 Hello and welcome to lecture 18 of this lecture series on Introduction to Aerospace Propulsion. Starting with the last lecture that was lecture 17, we had started discussions on application of thermodynamic principles to different engineering cycles. And we had discussed about three different cycles in the last lecture, which were the auto cycle, diesel cycle and the dual cycle. And we shall continue our discussion on different power cycles in today's lecture. And what we shall be discussing are two cycles, we will start our discussion with on two cycles, which are having efficiencies close to that of or equal to that of the Carnot cycle or Carnot efficiency. These are the Sterling and the Ericsson cycles. We shall then discuss about the Brayton cycle, which forms the ideal cycle for gas turbine engines. We shall discuss some variants of the Brayton cycle, that is Brayton cycle with regeneration, Brayton cycle with intercooling and reheating and regeneration. And then we will have some discussion on the one of the vapor power cycles, which is known as the Rankine cycle. Rankine cycle is the basic thermodynamic cycle for steam engines or steam power cycles. Now, let us start discussion on two of conceptual cycles, which are supposed to have a very high efficiencies. Now, we already know that the ideal auto and diesel cycles are not totally reversible. They are internally reversible, but it is possible to have irreversibilities outside the cycle, which means that ideal auto and diesel cycles are not totally reversible. This means that the efficiency of the ideal auto and diesel cycles can never be equal to the Carnot efficiencies, because they are not totally reversible. And so, if a cycle has to have efficiencies, which are close to that of a Carnot cycle efficiency, then they need to have, they need to be both internally as well as externally reversible. And for a cycle to become externally reversible, it is important that the temperature difference between the source and the sink, that is heat transfer takes place at a differential temperature D T. And this does not happen in the auto and diesel cycles, because we have seen that heat transfer takes place through a temperature of well through a heat transfer rate of Q in and heat transfer out at Q out. And therefore, as long as there is this temperature differential, the cycle becomes internally, it is still internally reversible, but not externally reversible. So, in order that any cycle becomes totally reversible, it is necessary that heat transfer takes place through a temperature difference, which is which does not exceed a certain differential amount of D T. So, how do you do this? How do you transfer temperature through a very small differential and still be able to have very high efficiencies? So, Sterling and Ericsson cycles are two cycles, which have isothermal heat addition. So, isothermal heat addition is something, which is there in a Carnot cycle, where you have a reversible isothermal heat addition as well as heat rejection. Sterling and Ericsson cycle comprises of isothermal heat addition and heat rejection through a certain process, which is known as regeneration. So, let us understand what is meant by regeneration. So, regeneration that is is a process, which is common to both Sterling and Ericsson cycles. Regeneration is basically a process, which involves heat transfer through a to a thermal energy storage device, which is a regenerator during one part of the cycle and transferred back, the energy is transferred back to the cycle during another part of the cycle. So, it is just conceptually shown in this diagram here. So, here we have a cycle, which consists of a certain working fluid, which has a certain energy associated with that and there is a thermal energy storage device, which is a regenerator, which is placed at the center here. So, during one part of the cycle, we transfer energy to this device and then this energy device transfers energy back to the working fluid during another part of the cycle. So, energy is transferred to the fluid during one part of the cycle and transferred back to the working fluid during another part of the cycle. So, regeneration is something, which is common to both Sterling as well as Ericsson cycle and this is why it is possible that Sterling and Ericsson cycles can have isothermal heat addition and isothermal heat rejection basically, because these cycles basically transfer heat or using what is known as a regenerator. So, regeneration process or a regenerator is something, which is common to both Sterling as well as Ericsson cycles. So, regeneration is something, which we have just understood that it is basically involves transfer of energy to a thermal storage during one part of the cycle and transfer that energy back to the working fluid during another part of the cycle. So, with this in mind that is regeneration enables a cycle to have isothermal heat addition and isothermal heat rejection, which is required if you want to have efficiencies, which are close to that of Carnot cycle efficiency. So, isothermal heat addition and isothermal heat rejection is an important aspect of a cycle, if it has to achieve Carnot efficiencies. So, let us look at what are the different processes that constitute a Sterling cycle. A Sterling cycle basically consists of 4 processes, which are totally reversible. Process 1 to 2 is isothermal process, T is a constant. So, we have an isothermal expansion process during which heat is added or heat addition takes place from an external source. The second process is a constant volume regeneration process, where there is internal heat transfer from the working fluid to the regenerator. The third process that is process 3, 4 is again constant temperature it is an isothermal process, compression process during which heat is rejected to an external sink. And the last process is a constant volume regeneration process during which there is an internal heat transfer from the regenerator back to the working fluid. So, these are the 4 different processes, which constitute a Sterling cycle that is there are 2 isothermal processes and 2 regeneration processes. So, on a PV and the TS scale or