 A piston-cylinder device contains 0.005 cubic meters of liquid water and 0.9 cubic meters of water vapor in equilibrium at 600 kPa. Heat is transferred at constant pressure until the temperature reaches 200 degrees Celsius. Determine the following. A. The initial temperature of the water. B. The total mass of the water. C. The final volume. D. The amount of work done in the process. E. The amount of heat transferred during the process. And F. Show the process on a PV diagram with respect to saturation lines. I will begin with a diagram of our problem. And then I will populate what I know about our beginning and end state points, which I'm calling state 1 and state 2 for now. At state 1 I know a pressure and a volume. Because I know that we have a combination of water vapor and liquid water, and they are coexisting. The only way that they can coexist in equilibrium would be for them to be at saturation conditions. Otherwise we wouldn't meet phase equilibrium. Which means that we must be at saturation conditions. And I will write that as we have a volume of saturated liquid and a volume of saturated vapor. Of 0.005 cubic meters and 0.9 cubic meters. Respectively. And then that means we have a total volume at state 1 of 0.905. Because the only things that are in our system right now are liquid water and vaporous water. Meaning at state 1 we have total volume and we have pressure. We also know that it's at saturation conditions. Then at state 2 we know that we had an isobaric process. So P2 is equal to P1, which we know. And we know T2, which is 200 degrees Celsius. So the logic here is this. The two independent intensive, I wrote P1 to too much weight. I'm going to back that off. I have to, it has to match guys. P1 was 600 kilobascals. Try that again. So the logic here is as follows. We have two independent intensive properties at state 2 that fully define state 2. At state 1 it's a little bit more complicated but not much. We know that we have a pressure and we know that it's at saturation conditions. And we know that we have a total volume, which is made up of a volume of a saturated liquid and a volume of a saturated vapor. We can use those two volumes along with the specific volume of a saturated liquid and a specific volume of a saturated vapor at 600 kilobascals at 6 bar to determine the mass of liquid and the mass of vapor that are in the problem. So this is a long-winded way of saying the two independent intensive properties that fully define state 1 are pressure and quality. Because once I know the masses, the mass of vapor divided by the mass of the mixture, which is mass of vapor plus mass of liquid, that's going to give me the quality at state 1. From the quality at state 1, I can interpolate any other properties that I need and we will be made in the shade. But to get there, we will first have to look up Vf at 6 bar and Vg at 6 bar. Then I can calculate quality. Then I can use quality to interpolate four properties that I need, including but not limited to temperature. So I will actually leave that quality calculation there and we will vaguely write get to mass and then quality here and then pressure and quality and get us to our lookups, which we want t1 and presumably other stuff too. Total mass of the water, we will have calculated once we figure out the mass of vapor and the mass of liquid, which I will write as Mf and Mg. Then Mf plus Mg, the mass of the vapor plus the mass of the liquid, is going to give us the total mass, which is part b. In part c, I can determine the total volume because I have enough information at state 2 to look up anything I want. I will be looking up specific volume 2, because at this point I will have mass 2, because mass 2 is presumably the same as mass 1. We can assume that if we call this a closed system. Then M2 being equal to M1 means we will have M2 and then M2 and specific volume 2 will allow us to calculate the total volume at state 2. So a is just looked up, b is M is equal to M1, which is equal to mass of liquid plus mass of vapor, which is Mf plus Mg. And then for part c, total volume at state 2 is mass 2 times little volume 2, which is equal to mass, which is what I'm calling the overall mass in the system, times specific volume 2. Okay. Then for part d, I want to know the total amount of work done in the process. Well, we only have one type of work occurring. It is boundary work, which means I'm going to calculate a boundary work. Remember that the total boundary work is going to be the integral of pressure with respect to volume. I know the pressure is constant because we have an isobaric process, so I will bring the pressure out of the integral and I will have pressure times. The integral of dv from v1 to v2, which is just going to be v2 minus v1. And I could write that in terms of other stuff, like for example mass times specific volume 2 minus specific volume 1, but at this point I will have the total volume at state 2 and I will also have the total volume at state 1 because I already know it is 0.905. So it's just as easy to leave it as total volume. Then for part e, I need to relate that work to heat transfer and the changes of energy inside of our system. For that, I need to perform an energy balance. So I'm going to define our system as the mass of water. It's both mass of vapor and mass of liquid water. That is