 A gas refrigeration system using air as the working fluid, or refrigerant, has a pressure ratio of 5. Air enters the compressor at 0°C. The high pressure air is cooled to 35°C by rejecting heat to its surroundings. The refrigerant leaves the turbine at negative 80°C and then absorbs heat from the refrigerated space before entering the regenerator. The mass flow rate of air is 0.4kg per second. Using that information, I want us to A. Determine the effectiveness of the regenerator. B. Determine the rate of heat removed from the refrigerated space. And C. Determine the coefficient of performance. So let's parse out that information. The problem tells me that I have a pressure ratio of 5. That pressure ratio is going to give me the difference in pressure, or the proportion of pressure, between the high pressure side and the low pressure side. Remember that all heat exchange processes in the Brayton cycle and therefore the Brayton refrigeration cycle and therefore the gas refrigeration cycle are isobaric. So P4 will have the same pressure as P3, which will have the same pressure as P2. Those three pressures collectively make up my high pressure side of this analysis. Furthermore, P5 will be the same as P6, which will be the same as P1. That makes up my low pressure side. The proportion of pressures between the high pressure side and the low pressure side is 5. We weren't told what the pressure is on the low pressure side, but if we imagine it was 1 bar, that means the pressure on the high pressure side would be 5 bar. Also note, it doesn't really matter what the pressure is. All that we actually care about is the proportion of pressures. We'll get into that a little bit more in a second. Next, we recognize that we were told three temperatures, 0°C, 35°C, and negative 80°C. Those three temperatures refer to state 1, 3, and 5 respectively. Lastly, I was told the mass flow rate of air, which because this is made up of steady flow devices, that mass flow rate is the same everywhere. m.1 is equal to m.2, which is equal to m.3, which is equal to m.4, which is equal to m.5, which is equal to m.6, all of which are 0.4 kg of air per second. The first thing I want to do before we get into analyzing what the problem is asking for is setting up my work-in, Q-in, workout, and Q-out relations. I recognize that because these are open systems, the turbine, the compressor, the heat exchangers, and the regenerator are all open systems that are assumed to operate steadily. My works and heat transfer terms are only going to be relative to delta H. So I can write the work-in to the cycle on a specific basis, the Q-in to the cycle on a specific basis, the work-out of the cycle on a specific basis, and the Q-out of the cycle on a specific basis, relative only to enthalpies in our state points. My work-in occurs in the compressor, for which the energy balance simplifies to H2-H1. My Q-in occurs in our heat exchanger on the cold side, and because heat is being added between 5 and 6, that means Q-in is going to be H6-H5. My work-out occurs in the turbine, where the energy balance would simplify down to H5-H6, excuse me, H4-H5, and my Q-out occurs in the heat exchanger on the warm side. Since I'm rejecting energy, a positive delta would be H2-H3. If I knew H1, H2, H3, H4, H5, and H6, I could calculate these six quantities from which I could determine anything else that I wanted. So my goal for the moment is going to be to determine enthalpy changes, and to do that, we have to consider two options. Option one would be to use our ideal gas properties of air. Remember that for an ideal gas, enthalpy is only a function of temperature. So I could use my temperatures to determine enthalpies. I could use my isentropic ideal gas relations to relate the pressure proportion to the proportion of reduced pressures. So for example, I could use T1 to look up an H1 and a PR1. I can multiply that number by 5 and use that result to look up T2 and or H2. My other option would be to assume constant specific heats. If I assumed constant specific heats, then I could rewrite all of these delta Hs as Cp delta Ts. In that case, I would only care about the temperatures. So for those of you watching at home, why don't you take a minute and think about which way you would prefer to work the problem? Do you want to assume constant specific heats or do you want to perform the table lookups? Well, you guys probably know what I want to do. I want to do both. And in order for us to do both, we need a