 Hello and welcome to lecture number 8 of this lecture series on jet aircraft propulsion. As we have been discussing in the last few lectures, there are different modes or forms of Brayton cycle which is the fundamental cycle based on which all the gas turbine engines operate. So, the Brayton cycle forming the basis of all gas turbine engines we have been discussing of the last couple of lectures on what are the different forms of Brayton cycle, the actual Brayton cycle and how is it that this Brayton cycle is used in an actual aircraft engine. And we have seen the ideal analysis or the ideal cycle form of the Brayton cycle used in the turbojet the simple turbojet engine which is the most fundamental form of a gas turbine engine. And then we have seen variants of turbojet engine like turbojet with after burning. Then we have also seen turbojet the turbofan engine which is a variant of the turbojet engine with a fan. So, as to have a lower effective jet exhaust velocity leading to better propulsion efficiencies and also fuel efficiency to some extent. Then we have also seen that there are different forms of turbofan engines like the single spool, the multi spool and so on. And then the other forms of the propeller forms of the jet engine the turbo prop the turbo shaft engine. And also a very simple form of a jet engine the ram jet engine which does not have any rotating components. So, these are the different types of jet engines that we have been discussing over the last couple of lectures. And also we have had some discussion on the basic cycle the ideal cycle of these different modes of jet engines. So, what we are going to discuss today are the components that constitute a jet engine. And how is it that we can evaluate the performance of these components? We are not going into discussion mode of the exact form of these components or what how is it that these components work and so on. We will just be treating these components as black boxes. But we will also look into a take into account the performance parameters which we can associate for each of these components. So, what we are going to discuss in today's lecture or the following we will be beginning our discussion with taking a look at the intake. Intake is one of the first components of a jet engine. Then following the intake is the compressor or the fan thermodynamically compressor and fan are the very same. And so the performance parameters that we define for compressor should also be valid for a fan. And then we will discuss about the parameters which are to be defined for combustion chamber, then the turbine, the nozzle and after burner and so on. So, these are some of the topics that we will be discussing in today's lecture. And we will as I mentioned we will not be discussing about the geometry and the design and other aspects of these components. These detail design and also the composition of these components will of course be discussed in detail in later lectures when we analyze each of these components one by one. Today we are going to just treat them as black boxes and that there is some performance penalty which we have to associate due to some irreversibilities and so on. So, let us discuss about what are the irreversibilities that could possibly be hampering the performance of these devices. So, we know that jet engine as we have seen has several salient components and one of the first components of a jet engine is the air intake or the diffuser. The basic perform the basic function of an air intake or a diffuser is decelerate the air incoming air and deliver it to a compressor. So, that is one of the basic functions of an air intake that it will decelerate the air incoming air directs it towards the compressor and in the process it also reduces this velocity increases the pressure a little bit and as we have seen in the ideal cycle compression begins with the air intake. So, part of the compression occurs in the air intake rest of it occurs in the compressor. The second component that follows an air intake is either a fan in the case of a turbofan engine or it could be directly a compressor in the case of turbojet engines or turboprops and so on. So, compressor or a fan as I mentioned are thermodynamically the same because both of these increase the pressure decelerate and so on. So, these are components which are thermodynamically the same and their function is to increase the pressure and in the process they also increase the temperature of the incoming air and then it is delivered to the next component which is the combustion chamber. So, combustion chamber is the component where fuel is added in the engine and combustion takes place releasing a lot of temperature and energy and at the end of the combustion chamber we have the air and the combustion products which are at high temperature and pressure at the turbine inlet. So, at the turbine inlet we would want combustion products or air to be at very high pressure and temperature. So, that this can be expanded through the turbine and in the process we can extract energy from such a system. So, combustion chamber delivers hot products which are at high temperature and pressure at the turbine inlet and then the turbine expands this and in the process it will deliver a shaft work and the function of a turbine in a pure jet engine is merely to drive the compressor. So, turbine does not have any other function it is meant only to drive the compressor, but whereas in certain other forms of engines like in a turboprop for example, turbine besides driving the compressor also drives the propeller. So, in such engines as we have discussed in the last lecture there is a separate turbine unit which is known as