 A two-stage compression refrigeration system operates with R134A between pressure limits of 1.4 and 0.1 MPa. The refrigerant leaves the condenser as a saturated liquid and is throttled to a flash chamber operating at 0.4 MPa. The refrigerant leaving the low-pressure compressor at 0.4 MPa is also routed to the flash chamber. The vapor in the flash chamber is then compressed to the condenser pressure by the high-pressure compressor and the liquid is throttled to the evaporator pressure. Refrigerant flows through the condenser at a rate of 0.25 kg per second. Assuming the refrigerant leaves the evaporator as saturated vapor and both compressors are isentropic, determine the compressor power, the rate of heat removed from the refrigerated space, the fraction of the refrigerant that evaporates as it is throttled to the flash chamber, the coefficient of performance, and the theoretical maximum refrigeration coefficient of performance in these circumstances. There is a lot to unpack in this problem. We were given a low-pressure, a high-pressure, and an intermediate pressure. We were given a single mass flow rate. We were told that the exit of the condenser is a saturated liquid, so it's not supercooled or compressed. Similarly, the outlet of the evaporator is a saturated vapor and both compressors are isentropic. Because we have two-stage refrigeration here with a flash chamber operating as an intercooler, we're going to have the same cycle diagram as from our previous video. I will identify these nine state points for this problem and try to determine enthalpies. The reason I need enthalpies for all nine state points is because I want to calculate the work in, the Q in, the workout, and the Q out. From those, I can determine a network, a net heat transfer, a coefficient of performance, and then from those and the mass flow rates, we can figure out the power input, the rate of heat removed from the refrigerated space, and so on. So I want nine state point properties. Because we're looking up enthalpy for R134A, we're going to need two independent intensive properties to drive that lookup in the tables. The first independent intensive property I have for all nine state points is the pressure. I was given three pressures, 1.4, 0.4, and 0.1 megapascals. Those correspond to the high pressure, the intermediate pressure, and the low pressure. So the first thing I'm going to do is populate all nine pressures. 4 and 5 are the high pressure. 1 and 8 are the low pressure. And then 7, 3, 6, 9, and 2 are all at the intermediate pressure. I'll refer to that as medium pressure, just to make abbreviating it a little bit more convenient. With those pressures in place, I have nine of my required independent intensive properties. I need nine more. Two of those nine come from the fact that I was told the quality at the outlet of the evaporator and the outlet of the condenser. Two more come from the fact that the compressors are isentropic. That means S4 is equal to S9 and S2 is equal to S1. Two of them come from the fact that we are assuming the expansion valves are operating isenthetically, meaning H6 is equal to H5 and H8 is equal to H7. Two of them come from the mechanical nature of the flash chamber because it takes a mixed stream at state 6 and separates it by phase. 7 is going to be entirely saturated liquid and 3 is going to be entirely saturated vapor. That leaves me with one independent intensive property unaccounted for. That comes from an energy balance on the mixing chamber. At this point in the analysis, I will know H2, M.2, H3 and M.3, meaning I can calculate the enthalpy at state 9. So at state 1, I'm going to want to look up the enthalpy and the entropy because that defines state 2, and which I can look up the enthalpy. At state 3, I want the enthalpy at state 4. I will have the entropy from state 9 and I can look up the enthalpy. At state 5, I want to look up the enthalpy and I actually don't even really need to look up anything at state 6 because I already will have had H5. At state 7, I want to look up an enthalpy, which gives me H8. And enthalpy at state 9 will come from an energy balance. So I have 1, 2, 3, 4, 5, 6, 7, 7 lookups that I can do right now. Before that, I'm going to want to go into my R134A tables. In my tables, R134A is going to be tables A10 through A12. I have a pressure and quality at state 1, which means I'm going to want to go into my saturation tables by pressure, which means that I want table A11. Moseying on over to table A11, I need to find my pressure. My pressure at state 1 is the low pressure, which is 0.1 megapascals. 