. The thing that I'm looking for, which I'm going to call x for now, minus the value at 0.06 bar, which would be 0.0010064. Remember that that column is not VF. It is VF times 10 to the third. Therefore, to get back to VF, you have to take this quantity times 10 to the negative third. Anyway, then I'm dividing by 0.0010084 minus 0.0010064. And I get a value for my V1 of 0.001008. So 0.001008 cubic meters per kilogram. One look up down. Do you want to go? Let's stay one. I apply the same proportion to my HF values. So I will jump back into the calculator. I will grab that same calculation and just replace the terms on the right. So this would be x minus 151.53 divided by 173.88 minus 151.53. And I get an H1 value of 168.293. That would be kilojoules per kilogram. And then my S1 value is going to be between 0.5210 and 0.5226. So calculator. We swap out those two values. I'm going to be taking x minus 0.5210 divided by 0.5226 minus 0.5210. And I get an S1 value of 0.5747. 0.5747 kilojoules per kilogram kelvin. Okay, stay one done. For state two, I have a pressure of 80. kilojoules per kilogram kelvin. Okay, stay one done. For state two, I have a pressure of 8 megapascals, which is 80 bar. And an entropy of 0.5747. So the first step is going to be to determine the phase at state two. Easiest way to do that would be to look up SF and SG at P2 and compare my entropy to those values. So at 80 bar, I'm going to be looking at an SF and SG value of 3.2068 and 5.7432. Because this first column is SF, this second column is SG. So at 80 bar, that's 3.2068 and 5.7432. And the logic goes, if my entropy is less than SF, I must have a compressed liquid. If it's greater than SG, I must have a superheated vapor. And if it's between SF and SG, then I must have a saturated liquid vapor mixture. So my value of 0.5747 is going to be less than SF, therefore I have a compressed liquid. So now I will jump over to my compressed liquid table, which is table A5. And I find the pressure subtable corresponding to 80 bar. And I see that alas, I don't happen to have a pressure subtable corresponding to 80 bar. Which means that I'm going to have to interpolate between 75 bar and 100 bar. Furthermore, I recognize that my value of 0.5747 does not appear on either one of these nice even rows, which means I'm going to have to interpolate between two rows and two sub-tables. Which means that I have a triple interpolation on my hands. And remember from thermal 1, there are two ways that we can go about doing that. Either interpolating between the rows first and then interpolating on the between the two sub-tables, between the columns essentially, or interpolate between the columns first to generate a little sub-table corresponding to 80 bar, and then interpolate between the rows. You can do whichever method you prefer, they both yield the same answer. But performing the interpolation by hand is a little bit easier to follow. If I start by interpolating between the two pressure subtables on the same row to essentially build a little sub-table of our own, that way you can more easily follow what I'm doing. So I'm going to jump back to the iPad. I'm going to bring us our own little sub-table corresponding to 80 bar. And I have a temperature column, an enthalpy column, and an entropy column. And my entropy is going to be between 0.5696 and 0.5686 and 1.0704 and 1.0688, which means that I'm going to be between the temperatures of 40 and 80. The 40, 80, and I'm going to interpolate for an h value at 80 bar and 40 degrees Celsius. And then I'm going to interpolate for an h value at 80 bar and 80 degrees Celsius. This would be, let's call this 1, 2, and then an entropy value at 40 degrees Celsius here and 80 bar, and then an entropy value at 80 degrees Celsius and 80 bar, which is here. And then I'm going to use my entropy, which is 0.5747, to interpolate for a value of enthalpy at that entropy. Does that all make sense? I mean, I could grab T2 while I'm here, but I don't actually need it. And it's kind of a waste of time to calculate yet another thing at this standpoint. So I will just leave it off for now. So we have four interpolations to do to establish the framework that we are actually using for our actual interpolation. Does that make sense? Okay, good. So jumping back to our tables, I'm grabbing the value of enthalpy at 80 bar and 40 degrees Celsius first. So that interpolation is going to be wakeup calculator. Solving between pressures would be 80 minus 75 divided by 100 minus 75. And that's equal to, and look, I'm going to be doing a whole bunch of interpolations here. So I'm going to make this a little bit easier on myself by plugging in variables. I can do that on my TI 89. If you have an 89 or a 92 or an inspire or equivalent, you can do the same thing. If you're working through this on a simpler calculator, you're probably going to have to just punch the numbers every time. And that's, of course, fine, but this will make it a little bit faster for us to work this on the example problem. So I'm going to say x minus a divided by b minus a. And then I'm going to evaluate it for x and I'm going to plug in an a value of, we are interpolating