 Refrigerant 134A is the working fluid in an ideal vapor compression refrigeration cycle that communicates thermally with a cold region at 0°C and a warm region at 26°C. Saturated vapor enters the compressor at 0°C, and saturated liquid leaves the condenser at 26°C. The mass flow rate of the refrigerant is 0.08 kg per second. Setting this up as a simple vapor compression refrigeration cycle, I want to determine the compressor power, the refrigeration capacity, here referring to the amount of heat absorbed from the cooling side, the coefficient of performance, remember that we have different coefficients of performance depending on if we are analyzing it as a heating process or a cooling process, and the theoretical maximum refrigeration coefficient of performance that could occur if everything were perfect in this scenario. To start, I'll draw a rough diagram and identify some state points. Using my model of simple vapor compression refrigeration, I have four processes, and those four processes occur between four state points. I have isentropic compression, unless I'm given enough information to deduce otherwise from one to two, I have isobaric heat rejection from two to three, I have an isenthalpic expansion process from three to four, and I have isobaric heat absorption from four to one. Since two to three and four to one are isobaric processes, I will denote the high pressure and low pressure sides, and then we can begin to identify state point property information. I was told that this device communicates thermally, which is a fancy way of saying exchange is heat with, a cold region at zero degrees Celsius, and a warm region at 26 degrees Celsius. That's going to give me enough information to come up with an approximate temperature at which my condenser and evaporator are operating. I was told that saturation, excuse me, I was told that saturated vapor enters the compressor at zero degrees Celsius. My compressor is from one to two, which means that I have a quality of one at state one, and I know that the temperature at state one is zero degrees Celsius. At state two, I have the same entropy as at state one. Because I assume that I have an isentropic compression process, unless I'm told enough information to deduce otherwise, like for example an isentropic efficiency. Furthermore, I know the process from two to three is isobaric, so if I had enough information to look up the pressure at state three, that would also fix the pressure at state two. The problem tells me that I have a saturated liquid leaving the condenser at 26 degrees Celsius. The output of the condenser is at state three. So at state three, I have a quality of zero and a temperature of 26 degrees Celsius. Lastly, at state four, I know that my pressure at four is the same as the pressure at one, because the process from four to one is isobaric. Furthermore, I know that as a result of the expansion valve occurring approximately isenthetically, I'm going to have the same enthalpy at four that I do at three. So at state one, I have a quality and temperature from which I can look up anything else I need. For example, I can look up the enthalpy. I can look up the entropy. At state two, since I know S1, that allows me to fix one of the two independent intensive properties, but not the other quite yet. So I'll step on to state three. At state three, I have a quality and a temperature that fixes state three, from which I can look up whatever I want. For example, I can look up the enthalpy and the pressure at state three, which would be the other independent intensive property, which fully defines state two. Using the entropy of state one and the pressure of state three, I can look up the enthalpy at state two. Once I have the enthalpy at state three, that's going to give me the enthalpy at state four, which is probably all that I actually care about anyway. If I wanted to complete the independent intensive properties at state four, I would need to look up the pressure at state one as well. So if I need to look anything up at state four, I would then have enough information to do it. So let's use our given independent intensive properties to perform our lookups for P1, H1, S1, H2, H3, and P3. To do that, we're going to go into our tables and we're going to look at the tables for R134A. Note that R134A is not water, so you're not going to use your water tables. At state one, I was told that I have a quality of one, or rather that it was a saturated vapor, which means that I'm going to get its properties from the saturation tables. I see that table A10 and A11 are both saturation tables. One is given an even increments of temperature, one is given an even increments of pressure, just like with water. Because I have a temperature, that means I'm going to want to use table A10. On table A10, I want to find zero degrees Celsius. Zero degrees Celsius occurs here. So the pressure at state one is 2.9282 bar, and then the saturation conditions at a saturated vapor for zero degrees Celsius are an enthalpy of 247.23 and an entropy of 0.919. So my pressure at state one is 2.9282 bar, 9282 bar, which would be just under 300 kilopascals. My H1 is 247.23, and my S1 is 0.919 kilojoules per kilogram Kelvin. Not too shabby. I already have three of my six properties. Since I don't have P3 