 Ideal air enters an adiabatic turbine at 1000 kilopascals and 500 kelvin. It is expanded to 100 kilopascals. The turbine has an efficiency of 75%. Assuming constant specific heats at room temperature determine first the temperature at the outlet in kelvin, the specific work produced by the turbine in kilojoules per kilogram of air, the specific entropy generated by this process in kilojoules per kilogram kelvin, and then I want us to postulate on how those answers change if the efficiency dropped to 65%, and I want us to plot these state points on a TS diagram. So an energy balance on my turbine is going to yield the specific work out is H1-H2. That's going to help me break the isentropic efficiency of my turbine into a more helpful quantity. Now in order to build this equation, instead of going to our equation sheet, let's think through this. Do you get more work out of a turbine in a perfect world where there's no friction, or in reality where there's friction and losses and stuff? You're right. The ideal work is greater, therefore it's in the denominator. Because of the simplification of my energy balance, that means the specific work actual is going to be H1-H2, and the ideal work is going to be H1-H2S. Remember, H2S does not exist. It is a stepping stone to get to H2 actual. Now I can take this one step further by recognizing that the problem specifically told me to assume constant specific heats at room temperature. For ideal air, remember that Cp is defined as del H del T, and Cv is defined as del U del T. So because I have a delta H, I'm going to be using Cp delta T in place of delta H. It is not a constant pressure process. Cp has nothing to do with that anymore. If it's an H, I use Cp. If it's a U, I use Cv. Ultraviolet, Harry Potter. Furthermore, because I was told to use specific heats at room temperature, that means that the Cp value I use is going to be the same regardless of what delta T I'm analyzing. Therefore they cancel. So this particular circumstance is going to yield a relationship of A to T is equal to T1-T2 divided by T1-T2S. That is only for this specific case where I've assumed constant specific heats at room temperature. Do not just write that equation down on your equation sheet and hope to remember when you can use it. I know T1, I can figure out T2S, and then I will have delta T. So T1 is 500 Kelvin. For T2S I'm analyzing an isentropic process. So how do I do that? Well, we just worked a problem with R134A. In that case we had looked up S1, used that equal to S2S to determine any other property that has 2S. At state 2S. Here though, I can't go look up the entropy. Because for my properties for air on table A22, I do not have entropy. I have this quantity which is different. This is a zero point entropy. We are going to be ignoring this column for the time being. Your solution manual will involve this entropy quantity in calculations. I don't want you to do that. What I want you to do instead is use the isentropic ideal gas equations. So we have this set of equations that are only applicable if I have constant specific heats for an isentropic process of an ideal gas. Are all three of those requirements met in this process? Only from 1 to 2S. Not from 1 to 2, only from 1 to 2S. So in the hypothetical analysis, from 1 to what the ideal output would be, I have an isentropic process. Therefore, I can use any of these equations. The one that is going to be most useful is the one in the upper left hand corner. So I'm going to say T2S divided by T1 is equal to P2S over P1 raised to the k-1 over k. Therefore, T2S is going to be equal to T1 multiplied by P2S over P1 raised to the k-1 over k. T1 is known, it's 500 kelvin. P1 is known, it is 1000 kilopascals. P2S is equal to P2, which is 100 kilopascals. I know everything in this equation except for k. For k and Cp, I'm going to look up the specific heat capacities for air at room temperature. For that, I'm going to go to table A20. We're going to use room temperature as 300 kelvin. At which point my Cp value is 1.005 kilojoules per kilogram kelvin, my Cv value would be 0.718 kilojoules per kilogram kelvin if that was helpful for this problem, and my k value is 1.4. So on table A20, k is 1.4, Cp of air is 1.005. Now, T1 was 500 kelvin, so I'm going to take 500 kelvin multiplied by P2S, it was expanded to 100 kilopascals, I believe. Yep, divided by P1, which was 1000 kilopascals, and then I'm raising that to the power of k-1, so 1.4-1, divided by k, which is 1.4. The T2S value that I get as a result is 258.974. Remember that because I'm using an absolute proportion here, T2S over T1, my temperatures have to be in kelvin, therefore T1 was in kelvin and T2S is going to be a result in kelvin. Now is that the temperature at the outlet? No, it is not. We have to use the isentropic efficiency in order to solve for the temperature at the outlet. So we can say T2S is equal to T1-A to T multiplied by T1-T2S. So T1 was 500, A to T was 75%, so 0.75 multiplied by 500 minus 258.97. Our temperature is going to be, excuse me, that was supposed to be T2 actual. Did I do the rest of the algebra right? Yeah, I did. So T2 actual is going to be 319.23. That's my actual temperature at the actual outlet, which is part A. For part B I was asked what the specific work produced by the turbine is. Well, for that I would normally take H1-H2, but because the problem told me to assume constant specific heats at room temperature I'm using Cp times T1-T2 and Cp we looked up as 1.005. Then for T1 I'm using 500 and for T2 I'm using 319.23 because that's the actual temperature at my actual outlet. So that's 181.674 kilojoules per kilogram. Note that I could have gotten to that answer by determining what the ideal work would be and then using this proportion or rather this proportion. Okay, the