for enthalpy first and 40 degrees Celsius. So 174.18 and a b value of. A b value of. That's, that's letter a again calculator and a b value of 176.38. So our first value here is 174.62. Okay. And then our second value is going to be enthalpy at 80 degrees Celsius. So a is. 340.84 b is 342.83 341.238 and then an entropy value at 40. So I'm using an a value of 0.5696 and a b value of 0.5686. Giving me an entropy of 0.5694 and I'm so confident that we're doing this correctly that I'm going to grab the other one as well so that I can just write them both down at the same time. That'd be 1.0704 and 1.0688. So those are our two entropy values forming the framework for our final interpolation here. So that was 0.5694 and 1.07008. And that seems very high. So I'm just going to go double check 1.07008. Yep. Okay. Now we have everything we need to actually perform our actual interpolation. So I'm going to say solve 0.5747 minus 0.5694. Divided by 1.07008 minus 0.5694 is equal to what we're looking for. X minus 174.62 divided by 341.238 minus 174.62. Calculator if you would please solve for X we get 176.384. And I'm going to write that up here 176.384. And while we're here I mean just for fun we can interpolate for temperature as well. I mean why not right? We did all this work already. Why not do a little bit of extraneous work? Look it's 40.42 degrees Celsius at state 2. Hooray. Anyway that process of going from state 1 to state 2 is a lot of interpolation. And it could have been worse. I mean imagine for a moment if we had ended up between two different rows and the two different tables and imagine if my state point was below the lowest pressure subtable then my interpolation would be between the lowest pressure subtable and the saturated liquid property. There is a shortcut that we can take to come up with our enthalpy value at state 2 and that shortcut is actually a navigation through fluid mechanics in order to get to a property. And that's by considering the work of a pump if we have incompressible flow. So we were to take our conservation of energy Reynolds transport theorem control volume equation and simplify it for a situation where we had incompressible flow undergoing steady state operation and we were to neglect friction we would end up with the work of a pump on a specific basis is approximately equal to specific volume times pressure change. And our specific work of our pump here is h2 minus h1. So if we make the assumption that the flow from one to two is pretty close to incompressible we can just take the specific volume at either state point because it's assumed to be the same because we're assuming that the density doesn't change times p2 minus p1 that gives us our difference in enthalpy. So I could say h2 is pretty close to h1 plus specific volume times p2 minus p1. We had h1 already it was 168.293 and we are adding to that our specific volume of state 1 which is 0.001008 cubic meters per kilogram multiplied by a pressure difference. Why don't we write that pressure difference in bar that would be 80 minus 0.075 bar and I recognize that a bar is 100 kilopascals and actually let me back up a second I want to get to kilojoules so best practice is to start a kilojoules and work backwards. A kilojoule can be written as a kilonewton times a meter and a kilopascal can be written as a kilonewton per square meter forgive me I'm running out of space here and everyone knows it's impossible to move when you're working digitally. Kilopascals cancels kilopascals bars cancel bars kilonewtons cancels kilonewtons cubic meters cancels square meters and meters and kilograms are left over. So if I were to take 0.001008 times the quantity 80 minus and 0.075 times 100 I would yield a quantity in kilojoules per kilogram which I could add to 168.293 to get an approximate value for h2 so let's try that out calculator if you would please wake up that's 168.293 plus 0.001008 times the quantity 80 minus 0.075 times 100 we get 176.349 so comparing contrast here with five interpolations we came up with a value that was 176.384 with zero interpolations and a single calculation we came up with a value that is 176.349 how much error is there in that calculation well let's find out 176.349 actually let me just grab the value so that we can keep the trailing decimals minus 176.384 divided by 176.384 that's what 0.02 percent that's probably an acceptable amount of error so here are my rules for that shortcut if you have a pump not a compressor but a pump and you're in the compressed liquid region and the saturated liquid line the whole time you can use this shortcut just be sure to note in your assumptions that you're assuming incompressible flow from one to two anyway we have the better value so let's keep the better value and move on to state three which is thankfully going to be just a matter of finding 80 bar in our saturation tables so back to table eight three we are looking for 80 bar which is down here I will grab the highlighter tool again and what I want at state three is an hg value this column here and an sg value this column here and I can't quite fit both of those on the same sheet of paper at any reasonable scale