 You have been tasked to analyze a gasoline engine with a compression ratio of 8 to 1. You know that 750 kJ per kg of heat is added during the combustion process, and you know that, prior to compression, the air is at 95 kPa and 27°C. Assuming the cold air standard, determine or complete the following. A. The maximum temperature and pressure occurring in the process. B. The network output. Note that that's not specified as to if it's specific work or total power. C. The thermal efficiency. D. The mean effective pressure during the cycle. E. The ratio of air to fuel burned during combustion, i.e. the air-fuel ratio. Assuming that the gasoline burned is pure octane. And lastly Part F, sketch a PV and TS diagram of this cycle. So first of all, let's reflect on the auto-cycle itself. In the auto-cycle, the process from 1 to 2 is our isentropic compression process. If the pressure and temperature are given prior to compression, that implies that 95 kPa is state 1's pressure, and 27°C is state 1's temperature. I know that 750 kJ per kg of heat is added during the combustion process, which is our process from 2 to 3. Therefore, I can say this is Q in from 2 to 3. Furthermore, since the auto-cycle only has one process with heat addition, that's going to be the specific Q in to the entire cycle. The compression ratio is given as 8 to 1. I'm going to note that as r is equal to 8. Then, because we are told to analyze this using the cold air standard, we are going to be using the properties for ideal air at 300 Kelvin. We jump into our appendices, what we're grabbing are properties from table A20, for ideal air at 300 Kelvin, Cp is going to be 1.005, Cv is going to be 0.718, and K is going to be 1.4. Furthermore, remember that the cold air standard means that we're going to be using the assumption of constant specific heats, so anytime we encounter a delta U or delta H, we are going to be substituting with Cp or Cv delta T. Furthermore, it means that we can use our isentropic ideal gas equations on any process that is isentropic. Now I'm going to ignore all of the statements that I actually want, and let's just start by approaching this in the same manner that we did our arbitrary power cycle. So ignore all of this for the moment. We'll come back to that in a second. Instead, I want us to calculate the temperature, pressure, and specific volume at all four state points. Again, this isn't necessarily the most efficient way to approach this problem, but I think it's useful in the learning process. We were told that the pressure prior to compression is 95 kilopascals. Note that that's going to be our ambient air because the process prior to compression is air intake in reality. That's why this is 95 kilopascals about atmospheric pressure in South Dakota, and T1 is about a reasonable temperature on a nice warm summer day when you would want to be driving around 27 degrees Celsius. Furthermore, 27 plus 273.15 is close enough to 300 Kelvin, and we could probably call it 300 Kelvin. But for the moment, let's leave it at 300.15. Just like with the arbitrary cycle, we can use our ideal gas law to calculate a specific volume since we know this temperature and pressure. For that, we will summon our calculator. And I'm going to gloss over the calculation of the specific gas constant for air because we've done that in previous problems. Remember that the specific gas constant for air is the universal gas constant, which is 8.314 kilojoules per kilo mole Kelvin, in the case of our metric unit system here. That number is retrieved from the inside of the front cover of your textbook. That's the bottom left corner of your conversion factor sheet. And you divide by the molar mass of the substance you're analyzing. In this case, ideal air has a molar mass of 98.97 kilograms per kilo mole. Therefore, we are taking 8.314 divided by 28.97, which is going to be 0.287-ish kilojoules per kilogram Kelvin. Then, since PV is equal to RT, V would equal RT over P. So I'm going to take 0.287 kilojoules per kilogram Kelvin. And I'm going to multiply by our temperature in Kelvin, which is 300.15. And then I'm going to divide by 95 kilopascals. And I walked through the unit conversions in that arbitrary cycle analysis that we had performed, so I'm not going to write it out in detail again. But we know we're going to have a result in cubic meters per kilogram. For state number 2, we don't have T2P2 nor V2 yet. The thing that gets us from 1 to 2 is recognizing that we are going from our big volume to our small volume. And we know that that compression ratio, which is a proportion of volume, remember, is 8. So the big volume is going to be 8 times larger than the small volume. Furthermore, because this is a closed system, that means that the proportion of specific volumes is also going to be 8. Therefore, V2 is going to be V1 divided by R, which is 8. So we can take that number we just calculated, divide by 8, and we get 0.11346. And with that, we actually have all four specific volumes. Because the process from 2 to 3 is isochoric, and again because it's a closed system, that means the mass is constant, that means V3 is equal to V2. And then we expand back to our initial specific volume, because the process from 4 to 1 is isochoric, therefore V4 is also 0.907. You could also think of it like the expansion ratio is the reverse of the compression ratio. So instead of going from big volume to small volume, you're going from small volume to big volume, and that proportion is still 8, that is big volume over small volume is still 8. Therefore, you could think of this like V4 over V3 as well. That gives us one independent intensive property at state 2. The other independent intensive property that we have is our specific entropy. Now we're not actually going to be trying to determine