 air enters the compressor of a regenerative stationary gas turbine engine steadily at 100 kPa, 27°C and 5 kM per second. The engine features a pressure ratio of 10, and is designed to induce a maximum temperature of 1400 Kelvin. Assuming the cold air standard and that the effectiveness of the regenerator is 80%, determine and complete the following and compare the results to Brayton cycle number one. First I want us to determine the specific network output of the cycle, and then the thermal efficiency of the engine. I will start with the diagram we had generated when we were talking about regeneration, and in order to complete the cycle here I will add in a cooling process between 6 and 1. That's the Q out box that we've been analyzing, which again doesn't exist. And then I recognize that my work in is only occurring in the compressor, my work out is only occurring in the turbine, my Q in is only occurring in the combustion chamber, and my Q out is only occurring in the Q out box. And to be consistent with the rest of the energy components I had written, I will write that as a specific Q out instead of a Q dot out. The regenerator itself is treated as being adiabatic, and all of the energy that is exchanged is exchanged internally. That means it doesn't cross my system boundary when I analyze the entire cycle. So when I'm calculating Q in, Q out, and Q out for the entire cycle, I don't include Q regen, because again it doesn't cross the boundary of the entire cycle. Anyway, I know that I'm going to have six state points here, so I will start generating a table. I know that I have an inlet temperature and pressure, that's state one's temperature and pressure, of 27 degrees Celsius and 100 kilopascals, and I know I have a pressure ratio of 10. And because all heat exchange processes in the Brayton cycle are assumed to be isobaric, that means that I actually know all of the rest of my pressures. Because P2 will be after the compression process, therefore it will be 10 times P1, and then P2, P3, and P4 are all the same pressure, and P5, P6, and P1 are all the same pressure. So this is a thousand, this is a thousand, this is a thousand, because 10 times 100. This is 100 and this is 100. Then I was told the maximum temperature was 1400 Kelvin, and I know that the maximum temperature is likely to occur after the heat addition process, so we will treat it as being T4. Remember it's always hottest after the fire happens, and we can come up with two of our state points at 3 and 6 by analyzing the regenerator because we know the of our regenerator. Then going from one to two is going to involve the pressure ratio and our isentropic ideal gas equations, going from four to five is going to use the pressure ratio and the isentropic ideal gas equations as well. So that should be enough to populate all the temperatures, which because I'm using the cold air standard is going to be enough to calculate the work in, the Q in, the workout, and the Q out, which will allow me to determine the network out and the thermal efficiency of the engine. So again, that's due to the fact that the problem told me to assume the cold air standard. If I wasn't using the cold air standard in this problem, I would have to look up PR1 and then use the fact that PR2 over PR1 is equal to P2 over P1, because that would be our isentropic ideal gas equation without the cold air standard. Then I would use PR2 in the tables to look up whatever I needed at state two, including but not limited to the enthalpy at state two. And I could calculate what my enthalpy would be at four by using the tables and the temperature. And then I would look up the PR4 as well, use PR4 over PR5 is equal to P4 over P5, again isentropic ideal gas equations and look up the enthalpy and probably the temperature corresponding to state five. And then I would set up my energy balance on the regenerator and use the effectiveness of the regenerator to determine H6 and H3, at which point I would have all of the enthalpies required to determine the work in the Q in the workout in the Q out, and therefore network out and thermal efficiency of the engine. Do you follow all that? Cool beans. So for my isentropic ideal gas equations from one to two, I'm using T2, that's a highlighter, it's unhelpful. I'm using T2 over T1 is equal to P2 over P1 raised to the K minus one over K. That again, came from our isentropic ideal gas equations when we've made the assumption of constant specific heats and other pressure ratios describing the pressure ratio across the compressor. Therefore P2 is 10 times higher than P1. So I'm plugging in 10 in place of P2 over P1. Therefore I can calculate T2 by taking T1 multiplied by 10 raised to the K minus one over K. We know T1, we know the pressure ratio, all we need is K and we can grab K from table A20 in our textbook. The