0.1 megapascals would be 1 bar. So at 1 bar, I can grab the specific enthalpy of a saturated vapor. And while I'm here, I can grab the specific entropy of a saturated vapor and write those down at state 1. That is 231.35 and 0.9395. From the entropy at state 1, I can look up the enthalpy at state 2. The first thing I have to do is fix the phase at state 2. I recognize that I have an isentropic compression process from a saturated vapor, which implies that I have a superheated vapor, but best practice is to check anyway. I'm going to find 0.4 megapascals, which would be 4 bar. And I'm going to compare my entropy to the saturated vapor entropy at 4 bar. So saturated liquid specific entropy at 4 bar and saturated vapor specific entropy at 4 bar are 0.2399 and 0.9145 respectively. The fact that my entropy at state 1 is higher than Sg at 4 bar implies that I have a superheated vapor. Now I can mosey on over to my superheated vapor tables and I look for a pressure subtable of 4 bar. At 4 bar, I see that I have entropies on either side of 0.9395, which I can use to interpolate for an enthalpy. For that, I need my calculator. So we're going to take 0.9395. Why did you not type 9? 0.9395 minus 0.9182 divided by, no, not divided by P. You can do a calculator. 0.9515 minus 0.9182. And then that is going to be equal to the thing that we're looking for, minus 253.35 divided by 262.96 minus 253.35, solving for x. We get 259.497. Let's make sure that makes sense. It does. So 259.497. At state 0.3, I have a quality of 1 again and the medium pressure, so I'm going to jump back into my saturation tables by pressure. And I'm looking for enthalpy. The enthalpy at 4 bar as a saturated vapor is 252.32. State 4, I have to skip for now. State 5, I have the quality of 0 and a pressure of the high pressure, which is 1.4 megapascals. So I find 1.4 megapascals, which would be 14 bar. And I see that the quality of 0 gives me an enthalpy of 125.26. So 125.26. Next up is state 0.6. At state 0.6, the enthalpy is the same as state 5, which is 125.26. Easy peasy. At state 7, I'm back to 0.4 megapascals, which is 4 bar and a quality of 0 this time. So my enthalpy will be 62 on the nose. 62. At state 0.8, I have the same enthalpy at state 0.7. So 62 on the nose. I can't determine 4 until I finish 9 and I can't do 9 until we figure out our mass flow rates and analyze the mixing chamber with an energy balance. So let's put the calculator away and step into an analysis on the mixing chamber. When I analyze the mixing chamber, I have two inlets, state 3 and state 2. One outlet, which is state 0.9. I'm assuming the mixing chamber operates adiabatically, that there are no meaningful changes in kinetic or potential energy, and I'm neglecting any work occurring in the process. So 1, 2, skip a few. We can jump all the way down to the sum in of m.h is equal to the sum out of m.h, which means m.2 h2 plus m.3 h3 is equal to m.9 h9. And then because I'm solving for h9, I'm going to be dividing everything by m.9. m.9 is the same as m.cycle. So I'm going to write this as m.2. I guess I could write m.9 still. Let's try to write that a little bit more neatly. m.2 over m.9 multiplied by h2 plus m.3 over m.9 times h3. I have h2. I have h3. m.2 over m.9 and m.3 over m.9 are going to come from the quality at state 6. Remember from the previous video, we had defined the proportions of mass flow rates within a flash chamber relative to the quality of the mixture's inlet. So at state 6, if we know the quality at state 6, that gives us the proportion of mass flow rate at 3 relative to 6. And then if we take 1 minus the quality at state 6, that gives us m.7 over m.6. So the quality at state 6 gives us m.3 over m.6. 