the power turbine or the free turbine which drives the propeller. So, such units exist only in those engines where we need let us say a propeller power output and so on. After the turbine depending upon the nature of the engine let us say if it is a turbo jet engine with an after burner then after burner would follow turbine that is the combustion products from the turbine would still have a lot of energy in it and also it is at high temperature and pressure at the same time there is substantial amount of oxygen still left in the combustion products. So, it is possible for us to add additional fuel after the turbine exhaust raise the temperature even further and therefore, the pressure as well. So, that there is an additional pressure available for expansion in the nozzle. So, this is used in engines where we need additional thrust for example, if we have to accelerate and cross the sonic barrier reach the supersonic speeds and also cruise at supersonic speeds then it is essential that turbo jet engines also have an after burner. So, that this can be achieved. So, nozzle is the last component of the jet engine combustion products are expanded through the nozzle and generates the required thrust that the engine is supposed to deliver. So, what we will discuss today are each of these components and how can we evaluate or judge the performance of these units? We will start with the diffuser and we will first take a look at what are the sources of losses that are possible in a diffuser or an air intake and how can we estimate the performance of such devices? So, let us take a look at the air intake performance first. Now, there are two sources of losses in an air intake one is valve friction which is present in all in fact, all the components and the second possible source of loss are shock waves, but this is valid only for a supersonic intake. So, a supersonic intake may also have shock losses in addition to valve friction losses and. So, how does it affect the performance? It obviously results in a reduction in the total pressure at the exit of the diffuser and also there is an efficiency penalty associated with this and. So, we will be defining what is known as an isentropic efficiency and the flow through intakes can always almost always be assumed to be adiabatic because there is hardly any heat transfer that takes place across the diffuser boundaries. So, it is it can be safely assumed to be adiabatic and. So, there are two parameters that will define the performance of air intakes one is the total pressure ratio which will basically tell us how much frictional losses has taken place this can also be expressed in the form of what is known as an isentropic efficiency of the diffuser. So, both these parameters are usually used to evaluate or assess the performance of air intakes and. So, how do we arrive at definition for let us say the isentropic efficiency? Total pressure ratio is straight forward it is the exit total pressure divided by the inlet total pressure and sometimes this is also referred to as pressure recovery that is total pressure ratio across the diffuser is it gives us an indication of how much pressure how much total pressure has been lost and. So, how much is higher the total pressure ratio better is the pressure recovery of such an intake. So, total pressure recovery is one of the parameters now to define the efficiency of intakes you will have to take a look at the temperature entropy diagram of the intake in a little more detail. So, that we are able to define a performance parameter which is basically the isentropic efficiency. So, what I have plotted here is a temperature entropy plot of an intake process, but unlike what we have been plotting earlier this is lot more details are shown here on what actually happens during an intake process. So, let me explain this in little more detail. So, we have ambient air which is at station a which is at an ambient pressure of p subscript a and then this is compressed all the way up to the total pressure at the exit that is p 0 2 and station one is the entry of the intake. If you recall the intake when we were discussing about the cycle analysis of turbo jet I had shown that there are two distinct points one is corresponding to the ambient condition and the second one that is station one which corresponds to the inlet entry the geometric entry of the intake. So, this means that there is already some compression which is taking place which is shown by the increase in static pressure here. So, there is an increase in static pressure, but because this is outside the intake boundary there are no losses associated with this process. So, the first compression process that is known as the external compression or pre compression as it is denoted by in some literature. So, pre compression does not suffer from any losses because there are no solid surfaces which bound pre compression, but from point one all the way up to point two there are sources of losses because of the frictional effects and so there is an increase in entropy as you can see it is no longer an isentropic process. Process from A to 1 is always isentropic process from 1 to 2 is not isentropic and so this is the static pressure at the exit of the diffuser that is given by point two. Whereas, 2 S denotes the temperature on the pressure line if the process were to be isentropic. Now, what are the corresponding total pressures? Now, the total pressure corresponding to point two which is the actual total pressure will be equal to P 2 plus V 2 square by 2 C p and that is the static pressure plus dynamic pressure that is P 2 is static pressure plus half rho v square will give us the total pressure that is P 0 2 