 Hello and welcome to lecture 6 of this lecture series on jet aircraft propulsion. The last few lectures you must have been introduced to different types of aircraft engines and also some of the basic performance parameters that are used in analysis of these jet engines. And in today's lecture what we shall do is to discuss about the basic thermodynamic cycles based on which all these jet engines the air breathing engines operate. And what we shall discuss in today's lecture would be the basic Brayton cycle which I am sure you would have learnt in your thermodynamics course or if you have done the first course on introduction to aerospace propulsion. And so we will first look at take a relook at the Brayton cycle, the ideal Brayton cycle, the variance of the ideal Brayton cycle that is Brayton cycle with reheating, regeneration and intercooling. And also we will understand about what is what are the actual Brayton cycle. So, what is the difference between an ideal Brayton cycle and an actual Brayton cycle. So, these are some of the topics that we shall be discussing in today's lecture. We will begin with the ideal Brayton cycle subsequently we will discuss about the variance of the Brayton cycle and then we will discuss about actual or real Brayton cycle. So, Brayton cycle was something which was proposed by George Brayton in 1870s and initially it was proposed to be used for reciprocating engines. But as we know it now that Brayton cycle forms the basis for all jet engines which are rotary in nature that these are not essentially reciprocating engines they are rotary engines. And so modern day gas turbines operate on Brayton cycle, but it basically works with rotating machinery. And one of the main differences between the gas turbines and the basic thermodynamic cycle is that the gas turbines operate in an open cycle mode. Whereas the Brayton cycle that we shall discuss today is basically a closed cycle system. And, but you can still model the gas turbine engines even though they are open cycle they can be modeled as closed cycle using the air standard assumptions that is we assume that the combustion products can also be modeled as air. And so using that assumption we can still continue to use the Brayton cycle analysis for gas turbine engines even though the gas turbine engines operate in open cycle mode. And in Brayton cycle analysis what we will see is that the combustion process which is what is there in an actual engine as well as the exhaust process is replaced by heat addition and heat rejection processes. We shall not be exactly modeling combustion or exhaust, but we will model them as heat addition process because thermodynamically that is much easier to analyze. So, heat addition is basically representing the combustion process in a gas turbine and the exhaust is modeled using a heat rejection process. And in one of the later lectures we will actually be using the Brayton cycle analysis with combustion and exhaust etcetera put into place. But in today's lecture we will actually be discussing them with these assumptions. And so Brayton cycle as we know it consists of four processes the ideal Brayton cycle consists of four internally reversible processes. And there are two isentropic processes and two constant pressure processes. So, the first process in the Brayton in the ideal Brayton cycle is the isentropic compression process which could be in a compressor. So, process 1 to 2 is isentropic compression in a compressor. The second process in the ideal Brayton cycle is the constant pressure heat addition which is from process which takes the process from state 2 to state 3. The third process which is between states 3 and 4 is isentropic expansion and in a gas turbine this is basically happening in a turbine. And the last process is the constant pressure heat rejection process between states 4 and 1. So, these are the four different processes which constitute a Brayton cycle an ideal Brayton cycle. And in an ideal Brayton cycle all these processes are internally reversible in the sense that there are no irreversibilities occurring within the system. There could be irreversibilities outside the system, but we are not really concerned about that. So, an ideal Brayton cycle consists of these four processes. The first process is an isentropic compression where during the compression process the pressure and temperature of air increases. And the second process is where you add heat to the system you add it is a heat addition process which in a gas turbine engine is basically the combustion process where fuel is burned and energy is released. And after the second process which is a constant pressure process because it is an internally reversible process. The third process is an expansion process where the high temperature and pressure air is expanded. And in a gas turbine engine it is partially done through a turbine and then through a nozzle which we will discuss later on. And after the expansion process heat is released and in a Brayton cycle it is basically a constant pressure heat rejection process. So, these are the four different processes which constitute an ideal Brayton cycle. So, let us take a look at how these processes look like on pressure versus volume graph as well as temperature versus entropy graph. So, the Brayton cycle on p v and T s diagrams are shown here. So, let us take a look at the p v diagram first. Here we have a process which starts at state 1. The