 Air enters a gas turbine with two stages of compression and two stages of expansion at 100 kPa and 17°C. This system uses a regenerator as well as reheating and intercooling. The pressure ratio across each compressor is 4, and 300 kJ per kg of heat are added to the air in each combustion chamber. Also, the regenerator operates perfectly while increasing the temperature of the cold air by 20°C. Determine this system's thermal efficiency. Assume isotropic operations for all compressor and turbine stages and use constant specific heats at room temperature. So the presence of the regenerator is going to be improving the thermal efficiency, and we are also trying to get a little bit more performance by breaking our compression and expansion stages into multiple stages. Meaning instead of compressing all at once, and then getting a lot of heat build up as a result, meaning instead of compressing all at once and increasing the temperature of the air quite a bit, we are compressing a little bit, cooling it off, and then compressing the rest of the way. And in the turbine, we are expanding, allowing the air to cool off, and then reheating it, and then allowing it to expand again to get more power. Now it doesn't explicitly say that we only have two stages of compression and two stages of expansion. Excuse me, it does. Two stages. And two stages. So what I really meant to say was it explicitly tells us that we have two stages of compression and two stages of expansion. I was about to say the general rule of thumb is that if you aren't told enough information to deduce otherwise, and you're told it has multi-stage compression and expansion, it's likely two stages. You reach a point of diminishing returns. Two stages is good, three might be okay under certain circumstances, but it's not as though you're going to have 50 compression stages because your economy of scale doesn't work out. Anyway, my system diagram then is going to involve a compressor, an intercooler, a compressor, a regenerator, a combustion chamber, a turbine, another combustion chamber operating as a reheating stage, and then another turbine. And then we will be closing the cycle through a Q-out box. So I will be drawing for a while. Okay, so I have 10 state points identified here. I have two compression stages. I have two expansion stages. The intercooler cools the air between the two compressors. The combustion chamber between 7 and 8 acts as a reheating stage. Then I have my regular combustion chamber prior to the first turbine, and then a regeneration process between 9 and 10 and 4 and 5. And I'm closing my loop with a Q-out box. Therefore, I have work in occurring between 1 and 2 and 3 and 4. I have work out occurring between 6 and 7 and 8 and 9. I have Q in occurring between 5 and 6 and 7 and 8. I have Q out occurring between 2 and 3 and 10 and 1. Also note, you could have 3 or 4 compression and expansion stages, and you would just stack compressors with intercoolers and reheaters with turbines like this. So imagine a particularly devious exam question that had like 7 compressors and 7 turbines, and you just be running through isentropic. I do gas equations all day. Speaking of running through calculations all day, I have 10 state points, so I will start by building a table here. I recognize that I was told to analyze the problem by assuming constant specific heats at room temperature. Therefore, this is a cold air standard analysis. I will point out that we are doing that in an effort to try to compare and contrast the different cycles against one another, and you should be just as proficient analyzing through without the cold air standard and acting as a crutch. So you could look up all your enthalpies by knowing your reduced pressures and calculating the pressure ratio across using PR1 and PR2, etc. That'll make a little bit more sense once we get through the analysis with the cold air standard. We'll talk again about how you would do it without the assumption of constant specific heats, but one thing at a time, table. And all I know so far is the temperature at state 1. So 17 degrees Celsius plus 273.15 means T1 is 290.15 Kelvin. I also know the pressure at state 1 was 100 kilopascals, so I'm going to write 100 here. Do I know any other pressures? You're right. I know enough to know all of them. The logic being, I was told the pressure ratio across each compressor. So that means the proportion of pressures from one to two is four, and the proportion of pressures from three to four is also four. Furthermore, all heat exchange processes in the Brayton cycle are assumed to be isobaric, therefore P2 is equal to P3, P4 is equal to P5, which is equal to P6, P7 is equal to P8, and P9, P10, and P1 are all the same. So I will be populating pressures for a while here, 100 and 100, and then two and three. And I should point out that we weren't explicitly told that each turbine expands the same proportion, but the rule of thumb is if you don't know enough to deduce, otherwise you assume that the turbines are trying to split the load evenly, and part of that is going to be expanding by the same proportion. So