and what should have been the total pressure if there were no losses it would be equal to the static pressure here P a plus V naught square by 2 and that is given by P 0 a. So, P 0 a corresponds to the total pressure of the intake if there were no losses at all. Whereas, P 0 2 corresponds to the total pressure in the presence of losses which means that P 0 2 is always less than P 0 a and for all engines P 0 2 will be less than P 0 a. So, how do we define an efficiency based on this? So, the efficiency that we are going to define will be based not on the pressures essentially, but it is on the temperatures or enthalpies corresponding to this line. So, we will define the difference in enthalpy for an actual process divided by an ideal process. So, based on what we have discussed now the isentropic efficiency of the diffuser can be defined as h 0 2 s which is the total stagnation enthalpy for the isentropic process minus h a which is the static enthalpy at the ambient condition divided by h 0 a minus h a. So, let us go back to the T s diagram and see that look at the definition once again. So, h 0 2 s is this point which is the stagnation enthalpy on the total pressure line actual if the process were isentropic. So, h 0 2 s minus h a divided by h 0 a which is the stagnation enthalpy on the process line if it were entirely isentropic. So, this goes all the way up to p 0 a. So, h 0 2 s minus h a divided by h 0 a minus h a. So, this is the isentropic efficiency of a diffuser if you were to assume that the static pressures are at the inlet and exit are the same then this can also be expressed in terms of the corresponding temperatures. So, we get T 0 2 s minus T a divided by T 0 a minus T a. So, enthalpy difference which enthalpy difference between those two points that is one corresponding to the total pressure line actual minus the ambient enthalpy divided by total pressure stagnation enthalpy for the isentropic process all the way up to p 0 a minus stagnation enthalpy minus the ambient enthalpy. So, this defines an isentropic efficiency of a diffuser based on temperature. So, what you should notice here is that the isentropic efficiency has been defined based on temperatures, but it is due to the pressure loss there is no loss in stagnation temperature occurring across a diffuser. So, this is one point which you have to keep in mind that there cannot be either an increase in stagnation temperature or a decrease in stagnation temperature unless there is either heat addition or heat removal. So, if you assume that the process is adiabatic which is true then there is no change in stagnation temperatures, but there is definitely a change in stagnation pressure because of frictional losses. So, this efficiency that we have defined is because of the loss in stagnation pressure and that we have expressed in terms of stagnation temperatures. So, adiabatic efficiency of a diffuser is basically t 0 to s minus t a divided by t 0 a minus t a. So, this can also be expressed in terms of the stagnation pressure ratios and so on. So, if you were to simplify this expression that we have seen we can relate the efficiency to the total pressure ratio that is pi subscript d and the Mach number. So, if you do a simple simplification of this what we have seen that is t 0 to s divided by t 0 to a is t 0 a is can be expressed in terms of the corresponding pressure ratios and so on because it is isentropic. So, if you were to simplify that what we get is eta d which is diffuser efficiency is 1 plus gamma minus 1 by 2 m square multiplied by pi d which is the stagnation pressure ratio raise to gamma minus 1 by gamma minus 1 divided by gamma minus 1 by 2 m square. So, here m is the free stream Mach number. So, what we can see is that the diffuser efficiency can be easily expressed in terms of the Mach number that is a flight Mach number and the stagnation pressure ratio. So, these are two parameters which based on which we can actually determine the diffuser efficiency. So, we have seen that diffuser efficiency is something that we can estimate based on the Mach number and the pressure ratio and remember that pressure ratio itself is one of the parameters or stagnation pressure ratio is in by itself one of the parameters which define one of the efficiencies of the diffuser. So, based on what we have discussed we should be able to now assess the performance of diffuser based on the pressure laws which is also expressed in terms of temperature difference and therefore, we get an adiabatic efficiency or isentropic efficiency of a diffuser. So, when we do a cycle analysis which is what we will do in the next lecture we will basically be using the efficiency definition based on what we the flight Mach number and what is seen is that the isentropic efficiency drops drastically as the Mach number increases. Because as we increase Mach number there would be the presence of shock waves and therefore, the resultant total pressure losses would be substantially higher and in fact, there are empirical correlations which are available for estimating the efficiency of a diffuser as a function of Mach number. So, there are military standards available which you can look up on in some textbooks which define the efficiency based on or as a function of Mach number and so some of these definitions could be used when we are doing a real cycle analysis of a jet engine. So, having discussed about intakes let us move on and take up the next component which is of interest that is the compressor or the fan. Now, remember I mentioned that compressor and fan are thermodynamically same. So, whatever efficiency definitions that we are going to define for the compressor will also be valid