 In this and probably in this lecture and in the next lecture, we will discuss the air standard auto cycle which is actually an idealization of the sequence of processes that are executed in a spark ignition engine ok. So, let us start with this cutaway view of a Mazda Sky active engine. So, this is a Mazda Sky active G engine which is actually a very interesting engine and we will talk a little bit more about this engine or at least the interesting aspects of this engine later on, but for now you can see that you know it is a four cylinder engine. So, here we see one piston, here we see a cylinder and another piston, one more cylinder and a piston and you can also see one more cylinder and a piston here. And the valves the intake and the exhaust valves are visible here, the camshaft is also visible here. So, that is the camshaft so connecting rod and this is the crankshaft. So, basically this is a gasoline engine and it executes the following sequence of processes air and fuel as I said gasoline vapor or taken into the engine. So, the fuel has to be in vapor form when it is taken into the engine. So, we are now focusing on one cylinder isolating one cylinder and looking at the sequence of processes that are executed in that cylinder ok. So, this is done during the so called intake stroke. Now, in the next stroke which is the compression stroke so stroke here refers to a travel of the piston from either bottom to the top or top to bottom ok. So, basically as you can see from this picture this is a reciprocating engine. So, the piston travels up and down. So, at the bottom most location of the piston is called the bottom dead center because the velocity of the piston there is 0 then the piston begins to move up. So, the top most location of the piston is called the top dead center ok. So, one stroke would correspond to movement of the piston from either top dead center to bottom dead center or bottom dead center to top dead center. So, during the intake stroke air and fuel vapor or taken into the engine and at the next stroke which is the compression stroke this mixture is compressed ok. Now, towards the end of the compression stroke and before the beginning of the next stroke the compressed mixture is ignited using a spark plug hence the name spark ignition combustion engine. So, the mixture is ignited using a spark plug and the combustion front travels outward from the spark plug and then into the mixture and it then pushes the piston down that is the next stroke which is called the power stroke. So, here the combustion products. So, as the combustion wave travels outward from the spark plug it consumes or combust the fuel and air mixture in its path and the gases then expand and generate power ok. So, this is the next stroke and in the exhaust stroke the piston moves from bottom dead center to top dead center and pushes the exhaust gases outside the cylinder. So, that the engine is then ready to receive the next or fresh mixture of air and fuel vapor. So, intake in the intake stroke the piston travels from the top dead center to the bottom dead center let me denote it like this. Now, in the compression stroke it travels from the bottom dead center to the top dead center. And the combustion stroke essentially occurs at constant volume which means that there is very little movement of the piston during the combustion process. So, that is not really a stroke, but it is actually a process in between the compression stroke and the power stroke ok. So, during the power stroke the gases push the piston downwards and generate power and during the exhaust stroke the piston moves upwards and expels the exhaust gases from the cylinder. So, that a fresh mixture of air and fuel vapor can be inducted into the engine. So, as you can see here this is the first stroke, this is the second stroke, this is the third stroke and this is the fourth stroke. So, the entire sequence is executed in four strokes and hence such an engine is called a four stroke engine. Now, we may represent the processes are gone by the air fuel mixture on a PV diagram like this. So, notice that the volume is bounded by the top dead center and the bottom dead center. So, the red chain line here denotes the actual the processes undergone in the actual SI engine. So, we start with the intake stroke and as you can see the piston moves from top dead center to bottom dead center where air is air and fuel vapor is taken in. It is then compressed as you can see here in the compression stroke, park then ignites the combustion mixture. So, that is essentially at constant volume. So, the entire mixture ignites and burns and the burning mixture then pushes the piston down like this. And then in the exhaust stroke, the exhaust valve opens and the gases are then pushed outside and the process is completed. Now, the cylinder is ready to receive a fresh mixture. Now, it may appear as though the process is a cyclic process, but in reality it is not because we taken a fresh mixture of air and fuel vapor in the intake stroke. It executes a sequence of processes and then at the end