 For our next example problem, I want us to consider the same auto-cycle, except this time I want us to work it without assuming the cold air standard. So this time we are not going to be assuming that the specific heat capacity of air is constant and evaluated at 300 Kelvin. That means we cannot use our isentropic ideal gas relations and we can't make the substitution of Cv delta T for delta U or Cp delta T for delta H. What we have to do instead is use our tables. So I'm going to start with the same approach we had last time. I'm going to draw a table, but I'm going to draw it with more columns. So I want us to come up with the internal energy because I know that's going to be useful for calculating our work in, our Q in, our workout, and our Q out for an auto-cycle. And in addition to pressure, I want us to look up the reduced pressure and in addition to volume, I want us to look up the reduced volume. So for an isentropic process, when we don't have the ability to assume constant specific heats, our isentropic ideal gas equations are this. We are saying Vr2 over Vr1 is equal to V2 over V1 and Pr2 over Pr1 is equal to P2 over P1. And Vr and Pr are a measure of reduced volume and reduced pressure. It is essentially a normalized parameter. You are talking about the volume relative to the critical point volume and the pressure relative to the critical point pressure so that they scale a little bit more easily. And for an isentropic process, we can say Vr2 over Vr1 is equal to V2 over V1 and Pr2 over Pr1 is equal to P2 over P1. So our isentropic ideal gas equations, these, are still useful from 1 to 2. They just aren't our temperature related functions that we had access to in the previous version. I will again populate what we know. We have 300.15 as a temperature at state 1 and 95 kPa for state 1. And using the ideal gas law, we can calculate the same value for a specific volume. That's cubic meters per kilogram, John. So to complete the other three columns in our table, I'm going to have to turn to our textbook. And what I want is going to be table A22. Note that A22 is going to have enthalpy, internal energy, this zero point entropy term, reduced pressure and reduced volume as a function of temperature. But these are only functions of temperature. They aren't functions of pressure. So we don't have to worry about multi-dimensional data like we do for the steam tables. We just have a one-dimensional table. As a result, it's smaller and they can actually fit two pages of content here within the same table. So it starts at 200 Kelvin, works its way down to 500, and then it starts at, excuse me, 440. And then it starts again up here. Once we get down to 740, we go on to the next page. So what I want to do is find 300.15. And 300.15 I can see is going to occur between 300 and 305. So to come up with an actual value for internal energy, I'm going to interpolate. Here we go. So I'm going to write this as a solve term just so that I don't have to rearrange it. And this makes it I think a little bit easier for people to follow what I'm doing. So 300.15 minus 300 divided by 305 minus 300 is equal to 214.07. Again, specific internal energy is the third column here. Whereas specific enthalpy is the second. I know that because internal energy is less than enthalpy because enthalpy is internal energy plus pressure times specific volume. So 214.07 minus, no, it's equal to x minus 214.07 divided by 217.67 minus 214.07. I'm solving with respect to x. So my internal energy value is going to be 214.178. And I can repeat that process for PR. That PR value is going to be between 1.386 and 1.4686. So I'm going to say 1.386 and 1.4686 and 1.43006 just for funsies. Then VR is going to be the same thing again, except this time I'm going between 621.2 and 596.0. I just, whoops, I forgot to subtract by 1.386. Interesting. Well, let's go correct that. Not 1.43, 1.38848. And then lastly, we can interpolate for our reduced specific volume, which I am going to change up my interpolation formula here a little bit. Since we're going to have to do a million of these, I find it more convenient to use variables for the lower and higher temperature values of the thing that we are interpolating for, just so that I only have to plug them in in two places instead of three. I know a lot of people will program a function to their calculator. That's fine too. But A, this calculator emulator that I'm using is not particularly good at storing any actual memory. And B, I also want to illustrate how you could do this without actually programming anything. So if you happen to have a TI-89 or newer, you can probably do something similar to what I'm about to do. I'm going to plug in a variable, L, let's say, for the value of the thing that I am interpolating for at the lower temperature. Not the lower value, the lower temperature. And then H for the higher temperature minus L. And then the vertical