the entropy because we don't necessarily need to. It's just the fact that the entropy is the same that fully defines our state 2. To actually calculate T2 and P2, we will go back to our isentropic ideal gas relations. I have the option of using any of these equations for the process from 1 to 2, but since I have a ratio of volumes, it's probably going to be most convenient for me to use this equation to come up with T2 and this equation to come up with P2. So what I'm going to write is T2 is equal to T1 multiplied by V1 over V2 raised to the k-1. And remember, if I multiply the contents of this parentheses by mass over mass, I could write that as specific volume over specific volume and the same proportion is true because the mass doesn't change. What I'm going to write is T2 over T1 is equal to V1 over V2 raised to the k-1. Therefore, T2 is equal to T1 multiplied by r raised to the k-1. Remember, r, our compression ratio, is big volume over small volume. So if you're trying to think through, should I plug in r or 1 over r? Just think through the proportions of actual volume here. Compression is big volume, too small volume, so it should be big number over small number. Therefore, r, not 1 over r. And then our T1 value was 300.15, so I'm going to write this as 300.15 multiplied by 1.4. 8 raised to the power of 1.4 minus 1. And I get a temperature at state 2 of 689.56. Then I follow the same logic with our pressure calculation, except this time I'm using k, not k-1. And I can write this as 95 multiplied by 8 raised to the power of 1.4 minus 1. And I get a pressure at state 2 of 218.253. And with that, we are halfway through our state point properties. Next, we need to consider the process from state 2 to state 3. I know that my ideal gas law would allow me to write pv is equal to rt. Since I have an isochoric process, I can write this as v over r is equal to t over p. Since this quantity on the left is the same at state 2 as it is at state 3, that means I can write t2 over p2 is equal to t3 over p3, which means that I can write t3 is equal to t2 over p2 times p3, or I can write p3 is equal to t3 over t2 times p2. However, neither one of those is helpful right now. I have one equation and two unknowns. I don't know t3. I don't know p3. So what is the thing that gets me from state 2 to state 3? It is the specific heat input. I know 750 kJ of heat is added between 2 and 3. So if I were to write this out as an energy balance, I'm saying we have an isochoric process. The piston isn't moving because we're assuming that the combustion happens instantly. Therefore, I can write this as delta u plus delta ke plus delta pe that's equal to little q in plus little work in minus little q out plus little work out. Note that I am neglecting the first several steps of the energy balance, jumping immediately to this form. I'm canceling the changes in kinetic and potential energy assuming that the changes in kinetic and potential energy of the system itself are negligible. Next I'm neglecting works because I have an isochoric process and q out because the heat is added. Therefore, the q in from 2 to 3 is equal to u3 minus u2. And then you know what's happening next, right? That's right. Because I have assumed constant specific heats, because I was told to use the cold air standard, which has as part of it assuming that the heat capacity is constant at 300 Kelvin, then I can plug in Cv multiplied by T3 minus T2. Therefore, T3 is equal to q in over Cv plus T2. So I'm going to say T3 is equal to 750 kilojoules per kilogram divided by 0.718 kilojoules per kilogram Kelvin. The kilojoules and the kilograms are both going to cancel, leaving me with Kelvin in the numerator. And then I am going to add T2, which is 689.56. And I get 1734.13. And then now that I have one of our temperature or pressure at state 3s, I can say P3 is equal to P2 multiplied by T3 over T2. So 218.25 multiplied by 1734.13 divided by 689.564 And just for good measure here, I'm actually just going to jump up and grab our P2 property that we calculated, because you know now is really the time to be a stickler about all those decimal places. Now we get a pressure at state 3 of 548.86 kilopascals. The last process we have to consider is the process from 3 to 4. That process is isentropic expansion. Because it's isentropic expansion and because we are assuming constant specific heats, we can use our isentropic ideal gas relations again. So we are essentially doing the same thing that we did here. We are just writing it as T4 over T3 is equal to V3 over V4 raised to the k-1. Because in our isentropic ideal gas relations equation sheet, one really represents the beginning of the process, two really represents the end of the process. Now do I plug in R here or do I plug in 1 over R? That's right, V3 is smaller than V4, therefore it's 1 over R. So I'm going to take T3 multiplied by 1 over R raised to the k-1. T3 was 1734.13 multiplied by 1 over 8 raised to the power of, I forgot, I have to push the carry button on the calculator, I can't just take care of the keyboard, raised to the quantity of 1.4 minus 1. And I get 784.823 kelvin. And then I'm going to repeat that process with k instead of k-1. For the pressure form of this equation, I'm going to take P4 is equal to P3 multiplied by 1 over R raised to the power of k. So P3 was 548.866 multiplied by 1 over 8 raised to the power of 1.4. So I've calculated a quantity here that is 29.864. Now I could write that down, however, here, let's just step through this. I'm writing this down, definitely going to leave it definitely permanent. Okay, now let's think through whether or not these state point properties make sense. We have, first of all, an isentropic compression process. We are decreasing the specific volume and as a result our pressure should