 A diesel engine has a compression ratio of 20 to 1 with an inlet of 95 kPa, 290 kelvin, and an initial volume of 0.5 liters. The maximum cycle temperature is 1800 kelvin. Assuming the cold air standard, determine or complete the following. A. The maximum temperature and pressure occurring in the process. B. The network output, again it doesn't specify if that's total work or specific work or rate of work, we have to use our best guess based on the information that we have. C. The thermal efficiency and D. I want us to sketch a PB and TS diagram of the cycle. We can parse out what we know. This 20 to 1 ratio is a compression ratio. The inlet pressure is going to correspond to the pressure of the air as it comes in prior to compression, which would be P1. Likewise, with the temperature, that temperature represents the temperature of the air as it comes into the engine. The intake occurs before compression, so even though we treat the diesel cycle as a closed system, it actually is the state point before compression. Furthermore we have an initial volume of 0.5 liters. Then we know the maximum temperature is 1800 kelvin. So quick question, knowing nothing else about the problem. Which state point do you think is going to have the maximum temperature? That's right, the one after the fire. Remember, the highest temperature usually occurs in the state point that occurs after the combustion process. It is hottest after the fire happens. Therefore, that Tmax is likely to be T3. Next we are analyzing the problem with the cold air standard. And I'm going to start this like we did our auto cycle. Even though it's not explicitly necessary, I want us to calculate the temperature, pressure, and specific volume at all four state points. Why you ask? Because it's a good character building opportunity. It's a good way to practice, it's a good way to learn. And that's what we're here to do, isn't it? Learn stuff. So I want temperature in kelvin. I want a very poorly vertical line. I want pressure in kilopascals. I want another poorly drawn vertical line. And I want specific volume in cubic meters per kilogram. Let's make this table nice and pretty. Just get rid of these edges. Yeah, that's beautiful. Beautiful table. Fill that in a little bit, you know, get nice and sealed. Aha. Okay. Now I can populate what I know. I know T1 was 290 kelvin. T1 was 95 kilopascals. And I'm not writing units because it's in the column header. Then we can determine the specific volume. Why? Because the cold air standard is the air standard assumptions plus the assumption of constant specific heats at room temperature. The air standard assumptions include treating the air in the system as an ideal gas. So I can use the ideal gas law to describe the characteristics of this gas. Before I can say pV is equal to RT, therefore V is equal to RT over P, which is the specific gas constant times temperature divided by pressure, so V is equal to R of air times temperature divided by pressure. Our R of air would be the universal gas constant, 8.314 kilojoules per kilo mole kelvin divided by the molar mass of air, which we can get from table A1, which is 28.97 kilograms per kilo mole. When I take 8.314 kilojoules per kilo mole kelvin divided by 28.97 kilograms per kilo mole, the kilo moles cancel and I'm left with approximately 0.287 kilojoules per kilogram kelvin. We've done that calculation together a million times, so I am going to just jump to the point where I'm running down 0.287 kilojoules per kilogram kelvin. And then our temperature was 290 kelvin and our pressure was 95 kilopascals. Kelvin cancels kelvin. To get the kilopascals and the kilojoules to cancel, or rather break apart and then cancel, I'm going to break them down to their components. A kilopascal can be written as a kilonewton per square meter and a kilojoule can be written as a kilonewton meter. Now we can play the unit cancellation game. Now we're left with square meters times meters divided by kilograms, which is cubic meters per kilogram of calculator. First if you would take 8.04 divided by 28.97 and get approximately 0.287 and then we can take that fine number multiplied by 290 and divide by 95 and we get 0.8761. Is that any other volume occurring in the cycle? Yes, it's also V4 because the process from 4 to 1 is isochoric. So I will just populate that down there. Next because we have a cold air standard analysis we are going to use our isentropic ideal gas equations to get from 1 to 2. And those isentropic ideal gas equations are these because I've assumed constant specific heats in the cold air standard, therefore I can use the form that involves our isentropic exponent K. So I'm going to say T2 is equal to T1 multiplied by V1 over V2 raised to the K minus 1 over K. 1 over V2 is our big volume divided by our small volume, so when I plug in compression ratio or 1 over compression ratio, that's right, I will plug in compression ratio. So I will take 290 multiplied by 20 raised to the power of K, which is 1.4 minus 1. We've gone through the lookup for CP, CV and K several times together, so I'm going to gloss over that, but just very briefly, if we open up our textbook, we can see that the CP, CV and K values for air at 300 Kelvin, which is what the temperature is in the cold air standard. We're assuming constant specific heats evaluated at room temperature, which is about 27 degrees Celsius or close enough to 300 to call it 300. So we're using a CP value of 1.005, a CP value of 0.718 and a K value of 1.4. Back to the problem, we are taking T1 multiplied by R raised to the 1.4 minus 1 and we get a temperature of state 2 of 961.19 and then we can repeat the process for the pressure. The only difference here is that we are going to take P1 multiplied