 Find the internal energy change between the following states of water. At state 1, we have water at a temperature of 200 degrees celsius and 100 kilopascals. At state 2, we have a temperature of 150 degrees celsius and 5000 kilopascals. As a general rule, you want to begin by trying to fix your state points. Remember that it takes two independent intensive properties to fully define a state point. At our state 1 conditions, our two independent intensive properties which fully define that state point are temperature and pressure. From those two properties, I can use the tables to determine whatever else I want, including, but not limited to, internal energy. At state 2, the two independent intensive properties which fully define our state point are temperature and pressure. From those two independent intensive properties, we can fully define state 2 and determine, among other things, the internal energy. So we have enough information to determine u1 and u2, and what the problem wants us to do is actually determine the change in internal energy, which is just going to be u2 and u1. Therefore, our approach will be use temperature and pressure to determine u1, use temperature and pressure to determine u2, and subtract them. And for the internal energy of our water, we are going to turn to the appendix of our textbook. In our appendices, we have properties for water in the liquid and vapor phases, but in order to try to present that information in a way that is somehow manageable in two dimensions, the tables for those properties are split by phase. On table A4, we have the properties of water when it is a superheated vapor. On table A5, we have the properties of water when it is a compressed liquid. On tables A2 and A3, we have the properties of the water when it is either a saturated liquid, a saturated vapor, or a mixture of the two, a saturated liquid vapor mixture. The only difference between A2 and A3 is one is listed by temperature in even increments of temperature, the other is listed by pressure in even increments of pressure, and that is just a matter of convenience based on what you have driving your property lookup. So it's going to be up to us to determine the phase of state 1 and state 2 before we can go look up the properties. And for that, we are going to turn to our PV and TV diagrams. Remember that on a PV diagram, the saturation lines form a dome shape, and a line of constant temperature on the PV diagram goes down into the right because of Kesha and Pitbull, or whatever it is that you use to remember which one is which. This line of constant temperature is an arbitrary line of constant temperature it could refer to any temperature, like for example 200 degrees Celsius. Then this horizontal region in the middle going across the dome is the saturation pressure corresponding to that temperature. And I'm going to draw this as a slightly better horizontal line. And we can look up that pressure corresponding to a temperature on our saturation tables. Remember that we have both a listing of those properties by pressure and a listing of those properties by temperature, and the only difference between the two is which one is most convenient. So if I wanted to, I could look up the saturation pressure corresponding to my temperature of 200 degrees Celsius and place my phase based on if my pressure 100 kilopascals is higher or lower than the saturation pressure corresponding to that temperature. Similarly, we can look at a TV diagram on which the dome appears approximately the same and a line of constant pressure goes up into the right because folks are always trying to turn the TV up or whatever and that horizontal jut across the dome represents our saturation temperature or boiling point. That saturation temperature can be looked up for a given pressure in the saturation tables and we can place our phase based on if our temperature is greater than or lower than that temperature. So we have two ways to approach fixing the phase for stay one. And in the interest of character building, let's actually try it both ways. So at state one, if I wanted to start by determining the saturation temperature corresponding to my pressure, I would have to go into my saturation tables in order of pressure. That would be table a three. So I jump over to table a three. And I see that I have saturation conditions corresponding to saturated liquid properties and saturated vapor properties for a given pressure. And those properties include specific volume internal energy enthalpy and entropy. The property that is not distinguished between a saturated liquid and a saturated vapor is temperature. Because the temperature corresponding to saturation for a given pressure stays the same the whole time. These lines going across my dome are horizontal. So the saturation temperature corresponding to a pressure is the same across the entire dome. So if I wanted to find a hundred kilopascals, I would have to convert 100 kilopascals to bar because this table happens to be in bar. The conversion between kilopascals and bar is conveniently up here on this little sticky note, which appears on each of these tables. 