 So, in this lecture, we will start discussing air standard cycles. And as I had mentioned earlier, we will look at three engines or two types of engines. One is the gas turbine engine and the other two are actually automotive engines. So, we will look at the spark ignition engine and the compression ignition engine under the category of automotive engines. Now, in all these three engines, atmospheric air is taken in as the working substance. Now, after it undergoes compression, fuel is then added to the air stream and in some cases air is taken in, fuel is mixed and then it undergoes compression, but it does not matter. Atmospheric air is taken in as the working substance, fuel is added to the atmospheric air and the oxygen in the air is used for the combustion of the fuel. So, when the fuel burns, the temperature of the mixture increases and it then undergoes further thermodynamic processes. So, the working substance is clean air until combustion takes place. After that, it is actually a mixture of combustion gases. So, for this reason, the process executed in these engines is not a cyclic process because we cannot take the combustion gases and send them back to the beginning of the process. So, they have to be exhausted into the ambient and fresh air has to be admitted into the engine. So, the important thing is the processes that take place in these engines do not constitute a cyclic process. So, then the question that arises is why then are we actually looking at cyclic process? If the actual process in the application is not a cyclic process, the primary reason is this. So, we can carry out the thermodynamic analysis of the cyclic process and we assume that air alone is the working substance when we do this. So, this is of course, an idealization and the cyclic process is usually called an air standard cycle. Although the practical realizations are quite different, the analysis that we do now with the air standard cycle gives us two important insights into the cycle. Number one, the parameters that control the performance of the cycle. That is number one and number two, how these parameters affect the performance of the cycle. So, it allows us to identify parameters that control the performance of the cycle. It also gives us insights on how these parameters affect the performance of the cycle. And these insights that we get from the air standard analysis can be carried over directly to the practical realization. So, the same parameters will control the performance of the device that we are talking about and their effect on the performance of the device will be the same as what we see in an air standard cycle. That is the motivation for actually doing an air standard analysis. Although it appears to be far from the actual realization, the insights that it provides carry over directly to the practical application. So, that is the motivation for doing air standard cycles. So, we start with the gas turbine engine first, then spark ignition and then compression ignition engine. Now, in all these cases, at least for the air standard cycle, the performance metrics are the same as what we used in the case of the Rankine cycle. So, specific power, first law efficiency based on energy based efficiency and then second law efficiency. So, these are the three performance metrics that we will be looking at in the case of air standard cycles also. The first one of course, is the gas turbine engine which operates on the so called, well it is the air standard Brayton cycle. The actual engine, as I said, does not execute a complete cyclic process, but all the other processes are executed. So, basically atmospheric air is taken inside the engine into a compressor section. So, as you can see, this is the compressor section in the engine, where the air is compressed through a pressure ratio, typically around 30 or 40 to 1. The air then goes into a combustor which is here, here fuel is sprayed into the air stream and the mixture of fuel and air then undergoes combustion in the combustion chamber. And the peak temperatures are usually seen in this part of the cycle and they can be of the order of about 1300 to 1400 degree Celsius, probably around 1350 degree Celsius or so. So, the air then expands in the turbine which is shown here. We have many rows of turbine blades. So, here only the rotor blades are shown, stator blades which are fixed to the casing have been removed. So, this is an inside view of the exposed view of the rotor of the gas turbine engine. So, there are many rows of turbine blades and you can see that, you know, the diameter of the turbine blade increases as the flow goes through because the flow undergoes an expansion and to accommodate the expansion, the height of the blades also increases in the flow direction. So, here as you can see in the case of the compressor, the height of the blade decreases because the flow undergoes a compression process. So, after