increase and our temperature should increase as well. That makes sense. Then we are adding heat to our process at the same specific volume. Therefore our temperature should increase and our pressure should increase. The fact that we ended with a higher temperature and pressure makes sense. Then we are expanding to drop power out of the gas so the temperature should drop and the pressure should drop. But our process from 4 to 1 is going to be an isochoric cooling process. So we would expect our temperature to drop and our pressure to drop. So the fact that we're going from 29.86 down to 95. Oh wait, no we don't. This is indicating that the pressure would rise during the cooling process that occurs at a constant volume. That doesn't match what I know would happen because the ideal gas law says that pressure and temperature are going to be directly related for a process at a constant volume. Therefore this implies I have a mistake. This is the value of thinking through your state points to double check that they make sense before you move on. So let's double check all of our pressure calculations shall we? P2 is equal to P1 multiplied by... There we go. So our mistake occurred back when we calculated P2. I'm sure that many of you were screaming at your screens indicating to me that I had made an error but I was just moving on with the calculation. As you know now that was totally on purpose and here to serve as a learning opportunity. So what we have to do is rework P2, P3 and P4 and get rid of the minus one because in our pressure equation here this doesn't have K minus one this just has K. There is no minus one so I should not have subtracted one from the exponent. So at state two I have a pressure of 1,746.02 and then I'm going to take that quantity I'm going to multiply by 1734.13 divided by 689.56 and I get 4,390.96 and then I am going to take that quantity and multiply by 1 over 8 raised to the power of just 1.4 and I get 238.909. Now let's think through that again. Our pressure goes up when we are isentropically compressing that makes sense. We have an isochoric heat addition process. The pressure increases as the temperature increases in a constant volume process that makes sense. Then the pressure drops as the gas expands that makes sense and then we have an isochoric cooling process where the temperature and pressure both drop again. I could go one step further in my number crunching here and recognize that the same ideal gas equation simplification that we came up for 2 to 3 is also going to be true from 4 to 1. So I should be able to write T4 over P4 is equal to T1 over P1. Therefore P1 should equal T1 over T4 multiplied by P4. So if I get the same pressure at state 1 that I started with that implies that I did everything correctly. So I am going to take 238.91 and I am going to multiply by 300.15 divided by 784.82 and I of course know that I should have a little bit of discrepancy here due to rounding errors but let's see what happens. We get 91.3695. I wonder if we can make that even tighter here. Hey look at that. When we grabbed all the decimal places we got 95.0006. That's a very strong indication that we at least used all the calculations correctly. Now that we have all of our state point properties the world is our oyster. So before we move on to calculating the stuff that I actually want us to calculate I would instead like us to calculate the specific work, the specific Q in, the specific work out and the specific Q out occurring in this cycle. I know that work in for an auto cycle is going to occur in 1 to 2 and because I have an isentropic process the heat transfers are going to disappear and I am left with U2 minus U1. For my heat addition term I know that that's going to be U3 minus U2 because of my above energy balance and we know that that's just going to be 750 kilojoules per kilogram. My work out is occurring between 3 and 4. Again I have isentropic expansion therefore the heat transfer terms are going to disappear and I'm going to be left with work out and delta U so I can write this as U3 minus U4 and to all of you asking why is it 3 minus 4 instead of 4 minus 1 remember that I have delta U is equal to work in minus work out or can disappears because the work is only out therefore it's delta U is equal to negative work out so I would write U4 minus U3 is equal to negative work out therefore the actual work out is U3 minus U4 and then similarly for Q out again because Q out is going to be in the energy out term therefore has negative in front of it this is going to be U4 minus U1 because it's beginning minus end then I'm going to substitute in CV delta T here so I have CV times T2 minus T1 CV times T3 minus T4 and CV times T4 minus T1 so by scrolling up to my temperatures here I can just crunch some numbers my specific work in would be 0.718 multiplied by 689.56 minus 300.15 279.596 just for good measure here let's calculate the heat transfer in and double check that we actually get 750 and we do which is good then our work out term would be 0.718 multiplied by the quantity 1734.13 minus 784.82 and we get 681.605 and then 0.718 multiplied by 784.82 minus 300.15 and we get 347.993 I bumped my iPad with my elbow and it scrolled up so sorry for all the nausea this is undoubtedly inducing but I had 279.596 kilojoules per kilogram can confirm that that is indeed 750 my work out term is 681.605 my Q out term is 347.993 then because this is a power cycle I should have net work in the outward direction which is going to be work out minus work in so I will write that as 681.605 minus 279.596 and I get 402 I guess we can write 402.01 I mean come on John be at least a little bit consistent and then just to check myself I'm going to take 750 minus 347.993 and I get 402 and I would bet if I use this 750.001 instead of the 