 Hello and welcome to lecture number 33 of this lecture series on Introduction to Aerospace Propulsion. We have been over the last lecture discussing about cycle analysis, the ideal cycle analysis for gas turbine engines. So, we started our discussion in the last lecture on trying to derive an equation for thrust and other performance parameters of jet engine like the fuel consumption, the specific fuel consumption, the thrust specific fuel consumption and the different efficiencies like the thermal efficiency, the propulsion efficiency and the overall efficiency. And so these are some of the performance parameters that we can define for evaluating the performance of air breathing engines. And in the last lecture we had our discussion on cycle analysis, the ideal cycle analysis for turbojet engines. So, turbojet engines are one of the basic forms of jet engines and we have done the cycle analysis for that which primarily involved trying to determine the performance across each of the components in terms of the temperature and pressure. And then subsequently for each of the components which constitute a turbojet engine and so once we reach the nozzle of the turbojet engine we can find out the exhaust velocity and from the exhaust velocity we can find out the thrust that the engine is going to develop under those circumstances. And once we calculate the thrust we can also determine other performance parameters like the fuel consumption as well as the efficiencies. So, that was the basic cycle analysis procedure that we had followed and we had discussed in the last lecture. So, in today's lecture we are going to continue with discussion on that, but we shall be taking up variance of turbojet engine. We shall be discussing about one of the most popularly used jet engines in current day civil aviation that is the turbofan engines. We shall be discussing in detail about turbofan engines and how we can carry out a cycle analysis for turbofan engines. And there are some different variants of turbofan engines as well that we shall be discussing about and some of those analysis I shall probably be leaving it to you for carrying out an exercise on the cycle analysis of some of the variants of the turbofan engine. And then we shall also move on to discuss some of the other types of jet engines which are used like the turboprop engines and the turbo shaft engines and also in very brief about the ramjet engines which we shall take up towards the end of this lecture. So, let us take a look at what we are going to discuss in today's talk. So, we shall be discussing about ideal cycle again we are not taking into account efficiencies of different components. We shall be discussing about the ideal cycles and we will begin with the turbofan engine and the turbofan engine which constitutes of different components which are some of most of which are similar to what is there in a turbojet engine. But, there are there is an additional component in turbofan engine which basically the fan and there are different configurations of turbofan engines which we shall be discussing in detail. And depending upon the particular configuration that the turbofan engine is operating on the cycle analysis needs to be modified slightly depending upon what configuration it is. We shall then be discussing about two other popular engines which are in use the turboprop and the turbo shaft engines. Turboprop and turbo shaft engines are also used primarily in of course, they are used in military as well as in civil aviation, but many of the civil aircraft which probably you might have seen or flown would be using the turboprop engines usually with those engines are used in smaller aircraft which have lesser seating capacity. Turbofan engines the high bypass turbofan engines are what are used in most of the large class jet engines and aircraft like some of the larger aircraft which carry about 300 400 passengers or even more all of them use the turbofan engines. And turbo shaft engines are used in helicopters and of course, they have both civil as well as military applications. And then we shall also be discussing about ramjet engines which are very simple form of a jet engine. In fact, it is even simpler than a turbo jet engine, but it is not being used for any civil application it is primarily for missiles and probably for some of the futuristic aircraft applications that ramjets might probably be used little later on, but currently it is primarily used in missiles. So, these are some of the topics that we shall be discussing in today's lecture and so, we will begin our discussion with discussion on the turbofan engines. Now, let us understand why we need a turbofan engine in the first place. Now, we have seen that in a one of the performance parameters that we can use to evaluate the performance of a jet engine is the propulsion efficiency. You already learnt in great detail about the propeller based engines and we have also seen that the propeller based engines have a very high propulsion efficiency as compared to the pure jet based engines, let us say the turbo jets. So, if you compare the propulsion efficiency of a jet engine a turbo jet engine with that of a propeller based engine, the propulsion efficiency is much higher for a propeller based engine as compared to turbo jet engines. So, if the propulsion efficiency is higher this also contributes to the overall efficiency of the engine. So, this is one of the parameters which we shall make use of in trying to modify the jet engine the basic jet engine. So, that we have a better overall efficiency as well. So, what we shall be doing is that if we can increase the propulsion efficiency of the jet engine in some way, then it means that we will also have a higher overall efficiency. So, in order to do that we need to increase the front stages of the compressor in terms of the diameter we have a larger diameter and therefore, the mass flow rate associated with that fan will also be higher, but it will exhaust at a velocity which is lower than that of the core engine and therefore, the propulsion efficiency can be increased. So, turbo fan engines are characterized by the presence of huge fans which are placed ahead of the compressor. So, if you have