by V1 over V2, which is still R raised to the K value, not K minus 1. If you make that mistake, it's understandable when you're practicing. I mean, it'd be perfectly reasonable for someone to accidentally use 1.4 minus 1 instead of 1.4. But make those mistakes now so that you can execute it perfectly when you calculate your quantities on the exam, 6297.46, 6297.46. And then what is the specific volume at that state point, you ask? Well, I'm glad you asked. We could do the same ideal gas law calculation that we did before. That ideal gas law calculation would just use our temperature that we just got 961.192 and divided by the pressure we just got, which is 6297.46, in which case we get 0.438. But there's another way to get to V2. That's right. We're recognizing that our compression ratio is equal to V1 over V2, which is also equal to specific volume 1 over specific volume 2. Therefore, I can say specific volume 2 is equal to specific volume 1 divided by R. So if I were to take 0.8761 and divide by 20, I should get the same quantity. The fact that I do is a good indication that I didn't make any sort of isentropic calculation error, 0.0438. Now is that specific volume the same as any other specific volume occurring here? No, it is not, because we have some expansion occurring between 2 and 3. 2 to 3 in the diesel cycle is isobaric, not isochoric. Is there anything else that I can populate at state 3? Yes, there is. That's 6297.46. That pressure at 2 is the same as the pressure at 3 because we assume that that expansion process occurs isobarically. And then we can also populate the temperature. We can deduce that by thinking through the expansion process from 3 to 4. We are expanding the volume and we are doing it ideally, which means that our pressure is going to drop and our temperature is going to drop. Our 3 is higher than 4 and from 2 to 3 we have a constant pressure heat addition process, therefore T3 is likely to be higher than T2, therefore T3 is going to be our highest temperature. T3 is equal to Tmax. So I'm going to write 1800 down here. And then now we have temperature and pressure so we can calculate a specific volume. Yay. That is going to be easiest to use this one. And swap out to 90 with 1800 and 95 with that same pressure from earlier. And we get 0.082. Put a 0 just so we have 4 trailing decimal places. Okay, we are 3 state points through this process. How do we get from 3 to 4? That's right, the isentropic ideal gas equations again. And when I jump to my isentropic ideal gas equations, we recognize that we are going to use this same proportion as earlier. We are going to say T4 is equal to T3 multiplied by V3 over V4 raised to the k-1. So we are going to take 1800 and multiply it by the proportion of specific volumes from 3 to 4. Is that R or 1 over R? That's right. That's your question. It's neither. Remember, V2 is not the same as V3. So when I write out my equation here, I'm saying T4 is equal to T3 multiplied by V3 over V4 raised to the k-1. I am going to use the actual specific volumes that we calculated earlier. Now what if we hadn't calculated those? Well, we could write this in terms of other things that we know. We could use the ideal gas law to write out V as R times T over P, in which case V3 over V4 is equal to R of air times T3 over P3 divided by R of air times T4 over P4. And in that proportion, the specific gas constant cancels. And I have T3 over T4 multiplied by P4 over P3. Now of course that doesn't work super great if you don't know P4. So we could try to rewrite P4 in terms of things that we know there. We could say P4 is equal to P3 multiplied by V3 over V4 raised to the k, not k-1, as per our isentropic ideal gas equations. And then we can write this also as T3 over T4 is equal to multiplied by P4 over P3. And because we know P3 and we know T3, we have two equations and two unknowns. And those two unknowns are P4 and T4. That being said, I don't even want to mess with that algebra. So I'm just going to use the actual quantities that we actually have. So I'm going to say T4 is equal to 1800 multiplied by 0.082029 divided by 0.87 and some stuff. And I'm going to find the carrot button on my calculator emulator raised to the 1.4 minus 1. That's minus 1. So we hit 697.98. And then for our pressure, we're going to repeat that process. I'm going to leave these specific volumes and just replace T3 with P3, 6297 and some change. And then I am going to get rid of the minus 1 and I have 228.65. Now what do you suppose I want to do next? And no, it isn't this stuff. That's right. I want to calculate the specific work in, the specific queue in, the specific workout and the specific queue out. For our work in, we would perform an energy balance on all of the processes where we have the opportunity for work in. Now the only place that we have work in is the compression process, which is 1 to 2. If we set up an energy balance from 1 to 2, we would have isentropic compression, which means that we get rid of our heat transfers. We have compression, which means that we have a work in term, not a workout term. We are neglecting changes in kinetic and potential energy, therefore the only terms left in our energy balance would be delta U and work in. So I would say work in is equal to U2 minus U1. And then because we're using the cold air standard, I can write Cv times T2 minus T1. Why is it Cv and not Cp? That's right, Cv is defined with respect to U, Cp is defined with respect to H. So when I'm writing this as du is equal to Cv dt and then integrating both sides and bringing out Cv because it's constant because we've assumed constant specific heats because that's baked into the cold air standard, then I'm left with delta U is equal to Cv times delta