750 that it should have been I would actually get the exact same number hey look at that it's the exact same number let's just round these to 0.01 that's easier all along cool we have a net work out we have a net Q in now let's start to actually consider what the problem statement actually wants first up I had maximum temperature and pressure occurring in the cycle well T max is going to be T3 I know that for a couple of reasons first of all it's the highest number it is the only one in fact that has four digits wow that's a big number I don't know why I'm running T it's 1734.13 the other reason I know that is because as a general rule of thumb in a power cycle it is going to be the hottest after the heat addition process i.e. it is going to be the highest temperature after the fire thing happens where there's fire there is often a high temperature so general rule of thumb your highest temperature is going to correspond to the state point after your heat addition process and our pressure is correlated so closely with temperature that our state point with the highest temperature is going to have our highest pressure as well so 4,390.96 then part B I wanted the network output so I have that already as a specific term it is 402.01 now do I have the ability to determine a power? I don't I have no indication of size there's no mass that I actually know therefore I cannot multiply by any sort of mass or size that would represent mass to come up with an actual power all I can quantify is the network out on a specific basis I'm giving it per unit mass that flows through the cycle that's the best I can do for part C the thermal efficiency of this power cycle is going to be the network out divided by heat transfer in which is going to be 402.01 divided by 750 and I get 53.6% for part D I am determining the mean effective pressure during the cycle well just a brief recap the mean effective pressure is the pressure that if held constant would yield the same specific network so if this piston cylinder arrangement was held with air at a constant pressure and push the piston down what pressure would yield the same specific network I'm describing a displacement here I can describe that as the piston moving from top dead center which I can call x1 down to bottom dead center which I can call x2 so the relevant work here is going to be the integral of force with respect to displacement and my force is going to come from pressure since pressure is defined as a force distributed across an area that force is going to be pressure times area which one combined into my integral gives me pressure times area times dx since my volume swept by the piston is a cylinder that has a constant area I'm going to group together these terms into dv the volume differential and I can write this as the integral of pressure with respect to the volume and then the pressure is constant because the hypothetical quantity that I'm calculating the mean effective pressure is an average pressure I'm assuming it's a constant so it comes out therefore I have pressure multiplied by change in volume so the net work out here would be the mean effective pressure multiplied by delta volume therefore the mean effective pressure which by the way the book abbreviates p-mef but I write it as MEP is net work out over delta volume if I were to divide the numerator and denominator by mass I could also write this as net work out over delta specific volume and since I'm talking about going from top dead center to bottom dead center that's going to be our smallest volume to our biggest volume that is going to be v1-v2 cool, now that I have that equation I can actually calculate a quantity so I'll pause here for a second actually I'll leave it I'll leave it mean effective pressure for this process is 402.008 divided by my specific volume is state 1 which was 0.9 or something I'm going to go on a scrolling adventure to find that in my calculator minus that same number divided by 8 note that because it's the same number divided by 8 I could factor out the same number and then multiply by 1-1 over 8 but at this point I'm already down this path so I'm going to take the actual number and I get 506.676 so what does that number represent? not a whole lot I mean it's very abstract it doesn't relate to any pressure in the actual cycle it's a pressure that if you were to hold constant and then sweep the piston through the same space you get the same network out and that's useful for a type of comparison between engines because generally speaking the one with the higher mean effective pressure is the one that is going to produce more power but there's so many other scale values and there are so many other ways of comparing engines to one another and it is not actually that tangibly useful anymore but we still calculate it because it is a quantity that you can calculate from the parameters of the engine so sometimes it's useful and in those very rare circumstances you should know how to calculate it or at least be vaguely familiar with what it represents next up the air-fuel ratio for that I am going to write down here again I was thinking about going to a new page I'll save that for the diagrams part E here I want the ratio of air to fuel which I'm going to abbreviate AFR that's air-fuel ratio and the AFR here is going to be the mass of air to mass of fuel in the engine now I know what you're thinking you're thinking but John we don't know the mass of air we know the mass of fuel so how can we possibly calculate this well instead of approaching it from its definition let's think about what's happening in the heat addition process we are adding heat to the engine by burning fuel and if we assume that all of the heat produced by the combustion process is going into the air which is a big assumption here then I