 Complete the following table for water. I have given us two independent intensive properties corresponding to five state points, and what I would like us to do is look up whatever else we need in order to complete this table. So at state point A, the two independent intensive properties that fully define our state point are temperature and pressure, and I want us to determine specific enthalpy, specific volume, and a description of the phase. We will do that for all five state points until we have a fully populated table. Each of these is going to begin by trying to fix the phase to determine which phase corresponds to our property. At state point A, what we have is a temperature and pressure. So the way that we will fix the phase of state A is to look up the saturation condition corresponding to one of our properties and to compare the other one to that saturation condition. We can look up the saturation temperature corresponding to a pressure of 10,000 kPa and compare that temperature to 140 to fix the phase, or we can look up the saturation pressure corresponding to a temperature of 140, then compare our pressure of 10,000 kPa to that pressure to determine the phase. That procedure would involve either table A2 or A3 depending on if we wanted to use the temperature to drive the saturation pressure lookup or the pressure to drive the saturation temperature lookup. In the interests of character building, I will show you both ways. The saturation pressure corresponding to a temperature of 140 degrees is going to come from table A2, which is the saturation tables listed by temperature. So once I find 140, I can see that the saturation pressure is 3.613 bar or 361.3 kPa. My pressure at state 1 is higher than that, much higher, which means that I must have a compressed liquid. I find the easiest way to do that, to consider what's happening on the PV or TV diagram, and to visually place the pressure corresponding to our saturation pressure on the PV diagram, because this line of constant temperature right now represents 140 degrees Celsius. So we have a higher pressure than PSAT, which means we must be to the left of the dome, which means we have a compressed liquid. We also could have gotten there by comparing our temperature to the saturation temperature corresponding to 10,000 kPa. To do that, we convert 10,000 kPa into bar. It is 100 bar, and I jump into table A3, wherein I can see that the saturation temperature corresponding to 100 bar is 311.1 degrees Celsius. The temperature 140 is below the saturation temperature of 311.1, which means that the water has not yet boiled. The only way that we can be less than TSAT on this TV diagram is to be to the left of the dome, which is our compressed liquid region. In either case, we conclude that the phase is a compressed liquid, so we can jump into our compressed liquid property tables, which is table A5. So on table A5, I need to find a pressure sub-table corresponding to my pressure, which was 100 bar. I have a pressure sub-table of 100 bar, and I can look up whatever I need because I have the temperature 140 right here. I don't even have to interpolate between these temperatures, nor do I have to interpolate between these pressures. It's just a matter of looking up the property. So I will start by populating the phase description column, and I will do that by running in compressed liquid. Then I will populate the enthalpy, the specific enthalpy at 100 bar, and 140 degrees Celsius is going to be 595.42. And the specific volume will also come from that table. It's this first column here. And the specific volume is a little bit tricky here because this column doesn't actually give us specific volume. This column gives us specific volume times 10 to the third. Does that make sense? This column is not specific volume. It is specific volume times 10 to the third. So in order to get back to a specific volume, we have to take this column times 10 to the negative third, which means our specific volume is 0.0010737. Basically the book is just trying to save a bunch of ink in space, corresponding to writing out a whole bunch of zeros. It's doing that by presenting this information in exponential notation. And the easiest way to remember which direction you have to go is to remember that this column is specific volume times 10 to the third, which means we have to take this quantity times 10 to the negative third and get back to specific volume. So we have 0.0010737. Let's see if I can remember that long enough to write it down. 0.0010137. Everybody says 0.0010737. Looks like I got it right. And that gives us row A. Woo, we are one fifth of the way there. For row B, we have a pressure and specific volume. And this time, we don't have the luxury of having two options to determine what our phase is. We have to look up the specific volume corresponding to a saturated liquid and a saturated vapor at 100 kPa that corresponds to this specific volume and this specific volume. And just like we did with temperature or pressure, we can deduce what the phase is based on our value compared to these specific volumes. If our specific volume is less than the saturated liquid specific volume corresponding to our pressure, then we must have a compressed liquid. If it's in between the two, we must have a saturated liquid-vapor mixture. If