undergoing expansion, the flow then leaves the engine. Now, as it undergoes expansion, power that is required to run the compressor is extracted from the turbine and the rest of the power is used to actually run the generator to produce electricity. So, this is a realization for a land based power plant. So, this is used for power generation, that is land based power generation. So, the output that we get from the turbine is partly used to run the compressor and the rest is used to run a generator which produces electricity. Now, notice that the air that is taken in the front undergoes all these processes and then it is exhausted to the ambient and not taken back into the compressor section because air is no longer clean. Here we are looking at an aviation application. This is the cutaway view of a General Electric GE-9X engine which is considered to be state of the art. This is the largest engine in production today and the most powerful engine in production today and as you can see the layout is very similar except for the fan in the front. So, this is the fan in the front, this is actually a turbofan engine. So, it has a huge fan in the front. And I say huge, the diameter of the fan is 11 feet. So, the diameter of the fan here is about the same as the fuselage of a 737 aircraft. So, the engine fan has the same diameter as the fuselage of a 737. So, you can imagine how big it is, it is 11 feet in diameter. But apart from that, you can see the core gas turbine engine inside and you see the compressor section here. So, there are multiple compressor sections, as you can see the compressor section extends all the way up to there. We then have the combustor and then we have multiple rows of expansion or turbine blades and as you can see the diameter of the blades or the height of the blades increases so that the diameter of the rotor also increases to accommodate the expansion of the fluid. So, the power that is generated by the turbine is partly used to run the compressors as well as the fan. The rest is actually, well let me put it this way, the power produced by the turbine is used to run the compressor and the fan. The remaining enthalpy in the gases is then converted into kinetic energy into the nozzle which is located downstream like this. So, the enthalpy of the gases is converted to power in the turbine which is used to run the compressor and the fan, the remaining enthalpy is converted to kinetic energy in the nozzle and used for propulsion purposes. In that case also the air that is taken in is mixed with the fuel so the exhaust gases are sent out into the atmosphere as they are not taken back into the engine. So, the cyclic so there is no cyclic process that is executed in this case also. So, both the land based gas turbine engine and the one used for aviation application use the Brayton cycle, but the process is not a cyclic process. So, what we will do in the air standard cases we will look at an air standard Brayton cycle which is an idealization of these types of applications. So, this engine as I said has maximum temperatures around 1350 degree Celsius and the overall pressure ratio in this case is about 60 or so. It is among the most powerful engines and this pressure ratio is extremely high and it has a bypass ratio of 10. Now, what this number means is that for every kilogram per second of air that goes through the core gas turbine engine 10 kilogram per second goes through the fan. So, it is a high bypass ratio turbofan engine which is why the fan diameter is so high. So, these numbers pressure ratio 60 maximum temperature 1350 degree Celsius these are typical of gas turbine applications although this is on the higher side. Typically we would be having pressure ratios around 40 or so. So, we will keep this in mind when we actually do the air standard Brayton cycle. So, we may draw or illustrate the Brayton cycle in block diagram form like this. So, we start with the air typically ambient air at 100 kilopascal 300 Kelvin temperature it goes to a compressor where it is compressed through a pressure ratio let us say 30 or 40 the air is then taken to a combustor notice that here this is an air standard cycle. So, the air is never mixed with the fuel. So, the combustor essentially is nothing but a heat exchanger. So, heat is added in the combustor to the air stream and it gets heated up. It then goes to the turbine where it undergoes an expansion process part of the power generated by the turbine is used to run the compressor and the rest goes out. If it is a land based application the rest will be utilized to run a generator for producing power. Otherwise the high enthalpy gases otherwise there is no work output or power output high enthalpy gases are taken to the nozzle for thrust generation. So, in this case after undergoing expansion the air comes out it is then taken to a cooler which plays the role of a condenser. Remember we had a