vertical line. On one side, I will have the analysis using the tables. On the other side, I will have the analysis using constant specific heats. And remember that these are independent of one another. We are working through the problem in two parallel universes, one where we've assumed constant specific heats, one where we haven't. We just happen to be doing them both simultaneously because, you know, parallel universes. So with my table lookup side, I want H1, H2, H3, H4, H5 and H6. For those, I will need T1 to look up H1 and I will also look up PR1 and then I will multiply PR1 by 5 because for an isentropic process like the compressor, PR2 over PR1 is going to equal P2 over P1, which the problem told us was 5. Therefore, PR2 will equal PR1 multiplied by 5 and then we can use that resulting PR2 to look up H2. We use T3 to look up H3. For the turbine, we have to work backwards. We can use T5 to look up H5 and PR5 and then we recognize that across the turbine from 4 to 5 and then across our turbine because we have an isentropic process, we can write our isentropic ideal gas relation again. We're going to say PR4 over PR5 is equal to P4 over P5 which again is going to be 5. This is the relation between the high side pressure and the low side pressure. Therefore, PR4 is going to equal 5 times PR5. Then once we have PR4, we can look up H4. So at this point in our analysis, we will have H1, H2, we will have H5, H4 and we will have H3. Furthermore, we will recognize that the heat exchanged across the regenerator is going from one side to the other. We're going from the hot side to the cold side. The amount of heat leaving the hot side is going to be equal to the amount of heat entering the cold side. So if we know H3, H4 and H1, we can figure out what H6 had to be. Again, the regenerator has entering energy in the form of 3 and 6, exiting energy in the form of 1 and 4 and the heat leaving the hot side is going into the cold side. So if we draw a control volume around the entire thing, we have m.1 H1 plus m.4 H4. That's our exiting energy because remember that we assume like in the Brayton cycle that the regenerator itself is adiabatic that all of the heat exchanged occurs internally. That will be equal to the entering mass flow rates times their enthalpies m.3 H3 plus m.6 H6. My next step is going to be divide by m. Because they're all the same and then solve for H6. H6 will be H1 minus H3 plus H4. If it makes it a little bit easier to relate to the logic we used earlier, I could write it in this order. So you could see that whatever energy exits the hot side is being added to the process from 6 to 1. So H6 minus H1 will equal H4 minus H3. But whatever the case, that energy balance will give us H6. Once we have all six H's, the world is our oyster. There's our outline. All we have to do now is actually do the look-ups and to make this a little bit easier to follow. I will rewrite this as a table. So first I will populate my temperatures. T1 was zero degrees Celsius. Zero degrees Celsius would be 273.15 Kelvin. T3 was 35 degrees Celsius. 35 plus 273.15 is going to be, thank you calculator, 308.15 and then T5 was negative 80. So T5 would be 293.15. Let me double check, negative 80 plus 273.15. Yes, 193.15. So I'm going to use T1 to look up H1 and PR1. Then I'm going to multiply PR1 by 5 to get PR2, which I'm then going to use to look up H2. I don't really need T2 nor T4 nor T6, but you know what? We have infinite time. Let's just determine them for funsies. And by funsies, I really mean, let's develop a full set of temperatures so that we can compare the table lookups to the colder standard analysis. But I've solved for about as much time as possible. Now it's time to interpolate for an H and a PR. So T1 is going to be between 270 and 280, which means that my enthalpy is going to be between 270 and 280. And my PR is going to be between 0.959 and 1.0889. So with my handy calculator, I will write 273.15 minus 270 divided by 280 minus 270 is equal to 270, excuse me, x minus 270.111. That was too many ones. I'm just going to hope you guys didn't notice. Divided by 280.13 minus 270.11. Note the correct number of ones. And that gives me an H1 of 273.266. And then I can do the same interpolation for PR, except I'm going to swap my enthalpies with PR terms 0.959 and 1.0889 and 0.959. And I get 0.99919. Naturally, I'm not going around that. I'm going to write all of them. 0.99919. Nice and arbitrarily precise. Let me fix that decimal point. Okay. Now, again, because