 R134A enters an adiabatic compressor as saturated vapor at 100 kilopascals at a rate of 0.7 cubic meters per second and exits at a one megapascal pressure. If the isentropic efficiency of the compressor is 87%, determine first the temperature of the refrigerant at the exit of the compressor and then the power input required to operate the actual compressor in kilowatts. So this begins much like a chapter 4 question. We have a compressor. We are assuming its adiabatic takes work to power the compressor and I will indicate its adiabaticness with the prerequisite fuzzy lines. So if I were to set up an energy balance on this compressor, that energy balance will simplify down to the work in is equal to H2 minus H1. If I neglect changes in kinetic energy and potential energy and if I neglect any work in the outward direction and any heat transfer whatsoever, I can define my inlet and outlet state points. I know that at the inlet I have a saturated vapor, which I will indicate by saying x1 is equal to 1 and I know a pressure. It's 100 kilopascals. Then I will write the total volumetric flow rate at state 1 is 0.7 cubic meters per second. At state 2, I know the exit pressure is one megapascal and the other thing I know at state 2 is that the efficiency of this compression process is 87%. So what that means is A to C is 0.87. So what I have to do is actually set up a third hypothetical state point. State 2S. State 2S represents what the outlet condition would be if everything were perfect. State 2S does not exist. State 2 exists. State 2S is just imagining what if everything were perfect? It would be like if I gave you a quiz that was out of 35 points and you were trying to figure out how many points did I earn and you didn't know that it was out of 35 points but you knew you earned 80% of the points. Well, you could look through the quiz and figure out how many points each of the problems was worth, figure out how many points that is in total and then multiply that by 80%. Same is true here. We figure out how much work is required by the compressor if everything is perfect and we multiply that by 1 over 87%. So at state 2S, I have the same outlet pressure because I'm assuming that the compressor is compressing it to the same pressure. I'm imagining a hypothetical situation where it's the same compressor, it just works perfectly. So the end pressure is the same. And because ideal compression is represented by an isentropic process, S2S is equal to S1. So my approach is as follows. I'm going to want to look up H1 and S1. Using S1, I can then look up what the enthalpy would be at the outlet if everything were perfect and then I can use the efficiency to place what H2 actual is. Once I know H2 actual, I can use that to determine T2. Then I know the specific volume and the total volume will, when combined, will give me the mass flow rate which I can use with the difference in enthalpy to figure out the total power. So here. A to C represents the proportion of actual and ideal work. A quick question. Does it take more work to compress something in an ideal world or in reality where there's friction and stuff? You're correct. It's reality where there's friction and stuff. So A to C is equal to H2S minus H1 divided by H2 actual minus H1. Therefore, H2 is going to equal H2S minus H1 divided by A to C plus H1. So X1 and P1 give me H1. S2S equals S1 and P2 gives me H2S. Once I know those, I can determine H2 which I can then use to determine T2. That's my structure. Okay. Now, H1 is going to be the enthalpy at a quality of one and a pressure of 100 kilopascals. For that, we'll go into our tables. Don't forget that we have R134A, not water. So if I go into the R134A tables and I want saturation by pressure because I have a pressure of 100 kilopascals which is one bar. So the row corresponding to my state one is this one. So at state one, my enthalpy is Hg at one bar which is 231.35. My S1 is Sg at one bar which is 0.9395. And my V1 is Vg at one bar which is 0.1917. Then at state two, I know that I have a pressure of one megapascal and an entropy of 0.9395. So with that entropy and that pressure, first question is what is the phase? I'm going to determine that by looking up Sf and Sg at one megapascal. One megapascal is 1000 kilopascals which is going to be 10 bar. 10 bar has an Sf of 0.3838 and an Sg of 0.9043. Therefore, if my entropy is greater than 0.9043, I have a superheated vapor. If it's less than 0.3838, I'm going to have a compressed liquid. My entropy is 0.9395, which means that I have a superheated vapor. So going into my superheated vapor tables, I can find the pressure corresponding to one megapascal which is here. And I see that my entropy of 0.9395 occurs between 40 degrees Celsius and 50 degrees Celsius. I'm going to be interpolating for enthalpy, which is the third column. So my enthalpy is going to occur between 268.68 and 280.19. So if I fire up my calculator, I can say 0.9395 minus 0.9066 divided by 0.9428 minus 0.9066 is equal to the thing that I'm looking for, H2S minus 268.68 divided by 280.19 minus 268.68. That's all equal to x, and I get 279.14. Therefore, H2S is 279.141. You know what? While we're here, let's look up T2S. I know we don't have to, but, you know, character building. So I'm going to use