my system, which is a control mass because it is a closed system. And that energy balance is going to start as they all do with delta e is equal to energy in minus energy out. I have a closed system, which means en could be q in and work in and e out could be q out plus work out. But for the left-hand side of the equation, we have to consider whether or not it's reasonable to assume steady state. Do you think we should treat this as a transient process or a steady process? We should treat this as a transient process. We can't neglect time because there are properties changing with respect to time. For example, the volume of our water changes from state 1 to state 2. There's going to be a change in volume that's a result of the fact that the temperature is changing. We have whatever the saturation temperature is at 6 bar. That's our T1, by the way, because it's at saturation conditions. So if I open up our tables and look up the saturation temperature at 6 bar, I can see that it is 158.9 degrees Celsius. Therefore we're going from 158.9 degrees Celsius to 200 degrees Celsius. That means at least one property of our system is changing over time, which means that we must treat this as a transient process. We can't neglect the effects of time. So if it's a transient process, that means the left-hand side of our equation could be delta u, could be delta ke, and could be delta pe. And I can neglect some of these terms by making some assumptions. First of all, I will assume that the only work is boundary work, is only work. Then I will assume q is only in. That is, we are only considering heat transfer in the inward direction. For there to be a heat transfer, there must be a temperature difference, which means that we must have a higher temperature on the outside than on the inside driving a heat transfer in the inward direction. For there to also be a heat transfer in the outward direction, we would have to have a complicated enough setup to warrant the temperature being some sort of complicated relationship on one side of the piston cylinder versus the other side of the piston cylinder, something like that. Whatever the case, we are assuming that there's only one type of heat transfer, it is q in. Then we can neglect q out and we've neglected work in because our boundary work is in the outward direction. I mean, we presumably have an expansion process because we are heating it up, but I will point out that even if we had a contraction process, as long as we plug in our boundary work as we calculated it, it is a workout because a negative workout is going to be a positive workout. Then I was told nothing about any significant changes in velocity or elevation. So I'm going to neglect changes in kinetic and potential energy. I will say changes in kinetic energy are small changes in potential energy are small. At which point we have delta u is equal to q in minus workout q in is what I'm looking for. So I'll write this as q in is equal to delta u plus workout and we had total u2 minus total u1 is our delta because we're analyzing from state one to state two and our boundary work was our workout put which was pressure times v2 minus v1. And just like in the previous problem, I could group together internal energy and pressure times volume two and internal energy plus pressure times volume one because p1 and p2 are the same. When I bring p inside the parentheses, it could be written as either state one or state two and substitute in h2 for u2 plus p2 v2 and h1 for u1 and p1 v1. This doesn't look much like a u. Try that again. u, that's better. And then I could factor out the mass because it's the same and write mass times little h2 minus little h1. I can totally do that. That's the convenience of enthalpy. It appears all the time. It allows us to just look up one quantity instead of looking up all three of those quantities or calculating a result from those quantities. But since we did this in the previous example problem and because I want you to get exposed to solving these problems in different ways, let's not do that. Not because we can't, not because we shouldn't, because we actually should. That'd be our best practice here. But because I want to do things a little bit differently this time in the interests of education. So let's instead write this as mass times little u2 minus little u1. And I can look those up, you know, for the same amount of effort as looking up h1 and h2 and then multiplying by, excuse me, and then adding pressure times v2 minus v1, which is honestly easier for me to just write the result of part d. Anyway, mass we will have because we calculated it up here. u1 and u2, we can look up. So I will add that to my list of stuff to look up u1 and u2. And I think that'll be everything that I need. Finishing the problem is just a matter of computation. So I've worked through my entire analysis symbolically without plugging in any numbers, without actually computing anything. I encourage you to get into this habit, because it allows you to simplify your analysis ahead of time. I mean, like this whole enthalpy substitution thing, it also allows you to only look up what you actually need to solve the problem. And if you make a mistake in your outline, I mean, if you had gone down a rabbit hole and realized, ah, that approach doesn't work, it's better to have realized that