our compressor is operating isentropically, which we are assuming because we don't have enough information to deduce otherwise, I could have given you an isentropic efficiency or I could have implied something about the efficiency. But right now, because we don't have any more information, we are assuming isentropic analysis. So PR2 over PR1 is equal to P2 over P1, which is equal to 5, which means PR2 is going to be 5 times 0.99999919. So if my calculator cooperates, 0.99999919 times 5 is 4.99996. Then we can use that PR to look up an H2. And again, for funsies, a T2, even though it's not explicitly necessary. So 4.9996 occurs between 4.915 and 5.332. So my enthalpy will be between 431 and 441. And my temperature will be between 430 and 440. So, time to interpolate again. Here we go. 4.9996 minus 4.915 divided by P, apparently. 5.332 minus 4.915 is equal to the enthalpy that I'm trying to find, which is x minus 431.43 divided by 441.61 minus 431.43. We're solving for x and we get 433.495. And then we can repeat the same interpolation for temperature as well. Swapping out 431.43 with 430 and 431.43 with 430 and 441.61 with 440 and we get 432.03. 432.03. Okay, one down. Three more to go. Our state 3 lookup is the temperature of 308.15. I don't need a PR at this time. All I need is H. So I am going to find 308.15. That's going to occur between 305 and 310. Therefore, I'm going to interpolate between 305.22 and 310.24. So, 308.15 minus 305 divided by 310 minus 305 is equal to the thing that we're solving for minus 305.22 divided by 310.24 minus 305.22. We're solving for x and we get 308.383. 308.383. Now, I'm going to use T5 to look up PR5 and H5 and then use my proportion of pressures from 4 to 5 to figure out what PR4 was and then use that PR4 to look up H4 and then for funsies T4. Then I can use an energy balance on the regenerator to fix state 6. So, 193.15 occurs between 200 and 210. I know what you're thinking. You're thinking, John. No, it doesn't. Well, remember that our extrapolation, which is what we have to do here because we're off the table, is set up using the same methodology as our interpolation. So I'm going to be extrapolating from the difference between 200 and 210 to what it will be at 193.15. So I'm going to take 193.15 minus the higher value, which is 200 and setting, dividing that by 210 minus, oh, come on calculator, 210 minus 200 and setting that equal to x minus 199.97 divided by 209.97 minus 199.97. And we get 193.12. Double-check that I typed the correct number of nines. Indeed, I did. So 193.12 is my enthalpy at state 5. And then we can repeat the process for PR. That PR is going to be 0.3363 and then 0.3987 minus 0.3363 and we get 0.293556. And again, naturally I'm going to use an obtuse number of decimal places because I think my tables look the best when they have a completely ridiculous number of digits. And then to go from PR5 to PR4, I'm going to multiply that by five. So all together now. Multiply it by 0.293556. We get 1.46778. And using a PR value of 1.46778, we can jump back to our tables. We can see that it lies between 1.4686 and 1.3860, which means that my enthalpy is going to be between 300.19 and 305.22. Furthermore, my temperature is going to be between 300 and 305. So if I take 1.46778 minus 1.3860 and I divide by 1.4686 minus 1.3860 and I set that equal to x minus 300.19 divided by 305.22 minus 300.19. Now before I hit the Enter button, I'm going to point out that because our value is so close to 1.4686, I'm expecting my enthalpy to be very nearly 305.22. The fact that it's so close to 305.22 as compared to 300.19 is a good indication that we set up our interpolation correctly. 305.17. And then I'll go swap my enthalpies for temperatures to solve for the temperatures, 305. 305 minus 300 and 300. And we get 304.95. Let me make that a little bit easier for you to see. And by easier, I mean possible at all. Okay, we have state points 1 through 5. Then again, we're going to use the result of our energy balance on the regenerator, which I will draw out again. So H6 is going to be H1 plus H4 minus H3. So calculator H1 273.266 plus 305.17 minus 308.383. We get 270.05. And for completion's sake, I will use that enthalpy to determine a temperature at state 6. 