 Problem 3. R134A flows steadily through an expansion valve and is throttled from 1 MPa to 0.6 MPa. The refrigerant is a saturated liquid at the inlet of the valve. Determine the temperature at the inlet, the specific volume of the inlet, and the temperature and specific volume at the outlet. So we have an expansion valve. So the purpose of the expansion valve is to allow the fluid to expand which is going to decrease the pressure and, unlike a turbine, there's not going to be any work out. So if I were to call the inlet state 1 and the exit or outlet state 2, then at state 1 I know I have a pressure of 1 MPa, which is 1000 kPa, which is 10 bar, which might be useful later on. At state 2 I know I have a pressure of 0.6 MPa. That's going to be 6 bar. You might be asking yourself, why does it say 0.6 MPa as opposed to 600 kPa? It's best practice here when you're directly comparing numbers to keep them in the same units. So here, because I'm trying to show the reader that the pressure is decreasing, it makes it a lot easier to see when you're reading if it's in the same consistent units. So going from 1 MPa to 0.6 MPa, I can see that it dropped by 0.4 MPa. But if I had just said the pressure changes from 1 MPa to 600 kPa, it would have been harder to recognize what's happening, especially if you were a reader that didn't understand what the units meant. You'll see that pretty commonly. People use different units of money to try to disguise the differences between the values. But I'm kind of getting off on a tangent here. What else do I know? The refrigerant is a saturated liquid at the inlet of the valve. Okay, so the quality at state 1 is 0. It's my other piece of given information. Before we deal with the temperature and anything else, actually let me back up for a second. At state 1 I know the pressure and the quality. That's going to be two independent intensive properties that's enough to fix the state. So already I can look up anything else I want to know. I can use that information to look up both the temperature and specific volume. And then in order to look up the temperature and specific volume at state 2, I'm going to need to know something else about state 2. So my analysis is going to be to try to get another property at state 2. So the best practice here would be to do an energy balance and a mass balance. So my mass balance, I'll be shorter, so let's start with that. Changing mass inside my expansion valve is going to be the mass entering minus the mass exiting. And now I'm going to have to consider the fact that this expansion valve can't really accumulate mass. So it's going to be reasonable to assume that my mass just goes from inlet to outlet. So I'm going to say that this is steady state. So the properties at the inlet are the same regardless of when you look at it. And the properties at the outlet are the same regardless of when you look at it. So if you looked at it 10 minutes after it started, that'd be the same as 25 hours after it started. What of the case? No properties change with respect to time. So if I were to use the time rate form here, dividing all the terms by dt, because nothing, no property can change with time in order for it to be steady, that means that this would be zero. So the mass flow rate entering would be equal to the mass flow rate exiting. m.in is equal to m.out. So my mass flow rate in is m.1, my mass flow rate out is m.2. So m.1 is equal to m.2, and I'm just going to abbreviate that as m. The same mass flow rate. Now on to the energy balance. So first question, is this an open or a closed system? I already know it's steady. Well, if I were to identify my system as being the valve itself, or rather the mass inside the valve, then I could see that I have mass crossing the boundary. Therefore, this is an open system. So my change in energy is the energy in minus the energy out. And again, I'm going to use the rate of change form here dividing all terms by dt. This is going to be dE dt is equal to the rate of energy in minus the rate of energy out. And because no property can change with time, if it's steady, this is zero. Therefore, E.in is equal to E.out. Now, because it's an open system, my energy could be entering as heat and or work, but now it's open, so I also have to account for the energy entering as mass. So I'm going to say the sum in of the mass entering and how much energy that mass brings within. And then the energy out would look the same. This could be Qout and or Workout. And then the sum out of m.theta. So my full energy balance would be Q.in plus Work.in. And then I only have the one inlet at state one. So the sum of m.theta.in is really just m.1.theta.1. And then my energy out could be Qout, Workout, and my only outlet is state two. So again, this is m.2.theta.2. So now I can start to reduce some of these terms. So what could I say about my expansion valve? Well, first of all, I see no obvious sources of work. There's no work going in. There's no work going out. So I'm going to neglect work altogether. If I wanted to be a good role model here, I would list that as an assumption. So I'm saying no work. What else? Well, generally we assume that