had a chance to take a closer look at the aircraft engine which is used in most of the commercial civil aircraft, you might have seen that the first component that you see is this huge fan and that is the main difference between a turbo fan engine and a turbo jet engine. And so this is something which you could see either if you are in an airport waiting for the plane or if you are boarding an aircraft then you probably will get a chance to see the engine in a little bit detail or at least the next time you board an aircraft make sure you can have a view of the engine. You will notice that there are these huge fans and the basic reason that these fans are present is that firstly they of course, give you a higher mass flow rate and we have seen that thrust is a direct function of mass flow rate and the other reason is that it will also have an effective lower exhaust velocity and therefore, the propulsion efficiency can be higher which means that turbo fan engines in general can have higher overall efficiency. But besides this turbo fan engines also have a higher thermal efficiency which we shall see little later, but the only issue is that because they have larger diameters the overall drag of the aircraft can be higher. But this can be compensated by the fact that turbo fan engines can generate larger thrust because of the presence of the fan which generates a lot of mass flow rate through the engine which is high in fact, higher than that of the core mass flow. So, if you take a look at the turbo fan engine characteristics. So, the first two points is what I was mentioning about that propulsion efficiency is one aspect which turbo fan engines will have higher than turbo jet engines because the effective exhaust velocity is lower and we can do that by increasing the fan diameter. And so, in a turbo fan engine a large diameter fan is used ahead of the compressor it generates a mass flow which is higher than that of the core mass flow. And so, this ratio of the fan mass flow to the core mass flow is called the bypass ratio. So, this bypass ratio for modern turbo fan engines can be as high as 6 or even 7 which means that the mass flow which is which is being generated by the fan is 6 times greater than the mass flow which is passing through the core engine. So, 6 times the mass flow actually goes through the bypass duct. And therefore, your overall thrust can be higher because the mass flow is much higher than what you get in the core which also contributes to thrust. At the same time you also have a lower effective exhaust velocity and hence an increased propulsion efficiency. So, turbo fan engines will in general have propulsion efficiencies which are higher than that of the conventional turbo jet engines. Now, let us take a look at a schematic of a turbo fan engine. We have already seen that for a turbo jet engine. Let us now see how we can analyze a turbo fan engine and what are the different types of turbo fan engines which are in existence. So, a schematic of a turbo fan engine is shown here. And what is shown here is a schematic of an unmixed turbo fan engine. Unmixed turbo fan engine means that there are two separate exhausts there is one exhaust from the fan and the other exhaust is the core engine that is the primary nozzle. So, let me explain what are the different components of a turbo fan engine. So, turbo fan engine like the turbo jet has a diffuser and immediately following the diffuser is a fan. And so, you can see that the fan diameter is much higher than that of the compressor. So, this is the compressor fan diameter is much higher than that of the compressor. And then downstream of the compressor we have the combustion chamber which is where the primary heat addition takes place. Then we have a turbine which could be in multiple stages we could have different stages of turbines. Then after the turbine we have the primary nozzle after the fan you might notice that there is a secondary nozzle. So, you could have exhaust velocities from the secondary nozzle as well as from the primary nozzle both of them contribute to the thrust. And if we were to carry out a cycle analysis we have already seen we normally follow a certain convention for station numbering. So, for the core engine it is pretty much the same as what we had done for the turbo jet engine that we have stations A for the free stream, one for diffuser inlet, two for the compressor inlet, three for compressor outlet, four for the turbine inlet, five for turbine exit, six is the nozzle entry, seven is nozzle exit. For the bypass duct we shall denote these numbers with a prime. So, 2 prime refers to the fan inlet, 3 prime refers to the fan exhaust which means 3 prime and 2 for the core stream are the same. And the secondary nozzle exhaust is 7 prime. So, these are different components of a turbo fan engine which is of an unmixed configuration that is these two streams do not mix within the engine valves. Now, it is also possible to have a turbo fan engine where these two streams mix and that is known as a mixed turbo fan engine which is what I shall show you in the next diagram. This is a mixed turbo fan engine where the core mass flow mixes with the cold mass flow before exhausting through a single nozzle. So, here there is only a single nozzle and therefore, this is very similar to that of a turbo jet engine and such engines usually have a lower bypass ratio as compared to the high bypass which I had shown earlier. So, such engines are also sometimes refer to as low bypass turbo jets because these bypass ratios are very small and they are mixed the mixing is done for other reasons as well for military advantages to reduce the infrared signatures and so on. So, these this is a schematic of a mixed turbo fan and so here the analysis would be very similar to what we had seen for a turbo jet engine just that we have to calculate the properties in the mixture and which is basically done by an enthalpy balance of these two streams. So, these are the different components of a turbo fan engine in two of its configurations. Now, there are multiple ways of operating a turbo fan engine. You could have a turbo fan engine with a single spool that is the compressor, the fan and the turbines are