so I apologize for these small numbers on your screen but if I write down 5.7432 for my entropy 5.7432 5.7432 and an enthalpy value of 2758 on the nose that gives me my lookups and for my lookup at state four I have generally the same process as state two that is I need to use my entropy at state four which is the same as state three and compare that to sf and sg at my pressure at state four and then use our value to determine the phase so I'm going to have an entropy of 5.7432 and if we scroll on back up to 0.075 bar and try to find the hand tool in adobe again I can scoot over zoom in I see that my entropy value is going to be between sf and sg I mean my sf value is going to be whatever the interpolation for three quarters of the way between 0.5210 and 0.5226 is I mean I guess we actually have that number it's like 0.57 whatever it was and three quarters of the way between 8.33 and 8.23 is going to be what like 8.25 8.26 I have an entropy value of 5.74 which is definitely between sf and sg therefore I have a saturated liquid vapor mixture so my interpolation for enthalpy at state four is going to come from interpolating between sf and sg at 0.075 bar and I need to apply that same proportion to the difference between hf and hg at 0.075 bar so lucky me I have more interpolation steps to do let's see if we can draw that out here on the iPad again so I am going to take at state four x4 which is equal to h4 minus hf at 0.075 bar divided by hg minus hf both at 0.075 bar that's equal to s4 minus sf at 0.075 bar divided by sg minus sf at 0.075 bar and thankfully hf and sf are already done because we did that for state one because we had 0.075 bar and a saturated liquid so all we really need to do to support this interpolation is going to be to figure out sg and hg so for sg at 0.075 bar I am going to take 0.075 minus 0.06 divided by 0.08 minus 0.06 and set that equal to the value that I'm looking for x minus 8.3304 divided by 8.2287 minus 8.3304 and I get 8.25413 8.25413 kilojoules per kilogram kelvin and for hg that's going to be the same proportion except using the values at hg that'd be x minus 2567.4 divided by 2577.0 minus 2567.4 and I get 2574.6 2574.6 and I should have mentioned this earlier but I'll mention it now when you are interpolating between values that are going the opposite direction as the value that you're using to drive the interpolation like in this case pressure is increasing from 0.06 to 0.08 and enthalpy is increasing from 0.06 to 0.08 bar but the entropy is going down it can be easy to accidentally switch the direction of your entropy interpolation you might have been tempted to write 0.075 minus 0.06 divided by 0.08 minus 0.06 is equal to x minus 8.2287 minus divided by 8.3304 minus 8.2287 because your brain wants you to write a positive quantity but it doesn't actually work because you aren't grabbing the value at 0.06 and 0.08 anymore you're switching them it's okay to have a negative difference because you have a negative number in both the numerator and denominator on the right hand side of that interpolation and that's fine just make sure you keep an eye out for it it's good to get into the habit of checking that the proportion of the way you are between the numbers actually makes sense so we got 8.25 I can see that 8.25 is closer to 8.2287 than it is to 8.3304 which should be expected because 0.075 is much closer to 0.08 than it is to 0.06 so just get into the habit of double checking the sanity of your interpolation to make sure that you didn't actually switch the direction and come up with a quarter of the way across as opposed to three quarters of the way across if that makes sense anyway I have everything I need to be able to solve for h4 now we could also solve for x4 I mean we don't need it but we already determined t2 I mean we could calculate it one thing at a time so s4 was 5.7 432 minus sf at 0.075 which was 0.5747 divided by sg which is 8.25413 minus sf which is 0.5747 is equal to the thing that I'm looking for minus hf which was 168.293 divided by hg which is 2574.6 minus 168.293 and I get an x value of 1787.81 and does that make sense we are closer with 5.7 432 to 8 than we are to 0.5 therefore our enthalpy should be closer to 2500 than it is to 200 and it is which makes sense so my h4 value is 1787.81 and with that enthalpy I have everything that I actually need to determine my work in my workout my q in and my q out which will get me thermal efficiency the back work ratio I can use the power along with my network out to determine the mass flow rate of steam I can multiply that mass flow rate of steam by specific q in and specific q out to determine parts d and e and then for part e I'm going to be relating q that out to m dot cooling water times the difference in enthalpy of the cooling water itself so I have everything that I need look up wise just for funsies here let's calculate a quality at state 4 just in case I decide to plot the ts diagram I have something to refer to so that is just going to be 5.7 432 minus 0.5747 divided by 8.25413 minus 0.5747 and look we get about 0.673 queens so next even though it isn't explicitly asked for directly let's calculate the