for a fan. So, in the case of compressor we are going to define what is known as an isentropic efficiency. We have already seen this in one of our earlier lectures for turbine and compressors we will take a relook at how these efficiencies are defined. Now, as we have approximated in the case of intakes compressors are also to a high degree of approximation adiabatic. And so how do we evaluate the performance? Compressor efficiency isentropic efficiency is basically defined as the ratio of ideal work of compression for a given pressure ratio divided by the actual work of compression for a given pressure ratio. So, this ratio that is ideal work divided by actual work for the same pressure ratio will define the compressor efficiency which means that ideal work of a compressor would be the difference between the ideal stagnation enthalpy at the exit of the compressor minus the inlet stagnation enthalpy divided by actual stagnation enthalpy minus the inlet stagnation enthalpy. So, compressor efficiency the isentropic efficiency of a compressor will be equal to h subscript 0 3 s minus h 0 2 divided by h 0 3 minus h 0 2. So, this basically defines the efficiency of a compressor based on what should be the ideal work that the compressor requires divided by what is the actual work that the compressor is needing. Which means that ideal work of the compressor obviously will be lower than the actual work that the compressor requires. So, to understand this let us look at the T s diagram of a typical compressor and take a look at how we can define an isentropic efficiency. So, what is shown here is the T s diagram for a compression process both the actual and ideal compression processes are shown here. So, the compression begins at 0 2 which is the inlet of the compressor P 0 2 is a constant pressure line corresponding to the inlet compressor pressure. The corresponding temperature is T 0 2. Now, an ideal process would be an isentropic process a straight line shown here going all the way up to 2 0 3 s. So, T 0 3 s corresponds to the stagnation temperature at the compressor exit and what is the actual temperature actual temperature is T 0 3 which is on this non isentropic line. So, you can immediately see that T 0 3 is greater than T 0 3 s and therefore, the actual work is greater than the ideal work. So, based on this if we define the compressor efficiency we have h 0 3 s minus h 0 2 divided by h 0 3 minus h 0 2 which can be approximated assuming that the specific heat at constant pressure is a constant then we get T 0 3 s minus T 0 2 divided by T 0 3 minus T 0 2. Let us simplify this now. So, this we can express as temperature ratios we get T 0 3 s divided by T 0 2 minus 1 divided by T 0 3 divided by T 0 2 minus 1. So, the first term that you see here that is the stagnation temperature ratio one of them is isentropic can be expressed in terms of the corresponding pressure ratios from isentropic relations that is T 0 3 s divided by T 0 2 is equal to P 0 3 divided by P 0 2 raise to gamma minus 1 by gamma and P 0 3 by P 0 2 is the compressor pressure ratio which is pi subscript c. So, isentropic efficiency of the compressor can be expressed or in terms of the stagnation pressure ratio that is the compressor pressure ratio pi c raise to gamma minus 1 by gamma divided by tau c which is the stagnation temperature ratio that in this case it is T 0 3 divided by T 0 2. So, based on the expression that we have seen here this basically relates the compressor efficiency that is isentropic efficiency related to two distinct parameters one is the compressor pressure ratio that is pi c and the other is the temperature ratio tau subscript c. So, isentropic efficiency can actually be expressed in terms of some of the design parameters in this case the design parameter happens to be the stagnation pressure ratio. So, depending upon what is the design parameter for the compressor for a compressor it is always the compressor pressure ratio you can express the compressor pressure ratio I mean you can express the efficiency as a function of the compressor pressure ratio and the temperature ratio. So, besides the efficiency definition that is the isentropic efficiency there are other efficiency definitions that are possible some of them are stage efficiency and poly tropic efficiency. And these are basically used when we take a look when we when we try to approximate or when we try to take a look at the actual nature of the compressor itself. Compressor as we know in a gas turbine engine is usually a multi stage compressor and therefore there is the the compression process is not a single process as we know it now it is it constitutes it is constituted of several infinitesimal compression processes. So, one of the ways of trying to estimate the efficiency of such a process is to use what is known as a poly tropic efficiency. So, we will discuss about stage efficiency little later when we take up the compressor analysis in detail and today we will discuss about the poly tropic efficiency what exactly is poly tropic efficiency. And it is to be remembered here that all these three definitions or efficiency terms are related to one another and today we will be relating the isentropic efficiency to the poly tropic efficiency similarly the other term that is stage efficiency can also be related to these two definitions as such, but that we will discuss later on. Now, let us take a look at what is poly tropic efficiency? Poly tropic efficiency is defined as the ratio of ideal work of compression for a differential pressure change divided by the actual work of compression for a differential pressure change. Now, isentropic efficiency if you remember was defined in the