before you wasted all the time calculating stuff that you might not need. So this is the way in which I would encourage you to work through these problems. And once you get experienced enough at this, you won't even need to write out the outline. You'll be able to just think through it and make sure that it's going to work before you even start putting your pencil to your paper. You'll see, you'll get very experienced at thermal problems, because you're going to get a lot of experience at thermal problems. It'll be a good time. But until then, let's calculate some numbers, shall we? First up, I want to look up VF and VG at 6 bar. And that is in the service of calculating a mass. I'm going to move those down there, and I'm going to go into our saturation tables by pressure because I'm driving the lookup with 6 bar, which is my pressure at state 1. So in our tables, we can see that we have VF, or rather this column, which is VF times 10 to the third, and VG. Therefore, VF at 6 bar is 0.0010, no, 1006. 0.001106. And VG, which was 0.3157. And then MF is equal to the volume of the liquid at state 1 divided by the specific volume of a saturated liquid, because this is saturated at state 1. And if I'm taking a quantity in cubic meters and dividing by a quantity in cubic meters per kilogram, I'm going to get a result in kilograms. I'm going to skip writing that out and canceling the units, because I'm assuming you guys are able to follow that, and I'm not going to waste the space on that. I will also point out the further we get into property lookups, the fewer of these steps I'm going to show you, because I will be assuming that you guys are getting pretty proficient at this. But I need my calculator, and I need to divide some numbers. That would be 0.005 divided by 0.001106, giving me 4.54 kilograms. MG, which is the volume of the vapor divided by the volume, specific volume of a saturated vapor at 6 bar, because our vapor is a saturated vapor at 6 bar. So this is going to be 0.9 divided by 0.3157, and we get 2.85. And when I add these together, I get the total mass at state 1, which is the mass everywhere, which is like 7 and change. Add a plus sign there, John. Getting a little bit ahead of yourself. 7.39378, 7.39378. Cool. Isn't that interesting? Even though we had way more volume of vapor than we had volume of liquid, we have more mass of liquid than we have mass of vapor. That's why it's important to remember that quality is a mass proportion. It's mass of vapor divided by mass of mixture, not a volume proportion. A very common mistake among Thermo1 students at this point in the analysis would have been to take 0.9 divided by 0.905, yielding a quality of like, what, 0.96, something like that? And that's very different from what we get if we take a mass proportion. The proportions of volume and the proportion of mass are only the same if all of the densities are the same. Anyway, from this, I can calculate a quality. So x1 is going to be the mass of vapor at state one divided by the mass of the mixture, which is just M1. And that would be 2.85 kilograms divided by 7.39378. And here let me use all of those trailing decimal places just to be consistent. There we go. So 0.385568. And I didn't necessarily need that number directly. I mean, we just really need the proportion to plug into our interpolations down the road or whatever. But you know, good character building. And hypothetically, this might be useful when we are graphing this, you know, in about 20 minutes. Now that I have the quality at state one, I have my other independent intensive property, from which I can actually proceed to interpolating whatever it is that I need at state one, which is U1. I could have grabbed T1 already, because I know T1 is just going to be T sat at P1, which is going to be the saturation temperature corresponding to 600 kilopascals, which was going calculator 158.9. I can write 158.9 degrees Celsius. And I can populate that down here. Did I want T1? No, I wanted T1. What am I doing? A. That was for part A. I didn't become temporarily Canadian. 158.9. Part eight done. Total mass of the water. We got that already. That was seven point. Come back calculator. 39378 kilograms. Part B done. Part C, I wanted the total volume at state two, which requires that we look up specific volume at state two. So let's just look up U1, V2, and U2 all at the same time, because we will need them all eventually. And if you are in the alternative universe where we had plugged in H1 and H2 and our energy balance instead of U1 and U2 and the boundary work, then you would look up H1 and H2 at this point. But I want U1. So that's going to be the quality of state one multiplied by UG minus UF, because quality is defined among other things as the proportion of the way you are across the dome, which we assume is the same for all of the properties. And that is U1 minus UF divided by UG minus UF. Again, this doesn't look much like a U. Let's try that again. U1. Therefore, U1 is going to equal X1 times UG. Man, this looks kind of like an H. What is with those U's? You know, the U's today. Always up to shenanigans plus UF. There we go. So quality at one we know UG and UF we can grab from this same table. Go ahead calculator. I see that at six bar UF is 669.9 and UG is 2567.4. So if the calculator comes back and I am going to highlight those numbers so that I don't lose track of which ones they are and that is not what I wanted to do. Oh gosh, oh gosh, oh gosh, right here. Okay, scroll that down just a little bit. UF and UG. There we go. So we are going to take our calculator again and take 0.385568 times the quantity 2567.4 minus 669.9 and add that to 669.9. And we get 1,401.52. 