work in the q in the workout and the q out I have an open system analysis for each of these devices and I recognize that I only have work in occurring in the pump so I'm just going to end up with h2 minus h1 and again that comes from an energy balance on the pump it's isentropic which implies adiabatic we're neglecting changes in kinetic and potential energy and we're assuming work is only in the inward direction therefore it simplifies down to h2 minus h1 energy balances on the boiler turbine and condenser respectively would yield q in workout and q out terms because only one thing is happening at a time so for q in I would end up with h3 minus h2 then workout would be h3 minus h4 and q out would equal h4 minus h1 so my enthalpy values we yield these four quantities and I will bring the calculator back up and just for the sake of not scrolling all over the place I'm going to calculate all four of them before I start running them down so 176.384 minus 168.293 is work in which is 8.091 and then h3 minus h2 yields q in so 2758.0 minus 176.384 and that's 2581.62 and then 3 minus 4 yields work out which is 2758.0 minus 1787.81 and 4 minus 1 yields q out which is 1787.8.81 minus 168.293 so my four quantities are 8.091 and then 2581.62 and then 970.19 kilojoules per kilogram and 1619.52 now you know what's coming next don't you it's right I want network out and net heat transfer in so the network out is going to be the workout minus the work in which is this number minus this number we get 962.099 and that should match the net heat transfer in the inward direction not sure why I wrote the outward direction the net heat transfer in would be q in minus q out which would be 2581.62 minus 1619.52 and just to be consistent here I will actually grab the actual numbers so we get the trailing decimals that my calculator decides we don't need just so I can be a little closer to the actual value that I'm expecting hey look we got 962.099 again calculating both is a good way to check that you built these equations correctly right now we are analyzing the simplest variation of the Rankine cycle that we will possibly consider just like with the Brayton cycle we're going to add stuff to it and as we start adding stuff to it our equations are going to become more complicated which means the probability of messing this up is going to increase so good practice is to check yourself now I can consider what I actually asked for in the problem which I believe started with thermal efficiency thermal efficiency would be the network out divided by the heat transfer in so I'm going to take our shiny new 962.099 and I'm going to divide by 2581.62 and I get 37.26 percent then I asked for the back work ratio the back work ratio is the proportion of work out of the turbine that goes back into the work in and in this case it should be a very small number because the pump takes almost no work at all so I'm taking 8.091 and dividing that by 970.19 which yields 8.3 percent excuse me 0.83 percent which means all of the rest of the workout is able to go on to network out which is actually the goal of this device so we have a tremendous proportion of our work of our turbine actually being the profit that we are trying to achieve this small back work ratio implies that not much work is being reinvested to get more proportion as profit if that makes sense then for part c I wanted the mass flow rate of the steam which is the working fluid in the cycle and to do that I'm going to consider the net power output of our cycle and recognize that that quantity can be expressed as mass flow rate times specific network out therefore mass flow rate of our steam is equal to the net power output divided by the net work output or more accurately the specific network output so I have 100 megawatts 100 megawatts and I'm dividing by 962.099 kilojoules per kilogram and for now let's calculate a quantity in the kilograms per second and then we can actually get kilograms per hour which is what the problem asked for so a megawatt is a thousand kilowatts and a kilowatt is a kilojoule per second so I'm going to yield kilograms per second as an answer so 100 times 1000 divided by 962.099 yields a mass flow rate of 103.939 kilograms per second and then I can get kilograms per hour which is 3,600 seconds in one hour so if I multiply that quantity by 3,600 I will yield kilograms per hour which is 374,182 kilograms per hour and then for part d and e I am going to need a new sheet of paper here so for part d I'm looking for q in I believe yeah and that's q dot in which is going to be mass flow rate times specific q in the mass flow rate in kilograms per second was 103.939 and if I multiply that by kilojoules per kilogram I will get an answer in kilowatts so multiplying by 2581.62 something broke yields total q dot in which is going to be 268332 and that's kilowatts so 268.332 megawatts and that's one way to get to that answer there is another way do you spot it well remember that our thermal efficiency represents among other things the specific network out over the specific q in and also the total net power output divided by the rate of heat input therefore I could have taken