same way, but that was not for a differential pressure change it was defined for just a given pressure ratio. So, this reduces to a differential work ideal divided by differential work actual which is equal to d H naught i which is differential enthalpy change ideal divided by d H naught which is the differential enthalpy rise actual. This is again approximated as d T naught i divided by d T naught. Now, for an ideal compressor we know that T naught i is related to P naught i raise to gamma minus 1 by gamma into a constant because temperature ratios can be related to pressure ratios using the isentropic relation. So, this we can by binomial expansion we can express this as d T naught i divided by T naught is equal to gamma minus 1 by gamma into d P naught i by P naught. This is from the first approximation using binomial expansion. So, if we use this in the polytropic efficiency definition then we get if eta poly which is the polytropic efficiency this is equal to d T naught i divided by d T naught which is again equal to d T naught i by T naught divided by d T naught by T naught. So, what we have defined earlier this can be simplified as polytropic efficiency is equal to gamma minus 1 by gamma into d P naught i divided by P naught divided by d T naught by T naught. We will simplify this further again that is polytropic efficiency we are now going to express that in terms of a parameter which we can very easily use in the isentropic efficiency terms. So, the if you rewrite the equation that we have just seen we get d T naught by T naught is equal to gamma minus 1 by gamma eta polytropic into d P naught by P naught. So, if we integrate this between states 0 2 and 0 3 which are the inlet and exit of the compressor then we get tau subscript c is equal to pi subscript c raise to gamma minus 1 divided by gamma multiplied by eta polytropic. Now this we can again use in our efficiency definition that we have just now seen. Therefore, isentropic efficiency of a compressor eta c is equal to pi c which is the compressor pressure ratio raise to gamma minus 1 by gamma minus 1 divided by tau c minus 1. This is also equal to pi c raise to gamma minus 1 by gamma minus 1 divided by since tau c is expressed in terms of pi c raise to gamma minus 1 by gamma into eta polytropic that we can substitute here. So, what we have here is a very interesting equation which relates 3 parameters it relates the isentropic efficiency to the stagnation pressure ratio and the polytropic efficiency. So, this means that if we know the polytropic efficiency and of course, the compressor pressure ratio we should be able to get a better estimate of the isentropic efficiency and vice versa. Suppose, we know the isentropic efficiency we also know the compressor pressure ratio we can find out what should be the polytropic efficiency for such a process. And this is something which we are going to use very frequently during the cycle analysis that is something we will take up in the next lecture during the tutorial. So, during the cycle analysis what we will discuss is how we can use some of these performance parameter terms in our analysis. So, that we get a realistic estimate of the performance of these gas turbine engines. So, compressor performance evaluation involves the compressor pressure ratio which is a design parameter then it involves the isentropic efficiency and the polytropic efficiency. We have also seen how we can relate the isentropic efficiency to the polytropic efficiency and the compressor pressure ratio. And the other term that we are going to use is stage efficiency as I mentioned, but this we will define little later because that requires little understanding of the working of these compressor systems that we will discuss when we take up the compressor analysis in detail. Now, the next component that we are going to discuss about is the combustion chamber or the combustor or the burner. So, different books call it in different ways. Now, in a combustion chamber there are basically two possibilities of losses. In a combustion chamber there are two possibilities of losses. One is the incomplete combustion and the other is total pressure losses. So, these are two different sources of losses that can affect the performance of combustion chamber. One is related to the combustion efficiency which states that given a certain amount of fuel how much of that has actually taken part in combustion whether the entire fuel has been converted to work output or so on or heat released or so on. The other parameter is the pressure loss because combustion chamber involves certain amount of total pressure loss and there are two again sources of total pressure loss. One is the viscous loss and the other is the pressure loss due to combustion occurring at finite Mach number something we have discussed you might have learnt about during your studies on heat addition on a constant area duct and so on. So, based on our understanding of the energy balance across the combustion chamber we should now be able to define what is the combustion efficiency because we know the combustion chamber outlet conditions we know the temperature there and so we know the enthalpy or energy content at the combustion chamber exit. We also know the energy content of the inlet because we know the compressor exit conditions. We also know what is the amount of fuel that is added. So, based on all this if we carry out an energy balance we will know what is the efficiency associated with this combustion products. So, in a combustion efficiency we can basically define by carrying out an energy balance and only thing that we have to remember here is that in a combustion chamber we will no longer be