it's greater than the specific volume of a saturated vapor, then we must have a superheated vapor. If it's exactly equal to the specific volume of a saturated liquid, it must be a saturated liquid. If it's exactly equal to the specific volume of a saturated vapor, it must be a saturated vapor. I will point out what we're here that the notational scheme we used indicate saturated liquid properties is a subscript f. So this would be specific volume f. And the notation we use for properties of a saturated vapor is subscript g. So vf is the specific volume of a saturated liquid, vg is the specific volume of a saturated vapor. And that's a very old fashioned way of writing things out back from when they referred to liquids as fluids and vapors as gases, thus f and g here. So we are comparing our specific volume to vf and vg and deducing the phase based on that property. Same is true on our plot of pressure with respect to specific volume. vf corresponds to this intersect, vg corresponds to this intersect. This abbreviation appears the same on the tv diagram as well. We have vf corresponding to the property here. We have vg corresponding to the property here. And we position our specific volume relative to vf and vg and deduce a phase from that comparison. So we want the vf and vg values corresponding to a pressure of 100 kilopascals, which would be one bar. So we will jump into our saturation properties by pressure, which is table a3. And I will find one bar. And I see that the saturated liquid specific volume is 0.0010432. And the saturated vapor specific volume is 1.694. Again, remember, this column is not vf, it is vf times 10 to the third, which means you have to take it times 10 to the negative third to get back to vf. Again, vf at one bar is 0.0010432, vg is 1.694. And I use my specific volume to compare to these two values and deduce a phase. If it's less than vf, it must be a compressed liquid. If it's greater than vg, it must be a superheated vapor. If it's between the two, it must be a mixture. If it's either this one or this one, then it is a saturated liquid or a saturated vapor, respectively. Our specific volume was exactly 0.0010432, which means we must have a saturated liquid. Then our other two properties will just be the property of a saturated liquid at 100 kilopascals. So the temperature would be T sat corresponding to one bar, which is 99.63. And our specific enthalpy will just be hf, which is 417.46. Because again, specific enthalpy of a saturated liquid at one bar is 417.46. Two rows down. In row C, we have a temperature of 20 degrees Celsius and a specific enthalpy of 2000. Our procedure here will be the same general procedure as for row B. We will look up hf and hg corresponding to a temperature of 20 degrees C. And then we will see how 2000 compares to those two values and deduce the phase. If 2000 is less than hf, we must have a compressed liquid. If 2000 is greater than hg, we must have a superheated vapor. If it's between the two, we must have a mixture. If it's exactly hf or hg, we have a saturated liquid or a saturated vapor, respectively. So we will jump into our saturation tables corresponding to temperature this time. We will find 20 degrees Celsius. And we will see that the specific enthalpy of a saturated liquid, hf, at 20 degrees Celsius is 83.96. And we see that the specific enthalpy corresponding to a saturated vapor, hg, at 20 degrees Celsius is 2538.1. My value is between the two, which means I must have a mixture of saturated liquid and saturated vapor, i.e. a saturated liquid vapor mixture. Let me point out while we're here that this column hfg just represents the difference between hf and hg that is useful for something later on. And the table is just helping us by not expecting us to subtract these two numbers every single time. So hfg is just specifically hg minus hf. You can ignore that column for now, unless you want to save a step on the interpolation process. Speaking of which, we are going to assume that the relationship between vf and vg under the dome is going to be the same as the relationship between hf and hg under the dome. i.e. if we have a specific enthalpy that is halfway between hf and hg, then we will assume that our specific volume is also halfway between vf and vg. And the proportion of the way we are between hf and hg and vf and vg has a special name. It is called quality. Quality represents how far across the dome you are. And it represents, say, h minus hf divided by hg minus hf, i.e. how far between hf and hg you are. It also represents v minus vf divided by vg minus vf. It also represents u minus uf divided by ug minus uf. It also represents s minus sf divided by sg minus sf. It is called quality because when we are talking about steam power, we want high energy water. If we have a liquid and vapor mixture that is more vapor than it is at some other point, then we have more quality. We have something more useful than at that other point. So the quality is a representation of how much vapor we have in our mixture, which is an indication of how useful it is in the steam power production process. Regardless of the reason behind the name, though, quality represents how far across the dome you are and in fact represents a mass proportion of the vapor in the mixture. So if you had half a kilogram