can write them as being equal to one another and I can write this as a magnitude as well which would just be across the process as opposed to per unit time and if I do that I can write this as mass of air times specific Q in is equal to mass of fuel times the heating value of the fuel that is since specific Q in is given per unit mass of air in the cycle we're describing the total heat added by taking that specific quantity and multiplying it by the mass of air then we are using the heating value of the fuel multiply by the mass of the fuel to come up with a total heat emitted by burning the fuel and now we can write this as the mass of air over the mass of fuel is equal to the heating value of the fuel divided by Q in and that would give us our air fuel ratio so Q in we know is 750 ish kilojoules per kilogram of air and our heating value for fuel is going to be the heating value of octane we were told to treat the fuel as pure octane so we go back into our appendices on table A25 we have a variety of heating values for a variety of substances including but not limited to octane now we have two different types of octane and we have two different types of heating values so we're going to be using one of these four numbers now which number we choose is based on a couple of assumptions that we need to make first of all do we treat the fuel as being a vapor G or a liquid L for our purposes right now we are just going to use the vaporized value for fuel we are treating it as a liquid that then evaporates that is a better way of modeling the engine if you are trying to be more accurate but then you have to account for the latent energy of the fuel and that energy is lost from the air and it becomes a bit of a more tedious calculation and since our analysis is an overview and is an idealized form of the model anyway we are going to neglect that if you want to get more information on accurate ways we are going to use an engine and the simplifications we make in the auto cycle and how we can adapt them to become a little bit more accurate you should take the internal combustion engines tech collective but for now we are grabbing from this row here then we have to consider whether or not we should treat this as a higher heating value or a lower heating value general rule of thumb if you don't know for sure use the lower heating value the higher heating value is more appropriate in situations where you are recovering every last bit of energy that you possibly can like if you consider a natural gas furnace the output of a natural gas furnace has some products that you can recover a little bit more energy from really really high efficiency natural gas furnaces can for example condense the steam that has been produced in the combustion process that vaporized water when condensed is going to dump a little bit of energy back into your system so in a very high efficiency circumstance we need to use a higher heating value your system would also have intricacies to be able to actually make that happen since we are just burning fuel within tiny piston cylinder arrangement with reckless abandon we are going to be using the lower heating value and neglecting this little bit of recovered energy that we could get if we had really really specific types of combustion processes so our value is 44,790 again vaporized octane already we aren't trying to account for how the liquid becomes a vapor and whether we should use the heating value of a liquid and not account for the latent energy or try to account for the latent energy or just treating it as a vapor for now we are using the lower heating value because we don't have the complexities that would recover that little bit of additional energy from the products ok that was a very long winded way of coming up with 44,790 kilojoules per kilogram of fuel then when I divide these two quantities I should get 1493,25 thank you calculator come on calculator about 59.72 so I kinda ran out of room here but there really should be units on that it's not just a unitless proportion even though it is kilograms per kilogram because they are different kilograms yeah that's not at all confusing this is kilograms of air per kilogram of fuel which could be grams of air per gram of fuel or pound mass of air per pound mass of fuel this is a proportion of the mass of air per mass of fuel and this is useful because it gives us an indication as to the head room that's left 14.7 per mass units of air per mass units of fuel so what this means is we have about 4 times as much air as we need to actually accomplish this combustion process so theoretically a person could throttle up this engine add more fuel to it and multiply the amount of fuel by about 4 times and still have approximately good combustion lastly we have part f which is to sketch a pv and ts diagram of this cycle so for that I'm going to jump over to a new page and I'm going to begin by creating some axes now remember move into the right on a pv diagram represents work out move into the left represents work in and those are reversed on the ts diagram move into the left represents q out move into the right represents q in and again just because we have actual specific volume numbers doesn't mean that I want us to use those I want to think about how the work is going to appear on the pv diagram so how many different opportunities for work do we have we have 2 the process from 1 to 2 is a work in therefore from 1 to 2 I should step from the right to the left and then 2 to 3 has no work occurring because it's isochoric then 3 to 4 is an expansion process so we have work out and then 4 to 1 is isochoric so we have no work so 1 to 2 2 to 3 3 to 4, 4 to 1 I should have 2 specific volumes big volume, small volume and I'm going to label those v1 and v2 and again we actually know them