cold air standard says to assume constant specific heats at 300 Kelvin. So we are using the properties at 300 Kelvin, which are CP of 1.005, a CV of 0.718, those are in kilojoules per kilogram Kelvin, and a K value of 1.4. So our temperature at state two, if the calculator would agree to help us out here, is going to be 300.15 times 10, 10 raised to the power of 1.4 minus 1 divided by 1.4. And we get a temperature at state two of 579.499. So I'm going to be jumping back to this page and writing down 579.499. And I could probably run that to 580 or 579.5, but we are going to be arbitrarily precise for the moment. So we got it from one to two and we can't get to three yet because we don't know five. So instead we have to jump to four and then analyze the process from four to five. Because I wasn't given an isentropic efficiency of our turbine, I'm assuming it's 100% efficient like the compressor. Therefore I'm using the isentropic adiogas equations again, which means we are going to be taking T5 over T4 is equal to P5 over P4 raised to the K minus one over K. We know P5 over P4 that's going to be one over 10 because we are going from the high pressure to the low pressure. So the after the turbine state point divided by the before the turbine state point is going to be a low number divided by high number, which is one over 10. So that would be 1400 Kelvin multiplied by one over 10 raised to the power of 1.4 minus one divided by 1.4. And we get 725.126. And again, arbitrarily precise my decimal points there. So we have four and five, we have one and two. In order to come up with three and six, let's write out what we know about our regenerator. I'm going to be saying the amount of heat exiting the hot side, key region is going to be equal to the amount of heat entering the cold side, which is also key region. And remember, from our discussion of the regenerator by itself, I'm going to be characterizing key region as H5 minus H6 from an energy balance on the hot side. And the amount of heat entering the cold side is H3 minus H2 from an energy balance on the cold side. And then because of the cold air standard, meaning that I'm assuming constant specific heats, I can plug in Cp delta T and place a delta H. So I can write Cp times T5 minus T6, and that's equal to Cp times T3 minus T2. And then because the Cps are evaluated at the same temperature, I can say T5 minus T6 is equal to T3 minus T2. I have one equation and two unknowns, because I don't know three and I don't know six. The other equation is going to come from our definition of our effectiveness of the regenerator. We know that that's 0.8 from the problem statement. And remember that that describes key region over key region max, which I can write out as H3 minus H2, or I could have written out H5 minus H6. And then I'm substituting in T3 for T5 because under maximum heat transfer conditions, the temperature at 3 and 5 is going to be the same, and the temperature at 2 and 6 is going to be the same. So instead of plugging in H3, I'm plugging in H5. Because remember, for an ideal gas, like in the air standard, the enthalpy is only a function of temperature. So the temperatures are the same, the enthalpies must be the same as well. And then because of the cold air standard, the assumption of constant specific heats, plugging in CP, delta D in place of delta H, and then they are evaluated at the same temperature, which is 300 Kelvin, so I'm canceling the same CP. Therefore, 0.8 is equal to T3 minus T2 divided by T5 minus T2. That gives me my other equation from which to calculate my two unknowns. And in fact, by writing it out this way, that is 3 minus 2 over 5 minus 2, I have managed to write it in terms of only one unknown, and that unknown is state 3. So I can use this to calculate T3, which I will do by performing algebra, hopefully anyway. So I am going to say T3 is equal to 0.8 times the quantity T5 minus T2 plus T2. So calculator. Here we go. Let's do this. 0.8. I'm just remembering that it said 80%. I'm going to double check that it is indeed 80%. Multiply by T5 minus T2, which is that 725 number minus that 579 number. And then I'm adding to that T2. Yes, indeed. And I get 696.001. And that was T3. And then I can solve this equation for T6, which would be T5 minus T3 plus T2. And I now know 3. So that's going to be this 725 number minus this 696 number plus that 579 number. Let's just double check that. 725, 696, and 579. Yep, it seems right. Therefore, T6 is 608.624. Another way of double checking that I did that correctly here, which probably isn't as relevant to you guys since you're presumably looking at a single sheet of paper all at once as opposed to a really tiny zoomed-in section. Another way to double check this is to double check that your delta T from 2 to 3 is equal to your delta T from 5 to 6. So