270 is going to occur between 260.09 and 270.11. So my temperature is going to lie between 260 and 270. So that interpolation will go 270.053 minus 260.09. And then of course, you have to remember to fix your parentheses. That is an important part of the interpolation. And then you divide by 270.11 minus 260.09. And that's equal to x minus 260 divided by 270 minus 260. And we get 269.943. Now that I have all six enthalpies, I can calculate the work in, the Q in, the work out, and the Q out. And then I can proceed from there to whatever else the problem wants us to do. So work in is H2 minus H1, which is going to be 433.495 minus 273.266. We get 160.229. And then Q in is 6 minus 5 to 70.053 minus 193.12. We get 76.93. For work out, we're taking 4 minus 5. And we get 112.05. And for Q out, we're taking 2 minus 3, which is going to be 433.495 minus 308.383 for 125.112. Then while I have these numbers, I recognize that for this entire cycle to operate as a closed system, I would expect for the net heat transfer in to equal the network out, or the net heat transfer out to equal the network in. Since this is a refrigeration cycle, we should have more net heat transfer out and network in. So I'm going to double check that those are correct. This being a refrigeration cycle, we would expect a network in and a net heat transfer out. I'm going to double check that we have the same number, 160.229 minus 112.05. It's 48.179 and the net heat transfer out 125.112 minus 76.933. And look, we got 48.179. The fact that those are the same means that I likely set up these equations correctly. So the network in is 48.179 and the net heat transfer out is 48.179. I'm going to clear a little more space so I can write this a little bit more legibly. Now let's see what the problem wants us to do. First up, we have the effectiveness of the regenerator. Well remember that the effectiveness of the regenerator represents how much heat is transferred divided by how much heat could be transferred if conditions were perfect. So I have to fight the urge to do my Jermaine Clement impression when I say conditions are perfect. Q-regen is going to be the heat exchange from one side to the other which could be H3 minus H4 or it could also be H1 minus H6. For Q-regen max, we are considering what would happen if everything were perfect and I think the easiest way to figure out what that is is by imagining a plot of temperature versus position. I'm going to define X as a position starting from the left. So the temperature at state 1 is supposed to be higher than state 0.6. The temperature at state 0.4 is supposed to be lower than the temperature at state 0.3. Theoretically the hottest temperature should be at position 3 and the lowest temperature should be at 6 and then we are adding heat from 6 to 1, moving heat from 3 to 4. So if we had the worst regenerator ever we would have horizontal lines between 3 and 4 and 1 and 6 and if we had a little bit of heat transfer that means 1 would scoot a little bit up relative to 6 and 4 would drop down by the same amount between 3 and 4. I guess they would drop by the same amount if we had a constant specific heat capacity here because there's a little bit of variance, it's not technically the same amount but pretty close to the same amount. Then as our effectiveness increases 1 goes up, 4 goes down until if we had an infinite amount of surface area 4 and 6 would be the same temperature and 3 and 1 would be the same temperature. Therefore, Q-region max is H3 minus H6. If you like to think about the effectiveness of the regenerator as a visual proportion it is the visual proportion of this distance here divided by this distance here. So whatever the case we can represent the effectiveness of the regenerator as H3 minus H4 divided by H3 minus H6. So 3 minus 4 would be 308.383 minus 305.17 divided by H3 minus H6 which would be 308.383 minus 270.053 and we get 0.083 or 8.3% effectiveness. So what that means is that the regenerator isn't exchanging very much heat at all and that makes sense. It's only dropping the temperature of the hot side stream by about 3 degrees and increasing the temperature of the cold stream by about 3 degrees. So low effectiveness matches what we're seeing in our state point data. Part A done. Part B asks us for the rate of heat removal from the refrigerated space. So that would be the amount of Q in. That's the heat exchanged between 4 and 5 and since we want the rate of heat removal we're going to take our mass flow rate which is 0.4 and multiply it by a specific Q in. So I'm going to calculate that first so that I don't have to scan up and down to see what numbers I'm typing in. I can just write them from the calculator. 