during the exhaust stroke, the combustion gases are sent out and a fresh mixture is taken in. Although at the same thermodynamic state, but the process executed in the engine is not a cyclic process that is very important to bear in mind. So, as the strokes are executed, the valves that are seen here are opened and closed appropriately. For example, during the intake stroke, the intake valve is open, but the exhaust valves are closed. That may be more than one intake and one exhaust valve. Typically, these are four valve engines, which means they have two intake valves and two exhaust valves. So, the camshaft ensures that the spring loaded valves, so you can see the, you can actually see the spring here. So, we can see the spring here. So, the camshaft ensures that the spring loaded valves or in the appropriate open or closed position as the processes take place. Now, although the sequence of processes appears to be cyclic on the PV diagram, the working substance does not execute a cyclic process that is very important to keep in mind. Now, in the equivalent air standard auto cycle, we assume that there is a fixed mass of air in each cylinder. There is a fixed mass of air in each cylinder and it executes a cyclic process. So, that is the idealization that we will make when developing the air standard auto cycle. So, we assume that there is a fixed mass of air M, which executes a cyclic process because the air inside the cylinder is fixed. We need not take in fresh air at the end of every, at the, I mean during the intake stroke or send any combustion products out because it is clean air, which means that the intake and the exhaust strokes are absent. So, the air standard auto cycle actually is a two stroke cyclic process. Whereas, the actual sequence of processes in an SI engine is a four stroke process as illustrated here. So, we have illustrated a four stroke engine here. So, that executes a, I am sorry, that executes four strokes. Whereas, the equivalent idealized air standard auto cycle executes only two strokes. Heat addition and heat rejection during the air standard cycle are assumed to take place at constant volume. And this is illustrated in blue in this, in this diagram. So, the air starts the cyclic process from state one. So, it is compressed from state one to state two. In the ideal case, we assume this compression process to be isentropic. Heat addition occurs at constant volume between state two and state three. Power stroke or expansion occurs between state three and state four. And then heat rejection at constant volume takes place between state four and state one and the cycle is repeated. So, notice that in going from state one to state two, the piston moves from the bottom red center to the top red center. And then in going from state three to state four, the piston moves from top red center to bottom red center. So, there is only two strokes. Now, if you go back to this actual engine that we have shown here, remember we fixed our attention on one cylinder and then we actually said that the four processes or four strokes are executed in one cylinder. The question of why have more than one cylinder than logically arises in the actual engine. This is required because as we discussed, power is produced only during one out of the four strokes in an actual engine, which means that the power generated is intermittent. It is not continuous. Unlike the case of a gas turbine engine, which executed a Brayton cycle, the power produced there is continuous whereas here the power produced is intermittent. So, to minimize the intermittency in the output power, we add more cylinders. So, we add as you can see here three more cylinders. So, when we have four cylinders, generally what is ensured is that at least one cylinder will always be executing a power stroke so that the power production appears to be continuous and not intermittent. So, at any given time, one of the cylinders is guaranteed to be executing a power stroke. So, the cylinder firing is sequence or the firing of the spark plug is sequence in such a way that one, at least one cylinder will be producing power at any given point in time. So, by having more cylinders, not only are we producing more power, we are also reducing the intermittency in the power generation. Of course, in the auto cycle, we focus on one cylinder only, multi cylinders extending this to multi cylinders is trivial so we need not discuss that here. So, we are focusing only on one cylinder and looking at this cyclic process. Now, since fixed quantity of air executes a cyclic process in each cylinder, unlike in the case of an air standard Brayton cycle, the processes here are not steady flow processes. These are actually non-flow processes. We have a fixed quantity of air in each cylinder and it executes the processes that we describe, compression, heat addition, expansion and then heat rejection. So, we cannot use steady flow energy equation, we use the non-flow form of first law. So, delta u is given as for process 1, 2 delta u is q minus w because it is an isentropic compression