because we calculated them but what I'm looking for on these graphs is relative positioning of the state points you wouldn't have needed to calculate this specific volume for this part of the analysis now we can do the same thing for pressure recognizing that we have the smallest pressure at state 1 and then we are increasing to a higher pressure at state 2 and then we are increasing again let's draw more like this we are increasing again to p3 and then we are cooling off between 4 and 1 so we should drop just a little bit so our expansion should go from 3 all the way down to 4 and then I can draw state 1 state 2 state 3 and state 4 and then I know that movement to the left is going to represent our work in so work in from 1 to 2 is going to be this region this curve and then movement from left to right from 3 to 4 represents our work out this quantity here is work out so when I take a work out minus work in what I'm left with is this quantity inside which means that this represents our network out now let's do the same thing for our ts diagram movement to the left represents q out movement to the right represents q in I have one process that has a q in term I have one process that has a q out term and it goes no heat transfer from 1 to 2 and then heat transfer in from 2 to 3 no heat transfer from 3 to 4 and then heat transfer out from 4 to 1 so 1 and 2 should be over here 3 and 4 should be over here so I will call this s1 and s3 and we know that we are increasing our temperature when we compress we are increasing it when we add heat to our process we are trying to get as much energy as we can which means it's going to drop a lot then we have cooling between 4 and 1 so it should drop just a bit so I'm going to draw this as t1 and then t2 and then t3 and then t4 back down here so state 1 is going to be here state 2 is going to be here state 3 is going to be here and state 4 is going to be here and state 4 is going to be here and just like with our pb diagram we could take the area under the curve from 1 to 4 and recognize that this is our q out the area under the curve from 2 to 3 and recognize that this area represents our q in therefore q in minus q out will leave us with this region here which is the net q in which is also equal to network out so these are our approximate shape pb and ts diagrams for this example problem these graphs are a little bit crude perhaps but they illustrate what's happening in the process if you understand how a horizontal displacement represents what can heat transfer on the pb and ts diagrams and you can reason through what vertical displacement implies about temperature and pressure changing over processes you can interpret a graph and have a pretty good idea of what's happening in a cycle or you can draw out the graph for a cycle without even having specific numbers if we wanted a better graph and better numbers I would recommend completing this on a computer that doesn't have the same inability to keep track of decimal places that I do or inability to draw lines of constant specific volume on a ts diagram or lines of specific entropy on a pb diagram for that I will turn to my friendly neighborhood matlab that's the calculator friendly neighborhood matlab that's the matlab anti-calculator well done OBS so in this matlab code I'm performing the exact same calculations that we did by hand we looked up a heating value of octane of 44,790 we looked up a cvncp value from the back of our textbook we have our given information and then we are just calculating the process from one to two using isentropic ideal gas relations two to three using an isochoric process three to four using isentropic ideal gas relations taking cv dot the t for all the things calculating a network out and then calculating a thermal efficiency mean effective pressure and air fuel ratio and note that what we get is approximately the same as what we had for our properties we have 1733.73 instead of our temperature which was 1734.13 and we have 4389.93 instead of 4390.96 of a network out that's about three kilojoules per kilogram higher and a thermal efficiency that is off by about 0.15% now in this class you are going to be expected to complete the work by hand but I would encourage you to get proficient at completing them in matlab as well first of all it allows you to spend less time on the things themselves and more time on the engineering side of things thinking through processes coming up with relationships between properties and even things like plotting out the pb and ts diagram are much faster and more accurate in matlab note here that I'm plotting out the pb diagram with two lines of constant specific volume and two lines of constant entropy and this is all graphed to scale properly also note that the pb diagrams are generally given with a logarithmic scale on the x-axis that's one of the reasons I'm not particular about using our actual specific volumes to place their positions on the pb diagram that we draw by hand note on the ts diagram we have two lines of constant entropy and two lines of constant specific volumes but the general shape of the plots is the same we were able to think through how the properties are affected by the processes and therefore what the general shape should look like furthermore just to elaborate again you should be able to look at a pb and ts diagram and deduce hey there's going to be work in from one to two there's going to be heat addition probably isochorically from two to three there's going to be work out from three to four and then there's going to be heat reduction I know that I'm decreasing my pressure which probably implies a decrease in temperature at a constant specific volume for an ideal gas and then looking at the ts diagram you can confirm ah yes we have compression from one to two