mounted on the same shaft, but that is a little unrealistic or an inefficient way of operating a turbo fan because if you mount all of them on the same shaft it means that all the stages have to run at the same speed. So, there are operational issues with regard to running the fan and compressor at the same speed. So, normally it is practiced to split the different turbo machines on and mount them on separate spools or shafts. So, you could have a turbo fan with two spools or a twin spool turbo fan where usually we have the fan and the low pressure compressor that is the initial stages of the compressor which is attached to later stages of the turbine and that is known as a low pressure turbine and the later stages of compressor which are known as high pressure compressor that is driven by the initial stages of the turbine which is the high pressure turbine. So, this is a twin spool configuration. You could also have a three spool configuration where the fan is driven by the LPT that is low pressure turbine, the LPC that is low pressure compressor is driven by an intermediate pressure turbine or IPT and the HPC that is high pressure compressor is driven by the high pressure turbine. So, these are different configurations of a turbo fan and these all these three configurations could either be operating in a mixed mode or an unmixed mode. So, you can see that there are different varieties of turbo fan engines or configurations possible and cycle analysis for each of them though the basic philosophy is the same, they could be slightly different depending upon which configuration of the engine is being analyzed at the moment. So, what we shall do is to analyze the turbo fan engine, the cycle analysis for an ideal turbo fan engine for one of these configurations and the others can be carried out in a very similar fashion and we shall only discuss that in a very short duration for that time. So, let us take a look at one of the configurations of an ideal turbo fan. Now, the different processes which involve, which are involved in an unmixed turbo fan cycle constitute the first process that is A to 1 which is air intake from the far upstream, which involves air from the far upstream brought to the air intake with some acceleration or deceleration. Well, you could also have intakes where there is either acceleration or deceleration depending upon the operating condition. We will not discuss that in detail here it is out of the scope of the current syllabus and the second process is 1 to 2 dash or 1 to 2 prime, where the air is decelerated as it passes through the diffuser. 2 prime to 3 prime is the compression process in the fan and 2 to 3 is the compression process in the compressor, which could be either an axial compressor or centrifugal. Most of the cases it is indeed an axial compressor which is used. 3 to 4 is the combustion process of the core engine, air is burned or heated in a combustion chamber. 4 to 5 is air is expanded in a turbine to obtain power to drive the compressor. So, this process that is 4 to 5 could happen in different stages depending upon the number of stages that are involved. 5 to 6 is there could be an after burner. Normally, in a turbofan it is unusual to have an after burner, but there have been engines which have low bypass ratios and still have after burners. And 6 to 7 is the nozzle, the primary nozzle where air is expanded through the primary nozzle. 3 prime to 7 prime is the air in the bypass duct is accelerated and expanded through a secondary nozzle. So, secondary nozzle is the nozzle which is present in the cold stream and does not take part in the combustion. Whereas, the primary nozzle is the one which passes the air or combustion products which is coming from the turbine after the combustion and then it gets expanded through the nozzle. So, there could be 2 distinct nozzles or you could have a single nozzle where which is what would be in a mixed turbofan where both these streams are mixed. So, you could have turbofan engines which are operating in different modes and what we shall do is take up one of these for our cycle analysis. And let us say we take up one case where we have the turbine, there are 2 stages of turbine a twin spool turbofan where the fan is driven by the one of the stages of the turbine and the compressor is driven by the another stage of a turbine. So, let us see the turbine consists of 2 stages, one stage drives the fan, other stage drives the compressor. We shall do a cycle analysis for this case and the other cases we will discuss in little bit detail. So, depending upon what configuration the turbofan is operating on, the cycle analysis can be slightly different depending upon what configuration it is. We will carry out an ideal cycle analysis for an unmixed twin spool turbofan engine. Then of course, there are also mixed versions of the same engine. So, like we did for the turbojet engine, we shall be we shall be carrying out cycle analysis by considering the components one by one. And then based on some of those parameters which are known a priori like the ambient conditions and the Mach number are fixed, the compressor pressure ratio, the fan pressure ratio and the turbine inlet temperature are also design parameters which are fixed or known a priori. So, based on these known parameters we can now carry out the cycle analysis. So, let us take up the first component that is the intake. So, for an intake we should be knowing the ambient pressure, the temperature and the Mach number and therefore, P a which is the ambient pressure, static pressure T a and the Mach number are fixed. Now, based on this the intake exit conditions which is T 0 2 prime is equal to T a into 1 plus gamma minus 1 by 2 m square. So, this is following the isentropic relations because the compression in the intake is assumed to be isentropic for an ideal cycle. T 0 2 prime is the stagnation pressure at the intake exit which is equal to P a into T 0 2 prime divided by T a raise to gamma by gamma minus 1 which is again as a consequence of the isentropic nature of this process. The second process or the component which follows the intake is the fan and for the fan we know that the fan pressure ratio is a design parameter which