the net power output divided by our thermal efficiency and gotten q dot in that would have also worked and that's going to be 100 megawatts divided by 0.37267 and we get 268.332 that would have been the faster way if we hadn't calculated kilograms per second as an intermediate step anyway part e is q dot out again there are two ways to do this the first would be m dot steam times specific q out kilograms per second times kilojoules per kilogram we yield kilojoules per second which is kilowatts so I could take 103.939 and multiply that by I forget what the q out number is 1619.52 and we get 168.332 megawatts do you spot the other way well the other way would be recognizing that our heat engine is taking in q in and outputting q out plus network out so an energy balance on the heat engine itself would say q dot in has to equal the net power output plus q dot out therefore q dot out could be written as q dot in minus whatever went to net power output which means that we could have taken 268.332 and subtracted 100 and gotten 168.332 that again would have been the faster method especially if I hadn't calculated m dot steam in kilograms per second then lastly part f we are considering how the cooling process actually occurs in the condenser so in our system diagram we just drew a circle and said condensation is happening here but we didn't actually elaborate on it so this is elaborating on it if I draw that circle again let's draw that as a bigger circle something like this yes that's helpful how about that yeah that'll work so this is my condenser and I have entirely too thick a line weight again make some more space I have my output of my turbine entering at state four and it is condensing it leaves at state one so the logic is we are spraying in our vapor in this case it's actually a liquid vapor mixture that is approximately 63% vapor and what 27% liquid so some amount of mixture as well and then we are running that through some coils where cooling water is passed and let's call this cooling water in and cooling water out it is gathering on those coils and it is dripping down and leaving as a freshly condensed saturated liquid so I could draw it something like this look all those droplets they're falling our gathering is a liquid and then leaving and then leaving so my question becomes if we have cooling water passing through this that's entering at 15 degrees Celsius and leaving at 35 degrees Celsius how much mass flow rate of cooling water do I need remember that the amount of heat rejected into this coil is queued out so what I'm saying is the heat rejected by the condenser is going into the cooling water and I could write that as the mass flow rate of cooling water times the specific heat absorbed by the cooling water which if I were to set up an energy balance would be m dot cooling water times h cooling water at the outlet minus h cooling water at the inlet now I'm in a situation where I am evaluating a delta h for water now I could look up the enthalpy of water if I knew two independent intensive properties but I don't I only know one so I could assume a pressure and then I can look up the enthalpy at that pressure but it might be better for me to recognize that I have a temperature difference of 20 degrees Celsius which means I have a relatively small change in temperature and my liquid is unlikely to evaporate as a result of going up to 35 degrees Celsius therefore it stays in the same phase the whole time that is the cooling water passing through the coils not the steam on the other side so if I were to assume the specific heat capacity of the cooling water was constant and again let's just remember I'm assuming the specific heat capacity of the cooling water is constant not the actual working fluid in my cycle then this would become m dot cooling water times cp of cooling water which would be cp of h2o times 35 degrees Celsius minus 15 degrees Celsius therefore m dot cooling water is equal to q dot out from my cycle divided by cp of water times 20 kelvin because the temperature difference in kelvin is the same as the temperature difference in degrees Celsius so that's going to be what was it 168.332 megawatts and I'm going to quickly run out of unit converter in the base so I'm going to bring this over here let's draw a big horizontal line and then we are dividing by the cp of water that is determined halfway between 15 and 35 halfway between the two would be 25 degrees Celsius 25 plus 273.15 is going to be about 300 kelvin so I'm grabbing the cp of water at 300 kelvin which from table 819 in my textbook means I am grabbing a number that is 4.179 4.179 kilojoules per kilogram kelvin multiplying by 20 kelvin so in order to get the energy inside of the heat transfer rate and kilojoules to cancel I need to break apart the megawatt let's write that as a thousand kilowatts and a kilowatt can be written as a kilojoule per second so kilowatts cancels kilowatts kill them excuse me megawatts cancels megawatts kilojoules cancels kilojoules and I want hours again so 3600 seconds in one hour will leave me with an answer in kilograms of cooling water per hour because we're playing the directly compare the