we have the Sterling cycle represented in the following way. So it starts with an isothermal process T is equal to constant, process 1 to 2 is isothermal process and process 2, 3 is constant volume regeneration process. And it is during process 1, 2 that is isothermal process that you have heat transfer from an external source to the cycle. Process 2, 3 is a constant volume process, it is a regeneration process during which energy is transferred from the cycle or from the working fluid to the regenerator which is a thermal energy storage device. Process 3, 4 is an isothermal heat rejection process during which heat is rejected from the cycle. Process 4, 1 is again a regeneration process during which heat is transferred back to the fluid from the thermal energy storage device. On a TS diagram it the Sterling cycle looks like this, first process is isothermal process. So, you have a horizontal line on the TS diagram, because it is isothermal heat addition. Process 2, 3 is constant volume process during which regeneration takes place that is energy is transferred to the to the regenerator during this process. Process 3, 4 is isothermal heat rejection process. Process 4, 1 is again a constant volume process, it is a regeneration process during which heat is transferred for our energy is transferred from the regenerator to the working fluid. And so, this is Sterling cycle as seen in PV and the TS diagrams. And so, there is another cycle which is very similar to a Sterling cycle and also has 2 processes which consist of regeneration processes. This is known as an Ericsson cycle. An Ericsson cycle consists of again 4 totally reversible processes. Here again you have an isothermal heat addition process wherein there is heat addition from an external source. Then there is a constant pressure heat regeneration. Unlike a Sterling cycle which had constant volume regeneration, Ericsson cycle has a constant pressure regeneration process. Then there is again an isothermal heat rejection through a compression process and finally, a constant pressure regeneration process wherein there is internal heat transfer from the regenerator back to the working fluid. So, the difference between the Ericsson and the Sterling cycle is only in the regeneration process. In the Sterling cycle, the regeneration was constant volume regeneration. In the Sterling cycle, well in the Ericsson cycle, it is constant pressure regeneration. The heat addition is isothermal in both the cycles Sterling as well as the Ericsson cycles. So, let us look at the PV and TS diagrams of the cycles. The PV diagram on the PV coordinates, we have the first process that is an isothermal process 1, 2 and the second process that is basically a constant pressure process which is a regeneration process 2, 3 is regeneration. Process 3, 4 is again isothermal heat rejection and process 4, 1 is constant pressure regeneration back to the that is heat transfer from the regenerator to the working fluid at constant pressure. So, processes 1 to 2 and 3, 4 are isothermal processes. Process 2 and 2 to 3 and 4 to 1 are constant pressure processes which are basically regeneration processes. On TS coordinates, the same cycle looks like this. We have isothermal heat addition, isothermal heat rejection and constant pressure regeneration during process 2, 3 and process 4, 1. So, this is Ericsson cycle on PV and TS diagrams. So, both these engines are totally reversible cycles because they have they are anyway internally reversible like auto and diesel cycles. They are also externally reversible because of the heat addition and heat rejection processes that are isothermal. That is heat addition and heat rejection are taking place at a constant temperature which means that they are also externally reversible. And so, since sterling and Ericsson cycles are totally reversible cycles, their efficiencies will be equal to that of the Carnot efficiencies. So, both sterling and Ericsson cycles have efficiencies equal to that of Carnot cycle because they are totally reversible. Now, there have been attempts to build sterling and Ericsson cycles, but they are extremely difficult to build and they are very cumbersome in nature. And therefore, though these cycles have been demonstrated in some way, of course there are some irreversibilities which will still take place, but these cycles did have very high efficiencies higher than the normal auto and diesel cycle efficiencies, but then they are very extremely cumbersome to build. And obviously, they still cannot achieve Carnot efficiencies, but the main point of the main aspect of sterling and Ericsson cycles is that regeneration increases efficiency. That is if you have a cycle which has some component of regeneration, it basically increases the efficiency of the cycle. And this is one aspect that has been implemented in many modern day cycles for improving the efficiency, which would mean that regeneration within the cycle will always improve efficiency and that is one of the aspects or one of the understandings that we get from sterling and Ericsson cycle analysis. And that is one aspect which is used in many of the modern day cycles. And so we will not discuss further on how sterling and Ericsson cycles can be implemented and so on. Let us take up another important very important cycle and at least as aerospace engineers, this is one cycle which we should be understanding in much detail and we shall be carrying out detailed analysis of the cycle lot more. This is known as the Brayton cycle. So, Brayton cycle forms the basic cycle, thermodynamic cycle for gas turbine engines, though it was developed initially for use in reciprocating engine. It was developed or proposed by George Brayton in 1870 basically for use in reciprocating engines, but modern day gas turbine engines operate on Brayton cycle but and work with primarily rotating machinery. Gas turbine engines operate in an open cycle mode and but we can model it as closed cycle using the air standard assumptions which we had discussed in the last lecture. And using these assumptions we can assume that the gas turbine cycle operate in an open though it operates in open cycle mode, we can approximate that