of vapor and half a kilogram of liquid in your mixture, then your total mass in your mixture is going to be 1 kilogram and the mass of vapor in that mixture is 50%. If you had 0.1 kilograms of vapor in a 1 kilogram mixture, then you have 0.9 kilograms of liquid and a quality of 10%. If you had 99% quality, then that means 99% of the mass in your mixture is vapor and 1% is liquid. Quality is useful for a number of things. It can be used as a stepping stone for interpolation. It also can be used as a visual reference for positioning state points under the dome visually. So you could say quality of 25% is going to appear here, quality of 75% is going to appear here, quality of 50% is here, and you can describe those relative positions by expressing a quality a little bit more conveniently. Whatever the case though, I am going to calculate a quality for row C just for fun. We don't need it to complete the table, but what are we here to do if not to learn things? So I am going to calculate a quality by taking our enthalpy value minus hf at our temperature, which was 20 degrees, divided by hg at our temperature minus hf at our temperature. And if you wanted to, you could just substitute hfg in the denominator and save yourself a subtraction step. So hf at 20 degrees Celsius was 83.96, hg was 2538.1. So if I pop up my calculator again, I can take 2000 minus 83.96 divided by, I lost it already, 2538.1 minus 83.96. And I get 0.781, which means that our quality is 78.1%. Now again, that wasn't actually explicitly necessary as a stepping stone. We could have just jumped from h to v, but now that we have it, we can also write it as v minus vf divided by vg minus vf. So I am going to say 0.781 multiplied by the quantity vg minus vf plus vf is my specific volume. So if I jump back into our tables and I say this number multiplied by the quantity 57.791 minus 0.0010018 plus 0.0010018 is our specific volume at state C here, which is 45.12. And I can also populate saturated liquid vapor mixture here. And then all we have to do to complete that row is look up the saturation pressure corresponding to 20 degrees Celsius, which is 0.02339 bar, which would be 2.339 kilopascals, 2.339 kilopascals. Row C is done. For row D, we recognize that we have a saturated vapor. So all of our lookups are just going to be the property of the saturated vapor corresponding to 30 degrees Celsius. Conveniently for us that is still going to be on table A2, we are just going to be looking at properties at 30 degrees Celsius instead of 20 degrees Celsius. So I will start with pressure. My pressure is 4.246 kilopascals, 4.246, 4.246. And then next is our specific enthalpy, which is here, hg, and hg at 30 degrees Celsius is 2556.3. There we go. 2556.3. By the way, the temperature, actually the temperature on A2 and the pressure on A3, both of those driving properties are on both the left and right sides of these tables. So if you're reading something from the right hand side of this table, sometimes it's easier to just grab the temperature over here. That way you don't have to try to scan across the entire column. I will also point out if you are looking at properties on physical tables, it is often really useful to use a straight edge, even if you don't think you need it, just to put it underneath the road that you're looking at, so you don't read off a wrong number. So 30 degrees Celsius has an hg value of 2556.3. So 2556.3, 2556.3. And then specific volume is just going to be vg, which is going to be 32.894. And that gives us rho d. For rho e, we have a temperature and pressure again, which means that we have two options to determine the phase that can look up the saturation pressure corresponding to our temperature or the saturation temperature corresponding to our pressure. We've done the whole, let's do it both ways thing a couple of times now. So I'm assuming you're getting relatively proficient at that. So I will just do it one way. I generally find it easier to look up the saturation temperature corresponding to a pressure because it's easier for most people to relate a temperature to the boiling point than it is to relate the pressure to the pressure at which the actual phase transition would occur at a constant temperature, because you naturally boil things at a constant pressure in your daily life. You boil water on top of a stove to make your ramen noodles or whatever. So you probably have a more intuitive understanding of how temperature changes between liquid and vapor for a given pressure. Anyway, 300 kilopascals would be three bar saturation temperature corresponding to three bar is on table A3. I see that it is 133.6. My temperature is higher than that, which means I must have a superheated vapor. And then I want specific enthalpy and specific volume. Those will come from our superheated vapor tables, which remember are in table A4 for us. And on table A4, I see a pressure sub tables corresponding to different pressures. And what I want is a pressure of three bar. I see three bar, which means I don't have to interpolate between pressure sub tables. So I'm just going to grab a value at 340 degrees Celsius. And oh no, I don't happen to have 340 degrees Celsius. So I will have to interpolate. Again, we are using linear interpolation for all of our interpolations for property