bar of the TI-89 is evaluated at, so I'm going to say L equals 621.2 And here is going to be the actual and. And H is equal to 596.0. So I get a reduced specific volume value of 620.44. So that gives us our values for state number one. Now how to get to two? Well, I get to two the same way that I got to two the previous time. I use the isentropic ideal gas equations. The difference here is I'm not using the form that has temperature in K. I'm using these two equations right here. These are still called the isentropic ideal gas equations. It's just not usually the ones that we refer to. So what I'm going to do is recognize that v2 over v1 is a parameter I know. It's 1 over r. Because remember, r was defined as the big volume over the small volume. Therefore, v2 over v1 is 1 over r, which means I can take vr2 is equal to vr1 divided by 8. Because our compression ratio is 8. Now, 620.444. Divide it by 8 is 77.555. I'm just going to leave three decimal places here. It hurts my soul a little bit to f556 instead of f5555. These are the sacrifices we have to make in the name of trying to keep our decimal places aligned. And then from that vr value, we can look up what we need for row number two. Because remember, table A22 in the textbook is one dimensional. So for a given temperature, I can look up everything and from one of those everything terms, I can look up temperature. So we will mosey back to the tables. I recognize that the vr column has specific volume decreasing as temperature increases. So value of 77.5555 is going to be at a higher temperature. I see that it's occurring right here between 78.61 and 75.5. I'm going to grab my highlighter tool again. I'm going to say our value for row two, state two, is going to be between 670 kelvin and 680 kelvin. And our interpolations are all going to be built around the value... The value 77.5555 minus the value at the lower temperature, which is 78.61, divided by the value with the lower temperature, excuse me, the higher temperature, which is 75.5, minus the value with the lower temperature, which is 78.61. And the first thing I'm going to interpolate for, let's say, is pr. So I'm going to go between 24.46, 54.46 and 25.85, 25.85, and I get 24.93, 24.93. And in the interest of keeping my decimal places somewhat aligned, they can tell relative scale. Next up, we're going to interpolate for internal energy, which is going to be this column here, because again, I guess you can't see what I'm doing, gesturing. This column here, because it's going to be less than enthalpy. So calculator, if you would please, let's go between. The value at the higher temperature is 496.62. The value at the lower temperature is 488.81. So my internal energy is going to be 491.458. 400 and, no, Pental, 491.458. And lastly, we can do temperature. Temperature is 16670 and 680. So higher is 680, lower is 670. And I get 673.391. That is row number two, about halfway done. To be a completionist here, I should fill out my pressure and my specific volume. Now, a specific volume is easy. I mean, it's just going to be v1 divided by 8 again. But I actually want to come about it the long way. Just like in the previous example, if I calculate v using the ideal gas law using my temperature and pressure values and I get the same as what I should get from the definition of our compression ratio, it's a good indication that I did these two values correctly. So pressure is going to come from our other isentropic ideal gas equation. I'm going to say P2 is equal to P1 multiplied by PR2 over PR1. And I am plugging in values of 95, 24 and change, and 1 and change. So calculator, that is 95 multiplied by 24.9313 divided by 1.38848. I get 1,705.8. 1,705.8. Cool. And then specific volume, again, we should expect to get about 0.113. Actually, let's just go check that actual value from our previous example. 0.113 is what we're expecting to see. 0.113 is what we're expecting to see. And here we go. Too far in, John. Slightly more out. You can do it. Good job. Nope, too far. Oh well. R times T is going to be 0.287 multiplied by 673.391 divided by P1,705.8. Show me 0.113. Hey, we got 0.113. It is accurate with an rounding error. I'm pleased with that. That's a good sign that we calculated our values for R2 correctly. At least, well enough to be able to move on. Now, true or false, we can still use these equations from 2 to 3. False, we can't. It's not isentropic. So to step from state 2 to state 3, we have to do what we did last time. Use our ideal gas law. We know Pv is equal to RT. And we know the quantity V over R is constant from 2 to 3 specifically. So we can say T over P must also be constant from 2 to 3. So I can say T2 over P2 is equal to T3 over P3. But again, I find myself in a bind here because I have one equation and two unknowns. I don't know T3. I don't know P3. How do we come up with T3, you ask? Well, for that, we've got our Q in term, which is 