 Our 134A at 800 kilopascals in 25 degrees Celsius is throttled to a temperature of negative 20 degrees Celsius. Determine the final pressure and internal energy of the stream. The way that we indicate a throttling valve in our diagrams is with this thing that looks like a bow tie. We can describe an inlet state and an outlet state. I'm going to establish a control volume and start a list of assumptions. First assumption I'm going to make is steady state operation. Next, I have an open system. I don't really need to list that as an assumption, but I will open system. And then if we consider what's happening in this throttling valve, we are trying to convert from within the enthalpy term to within the enthalpy term. And we typically neglect things like work and heat transfer. That's less to do with any insulation and more to do with the fact that the valve operates quickly. We say that a quick process does not have any opportunity to exchange heat because it doesn't have any meaningful time to exchange that energy. Sort of like if I lit a lighter and passed my hand through the fire, my hand doesn't get burned. It's not because I'm invulnerable, but my hand wasn't in the fire for long enough to exchange any meaningful amount of heat. So I can add adiabatic and even no work because the expansion valve has no opportunity for work to occur. While I'm here, I'm going to recognize that the goal of the expansion valve is to drop the pressure and it pushes that energy into other places within the enthalpy. Therefore, I can neglect any changes in kinetic nor potential energy. And with that, I have enough to start an energy balance and a mass balance. Because I have studied state operation, there are no opportunities for mass nor energy to change within the control volume with respect to time. As a result of that, whatever mass flow rate enters has to leave. I only have one inlet, it's state one. I only have one outlet, it's state two. Therefore, m dot one has to equal m dot two, both of which I'm just calling m dot for simplicity's sake. Similarly, the energy of the control volume can't change with respect to time, so whatever energy enters also has to leave. Therefore, e dot in equals e dot out. Because it's an open system, e dot in could be work heat transfer or the energy associated with a moving mass. Similarly, e dot out could be work heat transfer or the energy associated with a moving mass. Because I'm assuming it's operating adiabatically, I'm neglecting heat transfer. And because there's no opportunities for work to occur, I'm neglecting work. Which means the sum in of m dot theta has to equal the sum out of m dot theta. I only have one inlet, it is state one. Therefore, this becomes m dot one times h one plus specific k e one plus specific p e one. The sum out of m dot theta becomes m dot two theta two, which is m dot two times h two plus specific kinetic energy two plus specific potential energy two. And then because the mass flow rates are the same, they cancel. Because whatever kinetic energy there is at the inlet is assumed to be the same as the outlet, I'm assuming no changes in kinetic nor potential energy, whether those terms disappear as well. And we're left with h one is equal to h two. For this reason, we typically refer to these expansion valves as being isenthalpic. They are isenthalpic expansion valves, which means no change in enthalpy, constant enthalpy process. That constant enthalpy allows me to fully define state two from the temperature and the properties of state one. Now that state point two is fully defined, I can actually look up whatever I need. So I'm looking for the final pressure and internal energy of the stream, which means that we're looking for p two and u two. And in order to get to that, I need to look up h one, which I can apply at state two. So for those three lookups, I can jump into my property tables. I want the properties of r 134 a, which means I'm going to be looking at tables a 10 through a 12 alike with water. I can position my state point by considering my saturation conditions. At state one, I have a temperature and a pressure. So I can either look up the saturation temperature corresponding to my pressure and compare that pressure to it. Or I can look up my saturation temperature at my pressure and compare my temperature to that saturation temperature. In this case, it'll probably be easier to compare my temperature to the saturation temperature at 800 kilopascals. So I will look at these saturation tables for arm with 34 a by pressure, which is table 11 800 kilopascals is eight bar. The saturation temperature at eight bar is 31.33 degrees Celsius. My temperature at state one is less than the saturation temperature at P one, which means that I must have a compressed liquid. So populating that information and say he sat at P one is greater than T one therefore compressed liquid. It has not yet begun to boil. So now all I have to do is find my compressed liquid tables for R 134 a. Surely that won't be a problem. I'll just scan down my table of contents and I see saturated R 134 a temperature table, R 134 a pressure table, superheated R 134 a and then ammonia, huh, interesting. I don't have a