first process is isentropic compression process because it is a compression process there is a decrease in the specific volume you can see that at the end of this process the specific volume is lower than what it was at state 1. And there is an increase in its pressure at state 2. So, there is a tremendous increase in its pressure from between states 1 and 2. At the end of this process wherein the state is 2 heat is added and heat is added at constant pressure. So, pressure is a constant between states 2 and 3. So, heat is added that is q in as shown here. At state 3 there is another expansion process which increases in during which the specific volume increases. And because it is an expansion process there is a drop in the pressure and this again is an isentropic process. The fourth process is a heat rejection process which is again constant pressure process. And therefore, there is no change in pressure between states 4 and state 1 and heat is rejected during this process which is represented by q out. So, all these 4 processes are shown here on a temperature versus entropy plot. The first process between states 1 and 2 the isentropic compression process is a vertical line because it is isentropic process. So, entropy at states 1 and state 2 are the same, but because it is a compression process there is an increase in temperature. Now, during the second process which is a constant pressure heat addition process there is an increase in temperature there is also an obvious increase in the entropy of the system. Heat is added during this process pressure remains a constant during process 2, 3. The third process is isentropic expansion. And thus this happens between states 3 and 4 again it is a vertical line because it is isentropic. The fourth process between states 4 and 1 is constant pressure heat rejection process during which heat is rejected from the system and therefore, it is an isobaric process. So, these are the 4 processes as represented on pressure volume and temperature entropy plots. And it is very important that you understand these different processes because we are going to use these diagrams very frequently during the cycle analysis which is what we will do over the next 2, 3 lectures. So, make sure that you understood the representation of the Brayton cycle on these different processes. What we will do next is to carry out an energy balance for this particular cycle the ideal Brayton cycle. And then we can derive an expression for efficiency of a Brayton cycle and what we will see later on is that the efficiency of a Brayton cycle is a function of just the pressure ratio, the cycle pressure ratio raise to some function of the specific heats ratio of specific heats. So, we will now carry out an energy balance across the Brayton cycle that we have just discussed from the steady flow analysis. Because the whole process Brayton cycle involves the use of steady flow devices. And therefore, from steady flow process analysis which you must have studied during thermodynamics course we can actually derive an expression for the Brayton cycle efficiency. So, energy balance for a steady flow process we can express in terms of what is shown here the heat input the work output and delta H. So, Q in minus Q out is the difference the net heat added to the system Q in is the heat added Q out is the heat rejected from the system. So, the net heat transfer to the system is Q in minus Q out and what is the net out work output from the system W in is the work input which is basically the work required for the compression process W out is the work output of the system which is basically the work done by the turbine during the expansion process. So, the net heat added is given by the first term that is Q in minus Q out net work output given by the second term which is W in minus W out. And some of these two therefore, should be equal to delta H which is the net change in enthalpy of the system. So, what are these terms Q in and Q out? So, Q in is the net heat transfer to the system which is basically the net change in enthalpy between states 3 and 2 because that those are the two states during which heat is added to the system. So, difference between the enthalpies at states 3 and state 2 is basically the net heat input. So, Q in is equal to H 3 minus H 2 because H 3 is the enthalpy at state 3 minus H 2 which is enthalpy at state 2 which is basically equal to C p times T 3 minus T 2 because we know that enthalpy is a function of the specific heat at constant pressure times the pressure temperature. And therefore, if we assume that specific heat at state 2 and 3 are the same for an ideal cycle that that should be fine and therefore, C p multiplied by T 3 minus T 2 will give us the net heat added during the second process that is the heat addition process. And similarly, the heat rejected during process 4 will be equal to the net change in enthalpies between states 4 and 1 which will be equal to H 4 minus H 1 and therefore, that is equal to C p times T 4 minus T 1 that is the product of the specific heat multiplied by the change in temperature occurring during that particular process. So, Q in and Q out can actually be expressed in terms of functions of the specific heat at constant pressure multiplied by the change in temperature during which this process occurs. So, Q in and Q out can be expressed as these functions. Similarly, what we will do is we can also do the same thing for