750 kilojoules per kilogram. Once again, we can answer what that is equal to by performing an energy balance in the process from 2 to 3. I'm going to go 1, 2, skip a few. Everything cancels except for Q and delta U. So in our energy balance, we would normally have delta E in the left and then energy in minus energy out on the right. We have a closed system here, so it's going to be energy in is equal to Q in plus work in. Energy out can be Q out and work out. It's the transient process, which means that delta E could be delta U plus delta KE plus delta PE. We're neglecting kinetic and potential energy changes because there's no meaningful opportunity for the kinetic energy nor the potential energy to cancel or to actually manifest in this problem. And then we are neglecting works because it's an isochoric process and Q is only in because we are treating it as a heat addition process. You can also think of it like a net addition process for ignoring Q out. So we're left with delta U is equal to Q in on a specific basis because we don't have mass. Then I can say Q in is equal to 750 kilojoules per kilogram and that's equal to, you got it, U3 minus U2. Why can we not use CV delta T? Because we can't assume constant specific heats. Instead, we have U3 is equal to U2 plus 750 kilojoules per kilogram. Luckily for us, we have U2. It is 491.458. So we take that number that we got earlier and we add 752 it and that fixes our row three. 491.458. Pop up my calculator so you can see what I'm doing. And then we are adding 750 and we get 1241.46. Actually, it's going to be 458, isn't it? Calculator, how come you're rounding to two decimal places? Who told you to do that? Calculator. You're at float six. 491.458 plus 750. This should be 1241.4. Is it like because I didn't include any decimal places on 750? No, it's just rounding to two. I don't get you calculator. I don't get you. Our relationship is very strange. This calculator and I, honestly, the fact that we can work together at all is a testament to both of our collective patients. I'm going to write 458 as our thousands here and not 460 because I'm the one writing, not you calculator. Anyway, now that we have internal energy and we have one dimensional table we can look up everything else. So shall we then? To the tables. 1241 appears way down here. You can do it tables. Move over. Yeah, you got it. Okay, 1241 down here. Between 1223 and 1242. So if my highlighter tool is going to work here, I can say between this row and this row. I'll reposition a little bit more. Hopefully those of you reading on this on a tiny screen can see a little bit better. I'll pop up my calculator. Speaking of tiny screens and your face calculator. Got you good. So our interpolations this time around are going to be built on the driving proportion of 1241.458. 1241.458. That's right, calculator. Throw some opposable thumbs if you want to participate in this value minus 1223.87 the value at the lower temperature divided by 1242.43 minus 1223.87 and we are interpolating between let's start with temperature, shall we? 1520 and 1540. So we get a temperature of 1538.95 unless there's, you know, more decimal places that my calculator is not telling me about. So that temperature value is going to be 1000. 1000, no, pen tool. 1538.95 maybe extra decimals and then after that we can do pressure if we want to but let's keep rolling in the tables. Next up in our tabular data is going to be PR which is going to be between 636.5 and 672. 672.8 and 636.5 the value of 670.899, 670.899 and then VR for good measure is going to be between 6.569 and 6.854 we had 6.58393 why five decimal places calculator? I never understand you. 5, 8, 3, 9 so the only thing left here is P true or false I can use this equation again no I can't because it's not isentropic what I've got is our ideal gas law so I'm going to write P3 is equal to P2 times T3 over T2 so I will take our P2 value which we know is 1,705.8 multiplied by 1,538.95 divided by 673.391 we get 3,898.39 check that I plugged that in correctly 2.5 ish multiplied by 1,705 3,900 seems right 3, 8, 9, 8.39 now that row 3 is complete I can step on to row number 4 and I'm going to do that by using this equation again so this time it's going to be VR4 over VR3 is equal to V4 over V3 and remember that V4 over V3 is going to be the big volume over the small volume which means VR4 over VR3 is going to equal R therefore VR4 is equal to VR3 multiplied by R so I am going to take our shiny new specific volume 6.5839 I'm going to multiply by 8 and I get 52.6714 52.6714 okay now with that value I can jump over to our tables and scan around until I find 52.67 which I see is right here so acrobat if you would move over please 52.67 occurs between this row and this row that's probably as zoomed in as I need to be so we'll build our interpolation as being between 52.6714 and the value at the lower temperature which is 53.39 