T. If all else fails, just remember ultraviolet Harry Potter. Next up I have Q in. For Q in, I'm going to be performing an energy balance on any process that has heat transfer in the inward direction. That's only going to be our combustion process. And since this is the new one, I'm going to actually step through this energy balance. Energy balance on two to three. Remember that in our process from two to three, the busting goes down a bit. We are adding heat to our process. We have a transient process of a closed system. So I'll start with delta U is the excuse me, I will start with delta E is equal to E in minus the out. And then delta U because it's a transient process could be delta U or delta K or delta PE. E in could be Q in or work in because it's a closed system. And E out could be Q out or work out. And I'm going to neglect changes in kinetic and potential energy. Furthermore, I'm going to neglect Q out because I'm considering heat transfer in as the heat added. And then I'm going to neglect work in. But can I neglect work out? Well John, I hear you saying to yourself there's work out occurring in the power stroke. But there's also work out occurring in two to three because the piston moves down. The volume expands a bit. We have to account for the boundary work associated with that expansion. So I can't just neglect work out. That is boundary work. So I can say delta U is equal to Q in minus boundary work. And then I'm going to divide everything by mass at which point I have delta little U is equal to little Q minus little WB. And I can write this as U3 minus U2 is equal to Q in minus little WB. But I need to write Q in. So I'm going to solve for Q in. So Q in is equal to U3 minus U2 plus the boundary work occurring. Now what is the boundary work for this process? Well, specific boundary work would be written as the integral of pressure with respect to specific volume. And this is evaluated from two to three. How does that integral simplify for this process? This isobaric heat addition process. That's right. Pressure comes out. And I'm left with pressure times delta specific V, which is going to be V3 minus V2. And look, we could stop writing now. We could say Q in is equal to U3 minus U2 plus P times the quantity specific volume three minus specific volume two. But I'm going to keep going. I'm going to actually write this as U3 minus U2 plus pressure times specific volume three minus specific volume two. And then I'm going to bring the parentheses inside this parenthetical statement. So is that pressure P2 or P3? That's right. The answer is yes. It is both. Because both state point two and state point three have the same pressure. So I could write it as U3 minus U2 plus P3 times V3 minus P2 times V2. Does that make sense? And then I'm going to group together U3 plus P3 V3, just grouping together the state three state points. And then I'm going to subtract U2 plus P2 V2. Just grouping together my state two state points. And look, we're left with internal energy plus pressure times specific volume. Hey, that's a quantity we've seen before. That's a quantity we see a lot. So frequently do we see that quantity that we have a special name for it? It is a logical property made up of the combination of those. It is specific enthalpy. Specific enthalpy is U3 plus P3 times V3 and H2 is U2 plus P2 times V2. Because enthalpy is defined as internal energy plus pressure times volume. Or specific enthalpy would be total internal energy divided by mass, which is specific internal energy plus pressure times volume divided by mass, which could be pressure times specific volume. Therefore, Qn is H3 minus H2. So Qn is H3 minus H2. And then I'm going to make the specific heat capacity substitution and I'm going to use Cp because it's enthalpy times T3 minus T2. And then I'm going to get rid of all of this energy balance so that I have space again. Goodbye energy balance, fun for a while. So the workout is going to be the energy balance where any and all workout terms appear. Where are their workout terms appearing? Where are workout terms occurring in both 3 to 4 and as we just discovered 2 to 3. So this is the workout from 2 to 3 plus the workout from 3 to 4. The workout from 2 to 3 is our boundary work of an isobaric expansion process. We just figured that out. I would scroll down at this point had I not just deleted all of it. And as you know, deleted things digitally are gone forever. So I'm just going to recall it from memory. The boundary work from 2 to 3, the workout from 2 to 3 is equal to the boundary work from 2 to 3, which is the pressure at 2 or 3. I'm just going to write 2 times the quantity V3 minus V2. Now for the workout from 3 to 4, we actually have to consider an energy balance on the process from 3 to 4. So this energy balance setup is just like it was in the auto cycle. We have an isentropic expansion process. It is a transient process of a closed system. Therefore, delta E is equal to En minus E out simplifies down to delta U plus delta Ke plus delta Pe is equal to the quantity Q in plus work in minus the quantity Q out plus workout. And then I neglect all the heat transfers because isentropic implies adiabatic. And then I get rid of the work in because I only have an expansion process. Therefore, work is in the outward direction. And then I'm going to get rid of Ke and Pe because I'm neglecting changes in kinetic and potential energy that leaves me with delta U and work out. Therefore delta U is equal to negative workout. So U4 minus U3 is equal to negative workout. Therefore, the workout from 3 to 4 is U3 minus U4. Did you follow that? Well, if not, maybe it would be a good opportunity to practice going through the energy balance. Therefore, I can say the