 So what we're going to do is analyze the same problem that we had just completed with the cold air standard analysis and I want to work it again except without the assumption of constant specific heats. So I'm going to get rid of our temperature information that we had determined and we are instead going to be looking for enthalpies. So temperature really only matters in as far as its ability to get us to an enthalpy. And since this is just asking for thermal efficiency everything we're doing is just a means to an enthalpy. We want an enthalpy so that we can take the difference in enthalpies to determine our work in Q and workout in Q out terms and we will only determine as much as is necessary to get us that far. Does that make sense? So I just want enthalpies and everything else is a stepping stone to get there. The stepping stone that will help us is going to be the reduced pressure. So I am going to remember that I have all of our pressures already and those will only serve to relate our reduced pressure proportions here. Okay. And then our best friend in this analysis is going to be table A21 or table A22 where we are going to be looking at the reduced pressure and enthalpy corresponding to our T1 and then motoring through that way. So we begin with T1 which was 290.15 Kelvin. The temperature is day one with 17 degrees Celsius can bring to Kelvin allows us to look that up in the tables. And I can see on my tables that 291.15 is going to occur between 290 and 295. So I am going to pop up the calculator. We're going to do a little bit of interpolation here for enthalpy and PR1. So I'm going to switch that down just a little bit so we can see it. Remember my enthalpy is the second column. I can always tell which one is which because enthalpy is going to be higher than internal energy. Okay. So we are taking 291.15 minus 290 and we are dividing by 295 minus 290 and we are saying that that is equal to the enthalpy that we are trying to determine which I'm going to call x for now minus 290.16 and then we are dividing by 295.17 minus 290.16 and I'm solving this equation for x. So I am going to add solve out front so my TI-89 can keep track. And we get an enthalpy at state 1 of 291.312 and then I want the reduced pressure so we're going to be taking this same relationship and plugging in our values for reduced pressure instead of enthalpy. So if I pop back over here to the S, reduced pressure is going to be our first column in the second section. So that's 130.68 minus 1.2311. So 1.2311 goes here, 1.2311 divided by 1.3068 minus 1.2311. And we get a reduced pressure of 1.24151, 1.24851. And then the process from 1 to 2 uses our isentropic ideal gas equation, which for the non-cooled air standard analysis is rp1 over rp2, which I could write as rp2 over rp1 is equal to p2 over p1. rp2 over rp1 is equal to p2 over p1 and that is equal to 4 for this problem. Therefore rp2 is going to equal rp1 multiplied by 4. So I'm going to take 1.24851 multiplied by 4 and I get 4.99404. Then from that rp I can look up an enthalpy. So I am going to find 4.994, which I see occurs between a temperature of 430 and a temperature of 440. So start a new interpolation process and I am going to say solve out front. I'm going to take 4.99404 minus 4.915 and I am going to divide that by 5.332 minus 4.915. And I'm saying that that is equal to the thing that we're looking for minus 431.43 divided by 441.61 minus 431.43. And I'm solving that for x and I am getting out as a result 433.36. Note I could interpolate for temperature but since I don't actually need temperature I'm going to skip that because I am just doing whatever is necessary to get to a thermal efficiency. So our enthalpy at state 2 is 433.36. Then our state points 3 and 4 are assumed to be the same as 1 and 2 for the same reasons as in the cold air standard variation of this problem. So we are assuming that the compressors are splitting the work evenly so we need the same delta H and we are using the same pressure ratio proportion of 4 to describe the difference in our RP values. So I'm assuming it goes back down to our initial temperature and therefore the same RP and H. And then state 4 is going to have the same properties as state 2 because it's the same pressure ratio. And with that we are 4 out of 10 state points through the problem. From 4 to 5 we have the same increase in temperature that we did earlier we are increasing by 20 degrees Celsius. So I need to interpolate for a temperature of 433.36 and for that I will jump back into my tables and I am going to be interpolating for a temperature. So you remember earlier when I said I could interpolate for a temperature but I don't actually need it haha. I know things I'm John. I was incorrect. It turns out that I do actually need it. I can't just add 20 to the enthalpy because it isn't necessarily corresponding to an enthalpy change of 20. I could assume constant specific heats and then figure out the enthalpy change corresponding to a temperature change of 20 Kelvin but then I would be defeating the whole purpose of working this problem without that assumption. Anyway I had 431 as a temperature at state 4 or rather as a temperature at state 2 and it's the same at 4. So it's also the temperature at state 4 and it's 431.895 plus 20 is 451.895 not to be confused with 450.895. That would be a very reasonable mistake. So I am going to find that on the table and wouldn't you know it? It's split across the two columns. That's annoying. Well I can interpolate for an enthalpy and we will see if we need an rp. So that enthalpy is going to lie. Okay here let's start with temperature at 440 and 450. So I am taking solve first and then 431.895 minus 450 divided by 460 minus 450. Hey I was wrong. It's not split across columns. Huh why did I think it was? It is equal to because it's a temperature in an enthalpy and I thought it was an enthalpy because I'm stuck in enthalpy mode. It is equal to the actual enthalpy that I want here which is going to be x minus 451.8 divided by 462.02 minus 451.8 and I get an x value of 453.737. So note I could have added 20 because