even though we weren't actually told that the turbine, the first turbine, the high pressure turbine expands until it reaches 400 kilopascals, and then the second one expands from 400 down to 100 that is what we assume is the case. So you might see some given information like the high pressure turbine expands to a pressure of 1000 kilopascals, the rest of the expansion is completed in the low pressure turbine, or something like that. With that we have all the pressures because of the assumption of constant specific heats evaluated at room temperature, for which I'm grabbing the CPCV and K values of air at 300 Kelvin, which is going to be 1.005 kilojoules per kilogram Kelvin, 0.718 kilojoules per kilogram Kelvin, and 1.4 respectively. I can start working my way through the state points. I begin with one to two. The process from one to two is an isentropic process. Because of the air standard, I'm treating the air as an ideal gas, and because the problem told me to use constant specific heats, I can use our isentropic ideal gas equations. Meaning I'm going to be applying T2 over T1 is equal to P2 over P1 raised to the K minus one over K from one to two. Therefore T2 is going to equal T1 multiplied by four raised to the 1.4 minus one over 1.4. So that's going to be 290.15 multiplied by four raised to the power of 1.4 minus one divided by 1.4. And we get 431.161. And if I can manage to switch back, come on. You can do it. There we go. So I'll write that down for T2. And then the process from two to three is the intercooler process. We don't know enough information explicitly to know what that delta T is, but we can figure out T3 by looking at our assumption that the compressors and the turbines are trying to split the load evenly unless we're told otherwise. So our compressors are trying to have the same work in. The work in to compressor one is equal to the work in to compressor two. And the specific work of that compressor is going to be delta H because there's no opportunities for heat transfer because they are isentropic and therefore we are implying adiabatic. So I have H2 minus H1 is equal to H4 minus H3. And then because of the assumption of constant specific heats, I have Cp delta T, so T2 minus T1 and Cp times T4 minus T3. They're evaluated at room temperature. So this Cp is the same as this Cp. So I have T2 minus T1, which I know is equal to T4 minus T3, which I don't know. But remember that it's the same delta. The other thing that we know about the processes at one to two and three to four is that they are described by the isentropic ideal gas equations. So I can write T2 over T1 as being equal to four raised to the k minus one over k. And I can also write T4 over T3 as being equal to four raised to the k minus one over k. Now that constant four raised to the k minus one over k is the same from one to two as it is from four to three to four. Therefore I can write T2 over T1 is equal to T4 over T3. Therefore I have two equations and two unknowns. And I don't even have to go through the algebra to figure out what the solution is to this. What temperature at state three would yield a state four temperature that is going to have the same delta and the same proportion as one to two? You're right. The solution to this is that if I'm assuming the compressors are splitting the load evenly, T3 is equal to T1. So the intercooler cools back down to where it started or rather where that first compressor started. And I will point out that that is made as a result of an assumption. I want to note that with an asterisk to point out that we are assuming that the compressors are splitting the load evenly. But that's how we handle multi-stage compression and expansion. Unless of course we were told enough information to deduce otherwise. Spoiler, like the turbines. Then I can analyze from three to four in the same way that we analyzed from one to two using our isentropic adiogas equations. This time it's going to be T4 over T3 is equal to four raised to the K minus one over K. So calculator, if you would come back please. I'm going to take 290.15 multiplied by four raised to the power of Pp, nope. Presley, Presley, 1.4 minus one divided by 1.4. And we get, hey, would you look at that? The same T4. Interesting. And with that, we've made it through our compression stages. We had multi-stage compression with two compressors and it explicitly told us we had two compression stages. Then how do we get from four to five? There is a key piece of information in the problem that tells us how to get from four to five. See if we can spot it right there. It looks like I'm circling the word combustion. I'm not trying to circle the word 20. I guess the number's 20. The phrase by 20 degrees Celsius. The problem tells us that the regenerator operates perfectly while increasing the temperature of the cold air by 20 degrees Celsius. So it's saying the difference in temperature between four and five is 20 degrees Celsius. And it's an increase in temperature because that's what the regenerator is trying to do. Therefore, T5 is going to be 20 degrees higher than T4. So I can take 430.161 and add 