compressed liquid table. I don't have a compressed liquid table because we typically don't care about compressed liquid properties for refrigerants. We are typically jumping back and forth across the saturation dome and looking at how much energy is associated with that phase transition. Since I don't have a compressed liquid table, the best thing that I can do is approximate the property by using the saturated liquid property. So we're saying HF is about the same as HF at our temperature. What we're saying is we're assuming the pressure has no effect in the compressed liquid region. Note that that's not the same as assuming the temperature has no effect in the compressed liquid region. I mean, if I look at water, for example, compressed liquid water is on table A5. Note how much the enthalpies are changing as a function of temperature, even relatively small changes like 80 degrees Celsius to 180 degrees Celsius, which is what, about 25%. Is a difference of twice as much enthalpy. Compare that to doubling the pressure at the same temperature. We go from 336.86 to 338.85. That's not much change at all. That is why we can get away with assuming that we don't have pressure effects in the compressed liquid region. Therefore, we're looking up the saturated liquid property at our temperature, not at our pressure. So HF of R134A at 25 degrees Celsius. For that, we jump back to table A11. Excuse me, table A10, and I find 25 degrees Celsius. I don't happen to have direct rows for 25 degrees Celsius, so instead I have to use 24 and 26 and interpolate between them. So that interpolation is going to be 25 minus 24 divided by 26 minus 24. Let me pop out my calculator here. 25 minus 24 divided by 26 minus 24 is equal to the thing that I'm interpolating for minus 82.9 divided by 85.75 minus 82.9. Looking for X and we get 84.325. So we're saying that instead of grabbing an enthalpy for R134A at 800 kilopascals and 25 degrees Celsius, we're looking up the specific enthalpy at 25 degrees Celsius and oh, about 660 kilopascals. And we're saying because pressure doesn't affect the properties within the compressed liquid region that that number is close enough. With our new enthalpy, we can fully define state two for that, the first question is going to be what is the phase at state two? We have a temperature and an enthalpy. So we go back to our saturation tables. In our saturation tables at negative 20 degrees Celsius, we see that the enthalpy of a saturated liquid is 24.26 and the enthalpy of a saturated vapor is 235.31. Our enthalpy is 84.325. 84.325 is greater than HF, but less than HG, which means that we must have a saturated liquid vapor mixture. First of all, we can conclude that our pressure must be the saturation pressure at our temperature, which means that the first part of the problem is pretty easy. P2 is just 1.3299 bar because we were given a pressure in kilopascals. Best practice is to express an answer in kilopascals, which means 132.99 kilopascals. For U2, we have to interpolate across the saturation dome to come up with our property at a temperature of negative 20 degrees Celsius. So that interpolation will go H2 minus HF divided by HG minus HF is equal to quality at state two and that's also equal to U2 minus UF divided by UG minus UF. We could interpolate for an X2 and then use that to step to U2, but because we don't actually care about X2, we can actually just jump all the way from H2 to U2. So my interpolation is going to go. H2, which is 100, excuse me, was 84.325. I guess it's still on the calculator so I can just use that. Minus HF at negative 20 degrees Celsius, which is 24.26 divided by, let me add a couple more parentheses for good measure, divided by 235.31 minus 24.26 and that's equal to X minus UF, which is 24.17 divided by 215.84 minus 214.17. We're solving for X and we get 78.719. So we have a pressure and internal energy at the end of the process. We have completed the problem. But you know what? I think we should do something else. We're all here anyway. Let's say that I asked you to plot this and let's say that I asked you to plot it on a PV diagram. For that, we need to indicate our saturation lines. It might be helpful to plot out a couple of lines of constant temperature. For the sake of drawing out this example, let's say the top line is 25 degrees Celsius and the bottom line is negative 20 degrees Celsius. With that information, I can plot state one and state two as well as the process path. State one was at the high temperature. We have 25 degrees Celsius. So I'm going to go into my compressed liquid region and I will indicate state one like this. State two is on the negative 20 degrees Celsius line and I know that it's under the dome. If I had calculated a quality at state two, that would help me position my state point across the dome. So since we're here, let's just compute that number. We get a quality of 28.5%. That means our state two is going to be 28.5% of the way across the dome, which I can draw as being right about here. And that's good enough for my purposes right now. In the future, it might be useful to recognize how a line of constant enthalpy looks on these diagrams, but for now, that's just fine.