it's different and if I assume that the specific heat capacity value was 1.005 I would have determined a change in enthalpy of 20.1 which one added to 433.36 does not get the correct value of enthalpy. Therefore I am determining a theoretically more accurate result without that assumption. Next on to state 6. For state 6 I recognize that I was given a Q in. I know the difference in enthalpy between 5 and 6 then is going to be 300 so I can add 300 to this number and I get 753.46 no 753.737. Don't just add 300 to the quantity that you had calculated to demonstrate that it was different huh? 433.737 so 453.737 plus 353.737 look we calculated it we wrote it down now we can proceed. The analysis from 6 to 7 is going to use the pressure ratio across the turbine which is 1 over 4 which means the proportion of rp is also going to be 1 over 4 which means I need to look up the rp corresponding to an enthalpy of 753.737 so jumping back into our tables and perusing through our enthalpy column I see that 753 is going to occur between 745 and 756 so I'm going to set up my scroll down for some reason protocol and also set up my solve on my calculator it's going to be 753.737 minus 745.62 divided by 756.44 minus 745.62 and that's equal to x minus 33.72 divided by 35.5 minus 33.72 and I'm solving for x should be a comma what are you doing to me calculator and I get an rp value of 35.0 this is 55 so 35.055 okay and then I know rp2 which in this case is going to be 7 over rp6 is equal to p7 over p6 I guess why am I writing this as this is an rp this is pr excuse me hope that isn't been confusing pr7 over pr6 is equal to p7 over p6 and I am saying that that is 1 over 4 which is an assumption that we're making so 35.0553 divided by 4 is 8.76383 8.76383 and then I am going to look up an enthalpy corresponding to that reduced pressure jumping back into my tables I see that that reduced pressure is going to occur between 8.411 and 9.031 so I am going to interpolate I am taking 8.76383 minus 8.411 divided by 9.031 minus 8.411 and I'm saying that that is equal to x minus the enthalpy value corresponding to 8.411 that's 503.02 and I'm dividing by the enthalpy value at 9.031 which was 513.32 minus 503.02 solving for x yields 508.882 and then I have the same 300 kilojoules per kilogram of heat addition in the second combustion chamber which means that I am going to be taking 508.882 I'm going to add 302 that will give me the enthalpy at state 8 which is 808.882 and I need to look up an a reduced pressure corresponding to that value and away we go 803 excuse me 808 gonna be on the second page and I see 808 occurs between 800 and 810.99 so that interpolation will go 808.82 minus 800.03 divided by 810.99 minus 800.03 and that's equal to x minus 43.35 divided by 45.55 minus 43.35 and I get a reduced pressure 45.1269 45.1269 and I am dividing that number by four for the same reasons as the process from 6 to 7 of 45.1269 divided by four yields 11.28 11.2817 and I can determine enthalpy value from that by jumping over into our tables finding 11.28 which is going to be on the previous page 11.28 is going to occur between 11.1 and 11.86 this should be the last interpolation I need to do so I am going to take 11.2817 minus 11.1 divided by 11.86 minus 11.1 and that is equal to 554.74 minus 5 excuse me x minus 544.35 because that was 11.1 right yeah divided by 554.74 minus 544.35 and solving for x yields 546.834 546.834 why am I repeating it I don't need to remember it it's on my calculator that is 546.834 cool so we could determine the temperature corresponding to that enthalpy and then subtract 20 from it but I could also recognize that all of the heat leaving the hot side of the regenerator is going to the cold side of the regenerator which means q regen is the same from both perspectives and because the mass flow rate is the same that means the delta h value is going to be the same so I can find h10 by figuring out the difference in enthalpy between four and five and then subtracting that from nine so I'm going to take 453.737 I'm going to subtract 433.36 I'm going to add that to rather that delta t 20.377 and I'm going to take 546.834 and subtract that number and I should get 526.457 and now that I have all of my enthalpies I can calculate our work and heat transfer so just like last time the work in is occurring between one and two and three and four so that's h2 minus h1 plus h4 minus h3 the q in is occurring between five and six and seven and eight that'd be h6 minus h5 plus h8 minus h7 and that should be 600 unless I made a mistake work out is going to occur between six and seven and eight and nine so h6 minus h7 plus h8 minus h9 and q out is going to occur between two and three and ten and one I have all those enthalpies now so I am going to take 433.36 minus 291.312 and I'm going to add to that 433.36 excuse me 36 minus 291.312 I'm going to get 284.096 okay work in check then q in I am going to take 6 minus 5753.737 minus 5 which is 453.737 I'm going to add to that 808.882 minus 508.882 and I should get a missing parentheses how where was their parentheses huh and I should get 600 look I did hooray then 6 minus 7 would be 753.737 minus 508.882 plus 8 minus 9808.882 minus 546.834 I'm going to work out of 506.903 I'll do q out while I'm here that would be 2 minus 3 which is 433.36 minus 291.312 plus 526.457 minus 291.312 so those two quantities are 506.903 kilojoules per kilogram and 377.193 and then from that I can determine a network out that network out is going to be 506.903 minus 284.096 and I get a network out of 222.807 and I'm going to calculate a net q in which is going to be 600 minus 377 and I get a net heat transfer in of 222.807 and then a thermal efficiency is going to be that network that we just determined divided by our q in which was 600 which I'm going to scroll up for for some reason and I get 0.3713 and that is how we would solve the problem without the assumption of the cold air standard so it's not that much more difficult conceptually the most difficult part is keeping track of what you need to look up in order to get to the next state point.