1,401.92. And that was in kilojoules per kilogram. And now that we are done with this, we don't need it anymore. I will also point out that this interpolation also works for V1 minus VF over VG minus VEF. And that is the same as H1 minus HF divided by HG minus HF and S1 minus SF divided by SG minus SF. But we don't need those. We have everything we need almost. We need V2 and U2. So for that we are going to use the fact that we have a pressure and a temperature to look up two properties. Luckily we are experienced at this because we've been working our property lookup drills. For a pressure and temperature we fix the phase first by looking up the saturation condition corresponding to one of our properties and comparing the other property to that condition. So we can look up the saturation pressure corresponding to 200 degrees Celsius or we can look up the saturation temperature corresponding to 6 bar. I am going to look up the saturation temperature corresponding to 6 bar because I've already done it. Look at that it's 158.9. I see that 200 is greater than 158.9. Therefore I must have a superheated vapor. So I'm going to jump into our superheated vapor tables and I'm going to find a pressure subtable corresponding to 6 bar. So 6, 1.53. I see that 6 bar does not have its own pressure subtable. Alas, instead I'm going to have to interpolate. So I'm going to be interpolating between 200 degrees Celsius at 5 bar and 200 degrees Celsius and 7 bar. So that interpolation is going to go 6 minus 5 divided by 7 minus 5 is equal to U2 minus 264 2.9 divided by 263 4.8 minus 264 2.9. Solving that for U2 I would get U2 is equal to the quantity 6 minus 5 divided by 7 minus 5 close quantity times quantity 263 4.8 minus 264 2.9 close quantity plus 264 2.9. And for V2 that would be quantity 6 minus 5 divided by 7 minus 5 close quantity times the quantity 0.299999. 0.2999 minus 0.4249 close quantity plus 0.4249. When I'm interpolating for things like this from now on I will actually write it in terms of the proportion that we're using to drive the interpolation and I will use my calculator to solve it. I think that will make it easier for you guys to follow when I'm not writing down the intermediate step. So 6 minus 5 divided by 7 minus 5 and we're saying that proportion is equal to the thing that I want which is actually U1 here but I'm going to plug in x because it's convenient on my calculator minus 264 2.9 which is the value at 5 bar divided by the value at 7 bar which is 263 4.8 minus 264 2.9. And just like the previous time we talked about this don't overthink the fact that the value at 7 bar is less than the value at 5 bar. It's okay if we write a delta as a negative quantity because we're going to have a negative quantity over a negative quantity which is still going to be a positive proportion. It's fine. Don't overthink it. If you overthink it you will often get it wrong. And I get 263 8.85 and that was U1 so I can write that down over here. U1 is 263 8.85 and for V2 I'm going to repeat that process solve that would be 6 minus 5 divided by 7 minus 5 which is halfway between 0.4249 and 0.2999 and that is going to be x minus 0.4249 divided by 0.2999 minus 0.4249 and we get 0.335614. And with those quantities I can calculate the total volume at state 2 so I'm going to take the mass we had just had which is 7.39378 kilograms multiply that by the specific volume at state 2 which is 0.335614. That's a quantity in kilograms times a quantity in cubic meters per kilogram. The kilograms are going to cancel leaving me with a quantity in cubic meters so that was 7.39378 which I'm going to grab for some reason with all the decimal places and then I'm multiplying by 0.33614 etc and I get a total volume at state 2 of 2.48146 cubic meters. And that volume at state 2 is larger than the volume at state 1 which makes sense because we would expect an expansion process because we are heating up water and things tend to expand as they are heated especially considering that we had a combination of liquid and vapor and we evaporated all the liquid and vapor tends to take up more space than liquid so we should have expanded. Anyway now that we have both volumes I can proceed to calculating a boundary work this is going to be 600 kilopascals multiplied by a difference in volumes which is going to be 2.48 ish minus the volume at state 1 which was 0.905 and that was in cubic meters and I presumably want a quantity in kilojoules indeed I do so I will write a kilopascal as a kilonewton excuse me I'm getting a little bit ahead of myself again I encourage you guys to start at your destination and work backwards a kilonewt is a kilonewton times a meter and then a kilopascal can be written as a kilonewton per square meter kilonewton cancels kilonewtons kilonewtons cancels kilonewtons not kilonewtons cubic meters cancels meters and square meters kilopascals cancel kilopascals leaving me with kilonewtons therefore my boundary work is just going to be 600 times this