lookups. When we're performing these lookups by hand. Furthermore, I'm going to take mental note that our first column is specific volume and our third column is specific enthalpy. So I want to interpolate between 320 and 360, which means I'm going to need our calculator again. And I will interpolate by taking 340 minus 320. I can't actually see. Yep, 340 minus 320 divided by 360 minus 320. Yeah, that's going to be annoying. Can I scroll down more? I can probably scroll down more. No, too much. Down. There we go. And then that is going to be multiplied by our specific volume at 360 minus our specific volume at 320, which is 0.969 minus 0.907. And then we are adding that result to 0.907. And just for good measure, I'm going to add a multiplication operator here. And I get a specific volume corresponding to 340 degrees Celsius of 0.938. So I can populate that here. Specific volume was this column. And that was 0.938 cubic meters per kilogram. And then we can interpolate for enthalpy in the exact same way. In fact, I will just scroll on up, grab this equation, and replace the specific volume at 320 and 360 with the specific enthalpy at 320 and 360. So 3192.2 minus 3110.1 plus 3110.1. And I get a specific enthalpy of 3151.15. So 3151.15. And with that, we have all of our property lookups. I will point out that you can interpolate between pressures in the same way that you can interpolate between temperatures. So let's just try one of those out ones. I'm going to get rid of my quality notes here. That was upon, it's both quality and of quality. Imagine if we had a row F. And row F had a temperature of 320, but a pressure of 200 kilopascals instead of 300. I mean, I made the table, I can add more rows to our table. Okay, we have an extra problem to do here at 200 kilopascals, which is 2 bar. I can look up the saturation temperature. Saturation temperature corresponding to 2 bar is 120.2 degrees Celsius, which means that I have a superheated vapor again, which means that I'm going to go back into table A4. And I'm going to find the pressure of subtable corresponding to 2 bar. I'm sure there will be a pressure of subtable corresponding to 2 bar. So confident, am I that I will just peruse on here? Oh man, that's not 2 bar. That's not 2 bar. That's not 2 bar. That's not 2 bar. Oh no, guys, I have a pressure of subtable at 1.5 bar, and I have a pressure of subtable at 3 bar, but I don't have 2 bar. Oh no, what do I do? I interpolate. So the interpolation is going to be at 320. It's just that we are interpolating between 1.5 bar and 3 bar instead of interpolating between two temperatures on a given pressure subtable. So I want specific volume and specific enthalpy. So I will interpolate between 1.819 and 0.907 and 3113.5 and 3110.1. By the way, when you're interpolating, it doesn't actually matter if the value is increasing or not. Just make sure that the order of your temperatures matches the order of your properties. So don't overthink if you need a positive delta or not, because that will often cause you to go astray. Instead, just keep track of which one you're doing first. So I would take 2 minus, excuse me, calculator. You're not responding. Try that again. 2 minus, oh, 2 parentheses. One more time. So I'm going to take 2 minus 1.5 divided by 3 minus 1.5 multiplied by the value at 3, which was 0.907 minus the value at 1.5, which is 1.819, quantity plus the value at 1.5, which is 1.819. And I get a specific volume of 1.515. And I can do the exact same thing with specific enthalpy. So I'm going to take 2 minus 1.5 divided by 3 minus 1.5 and multiply by the value at 3, which is 3110.1 minus the value at 1.5, which is 3113.5. Pay no attention to the fact that that's going to produce a negative result, because when we add it to 3113.5, it will just produce something that's lower than that, which is what we want. Many, many people would write this as 3113.5 minus 3110.1, which is incorrect, because we wrote 3 minus 1.5. Does that make sense? Anyway, then we are adding to that 3113.5, which gives us 3112.37. 3112.37. So we completed row F in the same way. We were just interpolating between pressures instead of interpolating between temperatures. And in fact, you're going to have to get proficient at interpolation between both and sometimes, even at the same time. To try that out, let's add a row G, shall we? On row G, let's imagine that I had given you a temperature of 340 and a pressure of 200. So if we were to go through the same procedure looking at the saturation temperature corresponding to our pressure, again, that saturation temperature corresponding to our pressure is going to come from double A3. I see that at a pressure of 2 bar, I have a saturation temperature of 120.2. Our temperature is higher than that. Surprise! It's a superheated vapor. So I'm going to jump back to our superheated water vapor tables. I'm going to scan for a pressure subtable corresponding to 2 bar. And alas, I do not have one. So I end up in the same situation as earlier. I have a pressure that is between 1.5 bar and 3 bar. But this time, I also have a temperature that is between 320 and 360. Meaning I have to interpolate for both temperature and pressure at the same time. So this is going to involve interpolating to produce values from which to interpolate between. I call this triple