the work done by the system. And we know that efficiency of any cycle can be expressed as the ratio of the net work output of the system divided by heat input of to the system. So, thermal efficiency of an ideal Brayton cycle if we were to assume the air standard assumptions will be eta Brayton that is eta thermal thermal efficiency of the Brayton cycle should be equal to W net by Q in which is equal to 1 minus Q out by Q in. And why is that for a cyclic process if you recall from thermodynamics we know that W net should also be equal to Q net and Q net is Q in minus Q out. Therefore, W net divided by Q in reduces to 1 minus Q out by Q in. We have already derived expressions for Q out and Q in Q out is equal to C p times T 4 minus T 1 Q in is equal to C p times T 3 minus T 2. Therefore, we can reduce this to 1 minus T 1 multiplied by T 4 by T 1 minus 1 divided by T 2 into T 3 by T 2 minus 1. Now, process 1 2 and 3 4 are isentropic that something we have already discussed that the first that is a compression process and the expansion processes are isentropic. And because they are isentropic as well as the fact that the second process and the fourth process are constant pressure processes where p 2 is equal to p 3 and p 4 is equal to p 1. We have T 1 by T 2 is equal to p 2 by p 1 raise to gamma minus 1 by gamma this is from the isentropic relations. Now, since p 2 is equal to p 3 and p 4 is equal to p 1 this should be equal to p 3 by p 4 raise to gamma minus 1 by gamma which in turn is equal to T 3 by T 4. That means, we can replace the temperature ratios T 1 by T 2 with T 3 by T 4 and if we do that what happens that if you were to substitute for T 1 by T 2 with T 3 by T 4 we can actually reduce the thermal efficiency of the Brayton cycle as 1 minus 1 divided by r subscript p raise to gamma minus 1 by gamma where r subscript p is the pressure ratio which is p 2 by p 1. So, what we see here is that the thermal efficiency of a Brayton cycle is something which is relatively simple that is in the sense that you can very simply by knowing the pressure ratio and assuming the ratio of specific heats one can very easily express thermal efficiency of a Brayton cycle in terms of two simple parameters that is one is the pressure ratio that is the cycle pressure ratio p 2 by p 1 and the other parameter that is involved here is the ratio of the specific heats that is gamma. So, thermal efficiency of a Brayton cycle of an ideal Brayton cycle is 1 minus 1 divided by r p which is the pressure ratio raise to gamma minus 1 by gamma and. So, using this simple expression one can actually estimate what would be the efficiency of an ideal Brayton cycle. In fact, what we will discuss little later is that even for real Brayton cycles one can get an estimate of how much the efficiency maximum efficiency can be because as you already know that actual engines will obviously have efficiencies which are lower than that of the ideal corresponding ideal cycles. And so if we know the ideal Brayton cycle analysis you can determine the efficiency of such a Brayton cycle and that will give us the maximum efficiency that the actual cycle is going to have. And that is where the significance of an ideal cycle comes into picture in the sense that it gives or it sets the standard for the performance of actual engines. And so what we will do now is to take a look at the different combinations or modifications that can be done to an ideal cycle and therefore, enhance the performance of an ideal cycle. One of the first modifications that we will be discussing is an ideal Brayton cycle with regeneration. So, regeneration as you might be knowing is a process during which one can transfer energy during one part of the cycle and to some storage device and then during another part of the cycle recover the energy back. Now, in a Brayton cycle regeneration can be carried out using the hot air exhausting from the turbine to heat up the compressor exit flow that is we or one could heat up the fuel that is going into the combustion chamber and so on. So, that we are recovering part of the heat which would otherwise be lost. So, using regeneration one can improve the efficiency of a Brayton cycle because we are reducing the Q out at the same time we are also reducing the Q in the sense that we can reduce the amount of heat that is to be added because we are using part of the heat which would otherwise be rejected. So, regeneration is a very clever technique of increasing the efficiency of systems thermodynamically of course, it is very easy to implement, but as we will discuss maybe later that practical implementation of regeneration for an aircraft engine has its issues and problems because anything that we try to do here on an aircraft engine might lead to increase in weight of the system which is something that an aircraft designer would never want to happen because weight of the aircraft is always to be kept as low as possible. And unless of course, that this increase in weight leads to substantial increase in efficiency and performance that it can outweigh the extra weight that the aircraft has to carry. So, basically regeneration would decrease the heat input of requirements for the same amount of net work output and therefore, we know that as you reduce the heat input with the same amount of work output the efficiency can increase. Let us take a look at regeneration cycle on a TS diagram again for an ideal Brayton cycle. So, this