 12 kilograms per second of water enters an adiabatic turbine at 600 degrees celsius and 12 megapascals, which expands it to a saturated vapor at 5 kilopascals. I want to know the temperature of the steam at the outlet, the power produced by the turbine, and the isentropic efficiency of the turbine. Well I'm going to start by identifying state points. I have state 1, which is the actual inlet conditions. I have a mass flow rate of 12 kilograms per second and a temperature of 600 degrees celsius. Furthermore I know the pressure at state 1 is 12 megapascals. Then I have state 2, where my pressure is 5 kilopascals, and I was told it expands it to a saturated vapor, so x2 is going to be 1. In order to actually answer the question, I need to establish a third state point, which is what the outlet would be if everything were perfect if I had an isentropic process, because remember, ideal compression and expansion are represented by an isentropic process. For state 2s, where everything is perfect, this hypothetical state point has the same pressure as p2 because I'm evaluating the same turbine, just at different operating efficiencies, then s2s is equal to s1. So at state 1, I'm going to be looking up an enthalpy. At state 2, I'm going to be looking up an enthalpy. And then at state 2s, I'm going to be looking up an enthalpy. And those three enthalpies are going to be what I use to determine the isentropic efficiency of this turbine. So quick question, before you go perusing your equation sheet, do you get more work out of a turbine in a perfect world where there are no losses, or in reality where there's friction and stuff? Right, it's an ideal world. The presence of losses decreases your work output from your turbine, therefore actual work is going to be lower than ideal work, therefore it's in the numerator of this proportion that should be less than one. From an energy balance on my turbine, for an adiabatic process, I can see that my specific work is going to simplify down to h1 minus h2, so I have h1 minus h2 divided by h1 minus h2s. So once I have all three enthalpies, I can determine the isentropic efficiency of the turbine. Then I also wanted the temperature of the steam at the outlet, so I'm just double checking that we know which state point is the real temperature, and then I have the power produced, which from the energy balance is going to become w.out, which is m.times specific work, which is m.times h1 minus h2. If it makes it easier, I will bring in an energy balance. We've done energy balances on turbines before, so I'm not going to go through it in complete detail, so the power output is m.times h1 minus h2. That will be how I answer part b. So all I have to do to finish the question is look up h1, h2, t2, and h2s. Looking up h2s requires that I know s2s, which requires that I look up s1 as well, so it's really five property lookups. And you guys know me, just for funsies here, let's look up t2s, just to compare and contrast the two. So first up, I have state 1, which has a temperature of 600 degrees celsius and a pressure of 12 megapascals. If I jump into my property tables, I recognize that the easiest way to fix state 1 is going to be to look up the saturation condition corresponding to either my temperature and pressure and compare the other to the saturation property. So I'm going to look up the saturation temperature at 12 megapascals, which means I'm going to go into my saturation tables for water by pressure. 12 megapascals is going to be 120 bar. At 120 bar, the saturation temperature is 324.8. Therefore, my temperature is greater than that, which means that I have a superheated vapor. So I'm going to go into my superheated vapor tables. I'm going to find 12 megapascals. And I see that my enthalpy at a temperature of 600 degrees celsius and a pressure of 12 megapascals is 3,608.3. Furthermore, my entropy is going to be 6.8037. So h1 and s1 are 3,608.3 kilojoules per kilogram and 6.8037 kilojoules per kilogram kelvin. Then next, I will do h2 and t2 because that should be a pretty straightforward lookup. I have a saturated vapor at 5 kilopascals, which means I'm going to go back to my saturation tables by pressure. 5 kilopascals is going to be 0.05 bar, which annoyingly is between 0.04 bar and 0.06 bar. So I'm going to have to interpolate for both my temperature and my enthalpy. That interpolation, fortunately, is going to be just halfway between 0.04 and 0.06. 0.05 minus 0.04 divided by 0.06 minus 0.04 is equal to x minus 28.96 divided by 36.16 minus 28.96, and that gives me a syntax error that gives me a temperature of 32.56, which represents my actual temperature at the actual outlet. 32.56 degrees celsius. The enthalpy will follow the same interpolation, just with the enthalpy values instead of the temperature values. Furthermore, note that I am interpolating for the hg values because I have a saturated vapor. So my enthalpy difference is going to be x minus the enthalpy at 0.04, which is 2554.4 divided by 2567.4 minus 2554.4, and I get 2560.9. 