yet, let's skip over state two and jump all the way to three. At state three, I have a saturated liquid this time, which means I'm still going to use our saturation tables. And again, I was given a temperature, which means that I still want to use table A10. This time, I'm going to grab our properties from the row that has a temperature of 26 degrees Celsius. So my pressure at state three will be 6.853 bar. My enthalpy would be 85.75. And my entropy, if I needed it, I guess, would be 0.3208. Since I don't need it, I will not highlight it. So let's see if we can write those down correctly. H3 was 85.75 kilojoules per kilogram. And P3 is 6.853. Now that I know P3 and I know S1, I have enough information to look up H2. For that, I first have to fix this phase at state two. I could probably deduce from the fact that I start at a saturated vapor and then compress that I'm probably going to end up in the superheated vapor region. But the best way to do this would be to look at our saturation tables and identify the saturated liquid and saturated vapor properties corresponding to our pressure. From how our S1 relates to those two properties, we can deduce what the phase has to be. So I'm going to jump into table A11 this time, and I'm going to try to find a pressure of 6.853 bar. Now because 6.853 is not an even increment here, it is unlikely to be directly on my tables. So from this, I could see that at 6 and at 7, my SF is about 0.3 and my SG is about 0.91. So from that, the fact that my entropy is higher than SG implies that I have a superheated vapor. You could also use A10 again, recognizing that this table is the same information as A11. It's just that one is given an even increment of temperature and one is given an even increment of pressure. And for this specific circumstance, recognize that we happen to have a row that is exactly what we need at state 3, and compare our S1 to the SG and SF at that pressure of 6.853 bar. I see that my SG here is 0.9082. S1 is higher than SG at P2, which means I must have a superheated vapor. So we can jump over to table A12 and look for a pressure sub-table of exactly 6.853 bar. Surely there will be a perfect sub-table for us. I have complete and utter confidence that there will be a... Oh, there isn't. That's sad. So we have 6 bar and we have 7 bar, but we don't have a pressure sub-table with the exact pressure of 6.853. That's inconvenient. In order to evaluate the enthalpy at state 2, I will have to interpolate. Better yet, I recognize that I don't have a pressure, and my entropy is going to lie between rows, which means that I have to interpolate between rows and between pressures all at the same time. I think the easiest way to do this is to essentially build our own pressure sub-table. We're using the properties at 6 bar and the properties at 7 bar to create a little table of our very own at 6.853. On that table, we will interpolate an entropy at 30 degrees Celsius, an entropy at the saturation condition, and then interpolate between those two to come up with our properties at a pressure of 6.853 and at an entropy of 0.919. So I'm going to be interpolating between the enthalpies and the entropies at saturation and at 30 degrees Celsius for a pressure at 6.853. So I will start the process of building my pressure sub-table at 6.853 bar. We want a column for temperature, which we don't really need, but it's a useful placeholder here. I have a column for enthalpy, which will be in kilojoules per kilogram, and I have a column for entropy, which will be in kilojoules per kilogram Kelvin. And on that table, let's make these straighter lines since I can. On that table, I will be filling in properties at saturation and at 30 degrees Celsius from interpolating between the pressure sub-table at 6 and the pressure sub-table at 7. And then I will interpolate between those two results for my entropy at state 2, which was 0.919. I don't really need the temperature at state 2, so I don't have to determine it, but if you wanted to, you could also interpolate for that. So we have to interpolate for an entropy on either side of 0.919 and an enthalpy on either side of the enthalpy at an entropy of 0.919, and then use the proportion of the way we are between 1 and 2 here to interpolate between 3 and 4, which gives us our enthalpy at state 2. So we have to perform five interpolations. Therefore, I call this a pent-touple interpolation. First up, we want the entropy at saturation and 6.853 bar. Then we want the entropy at 30 degrees Celsius and 6.853 bar. And then we want the enthalpy at saturation and 6.853 bar. And the enthalpy at 30 degrees Celsius and 6.853 bar. And then we can finally get to H2. Not too shabby. Five interpolations for that, but we'll use our calculator. On that first interpolation, we're going to be interpolating between 0.9097 and 0.9080, and we're going to be using pressure as our driving property for that interpolation. So to make it a little bit easier to follow on the calculator here, I will use the solve function so that it appears in the way that I would encourage you to write out your interpolations. Of course, if you wanted to solve that and write it as the entropy that we're looking for is equal to 6.853 minus 6 divided by 7 minus 6 is equal to the value that we're looking for here minus the entropy at 6 bar and saturation, which would be 0.9097 divided by 0.9080 minus 0.9097. Therefore, rearranging that algebraically, we get entropy at saturation and 0.6.853 is equal to 6.853 minus 6 divided by 7 minus 6 multiplied by the quantity 0.9080 minus 0.97 close quantity plus 0.9097. But one thing at a time. So we have 0, excuse me, we have 6.853 minus 6 divided by 7 minus 6. And then that is equal to x, the thing that we're looking for minus 0.9097 divided by 0.9080 minus 0.9097. And then we are looking for x and we get 0.90825. So that first value that we're looking for is 0.90825. I will use blue here to try to keep track of what we've interpolated for and what we're using as our given properties from the previous interpolations. 