20 to it and I get 451.161. Then how do we get from five to six? Well, again, there is a piece of information that allows us to calculate what T6 is. See if you can spot it. That's right. It is this. I was told that 300 kilojoules per kilogram of heat are added to both combustion chambers. That is, each combustion chamber is increasing the energy of the air by 300 kilojoules per kilogram. So Q in for the combustion chamber between five and six is going to be 300 kilojoules per kilogram. And if I set up an energy balance on that combustion chamber knowing that I have no opportunity for work, it's steady state operation. I have neglected changes in kinetic and potential energy. Therefore, I'm left with Q in is equal to 300 kilojoules per kilogram, which is equal to H6 minus H5. And you know what's coming next. We are told to assume constant specific heats so we can plug in CP times T6 minus T5. Therefore, if I were to take 300 kilojoules per kilogram divided by CP of air evaluated at 300 Kelvin and add that to T5, I would have T6 at CP of air evaluated at 300 Kelvin is 1.005. We know T5 is 450.161. So in order to come up with T6, all we have to do is take 300 divided by 1.005 and add it to that 451 number and we get 749.669. We made it through the high pressure side. Now we can start our expansion process. We are assuming that the turbines are splitting the load as evenly as possible, meaning that they are assumed to both have the same expansion ratio that is equivalent to the compressor's compression ratio. And I know that I have an isentropic process because our turbines have no isentropic efficiency listed. So we're treating them as being perfectly efficient, meaning we can use our isentropic ideal gas equations to calculate the temperature after the turbine. So we're saying T7 over T6 is equal to P7 over P6 raised to the K minus 1 over K. So I'm going to take that same number we just got our inlet condition. I'm going to multiply by 1 over 4 this time and I'm going to raise that to a power of, okay, let's try that again, multiplied by 1 over 4 and then raise to the power of 1.4 minus 1 divided by 1.4 and we get 504.49. Now you remember earlier, because we didn't know enough about the intercooler to determine anything else, we assumed that the compressors have the same work, therefore T3 is equal to T1. We can't do the same thing with the process from 7 to 8 because we have a piece of information that supersedes that assumption. It is the fact that that combustion chamber is also adding 300 kJ per kilogram. So in order to come up with T8, we are going to have to use 300 divided by 1.005 plus T7. So I pop my calculator back open and I take T7, the quantity that we just calculated and I add 2 at 300 divided by 1.005 and I get 802.997. Then to get from 8 to 9, I'm going to use my isentropic ideal gas equations, meaning I'm going to take T8, I'm going to multiply by 1 over 4 raised to the power of 1.4 minus 1 divided by 1.4 and I get 540.377 and then for 10, I'm going back through the regenerator and remember that the heat rejected from the hot side is going into the cold side, meaning that the change in enthalpy between 9 and 10 is going to be the same as the change in enthalpy between 4 and 5 that delta H is going to be represented as Cp delta T because I've assumed constant specific heats, meaning delta T between 4 and 5 is equal to the delta T between 9 and 10. Therefore, T10 is going to be 20 degrees less than T9, which is 520.377. And with that, I have all 10 temperatures, which means that the world is my oyster. I can calculate whatever else I want, including but not limited to work in, queue in, work out, and queue out. So we've done this a couple of times now. Why don't you pause the video here and see if you can generate work in, queue in, work out, and queue out. I'll give you about five seconds. The work in occurs between 1 and 2 and 3 to 4. That energy balance will simplify to H2 minus H1 plus H4 minus H3. Queue in occurs between 5 and 6 and 7 and 8. So I will write that as H6 minus H5 plus H8 minus H7. The work out occurs between 6 and 7 and 8 and 9. So I will write that as H6 minus H7 plus H8 minus H9. The queue out occurs between 2 and 3 and 10 and 1. So I'm going to write that as H2 minus H3 plus H10 minus H1. Did you get it right? Of course you did. Then I'm going to be assuming constant specific heats here and plugging in CP delta T in place of my delta H's. So I really have CP times T2 minus T1 plus CP times T4 minus T3 for work in, CP times T6 minus T5 plus CP times T8 minus T7 for queue in, CP times T6 minus T7 plus CP times T8 minus T9 for work out, and CP times T2 minus T3 plus CP times T10 minus T1 for queue out. And because it's the same CP, I can factor it all the way out. Therefore, I can jump all the way to the work in is CP times the quantity. That's not what I wanted iPad. Come on. Always be scrolling up T4 minus T3 plus T2 minus T1. And then queue in is going to be CP times that was 6 minus 5 and 8 minus 7, 6 minus 5 plus 8 minus 7. And that should be 600 if everything goes according to plan. And work out occurs between 6 and 7 and 8 and 9. So I will write that as 6 minus 7 plus T8 minus T9 and queue out occurs between 10 and 1. That would be T10 minus T1 and 2 to 3, so T2 minus T3. The CP value I'm using is 1.005. So we have everything we need to just start rolling here. I'm going to be using 1.005 multiplied by T4. That's 1.05, John. Try that again. 