10 to the second kilopascals or 100 kilopascals is equal to one bar. So how many bar are in 100 kilopascals? That's right, one. So if I find one bar here and I look up the temperature corresponding to that pressure, I can see that the saturation temperature is 99.63. So I will write that as follows. The saturation temperature corresponding to P1 is 99.63 degrees Celsius. So that means if this line of constant pressure represented 100 kilopascals, then the horizontal section of this line that is the line going across the dome occurs at 99.63 degrees Celsius. My temperature at state one is higher than that. Therefore, I have to be in the superheated vapor region because the part of that line of constant pressure that is higher than the saturation temperature is up here in the superheated vapor region. So the way I would write that is T1 is greater than T sat at P1. Therefore, I have a superheated vapor. And if I were to approach this the other way, I would look up the saturation pressure corresponding to my temperature. That would be a saturation pressure PSAT corresponding to a temperature at T1. And I would get that from table A2 because table A2 contains the properties of saturated water in order of temperature in even increments of temperature. So if I jump into table A2 and I find our temperature, which is 200 degrees Celsius, I can see that the saturation pressure corresponding to that temperature is 15.54 bar. Now how many kilopascals is 15.54 bar? That's right. It's 1554 because there are 100 kilopascals in each bar. So we multiply by 100 and we get 1554. So the saturation pressure corresponding to our temperature as per table A2 is 1554 kilopascals. That means if this line of constant temperature represented 200 degrees Celsius, then the horizontal section of that line, the saturation pressure occurs at 1554 kilopascals. Our pressure is lower than that, which means I have to be in this region over here. The region to the right of the saturation dome is our superheated vapor region. Therefore, I can write T1 is less than PSAT at T1. Therefore, we have a superheated vapor. Either route would have worked. And in fact, you should be familiar with both, because in some cases, you won't have the option or the luxury of using either or. You will only have one. And you have to make sure that you are proficient at both strings of logic that allow you to conclude that it was a superheated vapor. Anyway, with either route, we know that it is a superheated vapor at which point we can go into our superheated vapor tables. Our superheated vapor tables occur on A4 for water, so I can jump over to table A4. And I see a whole bunch of smaller tables. We liked tables so much that they gave us tables within our tables. These tables are cross sections of a three dimensional surface. We can't present three dimensional data on a two dimensional sheet of paper. So we cut it into cross sections and each of them gets their own little sub table. And each of these sub tables is for a certain pressure. So what we have to do is find the pressure sub table corresponding to 100 kPa. Remember that 100 kPa is equal to one bar, so we need to find the one bar sub table. I see that it is right here. From the pressure sub table, we can read off specific volume, specific internal energy, specific enthalpy, and specific entropy corresponding to certain temperatures. Note that the top row here is the property at saturation, which is the property right here. So if you imagine that this line of constant pressure represented one bar or 100 kPa, this table is giving me properties within this region in even increments of temperature. So it's telling us the property here and the property here and the property here and it goes all the way down to where it intersects the saturated vapor line. So that's the lowest value we have. Also remember that that's going to be the same property as if we look up the saturated vapor properties on our saturated liquid vapor mixture table. Even though the data is split across multiple tables, it's all the same. It's all properties of water. They aren't as distinct and far away from one another as they appear to be. So the property here could be determined from either table. Does that make sense? Okay. I'll get rid of all those drawings and we will jump back into this table. Again, the pressure sub table corresponding to one bar is what I want and I want internal energy, which is the second column and I want it at 200 degrees Celsius. Therefore, the internal energy at state one is 2,658.1 kJ per kilogram. And with that, we are halfway there. Next, we have to repeat the process for state two and the general principle is going to be the same. We are going to look up the saturation temperature corresponding to our pressure and use that saturation temperature as compared to our temperature to fix the phase or we could look up the saturation pressure corresponding to our temperature and compare our pressure to that saturation pressure to fix the phase. Which way would you like to do it? I agree. Both is better. Let's start by looking up the saturation temperature corresponding to our pressure. Our pressure is 5000 kilopascals. How many bar is 5000 kilopascals? Correct. It is 50. 50 bar because 100 kilopascals is equal to one bar. Therefore, I want to find the saturation temperature corresponding to a pressure of 50 bar, which I will do by jumping back to our saturation properties listed by pressure, which was on table A3. So 50 bar is down here and I see that the saturation temperature corresponding to that pressure is 264. My temperature at state two is 150 degrees Celsius. That's less than the saturation temperature corresponding to our pressure. What does that tell us about the phase? Correct. It is a compressed liquid. We know that because if we zoom in here onto a line of constant pressure on our TV diagram, that line of constant temperature across the middle of the dome is occurring at 264 degrees Celsius. The only way that we can be on that line and at a position that is lower than that temperature is to be to the left of the dome. Therefore, our phase is a compressed liquid. We also could have gotten there by looking up the saturation pressure corresponding to our temperature. That temperature is 150 degrees and I can find that on table A2. On table A2, I see that the saturation pressure corresponding to 150 degrees Celsius is 4.758 bar. 4.758 bar would be 475.8 kilopascals and I see that my pressure is