1 minus x6 gives us m.7 over m.6. Looking at the mass flow rates, I see 2 is the same as 1, which is the same as 8, which is the same as 7. I see m.cycle is the same as m.9, which is the same as m.6. So instead of writing m.2 over m.9, I can write 1 minus x6. Then m.3 is the same as m.3. m.9 is the same as m.6, which is the same as m.cycle. So this can be written as the quality at state 6. And now I can determine h9 if I can look up 8x6. Fortunately for us, we have everything we need to determine a quality at state 6. So we go back into our property table lookups, and we figure out a quality at state 6. So I'm going to jump into my saturation tables at the medium pressure, which is 0.4 megapascals, which is 4 bar. And I'm going to be interpolating between the hf property, which is 62, and the hg property, which is 252.32. So quality at state 6 is going to be x6. Excuse me. I mean, it is going to be x6. h6 minus hf divided by hg minus hf. So the quality at state 0.6. Pop up our calculator, see if we can figure out which button is clear everything. Hey, look at that. Nope, much more better. And the quality at state 6, which was 125.26 minus 62 divided by 252.32 minus 62. And we get a quality of 0.3323. Let's call it 0.3324. OK, now up in our energy balance in the mixing chamber, we're going to take 1 minus that number. So 1 minus that number. Johnny got to use the correct minus sign. Multiply by h2. h2 was 259.497, 259.497. And then we are adding to that that same number again. Multiply by h3, which is 252.32. And we get 257.111. So I can write that up here for consistency. More importantly, I can write that down here. I guess this should be more accurately written as h9 is equal to 257.111, from which we can interpolate an S9. OK, now at state 0.9, we have an enthalpy and a pressure. That pressure is the medium pressure, which is 0.4 megapascals. I jump back into my saturation tables and I compare h9, which is 257.111, to 62 and 252.32. I observe that because h9 is between the two, I must have a saturated liquid vapor mixture again. Furthermore, I can interpolate directly for the entropy at state 9. The entropy at state 9 minus SF divided by SG minus SF is going to equal the enthalpy at state 9 minus 62 divided by 252.32 minus 62. So let's see if we can clear the calculator. Aha, and then we're solving. So this is going to be 257.111 minus 62 on the nose divided by 252.32 minus 62. And that's equal to the thing that we're looking for minus 0.2399 divided by 0.9145 minus 0.2399. Close parentheses looking for x. And we get 0.931484. Let's sanity check that before we move on. We have an enthalpy of 257.111. I see that that's what am I doing? That's not between SF and SG. That's bigger than SG. Therefore, we're going to be in the superheated vapor region. That makes sense. That's how we drew it on our TS diagram. What am I doing? Ah, I know what I'm doing. I'm demonstrating the way not to do it. Don't get too callous in your property table lookups. Remember to sanity check your numbers. Otherwise, we would have written down an incorrect entropy. And that would have been a mistake. Luckily for us, that was a learning opportunity, not a mistake. OK, now I'm going to look in my superheated vapor tables. I'm going to find 0.4 megapascals, which is 0. Excuse me, is a 4 bar. And I see that at 4 bar, I have enthalpies on either side of 257.111. Interestingly, they're the same two enthalpies as last interpolation. And our interpolation for the entropy is going to be between 0.9182 and 0.9515. So let's try that interpolation again. 257.111 minus 253.35 divided by 262.96 minus 253.35. And that's equal to the thing that we're looking for minus 0.9182 divided by 0.9515 minus 0.9182. Looking for x. And we get 0.931232. See, that's why we interpolated. We got a better number. It's better by 0.002-ish. Yeah. With 0.9312, I can perform the last interpolation for stay 4. I'll write that down. S9 is 0.0. 0.9312 kilojoules per kilogram Kelvin. Let's sanity check just to make sure that we didn't make any mistakes. 257.111 is between 253 and 262. And it is closer to 253.35. Therefore, our entropy should be closer to 0.918, which ours is, which is a good sign. Now, at stay 0.4, we have an entropy and a pressure from which we can determine an enthalpy. So the first thing we have to do is jump into our saturation tables by pressure again. We're going to find our high pressure, which is 1.4 megapascals, which is 14 bar. And we're going to compare our entropy of 0.9312 to the SF and SG at 14 bar. If my entropy was lower than SF, I know that I must have a compressed liquid. If it's greater than SG, that means I must have a superheated vapor. Mine is greater than SG, which means I have a superheated vapor. So we go back into our superheated vapor tables and we find 14 bar. And then we're going to be interpolating between an entropy of 0.9297 and 0.9658 for an enthalpy, which is going to be between 283.1 and 295.31, if we take 0.931232 minus 0.9297. And divide that by 0.9658 minus 0.9297. And we see that's equal to x minus 283.1 minus 295.31 minus 283.1. Wrap that in the correct argument and parentheses. 