quantity minus 0.905 and that's 945 work out 945.876 kilojoules then for part e I'm going to take our mass which we know and I'm going to multiply by a difference in internal energies and I'm going to add to that the boundary work we just calculated so in this case it doesn't make sense to plug it in as pressure times difference in volumes because we already computed that we'll scooch this down a little bit beginning with our mass which was 7.3937 kilograms and then we are multiplying by u2-u1 which is 2638.852638.85 minus 1401.92 92 he said hoping he remembered correctly and he did and that would be kilojoules per kilogram kilograms will cancel kilograms leaving me with kilojoules and then I'm adding to that the work out term which was 945.876 so what we're saying here is the heat end is being split some of it is going into the internal energy change of our water the rest of it is going into moving the piston up so if we add those two requirements together we will figure out how much total heat had to have been added so 7.3937 again I will scroll up for pointless precision on that number multiplied by 2638.85 should I grab the number with all the decimal places I mean probably not it didn't matter because it was just two decimal places anyway minus 1401.92 I'm not even going to scroll up for that and then we are adding to that quantity the boundary work we just calculated and we get 10,091.5 so we require 10 megajoules to accomplish this process that is 10 megajoules of heat has to be added to our piston cylinder arrangement to accomplish the process about 9 megajoules are going into the internal energy of the water about one is going into moving the piston up the last thing I asked for hold on let me check off these boxes to be consistent the last thing I asked for was a plot of this on a PB diagram relative to the saturation lines for that I will open up a new sheet of paper and remember that the dome looks like this on the left this on the right that's not exactly to scale but it's going to be good enough for now because all I'm trying to do is indicate approximate positions and I'm going to go about plotting this by first starting with our two temperatures we started at a temperature of was it 158.9 and ending at a temperature of 200 so by drawing our lines of constant temperature on the tv diagram I can define approximately where the state points are going to be because I know it has to be on that line of constant temperature and then I can use the relation to the saturation lines to indicate approximate position on a pv diagram remember that a line of constant temperature goes down because we're yelling timber because of pit bull or whatever mechanism you use to remember which direction the lines go on which diagrams furthermore remember that the saturation pressure increases as a function of temperature so the horizontal bit here where the line of constant temperature crosses our saturation lines is going to be at p sat for our two temperatures therefore the higher of the two must be our higher temperature so this is going to be p sat at what was that 200 and the saturation pressure corresponding to 158.9 degrees celsius which we already know is going to be 600 kilopascals I mean if we wanted to be real consistent about this I could go look up the saturation pressure at 200 degrees celsius that is the thing that we can do in our saturation tables we go to these saturation tables by temperature and that would be here and I'm looking for 200 degrees celsius I see that the saturation pressure is 15.54 bar which is going to be 1554 kilopascals so in my plot if I wanted to label this I could say 1554 kilopascals I don't actually need it to be labeled for the purposes of indicating the process in the diagram but it is kind of helpful anyway and I know where state one is going to be about because I know the quality at state one we calculated the quality at state one as 0.385568 so we are going to have a state point right about there 38 percent of the way across the dome that's what that means then we know that we have a constant pressure process from one to two so one and two are both going to be at 600 kilopascals they're going to be on this horizontal line so state two must be on the horizontal line and at the higher temperature so state two is going to be over here therefore my process looks like this I start here and I end here and the process appears as a horizontal line furthermore remember that horizontal displacement on the bb diagram represents a boundary work and that boundary work is going to be outward when we have movement to the right and inward that would be a specific work not just a squiggle when I move into the left therefore our area under this curve is going to be a workout term and I could shade that in this quantity here is our workout which as a result of the simplification of the integral is just this horizontal distance which was v2 minus v1 times pressure so specific boundary work would be specific volume at state two minus specific volume state one times the height here which is p and that would give us specific work our total work then would just be that times mass which would be pressure times total volume two minus total volume one which is what we had anyway these are the sorts of useful things that you can discern from looking at a pb diagram