process, process 1, 2 is in the ideal case it is an isentropic compression process. So, q is 0 and delta u is simply w 1, 2 so w 1, 2 during the cycle is m times u 2 minus u 1. Just like what we did for the Brayton cycle without any loss of generality, we will assume the isentropic efficiency of relevant work absorbing or work producing processes to be 100%. If the case when the isentropic efficiency is less than 100% can be handled without any difficulty in the framework that we are going to develop. Now, q 2, 3 is heat addition at constant volume and this is equal to m times u 3 minus u 2 and just like what we did for process 1, 2 the work interaction doing during process 3, 4 is actually m times u 3 minus u 4 and heat rejection at constant volume takes place during process 4, 1 and we may write the heat rejected to be equal to m times u 4 minus u 1. Now, the network produced during the cycle is the work that is developed during the power stroke minus the work that is absorbed during the compression stroke. So, that the thermal efficiency of the cycle may be written like this 1 minus u 4 minus u 1 over u 3 minus u 2. Now, if you assume the working substance which in this case is air to be calorically perfect such analysis is called the cold air standard analysis because we assume it to be calorically perfect then we may rewrite these expressions before both for the efficiency as well as the network that is produced. This may be rewritten by using the fact that u is equal to Cv times T. So, we may rewrite the specific power expression for specific power like this and the expression for efficiency like this. Here the quantity R is called the compression ratio and it is the ratio v 1 over v 2. What is that v 1 over v 2 is equal to v 4 over v 3 because 2, 3 and 4, 1 are constant volume processes. So, this v 1 over v 2 is called the compression ratio. In the case of the Brayton cycle we had a parameter called the pressure ratio which was the ratio of pressures at the end of the compression process or expansion process appropriately. Here we are actually utilizing volume. So, it is the volume at the beginning of compression to the volume at the end of compression and that is called the compression ratio. So, you can see that the cold air standard analysis brings out the fact that there are two parameters in the cycle one is T 3 over T 1 and the other one is R which is the compression ratio. Notice that v 2 is called the clearance volume. So, this is v 2. So, when we have a cylinder like this, so this would be the BDC and this could be the TDC. Notice that the piston for obvious reasons does not run all the way up to the top of the cylinder. So, it stops at some location here called the TDC. So, this volume which is v 2 in this figure is called the clearance volume and the distance between the or the volume between the TDC and the BDC is called the swept volume or stroke volume. So, v 2 is called the clearance volume and v 1 minus v 2 is called the displacement or the swept volume. So, when someone says that the engine has a displacement of 1 liter, let us say it is a four cylinder engine and it is stated that the engine has a displacement of 1 liter. That means that the displacement of each cylinder is 250 cc. So, both the efficiency and the specific work are plotted here against compression ratio. What is that just like what we did for the Brayden cycle? The performance metrics of the auto cycle are also specific power efficiency, first law efficiency and second law efficiency. So, you can see that the efficiency of the auto cycle depends only on the compression ratio and it increases and sort of tapers off after a compression ratio of about 10. We have not really plotted efficiency beyond a compression ratio of 10 either for the specific power or for the efficiency because spark ignition engines typically do not operate at a ratio higher than 10. Now, why not? The question that arises next is why not? Because in an actual engine, what is compressed as we can as we mentioned earlier, what is compressed is a mixture of air and fuel vapor or gasoline vapor. So, in this engine, if you try to compress the air and fuel vapor beyond a compression ratio of let us say 10 or so, what would happen is at the end of the compression stroke, the temperature would be higher because the air and fuel vapor are being compressed. And it is quite possible that at some spots in the engine, the temperature of the mixture can go above the auto ignition temperature of the mixture, in which case the mixture will self ignite even before the spark, even before the spark is given in the engine or just as the spark is given in the engine. Now, the entire power stroke relies on the fact that the location of the spark plug is such that when the spark is initiated, it initiates a combustion wave which then travels outward from the spark plug towards the piston. And it has to be done in a very specific manner. It cannot be done too early because if it is done too early, then if the mixture is ignited as the piston is still moving out, then the down coming