295.31 minus 293.1. Yes, yes. And then we're going to go back into our superheated parentheses, 295.31 minus 283.1. Yes, yes. Comma x, we get 283.618. So 283.618. And with that, we have all of the enthalpies at all nine state points, which means that we can proceed on to calculating our work in, our Q in, our workout and our Q out. For that, I'm going to need a little bit more space. So I'm going to shrink down my state point properties a little bit. Look at that, their travel size. First up, I have work in, which I recognize is going to occur in the compressors, since I'm writing this with respect to the mass flow rate of the cycle, I'm going to want to write specific work as power input divided by mass flow rate through the cycle. The total power input is going to be the power into the low pressure compressor, low pressure compressor, plus the power input to the high pressure compressor. Cannot believe that I said that accurately the first time. Power input to the low pressure compressor is going to be m dot one times h two minus h one. m dot one times h two minus h one. That comes from an energy balance on the low pressure compressor. I recognize that there's no opportunities for heat transfer because it's isentropic, which implies adiabatic. There are no meaningful changes in kinetic nor potential energy. I'm neglecting any work in the outward direction. I only have one inlet, I only have one outlet. It's steady state. Therefore, power input is going to be m dot one times h two minus h one, plus the power input to the high pressure compressor, which is going to be m dot nine times h four minus h nine. m dot nine times h four minus h nine. And we are dividing that by m dot cycle. So first proportion we have is m dot one over m dot cycle, is the same as m dot eight over m dot cycle, which is the same as m dot seven divided by m dot cycle, which is the same as m dot seven divided by m dot six, which is how we had defined one minus x six. So in my goal to write this in terms of as few dimensions as possible, I'm going to write that as one minus x six times h two minus h one. I'm running out of space. I'll move this a little bit more to the left. Hopefully I remember to speed up this part. If you still hear this, that means I forgot to speed up this part. Okay. Then we have m dot nine divided by m dot cycle, which is the same as m dot six divided by m dot six, which is just m dot one, excuse me, which is just one. So I'm going to add to this h four minus h nine. Cool. In terms of x six and enthalpies, one down, three more to go. For Q in, I recognize that the only opportunity for heat transfer to enter my control volume around the entire cycle is in the form of heat transfer into the evaporator. Therefore Q dot in is just going to be m dot eight times h one minus h eight. So Q dot in divided by m dot cycle, which is equal to m dot eight times h one minus h eight divided by m dot cycle. m dot eight divided by m dot cycle is the same as m dot seven divided by m dot cycle, which is the same as m dot seven divided by m dot six, which is one minus x six. So one minus x six times h eight minus h one. Two down, two more to go. Specific workout from this cycle is just going to be the power output of the cycle divided by m dot cycle. The power output of this cycle is zero because there's no opportunities for it. Zero divided by a number is zero. Three down, one more to go. Q out, there's only one opportunity for heat transfer to be rejected by my cycle when I draw a control volume around the entire cycle and that is in the condenser. Q dot out is going to be m dot four times h four minus h five. Again, that comes from an energy balance on the condenser. So m dot four times h four minus h five. Easy way to remember which way that is by the way is to consider the fact that the condenser is rejecting energy because it's rejecting energy. I should start with more energy than I finish. Therefore, a positive quantity would be beginning minus end, h four minus h five. Then I'm dividing that by m dot cycle. m dot four divided by m dot cycle is the same as m dot cycle divided by m dot cycle, which is just one. So this is h four minus h five. So we have one minus x six times h two minus h one plus h four minus h nine for work in. We have one minus x six times