interpolation because you are doing three interpolations. You can either do it between pressure first and then between temperature. I call that the I method because it makes the shape of a capital letter I. Or you can do it between temperatures first and then between pressures, which means that you are producing a letter H. In either case though, you're going to have to produce two intermediate interpolations from which to interpolate between to produce your third interpolation, which is what you actually want. So again, you could do pressure first and then temperature or temperature first and then pressure. Pressure first and then temperature is easier to draw. So I will do that. So I'm going to interpolate for a value at 320 and 2 bar and a value at 360 and 2 bar. And then we are going to interpolate between those two interpolations to come up with our actual value. So you could think of this as we are building a temporary table. And this is our temporary table, which is at 2 bar. And on our temporary table, we have a column for temperature, a column for specific volume, and a column for specific enthalpy. And what I want are results at 320 and then 360 so that I can then interpolate for 340. So we have three interpolations for specific volume and three interpolations for specific enthalpy. Does that make sense? I'm doing this one first and then this one so that I can then do this. I'm going to do this one and then this one so that I can then do this. Okay, so our interpolation for 320 and 2 bar is going to be as follows. We are going to have 2 minus 1.5 divided by 3 minus 1.5 times the value at 3 bar for 320 degrees, which is 0.907 minus the value at 1.5 bar and 320 degrees, which would be 1.819. And then we are adding to that 1.819. And we get a specific volume of 1.515. Note that we are doing the same thing that we had done earlier, because we happen to have the same property as our intermediate stepping stone for this hypothetical problem that I just made up. So 1.515 is the value that we are slotting in here, 1.515. That's the value at 2 bar and 320 degrees Celsius. I have to repeat this process for 360 degrees Celsius. So I'm going to take 0.969 minus 1.943 and add that to 1.943 to come up with our value at 360 and 2 bar, which is 1.61833. And now on my little pressure subtable that I've created, I have enough to interpolate for a value at 340. So I can take 340 minus 320 divided by 360 minus 320 and multiply that by 1.61833 minus 1.515 and add that to 1.515. And I get 1.56667, 1.56667. One triple interpolation down, one more to go. I start by interpolating for a specific enthalpy at 2 bar and 320 degrees Celsius. So I will jump over here and it'll be as our third column. So we are taking 2 minus 1.5 divided by 3 minus 1.5 multiplied by 3110.1 minus 3113.5 plus 3113.5. And I get 3112.37, 3112.37. And then I repeat the process for a value at 360 degrees Celsius. So I'm going to take this relationship and plug in the values at 360 degrees Celsius, which is 3192.2 minus 3195.0 because again I'm grabbing 3 and then 1.5 regardless of the direction of the property. And then I'm going to add to that 3195.0. And I'm going to double check that I add plus up here and not minus because I said minus and I wasn't sure if that was a reason. So I get 3194.07. That's 3194.07. And again, I have created enough stepping stones to produce the interpolation that will actually give me what I want. I'm going to take 340 minus 320 divided by 360 minus 320. I'm going to multiply that by 3194.07 minus 3112.37 and add that 2 3112.37. And I get 3153.22, 3153.22. And these are the two values that I actually populate into my answer 3153.22, 3153.22 and 1.56667, 1.56667. And again, that was a superheated vapor. So that's triple interpolation. That is among the more difficult interpolations that you will have to be able to perform. And I will point out that there are in fact instances where it's actually worse than that. Like for example, you might be interpolating, well, let's see if I can come up with an example. What if you had a pressure of, let's try to compress like this instead. Okay. Let's imagine that you had 100 degrees Celsius and 10 bar. Okay. I don't think that I'm actually going to make us do this together right now because I'm going to probably save that for a separate time. But if we had 10 bar, which is 1000 kilopascals and 100 degrees Celsius, that state point is actually between our compressed liquid tables and our saturation tables. Let me show you. So if we look up the saturation temperature corresponding to our pressure, which would be the saturation temperature corresponding to 10 bar, I can see that our temperature at saturation is 179.9. And I'm going to compare that to the temperature that we just arbitrarily wrote down. That temperature that we have on row number H is 100. 