process between 1, 2, 3 and 4 represents an ideal Brayton cycle on a TS diagram where process is 1, 2 and 3 for a Rycentropic 2, 3 and 4, 1 are constant pressure processes. Now, we know that heat addition takes place during this process that is process 2, 3 which means that the amount of heat that would have to be added that is Q in should be between states 2 and 3. Similarly, heat rejected for a simple ideal Brayton cycle would be between states 4 and 1. Now, regeneration is a process by which we can transfer a small part or a part of this heat which is rejected back to this process 2, 3 that is this heat between states 4 and 1 is something which is rejected. So, which means that if you can reduce the amount of heat rejected here and transfer it back here into Q in Q in also will reduce. Therefore, the net heat input will be reduced. So, regeneration as is shown here can take place between these 2. Let us say for an ideal cycle the amount of heat that has been transferred or recovered or saved is between 4 and 6 that is this much amount of heat has been transferred to this process 2, 3. We know that you cannot have 6 below this level because that would mean that the temperature is lower than that of the compression which is something which is not possible thermodynamically. So, this is the maximum that we can save or that can be regenerated transferred back to the cycle. And so that means that would be the ideally Q regenerated would be between 2 and 5 that is we can heat it all the way up to a temperature which is equal to the expansion temperature. And therefore, Q in also has reduced now. So, this is the new Q in that is because of the amount of heat that has been transferred back to this heat addition process. So, as a result of heat regeneration basically we have been able to recover some amount of heat. Now, the extent to which regenerator can approach an ideal regenerator if you have an ideal regenerator that means amount of heat that can be transferred is between states 4 and 2 that is H 4 minus H 2 would be the maximum regeneration that is possible that is this much amount of heat can be recovered. Well as we know it in actual practice that will not be possible actual heat regenerated would be equal to this that is it would be up to a temperature lower than the maximum temperature. So, Q regeneration actual would be H 5 minus H 2 the maximum regeneration possible is H 4 minus H 2 ratio of this is given by what is known as the effectiveness of the regenerator this is something probably we will discuss a little later when we take up the real cycle analysis later on. So, effectiveness will tell us the how much amount of regeneration has actually been carried out as compared to what is maximum possible. Now, if you use the cold air standard assumptions the thermal efficiency of an ideal Brayton cycle with regeneration we can derive an expression which will be what is shown here thermal efficiency of a regenerating Brayton cycle is 1 minus T 1 by T 3 into R subscript P raise to gamma minus 1 by gamma. So, what you can immediately see here is that the thermal efficiency depends upon yet another parameter in a simple Brayton cycle we have discussed that the thermal efficiency is a function of the pressure ratio as well as of course, the ratio of specific heats, but in a cycle with regeneration thermal efficiency also depends upon the temperature ratio. So, I leave it as an exercise for you to derive this expression it can be derived in the same way as we have derived the simple Brayton cycle efficiency only thing is here the heat added and heat rejected is different from what it was for the simple Brayton cycle analysis. So, from our discussion on regenerating Brayton cycle you should be able to derive an expression for the thermal efficiency of a regenerating Brayton cycle. Now, I mentioned in the beginning that there are other modifications possible for a simple Brayton cycle like inter cooling and regeneration is something we have discussed as well as reheating. Now, inter cooling is something which occurs during the compression process reheating is something that can be applied for the expansion process and regeneration as we have seen is happening between the heat addition and heat rejection processes. So, these are other modifications that we are going to discuss that are to do with inter cooling and reheating. So, let us take up look at what we mean by inter cooling and reheating. So, we know that the network of a gas turbine cycle is basically the difference between the turbine work output and the compressor work input. So, this can be increased by either decreasing the compressor work or increasing the turbine work or both that is you could either reduce the work required by the compressor or you could increase the work required or generated by the turbine or we could do both to get an improved efficiency. And so, let us take a look at the compression process first. The work required to compress a gas between two specified pressures can be decreased by carrying out the compression in different stages and then cooling the gas in between. So, compression process can become more efficient if we were to cool the compression processes or we split the compression process into different stages and in between these stages we apply cooling and this is known as inter cooling and if it is done for multi stages of the compressor it is known as multi stage compression with