0.90825. One interpolation done. The next interpolation is going to be using the same left hand side of the interpolation, but this time using the properties at 30 degrees celsius. So instead of 0.9097, I'm going to be using 0.9388 and 0.9197 and we get 0.922508, 0.922508. Two interpolations down, three more to go. Next, we have the same left hand side of our interpolation, but we replace 0.9388 with 259.19 and we replace 0.9197 with 261.85 and we get 261.459. Then replacing 259.19 with 267.89 and 261.85 with 265.37. We get 265.74. Now we have all the legs we need for the actual interpolation we want to do and that actual interpolation is going to be 0.919 minus 0.90825 divided by 0.922508 minus 0.90825 is equal to x, the thing that we're looking for, minus 261.459 divided by 265.74 minus 261.459 and we are solving that for x and we get an h2 of a number surely and that number is 264.687. So our h2 value is 264.687. Note that when we were in the Rankine Cycle Analysis we had used a shortcut to get from state one to state two that was predicated on the fact that we were analyzing a pump operating with a compressed liquid, we cannot use that shortcut here. A, we have a compressor, not a pump, because it's working primarily with vapor, not a liquid, and our vapor is certainly not incompressible. So we cannot use that shortcut. Now that I have h1, h2, and h3, which also gives me h4, because h4 is equal to h3, I can determine my work in, Q in, work out, and Q out. From that I can determine the rest of the analysis. So work in here is only going to occur in the compressor. That work in comes from an energy balance on the compressor. The compressor is an open analysis, which is steady state. As a result of that I have e.in is equal to e.out, e.in could be Q.in, it could be work.in, it could be the sum in of m.theta, e.out could be Q.out, it could be work.out, it could be the sum out of m.theta. Because I have an isentropic compression process, isentropic implies adiabatic, both heat transfers disappear. Because it's a compression process, I have a work input, and I assume not a work output. Furthermore, I'm assuming changes in kinetic and potential energy are negligibly small, which means my work in collapses down to just h2 minus h1. Q.in then would be between states 4 and 1. I'm neglecting changes in kinetic and potential energy. I have no opportunities for work. I assume heat transfer is only in the inward direction. I have steady state operation with one in one outlet. Therefore, Q.in becomes h1 minus h4. Work out occurs nowhere, because there's no opportunity for work out. Therefore, work out is zero. Q.out occurs in the condenser, and I recognize that I have no opportunities for work. I'm neglecting changes in kinetic and potential energy again. I only have one inlet, one outlet, it's occurring steadily, which means that I'm left with h2 minus h3. So using h1, h2, and h3, I can calculate a work in, a Q.in, and a work out term. So first up, h2 minus h1, that would be 264.687 minus 247.23, 17.457, 17.457 kilojoules per kilogram. Q.in is h1 minus h4, h4 is equal to h3, which means that I'm taking 247.23 minus 85.75, which gives me a Q.in of 161.48. Q.out occurs between 2 and 3. I have 264.687 minus 85.75, which gives me 178.937. From this information, I can calculate a net heat transfer and a net work. The net work would be in the inward direction. It would be work in minus work out. Therefore, I'm left with 17.457. The net heat transfer is going to be in the outward direction. That will be Q.out minus Q.in, which would be 178.937 minus 161.48, and I get 17.457. Because these are the same, that's an indication that I probably built these three equations correctly, which makes sense because they're very simple equations and is a generally useful way of checking your calculations before you proceed. Those two numbers being different implies that you've neglected some changes in energy. Remember that when I wrap this entire cycle in a control volume and analyze an energy balance on the control volume, I have steady state operation of a closed system, which means that I must have Q.in plus work in equals Q.out plus work out. The energy entering has to go somewhere. The energy entering is heat transfer in and work in. The energy exiting is only Q.out. Now that we have that set up, we can actually consider what the problem asks us for. The first thing it wants is compressor power. For that, I take my mass flow rate and I multiply by our specific work in. The mass flow rate is 0.08 kilograms per second. The specific work in is 17.457. Kilograms cancels kilograms, leaving me with kilojoules per second, which is kilowatts. 