from 2 to 3, we are increasing the temperature by, what, about 120 degrees? That's 120 Kelvin as well. So from 5 to 6, the temperature should drop by about 120, and it looks like it is. So that's a good sign that we use the right state points. Now armed with our temperatures, we can calculate our work-in, Q in, workout, and Q out. And then from that, we can determine a network out and a net heat transfer in. We can look at them and go, ah, they're the same number. Hooray, we did it correctly, and then determine a thermal efficiency. So work-in is occurring between 1 and 2, so an energy balance on the compressor is going to simplify down to specific work-in is equal to H2 minus H1. Q in is occurring between 3 and 4. Remember that the heat exchange within the regenerator does not cross the cycle boundary. So an energy balance on the combustion chamber is going to simplify down to specific Q in is equal to H4 minus H3. Work-out occurs in the turbine, and energy balance on the turbine is going to be H4 minus H5 for workout. And energy balance on the Q out box yields H6 minus H1 for Q out. With each progressive example problem that we complete together, I'm going to be going into less and less detail about where these equations are coming from. If you've been following along, you probably have a good idea of how to come up with these. And in fact, you should be pretty good at it by now. It might be worth just practicing some drills, looking at a cycle, and trying to write out these equations before you even get to the state points themselves. It's a good skill set to have. Then because of the colder standard, I'm plugging in Cp delta T. So this would be Cp times T2 minus T1, Cp times T4 minus T3, Cp times T4 minus T5, and Cp times T6 minus T1. And I'm starting to run out of space here. So I will make this entire thing just a little bit smaller. So Cp, again, from table A20 is 1.005. So we are using 1.005 multiplied by, calculator comes back, cswc. Thank you for your input calculator. We are using 1.005 times 579.499 minus 300.15 for work in. So work in is 280.746. And then 1.005 times T4, 1400 minus T3, which was 696.001. And that's 707.519. I'm going to write those down before they get lost. 280.746 and 707.519. And then I repeat the process for work out in Qout. It's 4 minus 5. So 1.005 times 1400 minus 725 ish. I'm going to work out of 678.248 and Qout is going to occur between 6 and 1. So 608 minus 1 yields 310. So my workout was 678.248 and Qout was 310.017. These are all kilojoules per kilogram. You know what's happening next. I want network out, which is going to be work out minus work in and net heat transfer in, which is Q in minus Q out. So I'm going to take 678.248 minus 280. And I get 397.5. And I'm going to take 707.519 minus 310. And I get 397.502. Parade and the same. That's a good indication we built those equations correctly. And we didn't plug in any incorrect numbers. At least if we did, we did it very consistently. Then I can determine my thermal efficiency by taking the network out over Q in, which is going to be 397.502, this number, divided by Q in, which was 707.519. And I get 0.5618 or 56.18%. And I was looking for this and this as my answers. Then I asked us to compare that to the first two brain cycle examples. So I will copy over those results. And then we can look at them at the same time and go, ah, look, two was lower, three is higher. Interesante. So we got 397.5. We got 56.18%. So in comparing these, I see that the presence of isentropic efficiency has decreased my thermal efficiency and my network out. And remember that the presence of the regenerator will decrease how much Q in is required for the same amount of network out. So it doesn't surprise us that we have the same network out because we have the same maximum temperature. So we're decreasing how much Q in was required. Therefore, we are increasing the thermal efficiency. So the presence of the regenerator is going to increase the thermal efficiency. And remember, even a terrible, terrible regenerator is going to be improving our thermal efficiency. So hopefully we can find a good sweet spot between spending more money on a heat exchanger that works pretty well, because theoretically to achieve 100% effectiveness, we would need an infinite amount of surface area, which is hard to do with just fins and more surface area. And that investment of money is going to mean that we're paying less money in fuel, and we could probably spend a little bit more money on our turbine and compressor to try to get a slightly higher isentropic efficiency. And it would be your job in this situation to calculate what the breakeven point is. Where is the threshold of spending more money on a regenerator or a compressor or a turbine isn't actually paying off in the long run using projections of fuel prices?