two numbers game so calculator I need you to get back on top of business let's take 168.332 and multiply by 1000 and then multiply by 3600 and then divide by the quantity 4.179 times 20 and I get 7.25 times 10 to the sixth so in the interest of direct comparison here I will write that as 7,250,480 kilograms of cooling water per hour and exaggerated commas so we figured out that the mass flow rate of steam required to operate this was 374,000 kilograms per hour does it make sense that we need way more cooling water to accomplish the cooling process yeah it does because remember that the working fluid here the steam is undergoing a phase change which means that latent energy is involved so in order to absorb that massive amount of energy with such a small sensible energy change here means we're going to require way more mass flow rate to do that so we should expect that the cooling water needs a much much higher mass flow rate than the working fluid itself so that does make sense and by the way maintaining a certain temperature difference in your cooling water is relatively common I mean most steam power plants don't actually just reject heat with some fins and a fan on the side of a condenser they usually run the working fluid through a heat exchanger pushing it into another fluid like for example cooling water and then dealing with the cooling water separately in some cases that might just be dumping the heat into cooling water in the form of a nearby river or stream in some cases they're pulling water out of a well and just letting that hot water back into the environment and if it's too hot then it's damaging in other cases they okay this is the heat engine okay steam power plant in other cases they build a big evaporative cooler and then run it through a heat exchanger and then spray the cooling water out into the evaporative cooler and then collect the condensed cooling water before pumping it back in that's a very easy way to get rid of that heat pretty quickly and then the resulting hot air is pulled upwards which means that you end up with this little plume of steam and water vapor leaving with the hot air up into the atmosphere so if you've ever driven by a power plant and you see a big cooling tower that looks like a nuclear silo it's probably not unless it's actually a nuclear station but that cooling tower is cooling the water that they are using to cool the condenser in the actual power cycle it's just a much more efficient way to handle that and the advantage of keeping your cooling water separate from your working fluid is you want your working fluid to remain pure you don't want there to be any sort of air mixed in with it you don't want there to be impurities you don't want there to be like a build up of any sort of minerals inside of your system whereas the cooling water you don't care so much about because you don't have to be quite as precious because it's a lot easier to maintain that system anyway that's enough rambling that's all the required parameters in this problem but just for fun here why don't we draw a ts diagram i think that'd be fun so i am drawing a ts diagram and because we are talking about a Rankine cycle it's going to be helpful if we draw and because we are talking about a Rankine cycle it's going to be helpful if we draw saturation lines in order to indicate our state points relative to so i'm going to draw go with a thicker line weight i'm going to draw a saturated liquid line here and a saturated vapor line over here it's not a particularly good drawing about that that's slightly better okay start over that'll work and i have two relevant lines of constant pressure here so i am going to draw those in black i'm gonna draw low pressure remember the line of constant pressure on a ts diagram goes up into the right high pressure up here and actually let's just draw this whole thing black so that our state points will be red poof so state one is a saturated liquid at the low pressure this is 0.075 bar and this is 80 bar saturated liquid at the low pressure would appear here then state two is on the high pressure line but that's an isentropic process between one and two which means we go straight up state three was a saturated vapor at the high pressure and state four is an isentropic process straight down the low pressure line therefore my cycle looks like this and note that if i had pictured my ts diagram when i was defining my state point properties i could have more quickly determined the phase of states two and four because going up from these saturated liquid line is going to yield a compressed liquid and going down from the saturated vapor line is going to yield a saturated liquid vapor mixture and just to be consistent with my thermal one drawings here i will draw a little arrow here and say this is 60 what was it 68 percent of the way over 67.3 let's be exact shall we okay then that region enclosed is my network out and just like when we talked about trying to improve our cycle in the Brayton cycle i think it's useful to consider what would happen on your ts diagram when you're thinking through the implications in terms of thermal efficiency so remember that the