combustion wave will tend to push the piston down during the compression stroke itself, which is detrimental to the engine as well as the process that we are trying to execute. Now, if it is ignited too late, that means that the piston would have started, would have already be on its downward stroke and the expanding gases will not be able to give enough of a kick to the piston to generate sufficient power. So, much of the energy of combustion is wasted. So, it cannot be too early, it cannot be too late. So, it has to be done in such a way that the combustion wave, the expansion of the combustion wave coincides with the downward movement of the piston, so that maximum kick is delivered to the piston and it can generate the maximum amount of power. So, it is a very carefully controlled sequence and if part of the mixture ignites early or just as the at some other location, just as the spark plug is ignited, then the combustion wave from this combustion process would actually also travel outwards and it will oppose the combustion wave that is coming down from the mixture that is ignited by the spark plug and this can actually cause unstable combustion inside the cylinder and unstable pressure variation inside the cylinder and the piston will not be kicked smoothly down, it will start to oscillate because of differing pressure waves, combustion waves hitting the cylinder at different instants in time. So, the carefully controlled sequence is destroyed and the engine then the piston begins to vibrate and the entire engine then begins to vibrate and this process is called knocking and this is due to auto ignition of the mixture in some parts in the cylinder away from the spark plug which is why the compression ratio of spark ignition engines are never taken beyond 10. So, you can see that the efficiency of the auto cycle more or less begins to taper off at about 60 percent around a compression ratio of 10 and if you look at the specific power for the auto cycle, notice that we have plotted three curves here each one corresponding to a particular value of T3 over T1. T1 is typically as you can imagine T1 is typically 300 Kelvin and peak temperatures in spark ignition engines, typical spark ignition engines or around 2100 Kelvin or so. This may seem higher than the temperatures that we saw for Brayden cycle, but we must bear in mind that the Brayden cycle as we already said executes steady flow processes which means that the temperature peak temperature that we are seeing there is sustained peak temperature whereas in the case of the auto cycle the peak temperature is seen only for an instant because it is a non-flow process and then the gases begin to expand and the temperature of the mixture decreases rapidly. So, the temperature at the instant of ignition can be as high as 2100 Kelvin, but it begins to decrease very rapidly afterwards. So, there is a big difference between the peak temperature in a Brayden cycle and the peak temperature in an auto and a diesel cycle which we will see next. So, this value of 7 is sort of representative of actual spark ignition engines and you can see that you know the specific power pretty much asymptotes after a compression ratio of 4 irrespective of the value for T3 over T1. The specific power increases with T3 over T1 as we can see from here, but for a given T3 over T1 the specific power sort of asymptotes after a compression ratio of 4. For the ideal air-sanded auto cycle or the cold analysis of the auto cycle, we can also derive an expression for the second law efficiency. So, the exergy supplied per unit mass during each cycle may be written like this. So, basically exergy is supplied during the compression stroke in the form of work and heat is supplied in the heat addition constant volume, heat addition process. So, the exergy supplied during the during process 1, 2 may be written like this and the exergy supplied through the heat interaction may be written like this. Now, we may rearrange this expression in the following manner by pulling out or T1 from this and then we can actually simplify and write this expression dimensionless exergy supplied. Notice that this is exergy supplied per unit mass whereas, this is the dimensionless exergy supplied made non-dimensional by using the quantity mcv T1. So, if we do this then the exergy supplied involves the compression ratio and T3 over T1. Similarly, exergy recovered may be written like this. So, this is basically the work done during the power stroke and this also may be simplified eventually to read like this. The dimensionless exergy recovered reads like this and again this involves the compression ratio R and the temperature ratio T3 over T1. So, now we are in a position to calculate the second law efficiency. Notice that we have taken Th to be equal to T3. So, we have assumed the reservoir to be at the highest temperature in the cycle when heat is supplied. So, the variation of second law efficiency for the auto cycle is shown here. Note that in contrast to the first law efficiency which involved only the compression ratio and not T3 