expansion valves are adiabatic. That's not necessarily because they're well insulated. It could just be that the fluid is passing through them very quickly. There isn't very much time to exchange heat. So unless I'm given enough information to determine otherwise, I typically assume that expansion valves are adiabatic. So I ignore both Q in and Q out. So that leaves me with m.theta. And remember that theta is the energy brought in by the mass or brought out by the mass. It's the energy that the mass carries, which consists of specific enthalpy plus specific kinetic energy and specific potential energy. So my energy balance becomes m.1h1 plus ke1 plus pe1. That would be equal to m.2 times h2 plus ke2 plus pe2. So I would calculate the specific kinetic energy as taking one-half times the velocity squared, and I would calculate specific potential energy as being the gravitational acceleration times the height. So this would be h1 plus one-half v1 squared over 2 plus gz1, and this would be h2 plus kinetic energy 2, one-half v2 squared, plus gz2. However, I don't see any significant changes in height, and I'm going to be assuming that the changes in velocity are negligible. And that really comes from the fact that my mass flow rate is the same, and these are probably going to have approximately equal cross-sectional areas. So if the mass flow rate is the same, the cross-sectional area is the same, velocity is going to be about the same. Whatever the case, I wasn't given enough information to determine the change in velocity, so it's reasonable to assume no changes in kinetic energy unless told otherwise. So I'm neglecting my kinetic and potential energies. And this isn't because this is zero. It's that the change in this is zero, so they would cancel. It's a pretty common mistake to see me canceling these and say, oh, kinetic energy is zero. It's not. It's the change in kinetic energy. That's why we write the change in kinetic energy is the change in potential energy is zero. But other things that cancel here, my mass flow rate, because m dot one is equal to m dot two, if I were to divide both terms by m dot, that would drop out. That means that h1 and h2 are the only terms left. So in an expansion valve, a general expansion valve is assumed to be isenthalpic, meaning that the enthalpy doesn't change. And that's just because there's no place for that energy to go. The energy entering has to be the energy exiting. But that makes it convenient because I can now look up h1 because I have two independent intensive properties. I can use the fact that h2 is equal to h1 as my other independent intensive property. And then I'll use h2 and p2 to look up anything else about state two. So now we can get to the fun part of this problem. Looking stuff up in tables. You just see how much room I have to work with just so I can kind of space this out. So let's start with state one at state one. I know the pressure and I know the quality. So I have a pressure of one kilopascal, which is 10 bar. And I have a quality of zero. That's right, right? 1,000 kilopascals. Yes, one megapascal, 1,000 kilopascals. So I don't have to determine the phase for this. I already know it. That's because it's a saturated liquid. The quality is zero. If it helps to picture this, if I drew a PV diagram, man, that was a terrible line. That's slightly better. If I drew a PV diagram, I know that my line of constant temperature is going to look like this. So I have two different states here. That's going to be two different temperatures. Probably meeting two different lines of constant temperature. So my quality here at state one is going to be zero. So I don't know if the temperature at state two is going to be higher or lower yet. It's probably reasonable to assume it's going to be lower. That's because I'm dropping my pressure here significantly. So if I drop my pressure, it's probably going to drop the temperature too. But anyway, I know that state one is going to be a saturated liquid. So I'll jump over to my tables and keep in mind that this problem... Let me go up a bit. This problem was R134A, not water. So I don't want to use properties of water. Actually, this is going to be pretty crazy properties for water. A lot of these probably won't even exist on the tables. So make sure you are looking at the tables for R134A. Anyway, jump over to my tables. I see that I have properties of saturated R134A on tables 10 and 11. So I want the pressure table because I have a pressure. So I'll jump to table A11. That's page 944. See if I can figure out how to change pages. 944. Here we are. Properties of saturated R134 by pressure. So I have a pressure of 10 bar, 1 megapascal, 1,000 kilopascal. So my temperature is going to be my saturated temperature of 39.39. So T1 is 39.39 degrees Celsius. And then 39.39 plus 273.15 is 312.54. So that's 312.54 Kelvin. Now what else do I want to look up about state 1 while I'm here? Well, the problem asked me for specific volume, I believe. Yep, specific volume. So I definitely want to look up specific volume. Now I could also look up the H1 because the H1 is going to be equal to H2, which is an important property to know for state 2. So back to the tables. Here at 10 bar, I could read off my saturated liquid internal energy. Not helpful. Saturated liquid. Specific volume. Definitely helpful. So that's 0.8695. But keep in mind, this is VF times 10 to the third. So in order to get this back to VF, I have to move the decimal place three places to the left. Remember, that's just for convenience for the table. So it doesn't have to have a whole bunch of leading zeros. So never neglect this times 10 to the third up over here in the table. 