the quantity h eight minus h one for Q in, we've worked out of zero and Q out is just h four minus h five. So I'll clear up a little bit more space and we can get down to calculating numbers. First up for work in one minus x six. So we have one minus that was 0.3324 times the quantity, times the quantity, not h. What are you doing calculator? Times the quantity h two minus h one, h two was 259.497 minus 231.35. We are adding to that, adding to that calculator. Quality of state six again. No, excuse me. It's just one, plus h four minus h nine, which is 283.618 minus 257.111. And we get 45.2979. So 45.2979 kilojoules per kilogram. And then Q in, just going to be one minus 0.3324 times the quantity h eight minus h one, h eight is 62 on the nose, h one, oops. H one is 231.35 negative 113.058. Why did I write that as h eight minus h one? Okay, to everyone who was screaming at their device that I wrote that backwards. Thank you. H one minus h eight, which is of course 231.35 minus 62. So for some reason I had reversed that when I wrote it over here for no reason other than lack of coffee, 113.058. Workout is zero, zero is zero. Hooray. And then Q out is four minus five, which would be 283.618. 283.618 minus h five, which was 125.26. I don't know why I'm adding parentheses, but never hurts to add parentheses I guess. 158.358. So with workout, work, excuse me, with work in Q in workout and Q out calculated, I can now calculate a network in and a net heat transfer out. Good practice to calculate both of those, even though we don't need both because they should end up being the same. And if they're not, that is a good indication that we did something wrong. But of course we would never do anything wrong. 45.2979. Remember you shouldn't assume you'll make zero mistakes. You should assume you'll make mistakes and build in mechanisms to catch them. Q out minus Q in 158.358 minus 113.058. So that number minus that number. 45.299. So they should be the same within a rounding error. They are the same. Let's assume that that happened as a result of rounding the enthalpies and the quality. I'm comfortable enough with those two numbers to move on. So we have A asked us for the power input. Power input would be the mass flow rate multiplied by the specific work in. Note that that's mass flow rate through the cycle. The mass flow rate through the cycle was given because we were given the refrigerant mass flow rate through the condenser. So the condenser is going to be the same mass flow rate as the cycle. Therefore the mass flow rate through the cycle is 0.25. So I'm taking 0.25 multiplied by 45.2979. 0.25 times 45.2979. And we get 11.32 kilowatts. Kilograms, cancels kilograms. We're left with kilowatts, 11.3245 kilowatts. Then we were asked to calculate the rate of heat removed from the refrigerated space. So removed from the refrigerated space means we're looking for Q dot in. Q dot in, again, would be the mass flow rate through the cycle times little Q in, which is going to be 0.25 kilograms per second, multiplied by 113.058 kilojoules per kilogram. And then we weren't asked for a specific unit, but when we're talking about cooling capacity, we could express it in kilowatts, or we could convert it to the imperial unit of cooling capacity, which you'll remember is tons. A ton of refrigeration is 211 kilojoules per minute. Now, I could express it in kilowatts, I could express it in tons. We weren't asked for a specific unit. So just for funsies, let's calculate it in tons of refrigeration. That'd be 211 kilojoules per minute. So one ton, 211 kilojoules per minute. I know that I always say best practice is to express answers in the same unit system as you were given. So really we should be expressing an answer in kilowatts as per that model, but we're just having a little bit of fun to explore that conversion a little further. So if we take 0.25 multiplied by 113.058 times 60 divided by 211, we'll get an answer in tons. Let's see if we can type that into the calculator. You can do a calculator times 113.058 times 60 divided by 211. We get 8.04. By the way, note that that's the mass flow rate of the cycle because that's how we defined little Q in. You could have also gotten there by recognizing m.8 times h1 minus h8 is the actual Q.in. So we could have calculated h1 minus h8 as the specific Q in per unit mass of mass flow rate through that device and then multiply by the mass flow rate through