condenser in the Rankine cycle here we have a cooler which is also a heat exchanger. So, heat is rejected to the ambient here. So, the temperature of the air is about 300 Kelvin when it comes out of the cooler and after expansion the turbine the pressure is also 100 kilopascal. So, the cycle can be repeated. So, if the cycle is illustrated on a TS diagram it looks like this. So, we have compression from this pressure to the higher pressure. So, basically this cycle also operates between two isobars just like the Rankine cycle. And we have compression from state one to state two as in case it is an isentropic compression process. In the case of an actual compression process the exit state at the exit of the compressor would be state two or the state at the exit of the compressor would be state two. Heat is then added to the working substance until it reaches state three and then the fluid undergoes expansion in the turbine from 3 to 4 S to the lower pressure. Heat is then rejected in the cooler as shown by process 4 S to 1 and then the cycle is repeated. In this case also like what we did with the Brayton cycle without any loss of generality we will assume the compressor and the turbine efficiency to be 100 percent in all our examples. Other values for actual or realistic values for this can easily be accommodated in our analysis. So, there is no loss of generality in assuming the compression and expansion process to be ideal. Now, since the compression and expansion process are ideal the area under the curve actually denotes the heat interaction. So, this is the heat rejected and this is the heat that is supplied is the heat that is supplied and the difference is shown here. So, this would be the difference let us just so that is the difference and as we know from our first law for a cyclic process. So, this is equal to Q H minus Q C every cycle and that is equal to W net and that is what is shown here. Now, normally when we look at air standard Brayton cycle we assume that the peak temperature in the cycle T 3 is fixed. So, we assume that the maximum temperature in the cycle is fixed that is a reasonable assumption to make because we cannot have any arbitrarily high temperature remember the 1350 degree Celsius that I mentioned in connection with this. This itself is a very high temperature this is about 300 to 400 degree Celsius above the melting point of most metals and alloys, but still in this application the blades do not melt because of the excellent blade cooling system that is used. So, it is reasonable to assume that the peak temperature is fixed and that is what we will actually use in the analysis. So, if I apply steady flow energy equation to each one of the component we get the power required by the compressor to be equal to this and if you assume air to be calorically perfect then we can actually write this as m dot C p times T 2 S minus T 1 assuming air to be colorically perfect that is why we are saying ideal here. We can do the same thing for the heat addition in the combustor and the power produced by the turbine. And since 1 to 2 S is isentropic we may write it like this T 2 S equal to T 1 times P 2 S over P 1 raised to the power gamma minus 1 over gamma and the same for T 4 S. Remember P 2 S and P 3 and P 4 S and P 1 lie in the same isobar. So, P 2 S and P 3 I am sorry state 2 S and state 3 and state 4 S and state 1 lie on the same isobar. So, we may write this and I can eventually write like this where R P is the pressure ratio. So, the performance matrix that we are looking for or W X dot specific power as well as thermal efficiency. So, we may write W X dot net equal to this after making use of the expression that we have derived so far and heat supplied may be written like this after substituting for the temperatures and we get thermal efficiency of the cycle to be equal to this. So, these expressions make clear that there are two parameters that control the performance of the cycle. Remember that was one of the insights that we actually wanted from the analysis from the air standard analysis. So, T 3 over T 1 and R P are the two parameters that affect the performance of the cycle. So, W X dot net depends on both of them although in the ideal case when we assume the isentropic efficiency to be 100 percent the efficiency does not depend on T 3 over T 1, but if you relax that assumption then definitely eta will also depend on T 3 over T 1. Nonetheless, W X dot specific power is metric performance metric that we are looking at. So, that depends on T 3 over T 1 and R P and efficiency in this case depends only on R P. So, we have plotted here in this graph W X dot net over M dot C P times T 1 on the Y axis and efficiency eta on the X axis for various values of T 3 over T 1. So, T 3 over T 1 equal to 6 for this curve, T 3 over T 1 equal to 5 for this curve, T 3 over T 1 equal to 4 for this curve. The solid line corresponds to compressor and