that same proportion and I'm going to say x minus the temperature at an enthalpy of 268.68, which is going to be 40 degrees Celsius. And I'm dividing by 50 minus 40 degrees Celsius and that gives me a temperature of 49.088. So let's think about this for a moment. Take a break out of our calculations and actually think about what this means. We are compressing R134A. At the beginning of the compression process, it has an energy of 231.35 kilojoules per kilogram. If everything went ideally, at the outlet of the compression process, it would have an energy of 279.14. Also, it would have increased from a temperature that was whatever the saturation temperature was at 1 bar to a temperature of 49.088 degrees Celsius. If my efficiency is 87%, would you expect H2 and T2 to be greater than these or less than those? You should have said greater. They are going to be greater because I'm putting energy into the substance. If I'm not doing it as effectively, then I'm going to have to put more energy into the substance to get to the desired pressure. That means my enthalpy will increase as a result, which is why it's in the denominator of this proportion. And my temperature will also increase as a result. So T2 should be greater than 49.088. General rule of thumb, the temperature at the end of the process is going to be higher if there's friction. Anyway, I know H2 then is going to be H2S-H1 divided by A to C plus H1. So calculator, if you would come back, that's going to be 279.141 minus the enthalpy of state 1, which was 231.35 divided by 0.87 plus 231.35. So my enthalpy of that state point is going to be 286.282. That'll be enough to allow me to calculate the power. Then the temperature is going to be whatever the temperature is at 1 MPa and 286.28 kJ per kg. So what is the phase at state 2 actual? Well, we can follow the same logic that we did earlier. It will lead us to the same result. The logic is if we look at the saturation tables for R134A and we find 10 bar, we can compare our enthalpy to the saturated liquid specific enthalpy and the saturated vapor specific enthalpy. Those were 105.29 and 267.97. If our enthalpy was less than 105.29, that means we have a compressed liquid. If it's greater than 267.97, that means I have a superheated vapor. My enthalpy is 286, which means that I have a superheated vapor, which means I go back to where I started. On my superheated vapor tables, I see that my enthalpy of 286.28 is going to occur between 50 and 60 degrees Celsius. I can interpolate for the actual value by taking my enthalpy, which is 286.28 minus 280.19 divided by 291.36 minus 280.19 and I'm saying that that's equal to x minus 50 divided by 60 minus 50, which gives me an x value of 55.45. Therefore, the temperature at the end of the process is 55.45 degrees Celsius. Using the logic that we applied earlier, if my efficiency were 70% instead of 87.5%, would T2 increase or decrease? You're right, it would be even higher. As the efficiency gets lower, the temperature is going to increase. The other aspect of the problem was, what is the required power input? Well, for that, I can set up a full energy balance, but we've done that on compressors a couple of times together. So we can now jump to the power input is equal to the mass flow rate times h2 minus h1 if I'm neglecting heat transfer. My mass flow rate can be determined because I know the specific volume and the total volume. So I'm going to say mass flow rate is equal to volumetric flow rate divided by a specific volume. So 0.7 cubic meters per second divided by my specific volume, which was 0.1917 cubic meters per kilogram cubic meters cancels cubic meters and I'm left with kilograms per second. And I could calculate that number, but I don't want to. So we're just going to plug it in symbolically. Pure opportunities for error with that. So we have 0.7 cubic meters per second divided by 0.1917 cubic meters per kilogram that I'm multiplying by the difference between the enthalpy at state two actual which is 286.282 minus the enthalpy at state one which was 231.35. My goal is to come up with an answer in kilowatts and I recognize that a kilowatt can be expressed as a kilojoule per second. So at this point, kilograms cancels kilograms, seconds cancel seconds, kilojoules cancels kilojoules leaving me with an answer in kilowatts. So calculator, I need you once again. That is 0.7 divided by 0.1917 multiplied by 286.282 minus 231.35. We get 200.586. So with a compressor that has an efficiency of 87%, it takes 200.586 kilowatts of power to compress R134A at 0.7 cubic meters per second from an initial pressure and saturated vapor to an exit pressure of one megapascal. We also determine that the temperature at the outlet would be 55.45 degrees Celsius. While we're here, let me pull you with the same question as earlier. How does 200.586 kilowatts change if my efficiency were to drop? What if the efficiency of the compressor were 70% instead of 87%? You're right, the power required would increase. I'm now having to invest energy to compress the substance and also I have to supply the energy required for the losses which means that if the efficiency dropped, the power required would increase. So the answer to A and B would both go up as a result of the lower efficiency.