over T1, the expression for second law efficiency involves T3 over T1 also. So, it has been plotted here for different values of T3 over T1, representative values of T3 over T1 same as what we have used here for the auto cycle. So, these are representative values of T3 over T1 and you can see that the second law efficiency generally increases with compression ratio. However, it decreases with the T3 over T1. So, as the maximum temperature in the cycle increases, the heat supplied or the reservoir temperature also increases. So, the loss of exergy due to heat supply across a finite temperature difference increases because this temperature difference keeps going up as the temperature of the reservoir keeps increasing. So, as T3 over T1 goes up, you can see that the second law efficiency comes down, which makes sense. In fact, later on when we discuss diesel cycle, here also you can see that as one of the parameters goes up, the important parameter goes up, the second law efficiency of the diesel cycle also behaves in a similar manner. But the important takeaway from here is that the second law efficiency is the highest corresponding to compression ratio about 10 or so. So, it is a second law efficiency can be as high as 75 to 80 percent, which is quite good. And it depends on both the compression ratio as well as the value for the parameter T3 over T1. So, these are the insights that we are able to get from the cold air standard analysis, cold implying that we have assumed air to be calorically perfect as well. Let us go through a worked example illustrating this. So, at the beginning of an ideal air standard auto cycle, air is at 100 kPa and 298 Kelvin, compression ratio is 9, peak temperature is 2200 Kelvin. We are asked to determine thermal efficiency, mean effective pressure and the second law efficiency. So, we assume the cycle to be ideal, it is given that it is ideal. Processes 1, 2 and 3, 4 are isentropic and we may write P2 using equation of state between states 1 and 2. We may write P2 as R times P1 times T2 over T1 where R is the compression ratio. And since the process 2, 3 is a constant volume process, we may write P3 as P2 times T3 divided by T2. And in the same manner as what we did for process 1, 2, we may relate P4 and P3 as P4 equal to P3 times T4 over T3 times 1 over R. So, let us walk through the states in the cycle. So, we start with state 1, temperature is given. So, we go to the table with temperature as 298. So, that is given here. And we retrieve the specific internal energy from the table S0. And now in this case, we retrieve the value for the relative specific volume that we mentioned earlier VR. Because this is a non-flow process and compression ratio is defined based on volume. So, we retrieve VR. So, we retrieve this, this and this from the table. Now, S at state 1 is the same as S0 because that is at the same pressure, ambient pressure and temperature. Now, 1 to 2 is an isentropic process. And we may evaluate P2 using the expression that we wrote down earlier using this one. And remember, in our earlier derivation, we mentioned that V2 over V1 in case of an isentropic process is equal to V2 R divided by V1 R. So, that means that V2 R is 631.9 divided by 9 and that is approximately 70 something, approximately 71 let us say. So, we go to the table with this value for VR and then retrieve other quantities of interest. So, VR 70 something. So, let us see where that comes. So, that is over here. So, that comes between these two entries. So, we interpolate and obtain U. We need not get S from the, we need not retrieve S from the table because 1, 2 is an isentropic process. So, we retrieve this value from the table and S2 is equal to S1 because it is an isentropic process. So, next process 2, 3 is heat addition process, constant volume heat addition process. The peak temperature is given. So, we use this value of the temperature to go to the table. Let us just quickly see 2200 Kelvin. So, 2200 Kelvin comes here. So, we retrieve U from this table and S0 and VR. So, we retrieve U, S0 and VR from the table using the known value for temperature. And notice that S has to be calculated from S0 by using the expression that we wrote down earlier. So, S is equal to, I am sorry, S0 of T minus R natural log P over VRF. So, S may be evaluated. 3 to 4 is an isentropic process. So, S4 is equal to S3 and again we use this fact to calculate VR at state 4 which works out to be approximately 18 or so. So, we can then retrieve this value. With this value of VR, we can retrieve this value for U from the air table. So, now we are ready to carry out the calculations. So, work supplied during the compression process as we wrote down earlier, specific work supplied during the compression process is U2 minus U1 comes out to be 298 kilojoule per kilogram. Work done during the expansion process is U3 minus U4 comes out to be 1013 kilojoule per kilogram. So, the net work produced per cycle, specific work produced per cycle is 715 kilojoule per kilogram. Heat added per unit mass during each cycle is U3 minus U2 that is 1361 