workout in my cycle here is P2 times V3 minus V2 plus U3 minus U4. So I can make the constant specific heats substitution at which point I'm writing CV times T3 minus T4. And you know how in the auto cycle it was like super convenient that we could just write everything in terms of temperature? I'm going to try to do that here too. And in order to do that here too, I'm going to look critically at this relationship. Pressure times volume minus volume. Hmm. Well, if I were to bring in my parentheses, I could write that as P3 times V3 minus P2 times V2. Now, is there any substitution I can make for pressure times specific volume when I'm talking about ideal gases? Hmm. Hmm. Hmm. Hmm. Hmm. Hmm. Hmm. Hmm. Hmm. Hmm. Hmm. Hmm. Hmm. Hmm. Hmm. Hmm. So I could just write this R times T3 minus R times T2, which I could then factor out the R into, the specific as constant for air times T3 minus T2 plus CV times T3 minus T4. Now look, of course, we actually have the specific volumes and the pressures. You could calculate it that way if you want. But I'm just trying to write everything out in terms of temperatures where I can. I think that that will be convenient. And then just like last time, I'm going to get rid of all of this so I can clear space for myself. That'll just be two equal signs that won't be confusing at all. There we go. Oh, okay, a little bit more. There we go. And then, okay, I gotta be kind of consistent here. So I'm going to try that again. Gonna erase this. I'm going to scooch this stuff over and I'm going to erase this. Pre-constant specific heats simplification and the post-constant specific heats simplification. And hopefully no one will conflate this substitution with the constant specific heats substitutions. And I could also get rid of these. I don't really need those anymore. You guys followed it, right? Okay. We are three quarters of the way through work in, queue in, work out and queue out. That leaves me with queue out. And again, because this is the same as it was for the auto cycle, I'm going to gloss right through it. That energy balance is going to be delta E is equal to E N minus E out and it's a transient process of a closed system. So delta E could be delta U plus delta K E plus delta P and it's a closed system. So E N is going to be work in plus queue in and E out is going to be workout plus queue out. Therefore my full energy balance could be delta U plus delta K E plus delta P is equal to the quantity queue in plus work in minus the quantity queue out plus workout. I'm neglecting all of the works because it's an isochoric process. I'm neglecting queue in because it's a cooling process. I'm neglecting delta K E and delta P E because they are negligible. That leaves me with delta U is equal to negative queue out. And delta remember is always the end of state point minus the beginning state point. So that would be U one minus U four because the process is from four to one. So U one minus U four is equal to negative queue out. Therefore queue out is equal to U four minus U one. And then once I make the specific heats substitution, I have T four minus T three multiplied by our lovely CV quantity. And look, this all breaks down to a constant that we have either CV, which is 0.718 kilojoules per kilogram Kelvin, CP, which is 1.005 kilojoules per kilogram Kelvin, or the specific gas constant for air, which is 0.287 kilojoules per kilogram Kelvin. You guys wanna know another cool thing? R of air, actually R for any ideal gas can be written as CP minus CV. I know, that is cool. So you could really get away with writing this entirely in terms of CP and CV if you wanted to. But we gotta stop going down the rabbit hole at some point. So let's just get back to the calculation that we were doing that we didn't even really need to do but we're doing just for character building. That's where we are arbitrarily defining the end of our rabbit hole. So CV remember from table A 19, I think it is, no, table A 20. From table A 20, CV is 0.718, CP is 1.005, those are both kilojoules per kilogram Kelvin, and R of air is 0.287 kilojoules per kilogram Kelvin. So we can summon our calculator and get back to calculating stuff. 0.718 multiplied by T2, which is 961.19 minus 290. Work in is 418.914. And then CP was 1.005 kilojoules per kilogram Kelvin multiplied by 1800 minus T2, which is 961.19. Then my Q in is 843.004. And I will point out while we're here that if you didn't wanna make the enthalpy substitution, you could also write that as U3 minus U2 plus the pressure times the specific volume of subtraction, P times of E3 minus V2. And if you were to do it that way, you could write R of air times T3 minus T2. And that would be just fine. You could say CV times T3 minus T2 plus R of air times T3 minus T2. And then if you wanted to, you can recognize that R for ideal gas is a CP minus CV and rewrite it as just CP times T3 minus T2. Isn't that neat? That's pretty neat. I mean, I just, I'm sorry, I can't help it. Like U3 minus U2 plus P times V3 minus V2. Alternative ways to get to the simplification. That would be CV times T3 minus T2 plus R of air times T3 minus T2. And then I factor out T3 minus T2 as you do. And I have CV plus R times the quantity of T3 minus T2. And that is CV plus the quantity of CP minus CV times T3 minus T2. CV cancels CP and we're left with CP. That's cool, right? I know. It is cool. But we can't spend all day looking at cool stuff. Gotta get back to math. R of air we know is 0.287. And just to be extra cautious here, I could log in our value from this proportion up here. Or remember, I can also take 1.005 minus 0.718. You do you. And I'm multiplying by 1,800 minus T2, which was 961.19. That's a three. Okay, good. So our workout is 240.727. That can't