using the same specific heat across that because the temperature differential between the inlet and exit is substantial and so we can no longer use the same specific heat at constant pressure for the inlet and exit. So, we are going to use one specific heat for the fluid which is upstream of the combustor and the other value of specific heat for fluid which is downstream of the combustor. Now combustion efficiency which is denoted by eta subscript b where b stands for burner is defined as the following. So, this will be the ratio of the difference in the inlet difference between the exit enthalpy minus the inlet enthalpy divided by the fuel added. So, this means that if the efficiency was equal to 1 then we have the exit enthalpy which is m dot which is mass flow rate of air plus mass flow rate of fuel multiplied by the enthalpy equal to inlet mass flow that is mass flow rate of air multiplied by the corresponding enthalpy plus the mass flow rate of fuel into the heat release q dot f is the heat release rate of the fuel or heating value of the fuel. So, combustion efficiency is equal to m dot plus m dot f into h 0 4 minus m dot into h 0 3 divided by m dot f into q dot f. This we can simplify as m dot plus m dot f into c p 4 t 0 4 minus m dot c p 3 t 0 3 divided by m dot f q f. Now c p 4 is basically the c p at station 4 c p 3 is c p at station 3. So, this we will denote by c p g which is c p for the combustion products of the gases. This is the average value of gases downstream of the burner c p a is the average value of air upstream of the burner. So, combustion efficiency is one of the parameters that we can define to evaluate or assess the performance of combustion chambers. Now, what is the other parameter we have discussed? The other parameter we have discussed is the total pressure loss. So, there could be some amount of total pressure loss taking place across the combustion chamber. And what causes these total pressure losses? One reason is of course, the viscous losses the frictional losses in the combustion chamber. The other reason is that when combustion occurs at a finite mach number that usually leads to some amount of total pressure loss. So, both of these parameters put together lead to loss in total pressure across the combustion chamber. So, loss in total pressure would be given by the ratio of the corresponding total pressures that is p 0 4 divided by p 0 3 will be less than 1. So, in an ideal cycle we had assumed that it is a constant pressure process p 0 4 is equal to p 0 3 that is no longer true. So, it would be less than 1. In a combustion chamber the combustion efficiencies are usually very high because we know that the air to fuel ratio in combustion chambers in gas turbine engines are very high. And so, the combustion efficiencies are usually very high it is of the order of 0.95 96 even as high as 0.99. We will see some estimates or values of these losses towards the end of this lecture. So, in real cycle analysis we are going to use both these parameters that is efficiency as well as the total pressure loss. Now, having looked at the performance analysis of 3 different components now we have looked at the intake, we have looked at compressor and the combustion chamber. The next component that we shall be analyzing is the turbine. Now, turbine will be performance will be defined very much the same way as we defined a compressor. And so, we will be defining an isentropic efficiency for the turbine in pretty much the same way as we defined for a compressor. We will also then be defining a polytropic efficiency for the turbine. So, in turbine a performance we will again be assuming that the flow through the turbine is adiabatic. So, in actual engines of course, it is not really adiabatic because there would be turbine blade cooling. And therefore, there is actually heat transfer taking place there, but we are going to assume that that is negligible here. So, isentropic efficiency of a turbine will define the same way as we defined a compressor. In the case of turbine the actual work of for a turbine would be equal to the actual work that is developed by the turbine divided by the ideal work developed by the turbine for the same pressure ratio. That is w t actual divided by w t ideal which is equal to h 0 4 that is inlet enthalpy minus h 0 5 that is exit enthalpy divided by h 0 4 minus h 0 5 s. This can be simplified the same way as we did for a compressor. This would be equal to 1 minus tau subscript t which is the temperature total temperature ratio across the turbine divided by 1 minus pi t which is the pressure ratio across the turbine raise to gamma minus 1 by gamma. So, the turbine efficiency can be defined as we have seen it is very similar to that of a compressor, but just that the efficiency definition is reversed that is in the case of turbine the actual work output is less than the ideal work output. Whereas, in a compressor the ideal work is actually less than the actual work and that is why the efficiency definitions are reversed as in the case of compressors and turbines. Now, we will take a look at the T s diagram of a turbine actual and real turbines. So, that we can understand the efficiency definition little better. Now, a T s diagram of a turbine is what is shown here. We have a temperature and entropy on x y and x axis respectively. So, turbine begins at the expansion process or turbine process begins at 0 4 which is turbine inlet 0 5 is the actual point corresponding to the turbine exhaust 0 5 s corresponds to the isentropic condition that is if the process were to be isentropic. P 0 5 is the pressure constant pressure line and so 0 4 to 0 5 is the actual process. So, turbine efficiency