inter cooling. So, multi stage compression with inter cooling is one of the options of improving the performance of an ideal Brayton cycle that is by applying cooling. Basically, we split the compression process into different stages and then between these different stages we apply cooling. So, that for the subsequent stage it has to do lesser work as compared to compression between a single compression between the entire pressure ranges. So, if you were to apply the compression for between states 1 and 2 as a single unit as compared to splitting that process into different stages and applying cooling then the work required in the first case would be higher than what it would be for the second case. So, let us take a look at how it works. So, as the number of stages of compression increases basically the process will approach an isothermal process and ideally if we were to extend a Brayton cycle an ideal Brayton cycle to have efficiencies approaching that of a Carnot cycle then we can we need to ensure that the compression process and the expansion process both these processes are isothermal and if we add regeneration then it becomes an Ericsson cycle which is something you would have discussed during your you would have learned during thermodynamics. So, infinite number of stages of inter cooling compression processes makes it an isothermal compression infinite number of heating reheating of the turbine makes it an isothermal heat expansion process and if you also have a regeneration taking place between the heat addition process and the heat rejection processes that is a cycle with number of inter cooling stages of compression number of stages of reheating and also regeneration taking place will represent an Ericsson cycle and if you recall Ericsson cycle has efficiency which is equal to that of a Carnot cycle efficiency which is the maximum efficiency that you can get and that will be obviously, higher much higher than what a simple Brayton cycle generates. So, that is the advantage of having inter cooling let us take a look at how inter cooling works. Now, traditionally the process of compression takes place between states 1 and C. So, this is the normal state of compression that we are familiar with and we would like the compression process to take this path which is shown here which is an isothermal compression process which is what would be the process which requires the least amount of work and the way to do that is to split the compression process into several processes and let us say we split it into two stage compression with inter cooling that is process between state 1 and A this is the first compression process then we apply an inter cooling for an ideal cycle that is at constant pressure and the second stage compression is between states B and D. So, what is shown here by the shaded area is the amount of work that we have saved as a result of inter cooling which means that if we had infinite number of these stages of inter cooling the area under this whole curve will be the amount of work that we would have saved as a result of inter cooling and this is a tremendous amount of work that work that can potentially be reduced by splitting the compression process into several stages and applying inter cooling between them. And so inter cooling is one way in which we can increase the efficiency of cycle by reducing the work required for a compression process by trying to approximate a compression process to an isothermal process as much as possible. Of course, it is also known that you know that you cannot have infinite number of stages of inter cooling due to practical limitations, but even if we were to have finite number of stages of inter cooling let us say one stage inter cooling two stage and so on it will still have work requirement which is lower than what an ideal simple Brayton cycle requires. So, inter cooling is one way in which we can reduce the work required for compression and that is one of the ways of modifying a simple Brayton cycle so as to get a better performance. What is the other way of doing it as we have discussed it is reheating that is at the end of the compression well the third process which is the expansion process is basically a an expansion in a turbine in a gas turbine engine. What we try to do is to split the expansion process into multiple stages we have multiple stages of turbines and in between these two stages or several stages we apply heating. So, reheating between different stages of turbines also gives us an improved work output. Let us take a look at how that works in a two stage gas turbine ideal Brayton cycle with inter cooling reheating and let us say regeneration. Let us take a look at the compression process first this is a two stage inter cooling compression process the compression has been split into two stages. So, this is state 1 and 2 after that it is cooled that is the inter cooling wherein heat is removed Q out now consists of this as well and the second stage of compression between states 3 and 4. After this there is a heating that is the heat addition process at the end of which there is expansion the first stage expansion taking place that is between states 5 that is between states 6 and 7. And then the second stage of expansion taking place between states 8 and 9 between these two stages of expansion we have a heat addition taking place which is the reheat process. So, we now have two new heat exchange processes