integral under the process from two to three on a ts diagram would represent the heat transfer in so if i switch to blue color here this region here is my q in so the thermal efficiency is going to be the red hatched area divided by the blue hatched area that's a visual representation of my thermal efficiency so if i asked you a question like what if the condenser was operated at 0.06 bar instead of 0.075 bar you could immediately determine how that's affecting the thermal efficiency by visualizing what if the bottom line went down you have the same q in but you have a larger network out or rather you have approximately the same q in because you're scooting a little bit to the left but that same area is being added to the network out as well so network out makes a larger proportion of q in therefore dropping the condenser pressure would improve thermal efficiency or if i were to ask a question like what if the boiler were taken up to 90 bar instead of 80 bar how would that affect the thermal efficiency well you're adding the same quantity to both network out and q in therefore you're going to be improving the thermal efficiency or if we were to super heat the steam in the boiler maybe bringing it out to like here and then we are adding much more area to the network out then proportionally then we are to q in therefore that's also going to be improving the thermal efficiency moving the right line for the right also improves thermal efficiency so generally speaking increasing our boiler pressure dropping our condenser pressure and increasing our maximum temperature are all ways to improve our thermal efficiency if we set this problem up in matlab we end up with something that looks like this here i am using my boiler pressure and condenser pressure as our given information and we are looking up all of the state point properties using the x team tool set and then determining the work in q in workout and q out and calculating a thermal efficiency i'm getting a thermal efficiency of 37.25 with matlab which is good because when we worked it by hand we got 37.26 percent and then matlab also has the convenience of being able to plot this on a ts diagram so here i'm drawing my saturated liquid and saturated vapor lines drawing a line of constant pressure for 80 bar and 0.1 bar and then plotting our four state points state point one is appearing here state point two is actually visually directly on top of it because in our hand drawing the lines of constant pressure in the compressed liquid region are a bit exaggerated in reality the line of constant pressure follows so closely to the saturated liquid line that you can't really discern the difference between state one and state two visually on this plot i mean consider the fact that t1 is at our saturation temperature at 0.075 bar and t2 was 42 degrees celsius i mean they're very very close to each other state three is right here and state four is directly below it which matches what we drew which makes sense working through this in matlab allows you the ability to quickly modify given parameters and see the change that it has on your calculation very quickly i mean for example if i were to change the boiler pressure from 80 to 90 i can rerun all the calculations immediately and get a thermal efficiency of 37.65 instead of 37.25 or if i were to change this back to 80 and change the condenser pressure to 0.065 instead of 0.075 i get a thermal efficiency of 37.63 and that brings me to my next comment if we had considered the isentropic efficiency of our compressor and turbine i mean imagine a hypothetical situation where i had given you isentropic efficiencies for this process and asked you to perform all these calculations with those isentropic efficiencies all that would be different would be how we got to states two and four we would have used s1 being equal to s2s to look up in h2s and then used our definition of the isentropic efficiency for a compressor or a pump to determine h2 actual from h2s similarly we would have used s3 is equal to s4s to look up in h4s so note h2s and h4s would still be 176.384 and 1787.81 and then we would have used our isentropic efficiency of a turbine to represent the difference in enthalpy and use that to calculate h4 actual so we would have just had an extra step to getting to h2 and h4 and the rest of the analysis would be the same we would have just used h2 actual and h4 actual instead of these values which are h2s and h4s but the advantage of setting up your matlab code with an isentropic efficiency even if it's 100% is it allows us the ability to change those very quickly so instead of trying to go back and modify all your code if you have an isentropic efficiency instead of not having it you can just change it here so if we had an isentropic efficiency of a pump of 0.8 and an isentropic efficiency of our turbine of 0.9 these are very respectable efficiencies 80% and 90% let's see how much that affects our thermal efficiency shall we look at that it dropped all the way down to 33.8% wow that's quite a big change just by changing our isentropic efficiency just a little bit and now i think we can consider this example problem done