is defined as h 0 4 minus h 0 5 divided by h 0 4 minus h 0 5 s. This also can be expressed in terms of temperatures that is T 0 4 minus T 0 5 divided by T 0 4 minus T 0 5 s. Now, we will also now define a point of polytropic efficiency very much the same way as we define for a compressor because turbine also involves multi stages or expansion process is occurring in multi stages. So, we can actually define a polytropic efficiency for a turbine as well exactly the same way as we define for a compressor with the corresponding changes that is because of the nature of the process itself. So, the polytropic efficiency for a turbine is basically defined by the actual work output of a turbine for a differential pressure change divided by the ideal work output for the process for a given pressure ratio or for a differential pressure change. We have defined the same way for a compressor as well for a differential pressure rise here it is again for a differential pressure drop. So, polytropic efficiency will be equal to d w which is the actual work for differential or which is the actual differential work divided by ideal work d w i. This is in turn equal to d h naught divided by d h naught ideal which is again equal to d t naught divided by d t naught ideal. Now, using the same simplifications that we did for a compressor we can and using the binomial expansion which wherein we define this ratio d t naught i divided by t naught in terms of gamma minus 1 by gamma d p naught i by p naught. So, we get the polytropic efficiency eta poly is equal to d t naught divided by d t naught i. This is again expressed in terms of the temperatures d t naught by t naught divided by gamma minus 1 by gamma into d p naught by p naught. So, if we integrate between states 4 and 5 which is corresponding to turbine inlet and turbine exit then we get definition of isentropic efficiency of a turbine in terms of the pressure ratio and the polytropic efficiency. So, we have eta t which is the isentropic efficiency of a turbine which is equal to 1 minus pi t which is the pressure ratio raise to gamma minus 1 into eta polytropic divided by gamma minus 1 divided by 1 minus pi t raise to gamma minus 1 by gamma. So, here we have an expression for the turbine efficiency the isentropic efficiency which is again expressed in terms of the pressure ratio and the polytropic efficiency. Very similar to what we had defined for a compressor as well where we had related the isentropic efficiency to the pressure ratio and the polytropic efficiency. So, this is how we would evaluate the performance of turbine considering the turbine as a black box and remember that all through these analysis that we have been discussing we have not really gone into the mechanical construction or the geometric details or the design of any of these components because thermodynamically all that does not matter what matters is what is it that goes into this component and what is it that comes out of this component and how we can evaluate the losses that are occurring within the component. We will take up detail discussion of the analysis of these components etcetera in later lectures. Now, the next component that we shall evaluate is the nozzle. Now, again as I think I have mentioned earlier that nozzle is thermodynamically very similar to a turbine because a turbine expands the flow nozzle also expands the flow. But the fundamental difference between the nozzle and the turbine is that the turbine generates a work output nozzle does not generate any work output in that way it generates a reaction to thrust. So, flow in a nozzle again we are going to assume to be adiabatic which is very much true for a nozzle because the enthalpy drop across a nozzle is much higher than enthalpy changes occurring due to any heat transfer. So, what are the sources of losses in a nozzle? Losses in a nozzle could occur due to one is incomplete expansion it could be either under expansion or over expansion. The second source of loss could be friction and friction also can reduce the efficiency of a nozzle. So, how do we define an efficiency for a nozzle? Nozzle efficiency can be defined by h 0 6 minus h 7 where h 7 corresponds to the nozzle exit divided by h 0 6 minus h 7 s which is the stagnation which is the enthalpy at station 7 for an isentropic process. So, this will be clear if you look at this enthalpy diagram enthalpy entropy diagram or the equivalent is temperature entropy diagram. So, t 0 6 corresponds or h 0 6 corresponds to the nozzle entry conditions that is right here and on the pressure line that is p 0 6 and. So, this expands all the way up to 7 that is the nozzle exit static condition p 7 is the nozzle exit static pressure line. If the process were to be isentropic then it would expand from 0 6 all the way to 7 s and. So, since the process is non isentropic expansion goes all the way only up to 7. So, the efficiency is defined as h 0 6 minus h 7 divided by h 0 6 minus h 7 s. So, this defines the nozzle efficiency. So, besides this some of the books also use pressure ratio across the nozzle as one of the other parameters, but in this course which is going to limit the nozzle performance based on the efficiency. So, efficiency of a nozzle is one of the terms that we are going to use for assessing the performance of a nozzle. Now, as we have seen the way we can define the efficiency is by looking at the stagnation conditions at the inlet that is h 0 6 minus the static enthalpy at the exit h 7 divided by h 0 6 minus h 7 s. So, based on this we can define an efficiency of the nozzle. Now, the other component that of course is not true for all the engines, but it is present in the case of let us say a turbojet