and then we are adding heat which we know because it's a horizontal displacement to the right from two to three and then we have an expansion process from three to four isochrophically because our entropy doesn't change and then we are going from four to back to one where they heat reduction they heat output from right to left to get back to our initial temperature I have posted this code on d2l so you can poke through it if you want another nice thing about this is that if you wanted to perform a parametric analysis say you wanted to change a thing and see how that affected your results you can do that very quickly on matlab like let's say that we were considering by hand how compression ratio affects the thermal efficiency of our auto cycle we jump back into our handwritten calculations I want to see how r affects thermal efficiency well to do that it's probably easiest to write out our thermal efficiency and look for ways that r can appear I have net workout divided by q in which is workout minus work in divided by q in which is q in minus q out divided by q in because remember the net workout and the net heat transfer in should be the same and then I'm going to write that as one minus q out over q in appears between four and one so I can write this as one minus cv times t4 minus t1 because it's an output that means there's a negative on the right hand side of the energy balance so it goes from end to beginning and then it's flipped so it's beginning to end and then in the denominator I have cv multiplied by t3 minus t2 because it's a q in term we are not having to deal with the negation of the delta u so it's just n minus beginning and then because of the cold air standard so I'm left with one minus t4 minus t1 divided by t3 minus t2 and I can factor out t1 from the numerator this is t1 multiplied by t4 over t1 minus 1 over t2 multiplied by t3 over t2 minus t over t would just be one again now to try to reduce this further I'm going to split this into two chunks first we'll consider the right then we'll consider the left on the right side we have t4 over t1 minus 1 divided by t3 over t2 minus 1 now to figure out how these are related or correlated I'm going to write out all of my processes in terms of temperatures and to do that I think I'm going to need a little bit more space so let's start a new page so the question for the moment is how are these related to one another are they directly proportional are they indirectly proportional are they can they be reduced to a function of r etc and I'm going to approach this by writing out how my pressures are all calculated because the pressures are going to be determined as a function of temperatures and if I make the substitutions I should be able to write p1 on one side and then all of the rest of the calculations in terms of temperatures on the other and with any lock I'll end up with p1 in the numerator on the right hand side and then I can reduce to just a function that has to equal 1 so let's try this out first of all I was given p1 then I know p2 was calculated by taking p1 raised to the r to the k value not to begin to use the k-1 in that totally on purpose mistake that we learned together from p3 was p2 multiplied by t3 over t2 and then p4 was p3 multiplied by the quantity 1 over r to the just k not k-1 and then lastly we got back to state 1 by taking p4 multiplied by t1 over p4 I think I'm pretty sure p1 over p4 equals t1 over t4 yeah that makes sense so if I start my substitution process by starting at the end and working backwards I should be able to write p1 is equal to p3 multiplied by 1 over r to the k that's this term here and then I multiply by t1 over t4 and I work backwards again I can say p3 is equal to p2 multiplied by t3 over t2 multiplied by 1 over r to the k multiplied by t over t okay and then p2 was p1 multiplied by r to the k and then all the rest of the things since p1 is equal to p1 times this quantity I can conclude that the entire quantity has to equal 1 so I can say r to the k times t3 over t2 times 1 over r to the k times t1 over t4 has to equal 1 because I know that I have to multiply something by p1 and get p1 so it has to be 1 then I know that if r is a positive quantity and I guess k is a positive quantity then I can say r to the k times 1 over r to the k must equal 1 I mean you could write that out as a series expansion if you wanted but honestly the way that I approach that kind of calculation is to recognize that I'm taking essentially x raised to a positive integer I could say like x squared for the purposes of this analysis multiplied by 1 over x squared and that's the equivalent of running x times x times 1 over x times 1 over x and if you rewrite that as x over x times x over x you just get 1 because x over x is 1 and x over x is 1 therefore 1 times 1 is 1 anyway if those terms multiplied together are 1 then that means t3 over t3 now t3 over t2 multiplied by t1 over t4 is equal to 1 now we are cooking with peanut oil so I can say t3 over t2 is equal to divide by t1 over t4 I'm gesturing at my iPad I realize that you guys can't see that so it's not helpful but I'm saying t3 over t2 is equal to t4 over t1 now if t3 over t2 is equal to t4 over t1 then that means t3 over t2 minus 1 must equal t4 over t1 minus 1 so back before we enter the algebra dimension if we are looking at this equation here I can say this entire block simplifies to just 1 so for the cold air standard the thermal efficiency of an auto cycle is going to be 1 minus t1 over t2 isn't that cool? I mean we could have calculated the thermal efficiency that's like part d of our question once we figured out t2 let's see if we can take that even further well I know t2 was calculated by taking t1 multiplied by r to the k minus 1 therefore t1 over t2 is going to be 1 over r to the k minus 