2560.9 kilojoules per kilogram. So now for state 2s, I have an entropy of 6.8037 and I know that my pressure is still 5 kilopascals. So I'm going to jump back to my saturation tables by pressure. I'm going to find the correct pressure, which, hey, happens to be a 0.05 bar at the same row that we were just at. And what I have to do is compare my entropy to sf and sg at 5 bar. Well, I don't know sf and sg at 5 bar, but I do know that they're going to be somewhere between 0.42 and 0.52 for sf and 8.47 and 8.33 for sg. So my s value is definitely going to be between sf and sg, which means that I'm going to have a saturated liquid vapor mixture at state 2s. So what that means, unfortunately, is I have to determine what the values are for hf, hg, sf, and sg at 0.05. So I am basically building my own little row full of what those values would be at a pressure of 0.05 bar. So on my table, I'm going to have tsat, I'm going to have hf, I'm going to have hg, I'm going to have sf, and I'm going to have sg. Well, what is the saturation temperature corresponding to 5 kilopascals? We've already determined that 32.56. In order to interpolate for the rest of the values, I'm going to use the same exact interpolation. That's 0.05 minus 0.04 divided by 0.06 minus 0.04, which means that every single one of these is going to end up being hf is equal to 0.05 minus 0.04 divided by 0.06 minus 0.04. That entire quantity is 1 half times hf at 0.06 minus 0.04 plus hf at 0.04. So I'm going to, for simplification sake, just write that as 0.0, excuse me, 0.5 multiplied by the difference between the properties that we are interpolating between plus the hf property. So 0.5 times 151.53 minus 121.46, that's the values for hf, plus 121.46, that gives me an hf value for my new row of 136.495. Once again, for hg, that's going to be 0.5 times the quantity to 567.4 minus 25 255.4.4 plus, nope, there we go, plus 255.4.4, and I get 2560.9, which honestly, I should have just grabbed from over here, but you know, practice makes perfect. 2560.9, the interpolation for hf is going to be 0.5 multiplied by the quantity 0.5 to, excuse me, 521 minus 0.4226 plus 0.4226, which is 0.4718. And then sg is going to go 0.5, 0.5 times 8.3304 minus 8.4746 plus 0., no, plus 8.4746, and I get 8.425.4.025. So this is now my row. This is the row corresponding to pressure 0.05 bar. And on my shiny new row, I'm going to determine h2s, which is going to be between 136.495 and 2560.9, and t2s, which is known, that is just 32.56. I will scooch my table down. So for h2s, I am just interpolating s2s minus sf divided by sg minus sf is equal to h2s minus hf divided by hg minus hf. And that gives me x2s. And you know what, just for fun, let's also determine x2s. I mean, when in Rome, right, we already populated this little row of the table. So we might as well, that would be s2s, which is 6.8, 037 minus 0.4718 divided by not the letter p calculator 8.425 minus 0.4718, which gives me a quality of almost 80%. And then my interpolation for enthalpy is going to go 0.798404 times hg minus hf, which is 2560.9 minus 136.495 plus 136.495. And I get 20,473. That cannot be right. What went wrong? 2560.9. And I get 2072.15. h2s is 2072.15 kilojoules per kilogram. So note that the chapter 6 math here is not actually that complicated. The difficulty in this problem is primarily doing the labor to look up those enthalpies, which is why it's so hard to do well at chapter 6 if you are not already doing well at chapters 3 and 4. This is a chapter 4 analysis. All of the property lookups were chapter 3. But now that I have all of those, I will first figure out the power output. And do you know what? I should really sort this better. A, T2, we determined already is 32.56. B is going to be the power output, which I will drag down here. Mass flow rate is 12. So I'm going to take 12 times my enthalpy difference. 12 times 3608.3 minus 2560.9. And I get 12568.8 kilowatts, which is going to be 12.57 megawatts. And for part C, I'm going to drag this down. We'll actually just copy and paste. So I have H1 minus H2 divided by H1 minus H2S. So that will be 3608.3 minus 2560.9 divided by 3608.3 minus 2072.15. So my isentropic efficiency of my turbine is 0.682 or 68.2%. Cool. Cool, cool, cool. Even though I didn't ask for it, let's draw this on a TS diagram. Let's add in a part D here. On the TS diagram, surprising no one, the saturation lines appear as a dome and lines of constant pressure go up and to the right. And just like they did with TV diagrams. So state one is going to be a superheated vapor. State two S is going to be directly below that. State two is going to be right here. So this is my actual process. One to two S appears as a vertical line because it's an isentropic process. Note that if I were drawing this to scale, the percentage of the way I am across the dome for state two S would be whatever this was, which is 80%. The quality of state two S was 79.8%. Scale might not be that bad actually. And two is directly to the right because it's under the saturation dome. And that's my TS diagram. I can label the pressures. That's probably good. D one is 12 megapascals. P two and P two S are both five kilopascals. And now I'm really done.