taking place here one is heat removal from this process during the inter cooling and heat addition taking place during reheat process. We have also shown a regeneration here an ideal regeneration wherein this much amount of heat between states 9 and 10 is transferred to this process between states 4 and 5. So, that this is the heat regeneration that is between states 4 and 5. So, h 4 minus h 5 minus h 4 will be the Q regenerated which is basically equal to h 9 minus h 10 heat input has now been reduced which is between states 5 and 6. So, that would be h 6 minus h 5 plus h 8 minus h 7. So, these are the two stages of heat addition and what about heat rejection? Heat rejection between states 10 and 1 that is Q out as well as heat rejection during the inter cooling process between h 2 or state 2 that and state 3 that is h 2 minus h 3. So, heat rejected would be h 10 minus h 1 plus h 2 minus h 3 heat added would be h 6 minus h 5 plus h 8 minus h 7. So, this is basically showing a Brayton cycle which has two stages of inter cooling and two stages of reheating that is reheating between two stages of expansion and cooling between two stages of compression. So, an inter cooling and reheating process with regeneration is one of the ways of increasing the performance of a Brayton cycle substantially that is by making these modifications. So, thermodynamically it is very easy to implement this in the sense that we can always split the compression process into several stages and say that the efficiency will be higher and so on. But there are practical limitations to how many stages can we actually do this because as we increase the number of stages of either compression or expansion the mechanical complexity also substantially increases. So, in terms of mechanical design and complexity weight and cost and so on it does not make an economic sense to have infinite number of compression and expansion processes. So, normally what is saying is that two to three stages of inter cooling and reheat is probably the maximum that one could be using these principles. So, that this whole thing is economically viable. So, what we have discussed so far is about the ideal cycles variants of ideal cycle which include regeneration inter cooling reheat and so on. Let us take a look at what would be an actual Brayton cycle or a real Brayton cycle how would be this different from an ideal cycle and how is it that we can estimate or carry out a cycle analysis for that. So, an actual Brayton cycle or actual Brayton cycles differ from ideal cycle in all these four processes. One of the basic assumptions of an ideal cycle is that all the four processes are internally reversible. Actual processes are not and for example, the compression and expansion processes in an ideal cycle are assumed to be isentropic in an actual cycle these processes are non isentropic. And the heat addition heat rejection processes were assumed to be constant pressure in actual processes there will be a pressure drop take which takes place during heat addition and heat rejection. So, presence of these irreversibilities will cause a Brayton cycle an actual Brayton cycle to be quite different from an ideal Brayton cycle. So, this is how an actual Brayton cycle would be on a T S diagram. So, what is shown here between states 1 to 2 s 2 s shows that the state represents an isentropic state. So, 1 to 2 s is isentropic process which is how it would have been if the process was ideal, but an actual process is non isentropic there will be an increase in entropy during the compression process. So, 2 a is for the actual state actual process takes place between states 1 to 2 a and during the second process there is as you can see there is a change in the pressure it is not a constant pressure process there is a pressure drop taking place during the heat addition process. And therefore, the temperature at the end of this process is 3 and the third process is an expansion process again it is non isentropic there will be an increase in entropy during this expansion process as well. So, this would have been the ideal temperature at the end of this expansion process given by 4 s actual temperature is 4 a again the last process is an expansion process is a heat rejection process during this which heat is rejected and there is a pressure drop taking place as well during this process. So, an actual Brayton cycle as you can see consists of all the 4 processes that is there is an expansion process there is heat addition and compression and heat rejection and so on. But all the 4 processes are different from an ideal Brayton cycle. So, basically the first 2 processes that is the compression and expansion processes are non isentropic heat addition and heat rejection processes do not take place at constant pressure. So, these are the fundamental differences between an actual and a real Brayton cycle. So, how do you account for these differences now we know that there are differences between an actual and an ideal Brayton cycle. Now, one of the ways of accounting for the differences between or the deviation of actual compressors and turbines is by estimating or defining the efficiencies during the compression and expansion process. Now, let us take a look at the compression for process first. Now, efficiency as we will define now as isentropic efficiency of a compression process is basically the ratio of the isentropic work to the actual work. Now, if you go back