1 neat so for the cold air standard only make sure that you write that in big bold letters only cold air standard because I had to assume constant specific heats to get this far way back over in this step I guess you could write that as only if you've assumed constant specific heats evaluated at the same temperature because I also need cv to cancel so any temperature would work potato potato thermal efficiency of an auto cycle could be reduced to 1 minus 1 over r to the k minus 1 that means that as I increase my compression ratio I am also going to be increasing my thermal efficiency furthermore I know that that line is going to look something like this I mean again forgive the very crude graph but as my compression ratio increases my thermal efficiency increases but we reach less and less improvement in thermal efficiency for each increase in compression ratio therefore increasing from say 7 to 8 is not going to be as much of an increase as increasing from 6 to 7 so we can confidently say increasing the compression ratio will improve the thermal efficiency of an auto cycle but the higher your compression ratio the less effective those gains are again going from 7 to 8 is not as much as going from 6 to 7 or going from 12 to 13 is going to improve it but not as much as going from 8 to 9 etc we got there again by entering the algebra dimension but we also could just look at our MATLAB code we could say change 8 to 9 how does that affect our thermal efficiency it goes from 56.45 to 58.45 that's pretty neat what if we keep going how do we go to 10 come on potato 60.16 no it went up 2% and then it went up 1 point like I'm not here to do math something less than 2 more than 1 it went up by a smaller amount the second time than the first time I could actually just run this in a loop and have it calculate the thermal efficiency for a whole bunch of compression ratios and graph them and in fact I have this code again available on D2L if you want to poke through it is determining the thermal efficiency for all values of compression ratio between 2 and 15 for this problem by increments of 0.05 so let's let the potato do its work so I'm having it plot at each step so if improving the compression ratio always improves thermal efficiency then why don't we just drive around gasoline powered cars and have a compression ratio of like 25 well that gets into another conversation about how the fuel is ignited versus how it would be ignited if it had too high of a compression ratio the temperature of the fuel mixed in with the air increases the temperature of the air increases and at a certain point the temperature of the air increases past the flash point of the fuel when that happens the fuel burns before it's supposed to it combusts during the compression process which is not what we want some of you may have heard that especially if you have a high performance engine and you're burning fuel with too low of an octane the octane of the fuel that you buy from a gas station to how much you can compress it before it begins to auto ignite so a higher octane fuel is going to have a higher flash point which means you can have a higher compression ratio before you begin to auto ignite your fuel before it's supposed to but even that the high octane fuel can only be compressed a little bit more in order to have an engine that can have a very high compression ratio without concerning itself with auto igniting the fuel we need to inject the fuel after the compression process and for that we are going to be turning to the diesel cycle which we will get to after we talk about the auto cycle a little bit more I will also point out that the compression ratio for most gasoline engines somewhere between about 8 and 12 whereas diesel cycles are typically closer to 18 or 20 or 24 in really high compression ratio cases so you'll forgive a little bit of a tangent about octane and fuels that when we pull petroleum out of the ground when we have crude oil that comes up it is a mix of hydrocarbons that is molecules that are carbons surrounded by hydrogen and we have a combination of short hydrocarbons long hydrocarbons and when we go through the refining process we are allowing them to separate so that we can grab out exactly what we want the other thing that's interesting about that is as the hydrocarbon lengths get longer their flash point temperatures increase their boiling point increases their freezing point increases and what you get are substances that are generally vapors at room temperature for very short lengths I mean you have your propane and your butane, etc then as you start to get longer in the about 7 to 11 carbon atoms length chain that's liquid at room temperature but just barely I'll give you ever noticed how quickly gasoline evaporates when you spill it on the ground at a gas station that's because it's boiling point is relatively close to room temperature when you compare it to say water that octane measurement on the fuel pump at the gas station is an indication of how much of that mixture is a longer chain if what you get is all 7's then that's going to combust at a lower temperature than if what you get is all 8's it's called octane for a reason how many of your molecules are at 8 or higher in some cases or just 8 and as we keep going up we start to get into diesel or burning fuel oil where you have say 11 to 14 kerosene in that range go a little bit higher and you have liquids that are almost solid at room temperature and that's where you get into lubricating oils so up a little bit further you have solids at room temperature so say paraffin wax it's a solid but just barely because it's freezing point is so close to room temperature you keep going up you get asphalt which is a solid at room temperature and takes quite a bit of heat before it liquefies