be right. Oh, excuse me. I forgot to add in CV times T3 minus T4. That's just the workout occurring from two to three. That's good. That was a fun aside that we did. Got all distracted by my CP and CVs. Plus 0.718 times the quantity T3 minus T4, which is 1,800 minus 697.98. And I get 1,031.98. That's a much more better number. Earlier, when I got a number that was less than work in, I was like, seriously, this diesel cycle has less work out than it has work in. It's a network in the inward direction. It's literally not a power cycle. That's cool. Well, we get 1,031.98 for our workout, which is higher than work in, which is at least a good sign. And then I will close this out by calculating Q out. So that would be 0.718 multiplied by 697.98 minus 1,800. And I get negative 791.25. What's wrong there? Oh, U4 minus U1, CV times T4 minus T1. I'm sure that you guys did that right. I just wrote it wrong. I can't talk and think at the same time. One of my biggest pitfalls. Okay, so let's try that again. We're looking for a positive number for Q out. Here we go. 697.98 minus 290, survey says 292.93. Okay, so 292.93, let's zoom in. So we have nice handling, 292.93. Yeah, that's clearly nice handling. And then above that was the workout, which was 1,031.98. That is kilojoules per kilogram. And then above that is Q in, which was 843.00, I guess. Are we rounding everything to do that some places, guys? What do you think? Yeah, I think so. And then work in was 481 and then some stuff, 481.91. Okay. And then what do you suppose I wanna do next? That's right, I wanna calculate the network out. Which is workout minus work in. And I also want to calculate the net heat transfer in, which is Q in minus Q out. I know that if this is a steady cycle and closed, they must be equal. So I'm looking for equal numbers. If we get something that isn't equal, that's a good indication that we made a mistake in this process. But I'm sure we did, I'm sure it's fine. 1031.98 minus 481.91 and we get 550.06, cool. And then if we take 843 and we subtract 292, we get 550.07. So let's all take a moment to consider whether or not being different by a hundredth here is a sign that we made a mistake or a rounding error. I'm going to go with rounding error. So now that we have that, it is easy enough to just calculate the thermal efficiency. Thermal efficiency is network out over Q in, which is going to be 550.06, let's call it, divided by 843. And we get a thermal efficiency of 0.6525. Now that we have all of that done, we can actually get to what it is that I asked for. So for part A, I asked for the maximum temperature and pressure occurring in the process. That would be the maximum temperature that we were given, easy. And the maximum pressure is both P2 and P3. Note while I'm here that in an effort to try to make exam questions faster, I will often ask for a proportion of a problem as opposed to the full workout analysis like we do here. And in those sorts of circumstances, it is a waste of time for you to calculate anything more than what I actually asked for. So if you imagine a hypothetical situation where I give you this given information and I just asked for the maximum temperature and pressure, you would only need to come up with P2. And that would just be P1 multiplied by 20 raised to the 0.4 power. That would be a very quick calculation, but if you approached it, like we've been approaching this example, where you draw out a table and you do character building exercises, you might waste a bunch of time with a bunch of extra parameters. So just bear that in mind. So I will say for part A, Tmax is given, so 1800 Kelvin and Pmax is P2, which was 6297.46. 6247.96, was it a nine in the 10th place? It was indeed not a nine. Bores are not nines. Okay, part A done, part B. Was it the thermal efficiency? It was not, it was the network out. So what do you guys think it's asking for? Do you think it's asking for the specific network or do you think it's asking for some sort of total magnitude of network? Well, it doesn't specifically ask for either. So let's do both, what do you think? Both specific work and total magnitude of work. So the specific network out, we calculated already, it was 550.06 kilojoules per kilogram. Now for total magnitude, big W, we're going to take mass times little w. So this is the mass of air multiplied by the specific work and the net outward direction. Specific network out, we know. Now where does mass of air come from? Well, we recognize that we have a closed system. We are treating the air at all four state points as being the same. So if we can calculate the amount of mass of air there is in the process at any of the four processes, we can answer the mass of air in the cycle overall. So M1, M2, M3 or M4 would give us the mass of air in the cycle. We have enough information to do that for one of our four state points. Which state point is it? It is state point one because we know the total volume. Note that this could be a total volume of the engine as a whole. It could be the total volume of an individual cylinder, but because we aren't given enough information to know otherwise, we're just going to say, this is the total initial volume. So I'm just going to say the mass in this volume is the mass in the entire engine. It's just a relatively small engine. Okay, now how much mass of air is there in a half a liter? Well, the most relevant parameter that we have already calculated is going to be specific volume of state one. Specific volume is total volume divided by mass. Therefore, mass would equal total volume divided by specific volume. I know both total volume of state one and specific volume of state one, so let's mosey on down here. And I will say mass of air is 0.5 liters, 