that device, but we can't mix and match. The way that we defined Q in from the perspective of the cycle itself is per unit mass of the mass flow rate through the cycle. Therefore, we multiplied by the mass flow rate through the cycle. Part C, the refraction, excuse me, the fraction of the refrigerant that evaporates as it is throttled through the flash chamber, that would be the proportion of m.6 that is leaving at state point three that's asking us for m.3 divided by m.6, which is how we had to find that proportion which is represented as the same quantity as the quality at state six. So what we're really asking is the quality at state six, which was 0.3324, yeah, 24. Part D, I believe, asks us for the coefficient of performance. So the first question we have to ask is, is this operating in a cooling mode or a heating mode? We can deduce that it's going to be a cooling mode because we were given descriptions of the refrigerated space. It implies to us that we are trying to refrigerate a space and as a result of that, we are considering the cooling, the desired effect of this device. That means we're going to be using COPR and not COPHP. COPR is the desired effect divided by what you have to put in to make that happen. So that'd be Q in divided by the network in. The network in for this cycle is going to be the same as work in because there is no workout. That means I have Q in divided by just regular old work in. Q in was 113.058 divided by work in which was 45.298, 113.058 divided by 45.298 divided by 45.298 is 2.4959. So our coefficient of performance is about two and a half. So just like in the previous example problem, we could pose the question, is that a good COP or not? And in order to answer that question, we have to consider how well is this operating within how well it can operate in theory if everything were perfect. So how well is it performing compared to what is theoretically possible if we have Carnot refrigeration? For that, we are considering the low side temperature and the high side temperature because those will dictate what is theoretically possible. For that, we are going to be assuming that the high side temperature is the same as our temperature at state point five. We couldn't use the temperature at state point four because as soon as the temperature of our refrigerant dropped, we wouldn't be rejecting heat transfer to the surroundings anymore. So what we're assuming is that T max is pretty close to T five. That's an assumption we're making because we have nothing better to operate on. For the minimum temperature, we're going to make a similar assumption because we have nothing else to work from or we're going to be assuming that our minimum temperature is T one or T eight. Because I was given one directly, I'm going to write it as in terms of a state one. So for T five and T one, I'm going to be looking up the saturation temperature corresponding to my high pressure and low pressure respectively. So back once again to our R134A tables. So in my saturation tables for R134A, I see that the maximum temperature, T five is going to be 52.43 and the minimum temperature is going to be negative 26.43. So for maximum coefficient performance conditions, we are going to be trying to write our COP in terms of just proportions of heat transfer. So that's going to be Q in divided by network in, which we can again write as Q in divided by Q out minus Q in because the net heat transfer in has to equal the net heat transfer out if we are operating steadily as a closed system. And then I can write this as one over one over Q in times the quantity Q out minus Q in. And then I can bring that inside of the parentheses and I get one over Q out over Q in minus one. Then we make the Carnot substitution that Q H over Q L is equal to T H over T L. I recognize that for a refrigeration cycle, Q out is going to be Q H because the heat transfer rejected from the cycle is going into the high side temperature. So Q out is Q H and Q in is Q L. Therefore, this would be one over T H over T L minus one, which is going to be one over 52.43 plus 273.15 divided by negative 26.43 plus 273.15. And then we are subtracting one in that denominator. So we have one divided by 32.43 plus 273.15 divided by negative 26.43 plus 273.15 minus one. We got a maximum COP of 3.13. That means our COP, which is less than our maximum COP is possible and actually performing pretty well given the circumstances.