is known and therefore, pi f which is the fan pressure ratio is equal to P 0 3 prime divided by P 0 2 prime where P 0 3 is the fan exit stagnation pressure. So, therefore, P 0 3 prime that is fan exit stagnation pressure is equal to pi f fan pressure ratio into P 0 2 prime P 0 2 prime has already been calculated. The fan exit stagnation temperature is T 0 3 prime which is equal to T 0 2 prime into pi f raise to gamma minus 1 by gamma where pi f is as we know is the fan pressure ratio. So, these are the first two components of a turbofan engine the intake and the fan. So, both these processes are isentropic for an ideal cycle and in fact the next process which we shall see that is the compressor will also be an isentropic compression process and so for the compressor the cycle analysis which involves isentropic relation is similar to what we did for the turbo jet. So, for the compressor the known parameter is the compressor pressure ratio which is fixed. Now, the compressor inlet pressure is known stagnation pressure is known because a fan exit stagnation pressure should be equal to the compressor inlet stagnation pressure and so P 0 3 prime should be equal to P 0 2 which is the inlet stagnation pressure of the compressor. Similarly, the stagnation temperatures so based on that since we know the compressor pressure ratio we can find out the compressor exit conditions the temperature and pressure and just the same way as we did for the turbo jet engine. So, for the compressor we have P 0 3 which is the compressor exit pressure stagnation pressure is equal to pi c into P 0 2 where P 0 2 should be equal to P 0 3 prime and P 0 3 prime is the fan exit stagnation pressure. Similarly, T 0 3 is equal to T 0 2 into pi c raise to gamma minus 1 by gamma here T 0 2 that is the compressor inlet stagnation temperature should be equal to fan exit stagnation temperature that is T 0 3 prime. So, these are the compressor exit conditions the stagnation temperature and the pressure and now we move on to the combustion chamber. Now, combustion chamber analysis is something you are already familiar with we did that in the turbo jet. What we do is to carry out an energy balance across the combustion chamber. So, at the inlet we have conditions of pressure and temperature coming from the compressor then fuel is added in the combustion chamber and the we find out the exit conditions. So, H 0 4 is the stagnation enthalpy at the combustion chamber exit there should be equal to stagnation enthalpy at the inlet which is H 0 3 plus the fuel flow rate that is f into q r which is the heat of reaction of that particular fuel. So, from this by simplification because we are assuming air to be ideal gas here with constant specific heats. So, H 0 4 should be equal to C p into T 0 4 which is equal to C p T 0 3 plus f into q r. So, from this we can find out the fuel to air ratio. So, that will on simplification we get the fuel to air ratio as T 0 4 by T 0 3 minus 1 divided by q r divided by C p into T 0 3 minus T 0 4 by T 0 3. So, from this we can find out the fuel to air ratio. So, the fuel to air ratio we can find because we have the conditions at the inlet and the conditions at the outlet of the combustion chamber because the turbine inlet temperature that is T 0 4 as I already mentioned is a design parameter and so that should be known a priori. So, once that is fixed and inlet conditions are obtained from the compressor analysis we get the inlet stagnation pressure and temperature. So, we can find the exit conditions and in an ideal cycle we assume that there is no pressure loss occurring in the combustion chamber therefore, P 0 4 should be equal to P 0 3. So, now we have the properties up to the combustion chamber inlet or combustion chamber outlet or the turbine inlet. Now, in the turbine the analysis is now going to be slightly different from what we did for the turbojet because here we have a twin spool engine where there are two stages of turbines one driving the compressor and the other driving the fan. Now, the first stages of the turbine first few stages or at least it could be one or two stages is known as the high pressure turbine because that is the turbine stage which is operating at the highest temperature and that is therefore, known as the high pressure turbine. High pressure turbine is going to drive the compressor because compressor requires more work than a fan and the later stages of the turbine which is known as the low pressure turbine or LPT will be driving the fan because fan requires lesser work than these compressor as a whole. So, we have to now split the compressor well the turbine stages into two and so, the high pressure turbine drives the compressor low pressure turbine drives the fan. So, if you have to do that now depending upon the type of a configuration we are going to use we have already assumed that the engine is going to have two spools. Now, if the engine has two spools the fan will be driven by the low pressure turbine and the compressor is driven by the high pressure turbine. So, from how do we calculate the properties across this the work done by LPT that is low pressure turbine should be equal to the fan work and work done by the high pressure turbine which is HPT should be equal to the compressor work. So, we are going to carry out a work balance between the fan and the low pressure turbine and then the compressor and the high pressure turbine. Now, since initial stages are the high pressure turbine and we know the turbine inlet temperature which is the high pressure turbine inlet temperature we will use that for carrying out the work balance. Now, so for the high pressure turbine which is what is the first component after the combustion chamber the work done by the high pressure turbine should be equal to the work required for the compressor. So, if we equate the two we have m dot t which is mass flow rate through the turbine which is the mass flow rate coming in from the compressor plus the fuel flow rate multiplied by C p into T 0 4 minus T 0 5 prime. So, here T 0 5 prime is the temperature at the high pressure turbine exit at HPT exit this should be equal to m dot a h which is the mass flow coming in from the compressor the subscript h denotes