 Air enters the compressor of a stationary gas turbine engine steadily at 100 kilofascals, 27 degrees Celsius, and 5 cubic meters per second. The engine features a pressure ratio of 10 to 1 and is designed to induce a maximum temperature of 1400 Kelvin. Assuming the cold air standard and that the operating efficiency of compressor and turbine are 90% and 85% respectively, determine or complete the following and compare the results to Brayton cycle number one. A, the specific network output, note that I'm asking for a specific network out and not the power output, and B, the thermal efficiency of this engine. So this setup is the exact same as our previous example. The only difference is that we have isentropic efficiencies for the compressor and turbine. Therefore, they are not isentropic anymore. So I will bring over my diagram because it's going to be the same engine. The only difference here is that I have to establish six state points. Which six state points I hear you ask? Well, we have one and two still and we have three and four still. But now we have to establish two s and four s as ideal representations of the state point. For the process from one to two, for example, we have an isentropic efficiency. Therefore, we establish two s as the state point that represents what the outlet would be if it were isentropic. And then two actual, which is just two, which is the actual state point two. And then the way that I get from, oh, looks like Q out and completely copy over. There we go. Then the way that I get from two s to two is by using the isentropic efficiency of our compressor. And the way that I get from four s to four is by using our isentropic efficiency of the turbine. And remember that the isentropic efficiencies represent the proportion of ideal work to actual work. For example, in the compressor, does it take more work to compress air in an ideal world where everything is perfect or in reality where there's friction and losses and stuff? That's right. It takes more work to compress in the actual case because you are having to overcome the friction and losses and stuff. Therefore, we write this as work in actual in the denominator. It's the bigger one and work in ideal in the numerator. And from our energy balance on the compressor from the previous version of this problem, we know the work in is going to be h2 minus h1. The only difference between these two equations is that in the numerator, this is h2 s minus one. That's what the outlet enthalpy would be if everything were perfect, if there were no losses. And then because we're using the Cold Air Standard Analysis, we can plug in Cp times delta T in place of delta H. And because these are evaluated at the same temperature, 300 Kelvin for both, I can cancel the Cps. Then on the turbine, let's think through it. Do you get more work out of an ideal turbine where everything's perfect or an actual turbine where you're losing some energy to friction and losses and stuff? That's right. You get more work out of an ideal turbine. The presence of losses decreases how much work you get out. And from our energy balance on the turbine, we know the specific workout is going to be h3 minus h4. And again, the only difference between the two is that in the ideal case, this is h4 s. And then I plug in Cp delta T because it's a difference in enthalpies. And I cancel the Cps because they're evaluated at the same temperature. So my isentropic efficiencies for both the compressor and the turbine are just a proportion of temperature differences. So I can get to T2 s in the same exact way I did in the previous problem because remember that in this first iteration of the problem, we had an isentropic process across the compressor. Therefore, our T2 in this problem is T2 s for the second problem. Similarly, T4 in this problem is T4 s in the next iteration of the problem. And then to get from T2 s to T2 actual, I use this equation, knowing T1, T2 s and eta c. And then to get from 4 s to 4 actual, I use this equation, knowing T3, T4 s and eta t. So since I know all of the temperatures and pressures for 1, 2 s, 3 and 4 s already, I can populate those. And my pressures are the same at 2 and 2 s and 4 and 4 s because that's the thing that I'm holding constant in my comparison of the different devices. I'm figuring out how much work is required to compress from 100 to 1000 by considering different compressors, an 85% efficient compressor or a 90% efficient compressor or a per 100% perfectly efficient compressor, an isentropic compressor. It would be like if I asked you to determine how much fuel is required to drive from New York to San Francisco in a 2019 Geometro versus a 2019 Ford F-150 versus a 2019 Honda Pilot. The thing that is constant between all three of those is the distance driven. You're just comparing how much fuel is consumed in getting there. Similarly, I'm figuring out how much work is required to compress from 100 kilopascals to 1000 kilopascals with different compressors. The 1000 kilopascal pressure at the end is the target that they all have to hit. Therefore, the pressures are the same. Now, I can solve this for T2. Everyone stand back doing algebra. But no, it is fraught with possibility of error. It'd be T2 s minus T1 divided by A to C