kilojoule per kilogram and heat rejected again this is the both these are constant volume processes as we mentioned earlier. So, heat rejected during each cycle is 646.31 kilojoule per kilogram. So, the first law efficiency of the cycle is 52.52 percent. The mean effective pressure is defined as W net divided by V1 minus V2. Remember, if you look at the indicator diagram, then let us see let me erase some of these things. The net work produced during the cycle, let me illustrate that here that is the area inside the cyclic process. So, that comes out to be so that is equal to this. So, the mean effective pressure takes this area, divides it by this volume this is V1 minus V2, divides it by this volume and gives us a value for pressure. So, it is as if we have a constant pressure and the area under that constant pressure line is equal to the W net. So, this is in many ways or this is in fact the average you may recall that we calculated the average temperature of heat addition, we calculated the average temperature of heat addition in the Rankine cycle. So, this and this is very, very similar to the average temperature of heat addition that we calculated from the Rankine cycle. So, this is equal to W net over V1 minus V2 which may then be rewritten like this using equation of state and the definition of the compression ratio and this comes out to be 940 kilopascal. What is that? The peak pressure in the cycle as we calculated here was 6,644 kilopascal whereas the mean effective pressure works out to only about 940 kilopascal. Now, exergy is supplied during the cycle, notice that exergy is supplied during the during the compression stroke and during the heat addition process. So, exergy is applied during the compression stroke may be evaluated simply by I mean simply I am sorry simply like this. So, this is the exergy specific exergy V2 minus V1 and the exergy associated with this heat addition may be evaluated like this. Here we have assumed that TH is the peak temperature in the cycle which is 2200 Kelvin. So, that heat is supplied from reservoir maintained at 2200 Kelvin. So, the exergy supplied comes out to be like this. Exergy recovered during the cycle is given by this expression. Remember heat rejection is to the ambient. So, no exergy can be recovered during that part of the cycle. The only exergy recovered is when power is produced or what is done during the cycle which we may evaluate like this to be 937.5 and so the second law efficiency comes out to be 67 percent. Because one interesting aspect about this engine whose picture I showed in the beginning of the lecture is the following. So, we have mentioned that you know compression ratios for spark ignition engines typically tend to be less than 10 and the reason for that is also apparent from this figure. And the next cycle that we are going to discuss is the diesel cycle and as you can judge from this figure, diesel cycles definitely have higher specific power and they also have higher efficiency and their compression ratios are higher than that of the auto cycle. So, this engine that you see here is a very interesting engine because this engine actually operates at compression ratios greater than 10 but less than that of a diesel engine. So, it typically operates at compression ratios around 11 to 12. So, you may then ask the question how does this engine void the issue with knocking. So, the engine has very very special sensors and other control equipments which measure the equivalence ratio at every instant that during the during the combustion process. In fact, during the entire cycle and then adjust the mixture equivalence ratio which is the amount of fuel that we are putting in continuously and constantly so that the combustion process is always stable. So, as a result the efficiency of the cycle is comparable to that of diesel engine specific power not comparable to diesel engine but still we actually get over the barrier that we mentioned earlier. Higher compression ratio definitely gives us higher power and the stability issue is addressed through the use of onboard diagnostics and direct computer control of the equivalence ratio that is being sent to the engine. This engine also uses additional interesting technologies like or additional interesting strategies like cylinder deactivation during low load operation so that when the load is low or when the power demand is low not all four cylinders need to generate power. So, couple of cylinders may be shut off. So, this engine has software and onboard diagnostics which continuously senses the load and then turns off cylinders so that you get higher fuel economy. So, the fuel is not wasted power is not wasted. So, these are some very revolutionary strategies that have been incorporated in this engine developed by BASDA which illustrates or underscores the fact that there is still a lot of scope for improving efficiencies of existing IC engines. In the next lecture we will look at the air standard diesel cycle and again do an analysis cold air standard analysis and then do a worked example using the air table.