higher than that. The only way that we can be at a higher pressure than PSAT for our temperature is to be to the left of the dome, which is a compressed liquid. So in either way, we end up with the conclusion that our phase is a compressed liquid. And when we use our compressed liquid property tables, we will see that we have, again, that's table A5. A5 is properties of a compressed liquid. On our compressed liquid tables, we see we have pressure sub-tables again in even steps by pressure. And what we want to do is find the pressure sub-table corresponding to our pressure, which was 50 bar. The 50 bar sub-table is up here. And what I want is the internal energy corresponding to a temperature of 150 degrees Celsius. So I'm going to be looking at this column here and I'm going to be finding 150 degrees Celsius. And I see 150 doesn't appear on this list. We have the property at 140 and we have the property at 180, but we don't have 150. So what we need to do is interpolate between 140 and 180. For our purposes, we are going to interpolate by using linear interpolation every time. It doesn't matter if it's the correct application for linear interpolation. When we are solving these properties by hand, we are using linear interpolation anytime we interpolate. Once you start getting into the habit of solving these problems with MATLAB, you can choose the interpolation style that seems to make the most sense and come up with a better curve fit than just connecting the dots with a line. But right now, connecting the dots with a line is way easier to do by hand. And let's take a moment to review what linear interpolation represents. I will do that on a new sheet of paper. So what we are doing is taking two known data points and connecting them with a line. So right now we are, let's say, using x to look up y. And we have a data point here and a data point here. That's x1 and x2, let's say. So in our case, that would be a temperature of 140 and 180. And what we actually need is something in between, say at 150 degrees Celsius. And we have y values corresponding to x1 and x2. But we don't have a y value corresponding to our x value. So what we do is we connect these two dots with a line. And we can use our understanding of that line to predict what the y value would be at that intersect. Now you could use y is equal to mx plus b, figure out the slope by using the rise y2 minus y1 over the run x2 minus x1, figure out the y intercept of that line and then use the equation of that line to determine a y value for a given x. But it's much easier to use similar triangles here. So I'm going to switch to blue. And we are saying this triangle here is going to have the same proportions as this triangle here. We are using similar triangles to make the math easier. What this means is this distance here, let's see, maybe green makes the most sense. This distance here divided by this distance here is going to be the same proportion as this distance here divided by this distance here, because the triangles have the same proportions. So I'm saying x minus x1, that's this distance here divided by x2 minus x1 is the same as y minus y1. This distance here divided by y2 minus y1 here. Again, what we're doing is we're saying the proportion of this region over this region is the same as this region over this region. Or we are saying this proportion of this distance is the same as this proportion of this distance. Either way, you end up with the same relationship. It's either x minus x1 divided by y minus y1 is equal to x2 minus x1 divided by y2 minus y1. Or x minus x1 divided by x2 minus x1 is equal to y minus y1 divided by y2 minus y1. This is a long-winded way of saying formally, if we are halfway between x1 and x2, we must also be halfway between y1 and y2. Or if we are a quarter of the way between x1 and x2, we must also be a quarter of the way between y1 and y2. That is the fundamental principle of linear interpolation. And using this information, we can approximate a y value by taking x minus x1 divided by x2 minus x1, multiplying it by the difference between y2 and y1, and then adding y1. So in our actual data, our x value is temperature and our y value is internal energy. So what I want to do is interpolate between 140 and 180 to come up with a y value and internal energy at 150. So let's try that with our calculator. What I'm going to do is take 150 minus 140 divided by 180 minus 140, and I'm going to multiply that entire proportion by the difference between y2 minus y1, which would be the property at 180, which is 759.63, minus the property at 140, which would be 586.76. And we are adding that result to 586.76. And we can include an approximate value for our internal energy at 50 bar and 150 degrees Celsius is 629.978. Therefore, y2 is 629.978. And from our earlier lookup, we had 2658.1. And now that we have both of those properties looked up, we can subtract in the internal energy. We can say 629.978 minus 2658.1 is negative 2028.12 kilojoules per kilogram. And that represents the difference in our specific internal energy, which presumably we would be plugging into our energy balance to determine something else about this process. I will point out that the property lookups are one tiny step in our workout problems. They are not going to be an actual workout problem in and of themselves, except for right now when we are learning how to look up properties. You should be good at the property lookup tables. You should be so good at them that you don't have to dedicate cognitive load to them. You should be so practiced at these that once you get to a point in a workout problem that you need a property, you can just go look it up without having to waste mental real estate on it. In order to get good at that, you are going to have to practice. So let's practice some more.