to the T S diagram I was showing in a compression process ideally if the process were to be isentropic the work required would be lower than what it would be for an actual process. Therefore, the efficiency the isentropic efficiency of the compressor would be equal to ratio of the isentropic work to the actual work. Now, for a turbine it is the other way around because the work output of an actual turbine will be lower than the isentropic work output. Therefore, the isentropic efficiency of a turbine will be equal to the ratio of the actual work to the isentropic work. It will be little more clear if we go back to the T S diagram and take a look at these definitions once again. Now, this is the compression process the ideal compression process between states 2 S and 1 the actual compression process between states 2 A and 1. So, this difference is because of the presence of irreversibility. So, H 2 S minus H 1 is the isentropic enthalpy difference H 2 A minus H 1 is the actual entropy difference enthalpy difference that is how it would define the efficiency of this process the compression process. Similarly, for the expansion process we have H 3 minus H 4 A the actual work divided by H 3 minus H 4 S which is the isentropic work. So, isentropic efficiency of a compressor would be equal to H 2 S which is the entropy at the enthalpy at the end of state 2 isentropic process minus H 1 divided by H 2 A which is enthalpy at the end of state the process actual process that is at state 2 minus H 1. For the turbine it would be equal to actual work by isentropic work which is equal to H 3 minus H 4 A which is the actual work or enthalpy at end of the expansion process at state A divided by H 3 minus H 4 S which is the enthalpy at state 4 for the isentropic process. So, here 2 A and 4 A represents the actual state which is corresponding to that of the compressor and the turbine exit and 2 S and 4 S are those corresponding to the isentropic states. So, using these isentropic efficiency definitions one can estimate the deviation that an actual cycle would have as compared to an ideal cycle. So, difference between an actual cycle and an ideal cycle can be in some sense by estimated using these efficiency definitions. Now, the other processes where there are differences between the actual and the ideal cycle is the pressure loss taking place during the heat addition process and heat rejection process. So, if we know the pressure loss that occurs between the heat addition process and the heat rejection process one could also estimate the losses that would be occurring during that process. So, as a result of this non isentropic compression expansion the compressor obviously needs more work than ideal cycle and the turbine generates less work than the ideal cycle. So, isentropic efficiencies as we have defined it now reflect the amount of deviation the actual or compression or expansion processes have from the ideal and the total pressure losses in which occur in the heat addition or rejection processes will also have to be considered. So, that we can have an analysis of the cycle which is different from that of an ideal cycle. Now, there are also some other differences which have which occur between ideal and actual cycles which are sometimes used in cycle analysis one is the change of specific heats with temperature. Now, we have assumed in ideal cycle for cold air standard assumptions that specific heats do not change with temperature which is not really true specific heat before combustion and after combustion could be substantially different. So, there is a change in specific heat which also could be considered other parameter that could be considered is the heat exchanger or regenerator effectiveness. If there is regeneration in the cycle then the third parameter that sometimes is considered is the mass flow rate of fuel that is often of course, it is much it is a negligibly small quantity for aircraft engines at least because mass flow rate of air is usually much higher many times higher than that of the fuel, but it is sometimes considered in analysis and also the efficiency of the combustion process itself given by the combustion efficiency is yet another parameter that is sometimes used. So, some of these parameters which also show the deviation of actual cycles from the ideal cycles are sometimes used in the actual or real cycle analysis. Now, we have already discussed about variance of the simple Brayton cycle the ideal cycle that could have reheating intercooling or regeneration. Now, actual cycles with the above will be different from the ideal cycles in terms of the irreversibilities that is an actual Brayton cycle with reheating will obviously be different from the real cycle because there could be losses taking place during the reheat process that could be pressure losses in the reheat reheating may not take place at constant pressure. Similarly, the isentropic efficiencies of individual stages of either reheating or intercooling then the heat exchanger effectiveness these are different parameters which will come into picture when we have actual cycles actual Brayton cycles which have some of these variance that is intercooling or reheat or regeneration which are put in. So, simple Brayton cycle with all these individual components or a combination of these modifications with reheat intercool or regeneration will need to have the isentropic efficiencies of individual components