plus T1 and the isentropic efficiency of the compressor was 90%. Therefore, I am going to take 579.499 minus 300.15 divided by 0.9 plus 300.15 and I get 610. By the way, a quick way of checking that your numbers make sense is by remembering that friction tends to make things hotter. So, does it make sense that the outlet of the less efficient compressor is a higher temperature? Yes, it totally does. And then I am going to be solving this equation for T4. Again, that concerns everyone. I'm doing algebra. T3 minus. So, put your to the right. T3 minus. There we go. Yeah. So, 1400 minus the A to T was 85. So, 0.85 times T3 minus T4 s, 1400 minus 725.126. And I get 826. And again, I will sanity check by comparing my 4s to my 4 and asking the question, does it make sense that the temperature increased as the result of friction? Yes, it does. Armed with these six temperatures, I can make the same calculations that I had done back over here. I am going to be taking Cp times delta T for everything. Now, I know what you're thinking. You're thinking, John. So, are these state points different? Are they T2 s instead of T2 or T4 s instead of T4? No. These are the actual temperatures. So, I am using the actual T2 and the actual T4 in these evaluations. So, Cp is still 1.005 as per table A20. And I am multiplying by 610.538 minus 300.15 instead of 579.499 minus 300.15. And I get 311.94. And then, instant repeat. So, next up was T3 minus T2, which would be 1,400 minus 610.538. And then after that was 3 minus 4, because that's worked out. So, 1.005 times 1,400 minus 826.357. And then was Qout, which is 4 minus 1, because remember, it's the process from 4 to 1, beginning minus end would be 4 minus 1. 1.005 times 826.357 minus 300.15. And armed with these four quantities, I can calculate the net workout and the net heat transfer in. The net workout would be workout minus work in, which is going to be 576.511. I'll use the actual number without the rounding that my calculator did behind the scenes. Jerk calculator. So, that I can have as close to the same number as possible. So, as to be able to point at them and go, look guys, they're the same. And then Q in minus Q out. So, 1,094 minus 528. You ready? Here we go. Look guys, they're the same. Hold up. That's not right. Okay. What did I do wrong? Hey, by the way, our numbers weren't the same. So, that's the sign that we did something wrong. That's the big hint to us that we made a mistake. That's why you calculate both net workout and net heat transfer in. If you don't check yourself, you will proceed direct yourself like we just did. Luckily, we caught it because of that built-in check and balance system. So, 826 minus 300.15. Yeah, that seems right. And then workout was 3 minus 4. Yeah, Q in is 3 minus 2. Ah, that would be my issue. That was 3 minus an entirely hypothetical temperature that didn't actually exist because I typed 310 instead of 610. Now, Q in is a much more reasonable 793.41. Okay, let's try this again. This number minus 528.838. Hey, look guys, they're the same. That's cool. And then with that new specific network out, we can calculate the thermal efficiency. And that thermal efficiency is going to be network out over Q in, which is going to be 264.571. I'll use this one instead of the other one. And we are dividing by 793.41. See if we had calculated the net Q in, even though we didn't have to. We wouldn't have caught that and we would have a very inaccurate number because we would be dividing by 1094 instead of 793. And we get 33.3%. And these were the two things we were comparing. So, I have this table down here so that if you were trying to work this problem by hand, you could populate over the properties. I didn't because I have the ability to just scroll over very easily. But it might have been useful for you to keep track while you're working through them. All I will write out is the thermal efficiency in the network out. That was 33.3% versus, I think that was 48.2. Am I remembering it correctly? No, I'm not. 48.2. And our network out here was 33, nope, 264.57 and 397.502. And then let's reflect on those numbers for a bit. So we went from an isentropic efficiency in our compressor of 100% to 90% and an isentropic efficiency in our turbine of 100% to 85%. And look how much of an effect that had. A 90% efficient compressor is still pretty dang efficient and yet it dropped our thermal efficiency from 48% to 33%. If you were operating this stationary gas turbine for power production for a region, the difference of 15% of your thermal efficiency is going to make a huge dent in how much fuel you have to burn in order to get a certain amount of network out. That is going to have huge implications on the operating costs of that facility. This sort of reduction in thermal efficiency is why we are going to spend the next few example problems talking about what seems like a very small modification just to try to get a little bit more thermal efficiency. Because every improvement we can make can make up some of that loss in thermal efficiency. And over the scope of 20 or 30 years of operation that can have a huge effect, it might be worth it to invest more money to have a slightly more efficient system, if it means we're spending less money on fuel, etc.