100 is less than 179.9, which means that we have a compressed liquid. So if I jump over to my compressed liquid tables, which is table A5, and I scan for a pressure sub table of 10 bar, I see that I don't have a pressure sub table. In fact, I don't even have two pressure sub tables from which to interpolate between what that means is I have to interpolate between the 25 bar table and the saturated liquid property. Does that make sense? I mean, if you think about this visually again, if you picture this on a PB diagram, then this line of constant temperature represents the line of constant temperature at 100 degrees Celsius. So we have properties corresponding to different pressures up here, 25 bar, 50 bar, 75 bar, and that's all well and good. But our state point is right here. Meaning we have to interpolate between this point and this point to come up with our actual state point property for row H. And in fact, let's just do that. So what I'm going to do is generate the same style of triple interpolation. It's just that my interpolations are going to jump between tables. So I'm going to have an interpolation between the saturation pressure at 100 degrees Celsius and 25 bar. So my interpolation is going to be actually here, let me just copy and paste that PB diagram. So we can reference it. So we are going to interpolate in the same general way, we are going to build a temporary table, which I will do in the same general ways earlier. And we are building a sub table for 100 degrees Celsius. And we have pressure and specific volume and specific enthalpy. So we have properties at 25 bar. And we have properties at PSAT. That's PSAT at 100 degrees Celsius. And we are using those to generate properties at 10 bar, the saturation pressure corresponding to 100 degrees Celsius. By the way, we can grab from table A two. So that would be a pressure of 1.014 bar. 1.014 bar. And that is our stepping stone to get to 10 bar. So 10 is between 1 and 25. So we can interpolate for specific volume between the specific volume at saturation at 100 degrees Celsius, specifically the specific volume of a saturated liquid at 100 degrees Celsius, which is going to be VF at 100 degrees Celsius. And the specific enthalpy of a saturated liquid at 100 degrees Celsius. So let's go grab VF and HF now. VF at 100 degrees Celsius is 0.0010438. So I will populate that 0.0010438. No, 10435. And HF is 419.04. And those properties are the saturated liquid properties because that is the row for this guy right here. The other one is going to come from the 25 bar table. So I'm going to go into our compressed liquid tables, find 25 bar, which by the way is only 25 because it was the lowest pressure subtable appearing on the compressed liquid water tables. I don't know if I actually said that aloud. And we're grabbing V at 100 degrees Celsius and 25 bar, which is 0.001042310423. And specific enthalpy at 100 degrees Celsius and 25 bar 420.85. 420.85. And now we have everything we need to actually interpolate for our actual property. So we are going to take 10 minus 1.014 divided by 25 minus 1.014. And we are going to multiply that by the property at 25, which was 0.0010423 minus the property at 1.014, which was 0.0010435. We are adding that to 0.0010435. And we get 0.001043. We can complete the same interpolation for enthalpy, probably faster for me to just start this over, as opposed to scrolling up 1.014 multiplying by. So 420.85 minus 419.04 plus 419.04 yields an enthalpy of 419.718. 419.718. That gives me enough to populate my specific volume and specific enthalpy for row F. What are we on H now? Row H, excuse me, 419.718 and 0.001043. And that was a compressed liquid. So that's probably the most difficult interpolation conceptually when you're going between the compressed liquid region and the saturated liquid properties. That's for when you have properties of a compressed liquid at a pressure that is lowest than your pressure subtable. And if I may make an observation while we are here, you'll notice that the properties don't change much. And if I may make an observation while we are here, it's relatively common practice to assume that in the compressed liquid region, especially for pressures that are less than the lowest pressure subtable, that pressure does not affect the property. So if you were to look up the property at the correct temperature with the wrong pressure, like for example VF at T, that that would be good enough to use, or HF at T. So if we looked up the VF property at 100 degrees Celsius, what we would have gotten is 0.0010435. And the logic here is that it is not uncommon for people to say, well, 0.001043 is just close enough to 0.0010435, then I can call them the same. And is that reasonable? I mean, depends entirely on the context. Probably because linear interpolation is still error prone, we are going to end up with a less accurate result than if we had used a better form of interpolation, like something that we generate with MATLAB. So it's probably not unreasonable to say, eh, we're approximating anyway, let's just approximate a little bit more by assuming it doesn't change at all with pressure and only changes as a function of temperature. So that would probably be good enough, but I expect you to be able to actually handle the interpolation if you need to. By the way, that assumption comes from if you look at the pressure subtables, you can see that the properties change a lot with temperature, they don't change much with pressure. So going from the 25 bar table at 20 degrees Celsius, compared to the 50 bar table at 20 degrees Celsius is not a huge change in properties when you compare 20 degrees Celsius to 40 degrees Celsius or 80 degrees Celsius on a single table. The 86.3 is very different from 169.77, but 86.3 is not very different from 