with after burning. So, what is an after burner and how can we define the performance of an after burner. Now, if you recall I had mentioned that after burner is basically a combustion chamber and therefore, the performance parameters that we have defined for a combustion chamber should also be valid for an after burner. So, combustion efficiency and total pressure loss across the after burner define the very much the same way as we have defined for the combustor should be valid for an after burner. So, if an after burner is used in an engine performance of the after burner will be evaluated the same way as we did for a combustion chamber. So, an after burner performance parameters would be the combustion efficiency and the total pressure loss. So, in those engines which have after burning the corresponding parameters for after burner that is the efficiency as well as the pressure loss will need to be taken into account when we are carrying out a cycle analysis. So, during the cycle analysis a real cycle analysis of an engine with after burner these two parameters also will come into picture. So, these are some of the components that we have defined the performance parameters associated with all these components the intake the fan or compressor combustion chamber turbine nozzle and after burner. So, is there any other source of loss in an engine well in some cases we would also consider losses due to the transmission of power through a shaft. So, mechanical efficiency is one more loss parameter that is used sometimes defined when we carry out a real cycle analysis. So, how do we define a mechanical efficiency well mechanical efficiency is basically defined as the ratio of the power leaving the shaft to the compressor divided by power entering the shaft from the turbine that is the power output of the compressor to the work output of the turbine. So, if these two are equal obviously the mechanical efficiency is equal to 1 and why is it less than 1 it is less than 1 because of loss of power due to the bearings that are present. So, in when a shaft that connects the turbine and compressor obviously, there are several bearings which hold the shaft and so there are numerous losses taking place through these bearings other source of loss is that some power is also used for driving accessories like fuel pump and oil pump and so on. So, that much power that is developed by the turbine does not actually reach the compressor because it is used by these accessories. So, these are two different sources of losses which may lead to some amount of mechanical losses or mechanical efficiency. So, having looked at all these different sources of losses you may now wonder how much would be a typical value for any of these efficiencies or pressure losses how do I know what loss to take. So, I have now listed some of the typical ranges of these efficiencies which are seen in a modern day jet engine and typical component efficiencies for all of them I have listed down here of course, this is just a range there could be engines which are out of this range, but this is typically for modern day jet engines. So, let us take a look at the typical values of component efficiencies. Now, diffuser is defined by the diffuser pressure ratio as well as the efficiencies and so if it is and these efficiencies can be different depending upon whether it is subsonic or supersonic. In supersonic intakes typically intake efficiencies are lower because of the presence of shock waves which is why you can see for subsonic intakes the values vary from 0.95 to 0.98 for supersonic the diffuser efficiencies vary from 0.85 to 0.95. So, it is lower than the subsonic component. Compressor efficiency the isentropic efficiency of a compressor typically varies between 0.85 to 0.9. For combustion chamber or the burner the burner efficiency combustion efficiencies are usually as I mentioned very high varies from 0.96 to 0.99. Burner pressure loss can vary from 0.9 to 0.95 depending upon the nature of the burner or combustion chamber. Turbine efficiencies can be different depending upon whether it is uncooled or cooled. For an uncooled turbine typically it varies between 0.85 to 0.92 for a cool turbine it slightly lower it varies from 0.84 to 0.9. For an after burner efficiencies are pretty much identical to that of a combustion chamber 0.96 to 0.99 burner pressure loss 0.9 to 0.95. Nozzle efficiencies usually vary from about 0.95 to 0.98. Mechanical efficiencies again vary from 0.96 to 0.99. So, these are some of the typical component efficiency values just to give you an idea as to how much can these losses be. It is as you can see varies between about 10 to 10 to about 1 to 10 percent depending upon what the nature of the loss is. And so, these are some of the component efficiencies which of course, we will be using when we carry out some cycle analysis later on. So, let me summarize today's lecture. So, today's lecture was devoted to taking a look at the different components that constitute a gas turbine engine and how we can evaluate the performance of each of these components by defining pressure loss and efficiencies. So, we had taken a look at the intake performance, the combustion chamber, the compressor or the fan, the combustion chamber, the turbine, the nozzle after burner and also the mechanical losses. So, having discussed about these losses and how we can evaluate these losses, we will devote the next lecture for a tutorial where we will discuss about the ideal cycle analysis a little bit and also how we can evaluate the component performance in some detail. So, that we will take up as a tutorial when we solve some problems on these topics during the next lecture.