0.08 times 17.457 is 1.39656. Let's call that 1.4 kilowatts. Part B asks for the refrigeration capacity. The refrigeration capacity here is actually asking for Q.in. It's asking how much heat can it pull from a space if it's being used to refrigerate that space. That would be mass flow rate times specific Q.in, which would again be 0.08 kilograms per second, multiplied by specific Q.in, which was 161.48 kilojoules per kilogram. Kilograms cancels kilograms, so I could express an answer here in kilowatts. So I could say 0.08 times 161.48 and get 12.9184. And since the problem didn't specify what units it wanted, that would probably be reasonable. However, in the United States, a common unit of refrigeration capacity is tons. And to see that conversion, you can jump into our conversion factor sheet. And on our conversion factor sheet, we see that one ton of refrigeration is equivalent to 211 kilojoules per minute. Why tons of refrigeration? Well, that has to do with how refrigeration used to occur with big old blocks of ice. A ton of refrigeration has to do with how much ice you needed in order to refrigerate a certain space. It doesn't really make sense with modern refrigeration, but then again, neither do most of the imperial units. So tons of refrigeration are a vestigial unit, which we still have to deal with. And as a result of that, we should convert this from kilowatts into tons. So I will actually express an answer in tons. And for that, I recognize that one ton of cooling is 211 kilojoules per minute. 211 kilojoules cancels kilojoules. In order to get tons, I have to convert one minute to 60 seconds. Then minute cancels minute, second cancels second, and I'm left with tons. 0.08 times 161.48 times 60 divided by 211 gives me 3.67 tons. Part C asks us for the coefficient of performance. So now we have to consider, are we analyzing a refrigerating process that is a refrigeration cycle operating with the intention of cooling something, a refrigeration cycle operating as a refrigerator, or are we considering a refrigeration cycle operating in a heating environment where the side that we care about is the queue out? Well, from context clues, I know that it is operating as a refrigerating process. And the big context clue for that is the fact that we wanted the refrigeration capacity. If we didn't know that, we would have to make an assumption and move on. So the fact that it was asking for the refrigeration capacity implies that we care about its ability to refrigerate, which means that we care about the cooling end of it. So for that, I have to express my coefficient of performance in terms of the side that I care about, which would be Q in. So what I'm doing is calculating a COP R. Remember that COP, like all of our forms of efficiency, is what we're trying to do, divided by what we have to put in to make that happen. What we're trying to do here is cool, which would be Q dot in or specific Q in. And what we have to put in to make that happen is network in. In this case, in the case of vapor compression refrigeration, network in is just work in because there's no workout. So I can write this as little Q in over specific work in or Q dot in over work dot in. So I could take two point, excuse me, 3.67 tons divided by 1.4 kilowatts and express that in tons per kilowatt or I could convert it into a unitless proportion, but it's easier to just use the specific quantities because those are already in the correct units. So 161.48 divided by 17.457 gives me 9.25. So the coefficient of performance of this system is 9.25. From that information, we could calculate a variety of other things. We could come up with a lot of different ways of evaluating how well this refrigeration cycle is operating, but COP is a good general metric to begin with. Lastly, what I wanted to know was what could the COP be if this were operating perfectly? So if this was operating as a Carnot refrigeration cycle, if everything was perfect and there were no losses, how high could the coefficient of performance be? That's useful because it gives us a benchmark to compare our COP against. If the maximum COP is 6, and we have a COP of 9.25, that implies that this refrigeration cycle is impossible. If the maximum COP is 15, a 9 is smaller than 15, so it's possible, and there's some headroom to improve. If my maximum COP is 9.5, then we can say that, well, this is theoretically possible, but it requires that it operate nearly perfectly. There's not much headroom to improve upon this. For COP R max, what I want to do is make the Carnot substitution and substitute the heat exchange on the hot side and heat exchange on the low side with the temperature distribution driving them in the first place. For that, I'm going to have to rewrite my COP equation in terms of a proportion of heat transfer in and heat transfer out. Remember that for a refrigeration cycle, the heat transfer out is the heat exchange with the high temperature side, and the heat transfer in is the heat exchange with the low temperature side. So QAH is Qout, and Qin is Ql. So I have to rewrite this proportion of Qin and work in in terms of just heat transfers. For that, I will consider the