0.5 liters divided by the specific volume of state one, which was 0.8761. And that is cubic meters per kilogram. And then we recognize that one cubic meter has in it 1,000 liters as per our conversion factor sheet. I think the conversion factor sheet actually gives us, says one liter is 10 to the negative third cubic meters. So I should have really written one liter is 10 to the negative third cubic meters. I just happened to remember that one cubic meter contains 1,000 liters. So use the reciprocal if you want. But I'm saying one cubic meter contains 1,000 liters, so liters cancels liters, cubic meters cancels cubic meters. I'm left with kilograms. And by the way, this should be a tiny amount because it's a gas and it's a small volume of gas at that. So calculator, you come back. 0.5 divided by the quantity. Now I'm gonna go on an adventure to find specific volume with all the decimal places, even though it doesn't matter. There we got it, multiplied by 1,000. And we get 0.000571. So this engine contains half a gram of air, neat. And then I'm multiplying that by our specific network out. So I'm saying 0.000571 kilograms multiplied by 550.06 kilojoules per kilogram. Kilograms cancels kilograms. And I'm going to be left with a quantity in kilojoules or possibly even joules. We'll see. We get 0.313 kilojoules. 0.314 kilojoules, let's call it. So again, it didn't specify what it wanted if it was total volume or specific volume. If this were an exam question, I would specify. Rest assured. Now, for part C, I am asked to determine the thermal efficiency. Luckily for us, we already did that. It's 0.6525. Therefore, I'm going to write that as 65.25%. And then for part D, it wants the PV and TS diagrams. So for that, I'm gonna go to a new page because I want us to have as much workspace as we need to do these graphs justice. I'll move this online. And you know what? Just for fun here, let's do them on top of one another. Yeah, that's fun. So John, what did you do for fun this weekend? Oh, I graphed my PV and TS diagrams on top of each other, as opposed to horizontally. It was very cool. More than being cool, it is illustrative. Remember kids, helpful graphs are cool. Okay, and again, I've said it before when I'm going to reiterate, I expect you to be able to think through these graphs and draw the general shape without having specific numbers based entirely on the processes. So remember that on PV and TS diagrams, the horizontal displacement means something. Horizontal displacement on the PV diagram represents work. Specifically, movement to the right on the PV diagram represents workout. Movement to the left represents work in. So if I were to recognize that I had work in, say from the process from one to two, I could draw that as a movement to the left. And then from two to three, I have some expansion. So I have some workout. So I'm moving to the right. And then from three to four, I have more expansion. So I move more to the right. And then from four to one, I have no work occurring. Therefore, I'm going to have no horizontal displacement. So one to two steps to the left, two to three steps to the right, three to four steps to the right and gets us back to where we started. Therefore, I can conclude that we have four different horizontal values here. One low one, one intermediate one, one high one. And the high one is state one and state four. The low one is state two, the middle one is state three. And yes, I know we have the actual numbers, but thinking through the graph is the important part here. Okay. And then we can think through the pressures. We can do that by recognizing that we have a compression process from one to two. So our pressure should increase. And then from two to three, we have an isobaric expansion process. That's the heat addition process, I guess I should say. So the pressure is constant from two to three. From three to four, we have some expansion occurring. So it should drop in pressure. And then from four to one, we are cooling back off, which should also yield a drop in pressure. So I'm going to identify three different vertical positions as well. One will be the low pressure. Then we jump up to our high pressure at two and three. And then four is below that. Therefore, my processes look like this. One is here. Two is here. Three is here. And four is here. And later on, you'll learn how these curves actually look. For now, we can just draw them as rough curves. Actually, you can get away with the straight lines, that's fine. And the region closed here is going to be the net workout because the integral from two to three represents the workout from two to three. The integral from three to one represents the workout from three to one, excuse me, from three to four. Actually, that's the same as three to one because the horizontal displacement is the same. So the workout is the total area under the curve from two all the way over to four. And then the work in is going to be the area under the curve from one to two because it's the integral of that pressure times specific volume, excuse me, the integral pressure with respect to specific volume from one to two. And because it's right to left, it's going to be in the inward direction. So when I take the workout minus the work in, I'm left with that region inside the cycle, which is the network out. One graph down, one more to go. On the TS diagram, not to be confused with the TV diagram. Horizontal displacement represents heat transfer. Move into the right, represents heat transfer in, move into the left, represents heat transfer out. And then I can think through my horizontal displacement by looking at the heat transfer opportunities. From one to two, I have an isentropic process which has no heat