hot exhaust or the hot core of the engine multiplied by C p into T 0 3 minus T 0 2. Now, from this we can find out the HP turbine stagnation temperature at the exit of the HP turbine. So, T 0 5 prime is equal to T 0 4 minus T 0 3 minus T 0 2 by 1 plus f. Similarly, P 0 5 prime is equal to P 0 4 into T 0 5 prime by T 0 4 raise to gamma by gamma minus 1 where P 0 4 is equal to P 0 3 for an ideal cycle there is no pressure loss in the combustion chamber. So, the stagnation pressures are the same. Now, after the high pressure turbine we now move on to the low pressure turbine here the low pressure turbine drives the fan and therefore, m dot t remains the same as that for the high pressure turbine m dot t into C p multiplied by T 0 5 prime which is the inlet stagnation temperature of L p t minus T 0 5 which is exit stagnation temperature of the L p t and this is equal to m dot ac which is mass flow rate required for the fan into C p into T 0 3 prime minus T 0 2 prime. So, here we have the exit stagnation temperature exit stagnation temperature which is T 0 5 that is what we need to find out. So, therefore, if you simplify this we get since m dot t is equal to m dot a h plus m dot f. So, if we divide throughout by m dot a h we get 1 plus f into T 0 5 prime minus T 0 5 is equal to b which is the bypass ratio into T 0 3 prime minus T 0 2 where b is equal to m dot ac by m dot a h that is the cold mass flow divided by the hot mass flow. So, from this we can simplify and find out the exit stagnation temperature of the low pressure turbine which is T 0 5. T 0 5 should be equal to T 0 5 prime minus b into that is bypass ratio into T 0 3 prime minus T 0 2 prime divided by 1 plus f and similarly the stagnation pressure which is T 0 5 which is equal to T 0 5 prime into the temperature ratio T 0 5 divided by T 0 5 prime raise to gamma by gamma minus 1. So, we have now carried out the analysis across the low pressure turbine. We already carried that out for the high pressure turbine and so at the LP turbine exit we now have properties which are basically corresponding to the turbine exit. Now, from the turbine exit we will now move on to the nozzle and the nozzle exit conditions that is the exhaust velocity can be determined which in turn we can use to find out part of the thrust which is generated by the hot nozzle or the primary nozzle. Similarly, we will also find out the thrust which is contributed by the secondary nozzle that is the bypass duct and so the total thrust will be equal to the sum of these two for this case that is the unmixed turbofan case. The mixed turbofan case of course, there will be only a single exhaust which would be a mixture of the bypass mass flow with the core mass flow. So, moving on to the primary nozzle the analysis is identical to what we did for the turbojet and so we have the stagnation temperature and pressure at the turbine exit and in an ideal cycle there are no losses occurring in that duct subsequently therefore, these temperatures and pressures will be the same as that of the nozzle entry. So, at the nozzle entry if we do not have any after burner we have T 0 6 is equal to T 0 5 and P 0 6 is equal to P 0 5. So, the nozzle exit kinetic energy will be equal to u e which is the same as we did for the turbojet is equal to square root of 2 C P T 0 6 into 1 minus P a by P 0 6 raise to gamma minus 1 by gamma. So, this is the exhaust velocity from the primary nozzle. Now, for the secondary nozzle we have u e f where f corresponds to that of fan we can derive an equation for the exhaust velocity from the fan from the secondary nozzle in the same way as we did for the primary nozzle. So, u e f is equal to square root of 2 C P T 0 3 prime which is the nozzle inlet stagnation temperature into 1 minus P a by P 0 3 prime raise to gamma minus 1 by gamma. So, now we have the exhaust velocities from both the nozzles primary nozzle as well as the secondary nozzle. So, how do we now calculate the thrust because mass flow rate through both these streams are different they are governed by the bypass ratio. So, now that we know the bypass ratio we can calculate the thrust which is the total thrust which will have two components one is from the core duct and the other is from the bypass duct. So, the thrust that is generated by this unmixed by turbofan engine is equal to the first part of it is the thrust by the core nozzle which is m dot a h which is mass flow rate through the core stream into 1 plus f into u e minus u. And the second term that is the thrust generated by the bypass duct will be equal to the bypass ratio into m dot a h which is basically equal to m dot a c into u e f minus u which is equal to the velocity differences through the bypass duct. Of course, there will also be a pressure thrust term if it is to be assumed negligible then this is the total thrust which is developed by this turbofan engine the ideal turbofan where the two streams are unmixed. Now similarly, we can calculate the specific fuel consumption or the thrust specific fuel consumption and the efficiencies in the same way as we did for the turbo jet case. And once we calculate thrust the other parameters are straight forward. Now if the turbofan consists of a mixed configuration that is the two nozzles mix and there is only a single exhaust nozzle if it is a mixed turbofan engine then there is only one nozzle which generates the thrust. So, in this case the difference in the cycle analysis comes only at the nozzle entry how do you find the temperature of the nozzle entry which is not equal to the turbine exit temperature because there is a bypass duct mass flow which mixes with the turbine exhaust mass flow. So, what we need to do there is to carry out an energy balance between the two streams which is the bypass stream and the turbine exhaust stream both of them mix then you have a final mixture enthalpy. So, final mixture enthalpy that is m dot into h 0 6 should be equal to m dot ac into the temperature and that there is enthalpy of the cold stream plus the mass flow rate of the hot stream multiplied by the enthalpy of the turbine exhaust. So, from that you can find out the nozzle entry stagnation temperature and therefore, thrust because the nozzle exhaust velocity