1.005 times the quantity. 431.161 minus 290.15 plus 431.161 minus 290.15. And note that I'm not scrolling up to grab the numbers. I'm just plugging in what I wrote down. So there might be a little bit of error when I'm comparing the network out. When I'm comparing the network out to the net heat transfer in. So if it's off by like a couple of hundreds, it's okay. So my work-in is 283.432. Then I do the same process for queuing. This time taking 749.669 minus 504.49 plus 8 minus 7. Oops, I subtracted the wrong thing. T5 was 450.161. Then I'm adding T8 802.997 minus T7, which was 504.49. I get a queue in of 601.005. And then I am taking T6 minus T7, so 1.005 times T6 minus T7, 749.669 plus T8 minus T9, which is 802, excuse me, I'll ahead of myself again, minus T7, which is 504.49 plus T8 802.997 minus T9 540.377. The workout of 510.339 and queue out is last but not least, I'd be 520.377 plus minus 290.15 plus 431.161 minus 290.15 373.094. So queue out was at the bottom. I'll start there because that seems to make the most sense. 373, try that again, 373.094. And my workout was 510.338. That is the original is per kilogram, not another decimal place. What am I doing? 601.005. And then work in was 283.432. Okay. That queue in being so far off from 600 is a little bit surprising. I'm just going to go double check that I had plugged in those numbers correctly. I mean, I wrote those numbers out to like three decimal places. I shouldn't be off by one. Okay, 749.669 minus T5, which was 450.161. Type that correctly. And then I'm taking 802.997, which is T8 minus 504.49. Yeah, D7. Did I write those correctly? Just mosey on through my calculations here and 540.377, 802.997, 504.49, 749.669, 5451.161. Guys, was that the one where I was adding 20? Guys, did I really not add 20 correctly? Oh my gosh. I am so sorry that you had to sit through that for the last like 15 minutes, because I'm sure you caught that right away. You're just like, what are you doing? How did you add 20 and end up with 450, John? How does that even happen? Well, that means anything with a 5 in it has to be recalculated, which would be queue in and that is it. So queue in. Let's try that again. The queue in calculation is going to be 1.005 multiplied by 749.669 minus 451.161 plus whatever. Hey, look, we got 600. And if I know my calculator, it's actually 600.005, but it's just a jerk who sometimes likes to arbitrarily hide decimal places at 0.000075. Wow, that's surprising. Okay, fine. I'll accept your rounding. And I'll change this into a two and less of a snake symbol. Then I want us to calculate the network out and the net heat transfer in the network out is going to be the workout minus the work in the net heat transfer in is going to be the queue in minus the queue out. Network out was 510.338. And then I'm subtracting work in, which was 283.432. And I get 226.906. Queue in is 600 minus 373. And I get 226.906. So 226.906. And 226.906. And then thermal efficiency is going to be the network out divided by the queue in, which is 226.906 divided by approximately 600. We get 0.378, which is equal to 37.8%. Then if we check our math with MATLAB, we see that the thermal efficiency without any rounding errors was calculated to be 37.81%. Additionally, MATLAB can generate a PV and TS diagram for us, which allows us to see that our process here looks just like a regular rate in cycle, except we are kind of stretching the compression process to the left and the expansion process to the right. But I like to think of that as instead of trying to hit a certain horizontal position when you're going from left to right by achieving a super high temperature, you are instead going up, down, and then up again, allowing you to get further to the right without going crazy high on your temperature. That temperature would likely yield a failure of some sort in the turbines, which is probably why you would want to expand across multiple stages, get more network, but without the downside of the crazy high temperature. Also note that the regeneration process just introduces a couple of state points here, and that state point system appears on the same lines. That's our high pressure line and low pressure line. So we are just indicating that the heat transfer from five to six is now no longer this entire distance from four to six, it's now just five to six. And this horizontal displacement was the heat transfer that we quote saved unquote by not having to burn fuel. That was the return on investment that we were getting for our regenerator. Now what I would like to do next is actually work through this problem again, this time without the cool air standard. And instead of talking through it, I'm actually going to do it so that you can follow along if you want. And I'm going to break this out into a separate video.