the value at the higher temperature which is 51.64 and the value at the lower temperature which is 53.39 and then we are interpolating first for PR let's say which is going to be 41.31 and 43.35 we get 42.1477 and then next let's do internal energy so that's going to be between 576.12 and 568.07 which is 571.376 71.376 and then temperature is the last one so that will be between 780 770 I don't have a letter out front because of my OBS shortcuts I do great that was exactly what I wanted guys calculator on the letter let me just clear everything it's not just very frustrating and we get 774.106 774.106 then to finish off our row number four here I need pressure first which I can get by using the second isentropic ideal gas equation I could say P4 is equal to P3 multiplied by PR4 over PR3 and then that's going to be 42.1477 divided by 670.899 add some parentheses for good measure multiplied by 3898.39 and I get 244.907 then my last measure here is going to be using the ideal gas law again to calculate this specific volume and double checking that it equals about 0.906 because that's what I'm expecting to see 0.287 multiplied by our temperature which is 777.106 divided by our pressure which was 244.907 and we get 0.911 so it's a little bit off from 0.906 but every calculation that we do incurs a little bit of rounding error especially when my calculator just drops decimal places in its display and each one of those steps compounds so going from 1 all the way to the 4 all the way to state number 4 and being off by about 5% is probably to be expected but this is the downside of doing these calculations by hand so I'm going to come back to that in a second so I'm going to say it's close enough to move on and I'll just write 0.906 because we know that E is what it actually is and it's not going to actually affect anything I just don't want to confuse anyone that looks at just this table without hearing the explanation behind it but John I thought you said 4 to 1 as I said of course, yeah it is this is the effect of compounding error the work in, the Q in, the work out and the Q out so for that we're going to step back to what we defined for the auto cycle which is work in occurs between 1 and 2 and because it's isentropic there is no heat transfer so it's just U2-U1 Q in occurs between 2 and 3 there's no work because it's isochoric work out occurs between 3 and 4 and it's 3-4 because the work out has a negative out in front of it and the energy balance same for Q out which means it's U4-U1 and then we are plugging in our 4 internal energies expecting to see 750 for Q in let's just calculate all of these 491.458-214.178 gives us 277.28 this is really tiny on my iPad so it might be terrible handwriting but what else is new and then 1241.458-491.458 750 on the nose turns out when you add 750 to a number and then subtract first number you get 750 and then 1241.458-571.376 and you get 670.0852 the joules per kilogram and then lastly we have 571.376-214.178 and we get 357.198 like this one beautiful yep thank you for your cooperation in this matter iPad okay here we go 35 excellent 1 no should be 2 because I'm rounding 2-0 the joules per kilogram just because the perfectionist in me can't handle these okay there we go and then network out and net heat transfer in that's going to be 670.08 minus 277.28 and we get 392.8 and doing the same thing with 750 minus 357.082 and we get 392.9 again we should see the same number we can chalk up not seeing the same number as a result of here just double check is 0 so again we should see the same number the fact that they're really really close implies that it's probably just a rounding error but let me double check that I didn't type anything incorrectly I did you see it let's queue out 357.2 see now that's way closer because I mean it's just going to be like we're talking about adding and subtracting these four numbers shouldn't be much opportunity for difference anyway the fact that the net heat transfer in equals the network out is yet another indication that we probably calculated these values correctly using U1, U2, U3, and U4 so what I would like to do next is calculate the thermal efficiency that's our network out divided by Q in and that's going to be 392.8 by 750 and we get 52.37% again let's try that a little bit better 52.37 okay now I know that a lot of this is extraneous to what I actually asked for in the problem statement but I'm hoping that by going through all of these properties even though many of them are not particularly useful to this problem is helpful I mean this is all about learning right and what is determining extra parameters if not a learning opportunity it also allows us to do more of a direct comparison for example we have all four pressures and we can compare those to the four pressures over here so I can say RP1 is 95, RP1 is 95 that makes sense because we had the same starting value as the Cold