is a function of the temperature and the pressure ratios. So, if the configuration is a mixed turbofan then we can find out these parameters by carrying out this energy balance. Now, similarly there could be other configurations of turbofan engines which are possible like you could have different number of spools we have now carried out the analysis for a two spool configuration. You could have three spool configuration or you could have the same configuration with mixed or unmixed exhaust and so on. So, the analysis the process procedure for analysis is very much identical to what we have already discussed and so I will leave it as an exercise for you to carry out an ideal cycle analysis for a turbofan with three spools that is three spool configuration with the low pressure turbine driving the fan and intermediate pressure turbine driving the LPC and the high pressure turbine driving the HPC and you could have two types of configurations either mixed or unmixed configurations. So, I leave it as an exercise for you to carry out a cycle analysis for these two cases. There are two cases here that is it is a three spool engine and one configuration is a mixed turbofan the other is an unmixed turbofan. So, based on the discussion we had you should be able to now carry out the cycle analysis for these two configurations of a turbofan engine. So, we have now discussed cycle analysis for turbofan engines and as I mentioned there are different types of turbofan engines that you could come up with and cycle analysis will be slightly different depending upon what type of configuration it is. Now, the next type of engine that we shall be discussing there are two set types of engines together we shall discuss these are the turboprop engines and the turboshaft engines. Now, these are two types of engines which are slightly different from the turbojet and turbofan in the sense that these engines also develop a certain shaft power to either drive a propeller in the case of a turboprop or you could be driving a main rotor blade as in the case of a turboshaft engine, but the difference between a turboprop and the turboshaft is that in a turboprop there is also thrust generated by the nozzle exhaust which is absent in the case of a turboshaft. Turboshaft necessarily generates only shaft power to drive the main rotor blade there is hardly any nozzle exhaust thrust which is developed by the turboshaft engine. So, in a turboprop or turboshaft engine in a turboshaft prop engine generates a substantial shaft power in addition to the nozzle thrust and turboshaft engines on the other hand generates only shaft power and turboshaft as I mentioned in the beginning they are used in helicopters and the shaft power is used to drive the main rotor blade. In a turboprop engines that the advantages and limitations are primarily to that of the propeller in the sense that the advantage of a turboprop engine is it is high propulsion efficiency. The limitation is that the presence of the propeller leads to limited speeds that the propeller based engine can operate and so the limitations as well as advantages are both on account of primarily the propeller and so both these types of engines turboprops as well as turboshaft engines are primarily limited to relatively lower speeds because at higher speeds the compressibility effects come into picture and that could lead to increase in losses because of compressibility effects and shocks which could occur at the tip of the propeller or the blades and so their applications are limited to relatively lower speed but they are still not very low speed but relatively lower than a turbofan or a turbojet engine. Now if you let us take a look at a schematic of a typical turboprop engine now so what we have here is a turboprop schematic of that so some of the components you are already familiar with which were discussed for the turbofan and turbojet engines. We have a compressor then a combustion chamber and turbines so I have shown two types of turbines of stages of turbines one is the compressor turbine which drives the compressor and we also have a power turbine which is meant to drive the propeller. So this is the propeller here and this is driven by the power turbine in a turboprop and since the propeller speeds are limited the rotational speeds are limited the invariably there is a gearbox which reduces the speed of the shaft from the turbine so that the propeller can run at relatively lower speeds. So, gearbox is also a constituent of this arrangement and you may also have a propeller pitch control mechanism for changing the pitch which you had already seen in some of the earlier lectures and there is also a nozzle at the exhaust that is generating some amount of thrust. So, the power turbine exhaust through a nozzle which also generates a certain amount of thrust. So, these are the different components of a turboprop engine and it is very similar for a turbo shaft engine also instead of a propeller we have the main rotor blade and the nozzle component of the thrust is very limited it is negligible in a turbo shaft engine other components are more or less the same. Now in both these engines turboprops as well as turbo shaft engines they usually have a free turbine or a power turbine which is used to drive either the propeller in the case of turboprops or the main rotor blade in the case of turbo shafts. So, the speeds at which the propeller can rotate is basically limited by as I mentioned compressibility effects and stress limitations and. So, they have to have a speed reducer so that the speed at which propeller rotates can be limited and. So, in the case of turboprop I mentioned there are two components of thrust the thrust because of the propeller itself and the thrust on account of the nozzle exhaust and in the case of turbo shaft it is just the primary the thrust or the entire power of the engine is used for driving the main rotor blade. So, the cycle analysis of both these engines are in most of the components being identical to that of the turbo jets or the turbo fan cycle analysis remains more or less the same except for the free turbine or the power turbine which is used for driving the propeller or the rotor blade. So, that is the only