0.4 times 76.933 gives us a Q dot in of 30.77 kilowatts. So way down here I can write Q dot in is equal to mass flow rate times specific Q in which is equal to 0.4 kilograms per second multiplied by the 76.933 kilojoules per kilogram and a kilowatt is defined as a kilojoule per second. So kilojoules, kilojoules, kilograms, kilograms. That can cancel seconds left with kilowatts. 30.77 of them specifically. Then part C asks us for the coefficient of performance. We have to figure out if this is operating in a cooling mode or a heating mode. We can deduce that it must be operating in a cooling mode because we're talking about the refrigerated space as opposed to that place where we're getting all the heat that we don't care about. That's an implication that we're using this as a refrigerator. We can also deduce the fact that a gas refrigeration system is really bad at moving heat and it's really only useful in a situation where you're trying to air condition or refrigerate something and you really, really want to conserve weight at all costs. So costly is the weight that you are willing to pay a bunch more work to do that. If you wanted a heating system it would be much more effective to heat it in other ways. As a result, we're going to call this a COPR which we can write as QIN over net work in. QIN was 76.933. So if I pop up my calculator, that would be 76.933. And our net work was 48.179. We get a coefficient of performance of 1.6. I'm leaving a bunch of decimals so that when we compare the cold air standard versus the not cold air standard we can see with a little bit more detail the difference between the two. With that, our problem in parallel universe number one is done. So all we have to do now is repeat the entire problem all over again except this time what if we had assumed constant specific heats and more specifically what if we had assumed constant specific heats at 300 Kelvin? As a result of that assumption we can replace all of our delta H's with Cp delta T's. So once we look up the Cp of air at 300 Kelvin and determine all six temperatures we can calculate the work in, the Q in, the workout and the Q out. Since the lookup is easier, let's grab that before we get into our temperatures. From table A20, the Cp of air at 300 Kelvin is 1.005. Next we need six temperatures. For that I'm going to build a little table. So 1, 3 and 5 are going to be the same temperatures and look we have three of our six temperatures. We are halfway there. To go from 1 to 2 we are going to deploy our isentropic ideal gas relations and when we deploy our isentropic ideal gas equations with the assumption of constant specific heats we have a lot of options to choose from. The option that's going to be most useful here is this one. T2 over T1 is equal to P2 over P1 raised to the K minus 1 over K. P2 over P1 is our pressure ratio which is 5 for this problem and if we knew K we could determine T2. K comes from the same table as our Cp value. We're going to use the K value of air at 300 which is going to be 1.4. Therefore T2 is going to equal T1 multiplied by 5 raised to the 0.4 divided by 1.4 power and T1 was 293.15. So 293.15 multiplied by 5 raised to the power of our K value which is 1.4 minus 1 divided by our K value which is 1.4. We get a temperature at state 2 of 464.297. I'll write that a little bit more explicitly here. Oh excuse me, T1 is not 293, it is just 273.15. So let's correct our math. T2 is 432.62. Let's compare our state 2 temperature to what it was without that assumption. It is a little bit higher when we assume constant specific heats. Interesting. Next up we can repeat our process from 4 to 5. This time around we have temperature 5 and we are trying to find temperature 4. Well we can write out our proportions the same way. I mean, in order to write this equation, I had T2 over T1 is equal to P2 over P1 raised to the K minus 1 over K. We could write T5 over T4 is equal to T5 over P4 raised to the K minus 1 over K. We recognize that we have T5. We know P5 over P4 is going to be 1.5 this time. So solving for T4, we have T5 is divided by 1.5 to the K minus 1 over K. So 193.15 divided by the quantity 0.2, excuse me, I'll write 1.5 just to be consistent, raised to the power of 1.4 minus 1 divided by 1.4 and we get a temperature of 305.915. Let's compare that to without the assumption of constant specific heats. And interesting. When we assume constant specific heats, we have a slightly higher temperature. Whatever the case though, we have enough information to be able to calculate state 