Air Standard we got about 1746 this value is 1706 as a general rule of thumb the Cold Air Standard will overestimate things so the fact that we are a little high here is true to that P3 is 4390 versus R3900 P4 is 239 versus 244 still overestimating the amount that it decreased leaving us with a lower number and we can do the same thing with temperatures we had 689.5 the Cold Air Standard versus 673.4 we had 1734 in the Cold Air Standard versus 1540 and 750 versus 777 so again the Cold Air Standard is overestimating the amount that it decreased neat we got 52.4 on our actual analysis versus 56.4 again the Cold Air Standard has a tendency to overestimate so which of these two numbers is closer to the actual thermal efficiency of this actual engine well it's likely the non Cold Air Standard version because there are fewer assumptions made in the generation of this number but this number is not all that accurate to what it would be in an actual real gasoline powered engine because there are a lot of assumptions and simplifications built into the auto cycle model in the first place the reason that it's useful to calculate this parameter is not to actually get a real thermal efficiency of the operating engine the reason it's useful is because we can use this analysis to discover how changing certain things would affect the engine's performance for example if we were driving it on a cold day versus a hot day and the incoming temperature was 217 Kelvin instead of 300.15 really really cold day I guess we can compare how that's going to affect our power output how that's going to affect our thermal efficiency etc and the difference in those two calculations in our model should be reflective of the differences in the actual operation of the actual engine even if these actual numbers don't represent the actual thermal efficiency that's the value of the model does that make sense ok now for the actual things I asked for maximum temperature and pressure those are going to be 1538.95 and 1241.46 here I'll just write these out again emax emax and we had 1538.95 terrible handwriting it's really small 3898 versus neat there was a network output we got that was 392.8 c was the thermal efficiency we determined that conveniently on the same page here don't even have to scroll around and then for part d I wanted the mean effective pressure which is going to be our network out divided by v1 minus v2 so we are going to calculate 392.8 divided by the quantity 0.906 minus 0. actually let's just go 0.906 divided by 8 because there's a lot of uncertainty in my memory of v2 so 495 kilopascals and it occurs to me that I never walked you through that unit how I know it's kilopascals so let's do that now we had 392.8 kilojoules per kilogram and I'm dividing by a quantity 0.906 minus 0.906 divided by 8 in cubic meters per kilogram so when I take a kilojoule and write that as a kilonewton times a meter and write a kilopascal as being a kilonewton per square meter kilonewtons cancels kilonewtons kilajoules cancels kilajoules cubic meters cancels meter squared and meters give me with kilopascals so when I take a quantity in kilojoules per kilogram and divide by cubic meters per kilogram the result is in kilopascals and with that we have completed our second auto cycle as a general rule of thumb for all of the problems in chapter 9 I'm expecting you to be able to perform the analysis both with the cold air standard and without the cold air standard there are advantages to both the cold air standard is faster and it also simplifies very neatly allowing us to make nice correlations like we did for the thermal efficiency the non cold air standard version is going to be more accurate generally speaking but it's a slightly more accurate calculation from an inaccurate model so the advantage to that extra accuracy is very dependent on the circumstances do you actually care about being slightly more accurate in an inaccurate model and just like in the previous example I have set this up in matlab note that the given information is the same the only real difference here is that instead of calculating the properties of state 2 from our isentropic ideal gas equations I'm using the interpolation function to grab values out of table A22 by the way matlab's ability to interpolate is so much more convenient than calculating them by hand yet another massive time saver when you're completing these problems in matlab and then when I'm calculating the work in key win workout and key out I'm not making the assumption of constant specific heats and simplifying it down to CB delta T pretty much everything else is the same so with our temperature of 300.15 pressure of 95 compression ratio of 8 and a heat addition of 750 kJ per kilogram we get approximately the same results as what we had gotten when we worked it by hand and again this is going to be posted on DC wall so that you can poke through it if you want