component where there could be a slight difference between the cycle analysis. So, cycle analysis for the turbo shaft is very identical to what we had discussed only thing is here the power output is the shaft power and there is no thrust generated by the nozzle in turboprops thrust comprises of two components the propeller thrust and the nozzle thrust the therefore, the total thrust will be equal to some of the nozzle and the propeller thrust. So, if you look at the power turbine I mentioned that is the major difference between a turboprop or a turbo shaft and the other forms of engines. So, this is the power turbine that is shown here. So, 5 is the main turbine or the compressor turbine exist exit 6 is the power turbine and 7 is the nozzle exit. So, if you look at this process on an H s diagram it is an expansion process. So, an ideal cycle the expansion process is isentropic. So, stagnation temperature at 5 is T 0 5 and pressure is P 0 5. So, it expands all the way to P 0 6 in the power turbine and then rest of the expansion takes place in the nozzle. So, 0 6 to 7 which is the ambient pressure is the nozzle expansion. So, let us say this total enthalpy drop across the power turbine and the nozzle together is delta H a fraction of that enthalpy drop occurs in the power turbine which is let us say alpha. So, alpha times delta H is the enthalpy drop which occurs in the power turbine. So, 1 minus alpha into delta H is the power is the enthalpy drop in which takes place in the nozzle. So, delta H is basically the enthalpy drop in the both the power turbine as well as the exhaust nozzle put together fraction of that delta H which is alpha would be used by the turbine and remainder is used by the nozzle. So, the propeller thrust power which primarily comes from the power turbine let us say is the thrust power thrust by the propeller is tau to subscript P R which is the propeller. So, thrust power is this multiplied by u. So, thrust power is alpha delta H which is fraction of the total delta H which is occurring across the turbine power turbine plus the nozzle multiplied by the mass flow rate. So, propeller thrust will be equal to alpha delta H into m dot divided by u and. So, the next component of that is the nozzle thrust. So, exhaust nozzle thrust is tau n. So, tau n is mass flow rate times the velocity differences. So, here velocity difference you can find by the same way as we did for the turbo jet and turbo fans which is square root of 2 into the enthalpy drop multiplied by the fraction which is 1 minus alpha. So, u e is equal to square root of 2 into 1 minus alpha into delta H. So, 1 minus alpha into delta H is the fraction of the enthalpy drop which occurs in the nozzle. So, therefore, the total thrust which is equal to thrust generated by the propeller plus the thrust generated by the nozzle which is equal to alpha delta H into m dot divided by u plus m dot into square root of 2 into 1 minus alpha delta H minus u. So, here I have not described the step by step cycle analysis because it is identical to what we have done for turbo jets and turbo fans and all you have to do is to find out the temperature at the power turbine inlet and therefore, across the power turbine you could determine the enthalpy drop and which in turn gives you the propeller thrust. And once you know that the power turbine exhaust conditions are known, you can also find out the nozzle exhaust conditions and therefore, the nozzle thrust can also be determined. So, the total thrust will be equal to sum of these two thrust components part of which is coming from the turbine, part of it is coming from the propeller and rest of the thrust generation is taking place across the nozzle. And so these delta H that we have talking about the total enthalpy drop should be coming from the cycle analysis of the preceding components that is starting from the compressor, the combustion chamber and the compressor turbine. So, from there we can determine the enthalpy drop that will be required for these two components together and that in turn gives us the total thrust which is developed by the turbo jet turbo prop engines. And obviously, I want to calculate the thrust we can also determine the fuel consumption and the efficiencies. We already discussed the efficiency definition for turbo props wherein we also have the shaft power which needs to be taken into account while calculating the efficiencies. So, this is the general procedure for carrying out the cycle analysis for the turbo props. It is identical for a turbo shaft as well just that in a turbo shaft you do not need to really calculate the nozzle thrust because there is hardly any nozzle thrust developed in a turbo shaft engine. Now, the last cycle analysis which we shall discuss is that for a cycle which is a simplest form of an everything engine which is a ramjet engine. We will again not going to details of the cycle analysis because we have already discussed that. In fact, a ramjet cycle analysis should be much simpler than what we have discussed for all the other engines because ramjet is the simplest form of an everything engine. And some of the components we have discussed like compressor and turbine is actually absent in a ramjet engine. So, a ramjet engine consists of three basic components it consists of a diffuser or an intake followed by a combustion chamber and the nozzle. So, there is no compressor there is no turbine as well. So, you may wonder if there is no compressor and the turbine how is it that this engine is going to operate. So, in a ramjet engine the basic idea is that if the engine itself is flying at very high speed then the compression across the intake itself leads to substantial increase in the stagnation pressure and temperature to the extent that would not need to have a compressor anymore because the intake or the diffuser itself is compressing the flow to such an extent that the pressures and temperatures are high enough that you can have combustion taking place at that kind of temperatures and pressures. So, once you do not have a compressor you no longer need a turbine to drive the compressor. So, once the compressor is not required you would not need a turbine as well. So, exhaust of the intake goes into