6 and that is going to start to the same way as it did when we used our enthalpies, which is we're going to perform an energy balance on the regenerator. The only difference is that we are using the assumption of constant specific heats so we can replace the delta H's with Cp delta T's. So writing this as a positive quantity, I would have H3 minus H4 is equal to H1 minus H6. And I assume constant specific heats and I get Cp times T3 minus T4 is equal to Cp times T1 minus T6. Canceling the Cps because they're evaluated at the same temperature. Therefore T6 is equal to T1 minus T3 plus T4. T1 minus T3 plus T4. Let's see if I can remember that. 273.15 minus T3 308.15 plus T4 305.915. We get 270.915 and then we will double check. We should be adding heat between 6 and 1. We should be increasing the temperature slightly because again, a real lousy regenerator and then we are increasing between 3 and 4, both of those match up. That's a good sign. Now that we have all six temperatures, we can calculate our work in, our Q in, our workout and our Q out terms. So Cp value of 1.005 multiplied by T2 minus T1, which is 432.62 minus 273.15. Our work in is 160.267. Q in is Cp times T6 minus T5 and we get 78.154. Workout is Cp times T4 minus T5 and 15 minus 193.15. We get 113.329 and then Cp times T2 minus T3 1.005 times 432.62 minus 308.15. We get 125.092. Then just like before, I'm going to calculate a network and a net heat transfer and I'm going to do that in this space here. So net work in is equal to work in minus workout, scooch to the left and then net heat transfer out is Q out minus Q in. So we have 160.267, 160.267 minus 113.329 and we are comparing that number to 125.092 minus 78.154. The fact that we got the same number is a good indication that we built our equations correctly. 46.938 and 46.938. Now we can proceed to the rest of the problem. So again, we're using the same equation for the effectiveness of the regenerator. The only difference here is that we are plugging in Cp delta T and we could build that with the plot that I had drawn earlier with the temperature distribution across the regenerator, which by the way I can get rid of this axis definition. We can use the same temperature plot to relate our difference in temperatures just like we did with our enthalpies. So we're going to take the effectiveness is equal to Q Regen divided by Q Regen Max. The only difference is that we are using Cp delta T's. So Cp times T3 minus T4 divided by Cp times T3 minus T6. So 1.005, actually, since they're the same Cp, we can just cancel the terms. T3 minus T4, so 308.15 minus 305.915 divided by T3 minus T6, 308.15 minus 270.915 and we get an even worse effectiveness. Well done regenerator, 0.06 or 6%. Let's double check. I grabbed 4 and 6, which are 05.915 and 270.915. Yep, it really is just that much worse. For Part B we want Q.in, I believe, which was mass flow rate times little Q in. We have the same mass flow rate as earlier, which is 0.4 and then we're multiplying by a specific Q in which was 78.154. 78.154 kilojoules per kilogram and a kilowatt is defined as a kilojoule per second. Kilograms, cancels, kilograms, kilojoules, cancels, kilojoules, seconds, cancels, seconds. Leaving me with an answer in kilowatts. So 0.4 times 78.154 yields a Q.in of 31.26 kilowatts. Lastly, I want to calculate COPR again, which is going to be Q in divided by the network in. Q in was 78.154 divided by the network in which was 46.938. We get a coefficient of performance of 1.665 and that completes our second version of the analysis. I guess I had squared the other terms. Let me do that here. I guess boxed would be a better word. I had boxed the terms. So not that it really matters within the context of this example problem, but comparing and contrasting the analysis without the assumption of constant specific heats versus adding in the analysis with the assumption of constant specific heats. We can see that the assumption of constant specific heats means that we typically overestimate things. We're overestimating how much heat is transferred, which means that in situations like this, we're actually decreasing the result because we aren't doing as well because we could theoretically be transferring a whole bunch more because we're overestimating that amount of heat transfer. But whatever the case, we end up with a coefficient of performance about 1.6 to 1.65 depending on how you analyzed it and an amount of heat pulled from the refrigerated space of about 31 kilowatts.