turbine isentropic efficiency is to be 100 percent. And along each one of this curve the pressure ratio R P varies from 5 to 40 that is typical. So, we have allowed it to vary from 5 to 40. So, you can see that if I draw a vertical line from any of these curves irrespective of T 3 over T 1 the efficiency is the same for all these values. So, the efficiency values are the same for all these three lines which clearly shows that eta does not depend on T 3 over T 1. However, you can see that the specific work generally increases with R P and then reaches a maximum then begins to decrease. Efficiency increases continuously monotonically with R P as you can see from here as I move along this curve efficiency continues to increase. However, WX dot net over M dot C P T 1 increases with R P reaches a maximum and then starts to decrease. Notice that these values of T 3 over T 1 that we have chosen are also very representative of realistic applications. As I said the maximum temperature in the cycle could be around 1350 degree Celsius and air itself which is nothing but which works out to may be something like approximately 1600 Kelvin and the air itself enters at about 300 Kelvin. So, a typical value for T 3 over T 1 would be between 5 and 6. So, that is what we are looking at here. Although this value of 1350 will be smaller for smaller engines. So, 4, 5 and 6 nicely encompass all the practical realizations that we can think of. Now, let us return to the variation of W dot net with R P. Let us see whether we can get some insights from the TS diagram. So, let me just erase some of these things. Now, assuming that the maximum temperature in the cycle is fixed. What is that? This one illustrates the same cycle, but for a higher pressure ratio. What is that the pressure ratio here is between this isobar and this isobar. So, that is higher. And if I have, if I draw the cycle for a lower value of pressure ratio compared to this, it will look something like this. Let me just use a slightly different color for this. So, that is for a slightly lower value of pressure ratio compared to this cycle. And you can see the W net ideal in that case would look something like this. So, as the pressure ratio is increased, this state 0.3 keeps moving to the left. And if you look at the area included inside the cyclic process curve, you can see that you know the area, it is clear that the area begins to increase. And then once the pressure ratio becomes really high, the area in the process curve begins to decrease. And that is the trend that we are seeing here. So, you can see that the net power increases initially with R p reaches a maximum and then starts to decrease. In fact, by differentiating this expression for R p, so by differentiating this expression for I am sorry for W dot net with respect to R p, we can actually show that the specific work is a maximum when R p is equal to square root of T 3 over T 1 raise to gamma over gamma minus 1. Now, for this case, T 2 S becomes equal to T 4 S corresponding to this pressure ratio, T 2 S becomes equal to T 4 S. So, this would be a cycle for which so T 2 S will be equal to T 4 S. It will not be exactly the same state, the state will slide down but for some pressure ratio, T 2 S will become equal. So, 2 S will move like this and 4 S will keep sliding down this isobar like this. So, for a particular value of pressure ratio, which is given by the expression that we derived, T 2 S will be equal to T 4 S and that is when the specific work is a maximum. Now, the next question is purely theoretical, purely academic. How high can we go in pressure ratio? Is there an upper value? Is there a limiting value for pressure ratio? Notice that when I keep increasing the pressure ratio like this, 2 S as I said keeps moving along this line and eventually we are going to come to a situation where 2 S becomes equal to the maximum allowed temperature in the cycle. So, 3 keeps moving like this eventually 2 S and 3 merge for that pressure ratio when the exit temperature from the compressor is the maximum allowed, which means no more heat can be added, which means that this is a motoring cycle where the air is alternately compressed and expanded and no net power comes out of the cycle. So, in fact, for this situation, so for this limiting situation T 2 S becomes equal to T 3. In fact, if you look at this expression here, if you set T 2 S equal to T 3 then we can actually evaluate R P corresponding to this case. So, if for example, you take T 3 over T 1 to be 5, the corresponding value of pressure ratio comes out to be something like 128. So, for T 3 over T 1 equal to 5, if the pressure ratio is 128 or thereabouts, then the exit temperature from the compressor is the same as the maximum allowed temperature in the cycle. So, you can see that that is a ridiculously high value for the pressure ratio, but it is good to know that is the upper bound on the pressure ratio.