the total pressure loss taking place during either the intercooling process or during the reheat process and in some cases there could be heat exchanger effectiveness and so on. So, these are some of the parameters which will have to be considered when we have an actual Brayton cycle with these cycle modifications. So, let us take for example, an actual Brayton cycle with intercooling now actual Brayton cycle with intercooling would have intercooling taking place at different stages. So, there would be isentropic efficiencies associated with each of these stages. So, efficiency of each of these stages of intercooling is something that will have to be considered the second parameter that will have to be considered is the heat exchanger effectiveness between these two stages of or different stages of intercooling. Let us say we have a multi stage compression with intercooling. So, each stage of compression will have its own isentropic efficiency and intercooling will have some amount of heat exchanger effectiveness that will have to be considered. Now, for a Brayton cycle with let us say reheating alone that means that the expansion process of the turbine is split into multi stages. So, multiple stages of expansion with reheat in between and therefore, each stage will have its own efficiencies isentropic efficiency of expansion for each of these stages with the amount of total pressure loss that is taking place during each of these stages of reheating. So, total pressure loss and combustion efficiency during reheat that would have to be considered for an actual cycle with reheat. For intercooling we consider isentropic efficiency of each stage and also the heat exchanger effectiveness of the intercooling duct. With reheat we have isentropic efficiencies of each of these stages of reheating and the second aspect would be total pressure loss and combustion efficiency which will have to be considered during reheat. Now, with regeneration let us say the above cycles are being used with regeneration. Regeneration as we know it is basically a heat transfer process. So, heat exchanger effectiveness will also have to be considered during an actual Brayton cycle which operates with regeneration. So, an actual Brayton cycle with all these three modifications will have to be considered or will have to be analyzed while considering all these above irreversibilities which pertain to efficiencies of the compression process and efficiencies of the expansion process and also pressure losses taking place during either heat addition or heat removal or during reheat process. So, in an actual Brayton cycle if all these irreversibilities are taken into account one can actually estimate the net efficiency of such a cycle compare that with the corresponding ideal cycle efficiency and then we could basically have an estimate of where the loss in efficiency is taking place because if we were to have a Brayton cycle an actual Brayton cycle which has efficiencies approaching that of an ideal Brayton cycle then basically the improvement in terms of reduced compressor work increased turbine work output and heat recovery from the exhaust these are some of the aspects that will have to be taken into account. So, as to improve the efficiency of an actual Brayton cycle. So, let me summarize what we had discussed in today's lecture we were discussing about Brayton cycles and we started our discussion with the ideal Brayton cycle the ideal simple Brayton cycle which did not have any modification. So, it just consisted of four processes a compression process which was isentropic a constant pressure heat addition process isentropic expansion process and a constant pressure heat rejection process. So, these were the four different processes which compare which constitute an ideal Brayton cycle. So, this Brayton cycle can be operated with certain modifications or variations you could have intercooling in the compression process that is split the compression process into multiple stages with intercooling. So, as to reduce the net compressor work you could split the expansion process into multiple stages and apply reheating in between. So, as to increase the expansion work and also use regeneration that is recover part of the heat that is rejected transfer it to the heat transfer heat input process. So, variations of the simple Brayton cycle and then we also discussed about actual Brayton cycles or real Brayton cycles where in none of the processes are idealized there are none of the processes are internally reversible. For example, compression and expansion processes are non isentropic heat addition and heat rejection process do not take place at constant pressure. So, if you include all these irreversibilities then the process becomes what is known as an actual Brayton cycle. So, these were some of the topics we had discussed during today's lecture and what we will discuss in the next lecture would be basically to do with jet engine cycles that is how a Brayton cycle can be used for a jet engine for aircraft propulsion. We will discuss different versions of the cycle we will start with a simple turbo jet engine then we will continue with turbo jet engine with after burning then we will take up turbo fan and its variants and also we will discuss about turbo prop and turbo shaft engines. So, the jet Brayton cycle for all these different forms of engines is something well that we will be discussing during the next lecture.