the combustion chamber and from the combustion chamber the combustion products are expanded in nozzle to generate thrust. Now, the only drawback that is there for a ramjet is that this can work only if the ramjet itself is operating at a certain speed that is ramjet will start generating pressure ratios only if the ramjet is operating at a certain speed which means that ramjet engines cannot really generate any static thrust that is if a ramjet is stationary it is not going to start on its own. Ramjet has to be taken to a certain reasonably high Mach numbers or speeds before which ramjet engines can start developing or generating thrust. So, once the ramjet reaches a certain power or certain Mach number then the compression in the intake is high enough that you can have a stable combustion in the combustion chamber and so the ramjet can start operating. So, let us take a look at some of the salient features of a ramjet. Ramjet as I mentioned is one of the simplest forms of everything engines consists of a diffuser combustion chamber and nozzle only the new compressors or turbines and obviously ramjets are most efficient when they are operated at supersonic speeds and in fact the peak efficiency of a ramjet occurs around Mach number of 2 to 0.5 and again beyond that Mach number the ramjet efficiency also starts dropping. So, when we have already seen that when air is decelerated from a very high Mach number to very low subsonic Mach number this leads to a substantial increase in the pressure and temperature. Now, if you are already having such a high increase in pressure and temperature without the need of using a compressor then you do not necessarily need a compressor and therefore, no turbines and that is why ramjet is considered the simplest form of everything engine because you do not have any turbo machinery present there you do not have a compressor you do not have a turbine. So, there are no rotating components there and so since there are no rotating components in a ramjet you can actually have very high cycle temperatures in the cycle as compared to a turbojet where these temperatures are limited the turbine inlet temperatures are limited because the turbine blades which are rotating cannot withstand very high temperatures and so you have limited temperatures there. In a ramjet you do not have any such limitation there are no rotating components and so you can actually go for higher temperatures in a ramjet which in some sense means that thermodynamically ramjets may essentially have a higher efficiency because you are adding heat at a higher temperature and so that is one of the advantages that ramjets have you can actually operate it at higher temperature. So, let us take a look at a schematic of a ramjet engine and see look at what are the different components of a ramjet engine. So, ramjet engine primarily consists of as I mentioned three components the diffuser the combustion chamber and the nozzle. So, ramjet engine begins with a diffuser it requires a very efficient diffuser because it is going to operate at supersonic speeds which means that deceleration in a ramjet occurs through shock waves. So, if you look at the schematic that is shown here we have the diffuser part of it then fuel is added in the combustion chamber and then at the exit of the combustion chamber we have a nozzle. Now, I have indicated that diffusion or diffuser consists of or the compression process consists of two distinct processes the supersonic compression and the subsonic compression. So, this is a supersonic compression part of it where there could be shock waves there would be shock waves for decelerating it from supersonic speeds to subsonic speeds and at the end of the supersonic compression we have a normal shock which takes it to subsonic speeds and the subsonic flow enters the combustion chamber fuel is injected here in the combustion chamber and there are flame holders for ensuring that there is a stable flame in the combustion chamber and the combustion products are expanded in the nozzle. So, these are the different components or that constitute a ramjet and so if you look at the T s diagram of the same it is the basic Brayton cycle which we have seen process A to 2 is the isentropic compression in the intake 2 to 4 is the combustion process which is at constant pressure 4 to 7 is the isentropic expansion through the nozzle. So, this is the basic Brayton cycle that is shown there ideal cycle of the ramjet obviously is a simpler version of what we have seen for turbo jets and turbo fans and since there are no compressors and turbines the analysis is a lot simpler and so as I have already mentioned since ramjets depend upon the ram compression and the intake without the use of compressors they cannot generate any static thrust ramjets have to be taken to a reasonably high Mach number before which ramjets begin to generate thrust and so they cannot really start from 0 Mach number and start operating they have to be flown to a certain Mach number before which ramjets begin to operate. So, I will not discuss the ramjet cycle analysis in detail because it is a lot simpler and something we have already discussed for lot of other types of cycles. So, let us take a relook at what we have discussed in today's lecture we were discussing about different forms of turbo fan engines how do you carry out cycle analysis for these turbo fan engines the variations of the turbo fan engines like different spool configurations the mixed and unmixed forms of turbo fan engines we also discussed a little bit about turbo prop and turbo shaft engines and what are the different forms of the thrust in the case of turbo prop it is the nozzle thrust plus the plus the propeller thrust and so on and towards the end we also had a very quick discussion on ramjet engines and what makes ramjet engines different from other forms of engines. Now, so this completes the cycle analysis discussion which we have had over the last two lectures and in the next lecture what we shall do is to try and solve a few problems pertaining to ideal cycles of everything engines we will try and solve problems from turbo fans turbo jets and turbo prop engines and this we shall take up during our next lecture.