actual work is equal to 0.75 times the ideal work. The ideal work would be 1.005 times T1 minus T2S. So I could take 500 minus 258.974, multiply that by 1.005 and I get 242.231 kilojoules per kilogram. If I multiply that number by 0.75 I get the actual work which is 181.673. Potato potato. Part C asks me to figure out the specific entropy generated by this process. Well that seems involved. Little s gen, or rather let me show you where I'm getting that. Big s gen is equal to delta big s of the system plus delta big s of the surroundings. If I divide all three terms by mass then I can say little s gen is equal to delta little s of the system plus delta little s of the surroundings. I have an adiabatic turbine which means that no heat is exchanged with the surroundings which means that I have no entropy effect on the surroundings. So s gen is just delta s of the system. Now, how do we determine delta s of the system? Well, if I had water or R134a in this problem I can look up s1 and s2 and subtract them. If I had an isothermal process I can use the definition of entropy to figure out heat transferred divided by temperature but I don't. I have an ideal gas which means that we don't have tables that will give us entropy. Do you remember what we do when we don't have tables? We use these equations. Delta s is equal to the integral from 1 to 2 of cv as a function of t times 1 over t dt plus R times the natural log of v2 over v1. I could also take the integral from 1 to 2 of cp as a function of t times 1 over t dt times R times the natural log of p2 over p1. Which of those two equations is more helpful for this particular problem? You're right, it's the bottom one. I'm going to say delta s of the system is equal to... So first simplification I make here is by recognizing that cp is constant because I'm assuming it's constant because I was told to assume it's constant. Therefore cp comes out of the integral and I'm left with cp times the natural log of t2 over t1. It's the natural log of t2 over t1 because I'm left with the integral of 1 over t dt which is the natural log of t evaluated at 2 minus the evaluation at 1 which simplifies as a result of my logarithm rules to natural log of t2 over t1. Then I'm subtracting specific gas constant times the natural log of p2 over p1. cp is known at is 1.005. t2 and t1 are both determined now. p2 and p1 are both known. The only unknown here is the specific gas constant for air which is the universal gas constant divided by molar mass for air. I will leave that symbolic for the moment. So I have 1.005 kilojoules per kilogram kelvin multiplied by the natural log of t2 which was a number I'm pretty sure and that number was 319.23 Then I'm dividing by t1 which is 500 then I'm subtracting 8.314 kilojoules per kilomole kelvin divided by 28.97 kilograms per kilomole and then I'm out of space. Kilomole cancels kilomole so I'm left with kilojoules per kilogram kelvin then I'm multiplying by the natural log of p1 excuse me p2 over p1. p2 was 100 p1 was 1,000 So calculator if you would please. It's going to be 1.005 times the natural log of 319.23 divided by 500 then we're subtracting 8.314 divided by 28.97 times the natural log of 100 divided by 1000 and I get 0.20987 kilojoules per kilogram kelvin So before we do part D let's actually answer part E first How are A, B and C affected if the efficiency drops from 75% to 65% Think through that on your own Well first of all my temperature is going to increase I know that for two reasons One the presence of friction tends to increase the temperature at the outlet but I can also reason through the fact that the whole goal of this device is to extract energy A turbine that operates really really well will extract a bunch of energy leaving me with a nice low temperature If I don't have good operation of my turbine then I will not extract very much energy at all which means that I have a lot of temperature left over in the outlet stream Theoretically if I had an efficiency of 0% I would extract no energy or worse yet I would actually increase the energy in an attempt to try to extract energy Then work out is going to decrease because I have a smaller temperature change The higher temperature at T2 implies that delta T is smaller which means that I get less work out The lower the efficiency of my turbine the lower the amount of work that I get out of my turbine is Then S gen is going to increase because I am less reversible Remember, S gen is 0 if it's reversible if it's perfect The presence of losses is going to increase S gen The more losses there are the higher S gen is as a result Now for part D TS diagrams are useful for keeping track of efficiency and how it affects our properties and the lines of constant pressure go up into the right like that and they go up as the pressure increases So I could say that this line here is my P1 and this line here is my P2 Stay 1 and stay 2S are going to be on top of each other because I have an isentropic process If the entropy doesn't change I have no horizontal displacement on the S axis Then state 2 actual is going to be a little bit to the right and my entropy hasn't increased a little bit I have 2 actual over here So state 2S is going to be directly below State 2 actual is going to be off to the right and because these lines curve as I go up into the right that means that I have a smaller delta T between 1 and 2 My drawing is awfully poor If this were to scale I would see that my delta T would decrease from 242 to 181 and on this drawing it looks like it decreases from a number to just slightly less than that number but potato potato So state 1, state 2 actual and as my efficiency decreases I'm going to go further and further to the right So if I had an efficiency of like 10% that would probably be way over here