88.65. That's the logic. And 25 bar is a huge pressure change. Okay. Now I realized that we have gone down a rabbit hole of stuff in this problem that I didn't actually ask for, but let me just add one more thing. What if I had asked you to plot these state point properties on, let's say, a PV diagram? And I wanted you to plot these relative to the saturation lines. So I am expecting state point positioning that is accurate relative to the saturation lines and relative to each other, but not necessarily perfectly accurate in terms of scale. So this is a PV diagram. And what I'm asking for is a plot of the state points A through H indicated on this PV diagram. So what I'm going to do is start by drawing a bunch of lines of constant temperature, a line of constant temperature on a PV diagram goes down, correct because of pit bull and Kesha and stuff. So I'm going to draw a line of constant temperature here and here and let's say here. And that's going to correspond to 340, 140 and let's say 30-ish. So this is 340, this is 120, let's say, and this is 20. And again, those are the lines of constant temperature corresponding to those temperatures. So state point A is at 140 degrees Celsius and 10,000 kilopascals. So state A is going to be at a temperature that is a little bit higher than our line of constant temperature for 120, but it's going to be like way up here. It's way up into the left relative to our dome. B is at 100 degrees Celsius and 100 kilopascals and is a saturated liquid. So 100 degrees Celsius would be a little bit lower than 120. So I'm going to indicate state point B here. State point C was at a temperature of 20 degrees Celsius, which we happen to have a line of constant temperature for and was under the dome because it was a saturated liquid-vapor mixture. So I'm going to place it on the 20-degree Celsius line of constant temperature and I'm going to place it somewhere under the dome. But where to place it? Well, we know where to place it. We place it 78% of the way across the dome because we know the quality of that state point was 78.1%. Then state point D is a saturated vapor at 30 degrees Celsius, which is going to be just a little bit higher. Call it here. So if I wanted to, I could draw a whole bunch more lines of constant temperature, but I think it would be a little bit overwhelming on the spot right now. State point E is a superheated vapor at 340 degrees Celsius and 300 kilopascals. So that's going to be on our top line of constant temperature and to the right of our dome. And we can position it relative to B, C, and D by looking at that pressure corresponding to those pressures. So those were at pressures of, say, 4 kilopascals or 100 kilopascals. This scale isn't particularly good. This would be like 3 kilopascals. This would be 4 kilopascals and this is 100 kilopascals, like 3, 4, 100. Great scale there, John. And I am saying E is at 300. Therefore, I will place E maybe over here because it should be higher than B because it's higher than 100. But it should not be as high as the saturation pressure corresponding to 340 degrees Celsius. And then F, G and H, I can follow the same logic. I have a slightly lower pressure at a slightly lower temperature for F. So I could place that, say, over here somewhere. And in fact, I can place it better by looking at the specific volume, which is our x-axis. F compared to E has a slightly higher specific volume, so I really should place it over here. And then G was right in that same neighborhood. Should be up and to the right, just a Scotch here. And H was 100 degrees Celsius and N bar, which is 1000 kilopascals, which should be 100 degrees Celsius. I'll put it right about here. And that's the relative position of all of these state points on a PV diagram. It helps me to try to visualize the state points relative to a PV and TV diagram because that allows you to deduce some things about what direction things are going to occur. I mean, remember that we had integrated pressure with respect to specific volume to generate specific boundary work. Therefore, the area under a curve as drawn on a PV diagram is going to indicate boundary work. And if it's a region from left to right that represents an expansion process, and if it's a region that goes from right to left, that would be a compression process. Therefore, I can say on a PV diagram, horizontal displacement to the right represents workout. And horizontal displacement to the left represents work in. So if you had a process that went from C to B, so for some reason, we had a process that went from 20 degrees Celsius and 2.339 kilopascals to 99.63 degrees Celsius and 100 kilopascals, you know that that's going to involve a work in term, because visually, you're going from the right to the left, meaning you have horizontal displacement to the left, which means that you must have a compression process. I don't know. I'm a visual learner. So I like diagrams. It helps me keep track of things a little bit more clearly. And it also helps when you have cycle analysis with multiple state points and multiple processes. You can look at a cycle and get a good idea of what's happening between each process because you can see horizontal displacement appearing on the PV diagram. I think that wraps up everything I wanted to pack into this example problem.