10 bar 0.0008695. V1 is 0.0008695. Keeping meters per kilogram. Back to the table. H1 is another one that I want. So the enthalpy of a saturated liquid at 10 bar would be 105.29. It's a nice and easy convenient unit, kilojoules per kilogram. Grab, I just forgot the number. 105.29. 105.29. Then for state two, the two independent intensive properties that I know are the pressure, which is 600 kilopascals, which is just six bar, and H2, which is equal to H1, which is 105.29. So now I have to fix the face. Let me jump back to my table. So what I'm going to want to do is look up six bar on the saturated R134A tables and then compare my enthalpy to these enthalpies. So if I considered the graph here, the graph that I made earlier, I know that as I'm going from liquid R134A to vapor R134A, it's going to be increasing an enthalpy. I need to add heat to make that happen. So the enthalpy should be greatest in the saturated mixture region at a saturated vapor. So this would be Hg and Hf, which is saturated liquid, would be less than Hg. So if I get an enthalpy that's less than Hf, I know that my phase is a compressed liquid. If I get an enthalpy that's greater than Hg, that means my phase is a saturated vapor. If it's equal to either of these, it's a saturated liquid or vapor respectively, and if it's between the two, it's got to be a mixture. So I'm just guessing that because I know that a line of constant enthalpy on this graph looks something like that, I'm probably going to be dealing with a mixture, but let's compare numbers. Six bar, Hf is 79.5, Hg is 259.2, and my enthalpy was 105.3. Therefore, I'm dealing with a saturated mixture. So I'm actually going to have to interpolate to get my properties. So saturated mixture, that's because my H2 is greater than Hf and less than Hg. Okay, so the first thing I'm going to want to do is calculate the quality. Once I know the quality, I know how far over I am, and that'll make interpolating the other properties a little bit more convenient. So I could calculate the quality at state two from the specific enthalpy because I have it already, but I could just as easily use the fact that I guess I'll just write all this out. That'd be H2 minus Hf over Hg minus Hf. It is also V2 minus Vf over Vg, that's an H, Vg minus Vf. It is also U2 minus Uf over Ug minus Uf. It's also the specific entropy, but neither that nor internal energy are particularly helpful here. So my quality at state two is going to be my specific enthalpy, which is 105.29 minus the value of a saturated liquid at six bar, which was 79.48. Yes, 79.48, so minus 79.48. Then divided by the value of a saturated vapor at six bar, which was 259.19, 259.19 minus 79.48. So the quantity 105.29 minus 79.48 divided by the quantity 259.19 minus 79.48. Gives me a number of 0.14362. So my mixture is 14.5-ish percent vapor, which means that my dot is going to be 14% of the way across from here to here. So probably more like here, that's my state two. But I'll use that with my specific volume relation here to calculate V2. So if I take this here and solve for V2, I would have V2 is equal to X2 times Vg minus Vf plus Vf. So my quality at state two was 0.114362 times Vg minus Vf, which for six bar was 0.0341, 0.0341 minus Vf, which would be 0.0008196. Again, remember, 10 to the third up here. So 0.0008196, 8196, 8196, 8196. That would be cubic meters per kilogram. Then plus Vf, which is 0.0008196. So 0. I guess I could just do this times parentheses, 0.0341 minus 0.0008196 plus 0.0008196. I get a specific volume at state two of 0.0005599, 0.0005599. So that's one of the things I was looking for. And what was the other? Got all distracted. The temperature at the outlet. So I know T1, I know V1, I know V2. The last thing is the temperature at the outlet. Well, keep in mind that I am a saturated mixture here. My quality at state two is going to be 0.15, which means I'm somewhere on this line. I am a mixture. So the temperature is going to be the saturated temperature corresponding to a pressure of 0.6 megapascals. So the temperature at state two is just going to be 21.58 degrees Celsius, 21.58. And then to be consistent, 21.58 plus 273.15 is 294.73, which gives me my other answer, Kelvin. I was asked for units in Kelvin, right? Yeah, look at that. Remembering the problem that I wrote. Okay, so I have all of my answers. It's just a matter of plugging them in. So 312.54, 0.0008693, 8693, 294.73, 0.005599. Notice how I kept consistent units again to compare. I could have written 5.599 e to the negative third. Anyway, that's problem three, which is the end of the exam. You just finished exam two from spring 2015 semester of thermal one.