fact that my net heat transfer out has to equal my net work in. So I can say work in is equal to Qout minus Qin. So I can write this as Qin over Qout minus Qin. And then I can rearrange that equation to write it just in terms of heat transfer proportions. For that, I will bring Qin into the denominator, which I will write as 1 over Qin times the quantity Qout minus Qin. And then bringing that inside of the parentheses, I get 1 over Qout over Qin minus 1. Now that I have a proportion of heat transfers, I can make the Carnot substitution and write TH in place of QH, which is Qout because it's a refrigeration cycle, and TL in place of QL, which is Qin, for a refrigeration cycle. So I can write this as 1 over TH over TL minus 1. And that temperature distribution is referring to the temperature of the environment on the warm side and the cold side. So our refrigeration cycle occurs in this circle. The entire device, compressor and heat exchangers and expansion valve all exist within this circle. What we're looking at is the temperature on the outside of that cycle where we are pushing and pulling heat. Because we don't know the temperature that this is operating within, the best thing that we can do is consider the fact that it communicates thermally with the cold region at zero degrees Celsius and the warm region at 26 degrees Celsius. So we take that to mean the temperature of the warm region is 26 degrees Celsius and the temperature of the cool region is zero degrees Celsius. So we're plugging in zero degrees Celsius for TL. We are plugging in 26 degrees Celsius for TH. Because we're writing a proportion of temperatures, we have to use absolute temperatures, which means I'm going to write this as 1 over 26 plus 273.15 divided by zero plus 273.15 minus 1. We can pop our calculator back up here and then definitely get the correct number of parentheses on the first try. Definitely. Okay. 1 divided by the quantity 26 plus 273.15 divided by 273.15 minus 1 is 10.51. So my maximum CLP is 10.51. I'll write that here. So our coefficient of performance is 9.25 and the theoretical maximum that could occur if everything were perfect is 10.51. That implies that our refrigeration process here is theoretically possible in that it doesn't violate the second law of thermodynamics and that there isn't a whole lot of headroom. This is probably a very well-performing refrigeration cycle. Maybe even suspiciously too well-performing, but it's still theoretically possible. The last thing I want to do here, even though it isn't explicitly asked for, is to draw a TS diagram. So for that, I will move our subtable out of the way. I will erase our intermediate steps and I'll even add that to the list of requirements. Draw a TS diagram. Hey guys, look at that. We have a part E to do. Oh man, swear it just appeared on a nowhere. Despite the fact that it looks so professionally written. For that TS diagram, we're going to want straighter lines than that. We're going to want to draw it relative to the saturated liquid and saturated vapor lines for R134A, which let's say look like that. That is a very well-drawn dome. Maybe rotate the iPad. There we go. That's a little bit better. Still not great, but better. Now on this TS diagram, I'm going to draw a couple of lines of constant pressure, isobars if you will, and I will. And those two lines of constant pressure refer to the pressure of the high pressure side and the pressure of the low pressure side. Since I wasn't given those pressures, I understand those to be 6.853 bar and 2.9282 bar, referring to the high and low pressures respectively. So I will draw 6.853 and 2.928, respectively. And then on those two lines of constant pressure, we can indicate our state points. At state one, I recognize that I'm on the low pressure line, and I know that it's a saturated vapor, which means state one is going to be right here. Then from one to two, I have an isentropic compression process, which means that I'm going to move straight up because I have a constant entropy, giving me state two way up here. Then from two to three, I have an isobaric heat rejection process, which means that our working fluid is cooling. So we're going to be moving to the left until we reach the saturated liquid lines, which means state three is right here. And then from three to four, I have an isentropic process, which looks like this. So state four must be down here. So if we wanted to draw this a little bit more accurately, we could calculate the quality at state four. We could figure out the temperature at state two to ensure that we had slightly more two scale data. But what really matters here is the relative shape of this diagram. We recognize that when we go from two to three, we must be rejecting heat. When we go to four to one, we must be absorbing heat from its surroundings. Move into the right on a TS diagram represents a heat transfer in, move into the left represents a heat transfer out. And then this region under the diagram represents the net heat transfer, which also must be the net work. So we could indicate that we had heat transfer in over here. It could indicate that we have heat transfer out over here. But I think that clutters up my diagram. So let's just leave it like it was. And there we go. That's our first example problem.