transfer. So one and two are going to be horizontally on top of each other on the TS diagram. Two to three, I have a heat transfer in the inward direction so I should move to the right. Three to four, I have an isentropic process which implies no heat transfer. So three and four are going to be horizontally on top of each other. And four to one, I go back to where I started. So one and two are going to be to the left. Two and three are going to be to the right. No, not two and three. Again, three and four are going to be to the right. Then, vertically, I can think through what's happening in our four processes. First up, I have a compression process and since I'm compressing, I'm going to increase the temperature. From two to three, I have fire happening. Fire's going to increase the temperature. So two to three should also increase. Then from three to four, we have an expansion process so it should drop. And from four to one, we have a cooling process so it should drop. So I'm going to call this one and two and three. And then I don't know the relative positioning of four relative to two. So if I were just doing this properly, I would actually have to determine those quantities. And since we already did, I can see that T four is less than T two. Therefore, I should really draw this as four under two. Does that logic make sense? You could get there as well by thinking through which is going to affect the temperature more, isentropic expansion or cooling process and isentropic compression or heating process. But whatever the case, we have our rough shape. If you didn't know and you just drew two and four at the same temperature, that would be close enough for my purposes here. So one is down here, two is above it, three is up here, four is below that. And we have ourselves a parallelogram with a slight curve to it because those isentropic processes are vertical lines and an isobaric and isochord process appears as a curve on a TS diagram. And just like before, I can recognize that the area under the curve from two to three represents our heat transfer in. The area under the curve from four to one represents our heat transfer out. Therefore the region enclosed is going to represent our heat net heat transfer in the inward direction, which should be the same as our net work out. And that was everything that I asked for for this problem. Now we could be done, but I wanna add another part to this. I wanna add a part E. And for part E, let's say how much power would this engine produce if it was operating at a rotational speed of, I don't know, 2000 RPM. And let's say that I asked for that in horsepower. Well, to get there, I recognize that every time a diesel cycle occurs in this engine, I'm producing 0.314 kilojoules of energy. So what I'm going to want to do is take our net work out and I'm going to multiply by 2000 revolutions per minute and I'm going to multiply by one half. Why am I multiplying by one half? Because I'm only getting one diesel cycle per every two revolutions. If you think about this in terms of quantity, I can reference that total net work, which is 0.314 kilojoules. Think of it like 0.314 kilojoules for every diesel cycle. And then there are one diesel cycle per every two revolutions. Does that make sense? I mean, every other revolution is the actual diesel cycle and every other other revolution is the exhaust and intake strokes, which we are ignoring in our analysis above. So we have diesel cycle, boring, exhaust and intake, diesel cycle, boring, exhaust and intake. So 2000 revolutions per minute means we have 1000 diesel cycles. That's 1000 combustion processes, 1000 power strokes, et cetera. Okay, you'd also think of that like four stroke cycle is going to have one power stroke and the four stroke cycle occurs over two revolutions. Whatever logic you use to come up with this relationship, this is how we get to an answer. So I have kilojoules per minute so far and then I'm going to write one minute is 60 seconds and that is kilojoules per second, which is a kilowatt and then the conversion from horsepower into kilowatts comes from our conversion factor sheet. One horsepower is 0.7457. One horsepower is 0.7457, kilojoules per second. Okay, kilojoules, cancels, kilojoules. Seconds, cancel, seconds, kilowatts, cancel, kilowatts. And there we have our answer for horsepower. So calculator and we take this thing we just calculated times 2,000 times one half times one over 60 times 0.7457. We get seven horsepower or so. So the net power output is seven horsepower. Cool, now is there anything else fun I can add to this? Well, we could work through it again using the non-cold air standard. That might be a little bit less fun. In that analysis, we would just have to look up VR and use the proportion for our isentropic and gas equations without the cold air standard wherein VR two over VR one is equal to V two over V one. So I can use the fact that I know the compression ratio and VR one to determine VR two. And then I can use VR two to look up everything I need at state two, figure out the pressure using PR one and PR two because I can look both of those up. I could say P two is equal to P one multiplied by PR two over PR one. Once I had P two, that would give me P three. Once I know T three and P three, actually I just need T three, honestly. I can look up everything that I need at state three included but including but unlimited to U three and H three and VR three and PR three. And then I can use the same isentropic and gas relations that VR four over VR three is equal to V four over V three. And then using our actual specific volumes to come up with VR four from which I could determine whatever else I needed. The fact that V one is equal to V four means I know V three and V four and I know VR three so I can determine VR four and look up whatever else I needed. So you need to look up U one, U two, H two, U three, H three and U four. And once you had all of those, you could calculate these quantities the same, determine your networks, your thermal efficiency and so on. Since we did all of that already in the auto cycle, I will leave that as an exercise to the viewer. For the purposes of our in-class examples like this, I'm going to be using the cold air standard more often than not just because it's faster and it gives us more time to talk about the actual cycle and spend less time on table lookups. But bear in mind that you should be proficient at solving through these cycles with the cold air standard and also without the cold air standard. I expect you to be approximately as provisioned at both methods. Now, before we move on, let's consider how this problem would look in MATLAB. If we solve the problem in MATLAB using the same relationships we did when we were solving it by hand, we can figure out the maximum temperature, the workout and the thermal efficiency all much more accurately than we did by hand because we aren't limited to the same rounding errors that this calculator produces. I'm looking at you calculator. But it also allows us to graph how the PV and TS diagrams actually look. So take a second here to consider how close our very rough drawings were to the actual work created by MATLAB. We'd got pretty close and remember that we are trying to draw the rough shapes. So the acrographs are more useful in practicality, but the important part is considering what's happening between the four processes. As we start developing more complicated cycles, it's useful to use what's happening on the graph to keep track in your brain of what's actually happening in reality. It also is a quick and easy performance metric. We can look at how changes in our analysis affect the graph and infer things about that. For example, if I were to ask a question like if I were to increase the pressure at state two, and therefore state three, I'm going to move that top line up on our TS diagram. And by moving it up, I am going to be increasing the area inside of that graph. Therefore I will have more network. So higher pressure is going to involve more work here. We can also consider the implications of changing other things. Like you'll remember when we talked about the diesel cycle, I had mentioned the fact that we want a small cutoff ratio. The smaller the cutoff ratio, the better performance we actually get because the more power is happening in the power stroke where it's supposed to. Well, I could adapt this MATLAB code to using cutoff ratio as one of our input parameters. Our cutoff ratio in this problem was two. I mean, actually, we could have just calculated that. We could pop up our calculator. We could take V3 divided by V2. And we could say the cutoff ratio is 1.87. So about two. So the piston moves down such that the volume almost doubles during the heat addition process. Well, in an analysis where we were trying to consider what happens as we change the cutoff ratio, it might be useful to have MATLAB compute the solution with a bunch of different cutoff ratios. And I have done that for you. So I've wrapped that in a loop using the cutoff ratio as a given parameter. And we are computing all of our same results for every cutoff ratio between one and 25. In our plot of thermal efficiency plotted against compression ratio for each cutoff ratio between one and 25, observe that as the cutoff ratio increases, we are moving down. Therefore, we have a lower thermal efficiency for the same compression ratio. So it's in our best interest to make an efficient engine by making the cutoff ratio as small as possible, meaning that we are injecting the fuel very, very quickly. Another thing to discern from this graph is the consideration between auto cycles and diesel cycles. I asked a question like, how does the typical thermal efficiency of an auto cycle compare to the typical thermal efficiency of the diesel cycle? Which one is more efficient? Gasoline engines or diesel engines? Well, you'd say the diesel engine, that's right. Diesel engines are way more efficient than their gasoline counterparts. But if we look at this graph, we see that a cutoff ratio of one, which is equivalent to an auto cycle, has the highest thermal efficiency for a given compression ratio. I mean, if we consider a cutoff ratio of one, which would be our auto cycle to our actual cycle, which was about two, that thermal efficiency is dropping, maybe five or 10% for a given compression ratio. How can the fact that diesel cycles are more efficient than auto cycles in reality? Work with this graph. The reason is that diesel cycles have a higher compression ratio. In a gasoline engine, you are limited by the auto ignition temperature of the fuel. If you get a compression ratio too high, you start knocking, which is kind of productive. In our auto cycle example, we had a compression ratio of eight. So that would be right about here on our plot here. In our diesel cycle analysis, we have a cutoff ratio of two, but we have a compression ratio of like 20. So we are comparing this thermal efficiency to this thermal efficiency. That's why it's higher. Yes, the auto cycle is more efficient for a given compression ratio, but the diesel cycle can be more efficient because it has a higher compression ratio. It doesn't have to worry about auto igniting too early because the whole point of the engine is to auto ignite. Therefore, we are injecting the fuel after the compression process. Does that make sense? And again, these MATLAB codes are posted on D2L if you want to reverse engineer them.