as being operating in closed cycle mode. So, the in a gas turbine engine we will replace the combustion and exhaust by constant pressure heat addition and heat rejection. So, Brayton cycle basically consists of 4 internally reversible processes and process 1, 2 is isentropic compression. It is basically compression in a compressor in the case of a gas turbine. Process 2, 3 is constant pressure heat addition. Process 3, 4 is isentropic expansion and in the case of a gas turbine engine it is in a turbine and process 4, 1 is constant pressure heat rejection. So, these are the 4 different processes all of them are internally reversible which constitute a Brayton cycle. So, on PV and TS coordinates the Brayton cycle looks like this. We have the first process which is a compression process, process 1, 2 is isentropic compression and the process 2, 3 is constant pressure heat addition. So, heat addition at constant pressure. So, Q n takes place during process 2, 3. Process 3, 4 is again isentropic process it is an isentropic expansion process and process 4, 1 is constant pressure heat rejection process or Q out taking place during process 4, 1. On TS diagram or TS coordinates the Brayton cycle looks like this. The first process is isentropic. So, we have a vertical line here. Process 1, 2 is isentropic. Process 2, 3 is constant pressure process during which heat addition takes place. Process 3, 4 is isentropic expansion again a vertical line. Process 4, 1 is a constant pressure heat rejection. So, these are the 4 different cycles which constitute or 4 different processes which constitute a Brayton cycle. So, process there are 2 isentropic processes the compression and expansion processes are isentropic and we also have 2 constant pressure processes during which heat addition and heat rejection takes place. So, as we did in the case of auto and diesel cycles we will now derive an expression for finding the efficiency of a Brayton cycle in terms of certain pressure ratios which we will define little later. So, to derive an expression for efficiency we will primarily be doing an energy balance. We can assume the different constituents of the Brayton cycle to be steady flow systems and for steady flow process the energy balance is basically Q in minus Q out plus W in minus W out is equal to delta H and that is if you were to calculate the heat transfer to and from the working fluid Q in is equal to H 3 minus H 2 because it is a flow process or constant pressure heat addition. Therefore, it is equal to C p times T 3 minus T 2 Q out again is constant pressure heat rejection and therefore, that is H 4 minus H 1 that is C p times T 4 minus T 1. Therefore, we can determine the thermal efficiency of the ideal Brayton cycle and thermal efficiency is W net net work out wood by heat input that is 1 minus Q out by Q in which is 1 minus T 4 minus T 1 by T 3 minus T 2 which is equal to 1 minus T 1 multiplied by T 4 by T 1 minus 1 divided by T 2 multiplied by T 3 by T 2 minus 1. We know that processes 1 2 and 3 4 are isentropic and process 2 to 3 and 4 to 1 are constant pressure processes. Therefore, P 2 is equal to P 3 and P 4 is equal to P 1. Therefore, from isent because these process 1 2 and 3 4 are isentropic we have T 1 by T 2 is equal to P 2 by P 1 raise to gamma minus 1 by gamma which is equal to P 3 by P 4 raise to gamma minus 1 by gamma because P 2 is equal to P 3 and P 4 is equal to P 1 which is in turn equal to T 3 by T 4. So, if you substitute these values in the equations we get an expression for the thermal efficiency. So, thermal efficiency of a Brayton cycle is equal to 1 minus 1 by R subscript P raise to gamma minus 1 by gamma where R P is basically the pressure ratio P 2 by P 1 and that is basically the pressure ratio of this cycle and so Brayton cycle efficiency is primarily a function of the cycle pressure ratio and also it depends to in some way to the ratio of specific heats as well. So, Brayton cycle efficiency is the expression at least is very similar to that of an auto cycle. In an auto cycle as well we have seen the efficiency was equal to 1 minus 1 by R raise to gamma minus 1 by gamma where R was the compression ratio. In the case of a Brayton cycle the efficiency is 1 minus 1 by R P raise to gamma minus 1 by gamma where R P is the cycle pressure ratio the max pressure by the mean pressure and so the efficiency of the Brayton cycle is primarily a function of the pressure ratio and it also depends on the ratio of specific heats in some way and which was also the case for an auto cycle as well. And so this was the Brayton cycle the simple Brayton cycle as we know it now which primarily consists of these four different processes two isentropic processes an isentropic compression and an isentropic expansion and two constant pressure processes that is one isentropic well constant pressure heat addition and constant pressure heat rejection. So, all these four processes together constitute the Brayton cycle and the simple Brayton cycle forms the basic thermodynamic cycle on which all gas turbine engines operate. Now, we shall now look at some of the modifications that can be done on the Brayton cycle which can primarily increase the efficiency improve the efficiency of the Brayton cycle. So, one such modification which we will discuss is known as the regeneration and regeneration is basically trying to use some of the energy which is getting wasted from the exhaust of the cycle that is we would like to use part of the heat rejected from the cycle and use it in some other part of the cycle. So, regeneration as we have seen for sterling and Ericsson cycle involves transfer of energy from one part of the cycle to another. So, in the case of a gas turbine engine regeneration can be carried out by using hot air exhausting from the turbine to heat up the compressor exit flow that is before the air enters the combustion chamber we could heat up the air using energy from the turbine exhaust. So, thermal efficiency of the Brayton cycle is likely to increase because the some part of the heat rejected is basically being reused that is you are increasing the net work output by reusing part of the heat which is otherwise getting rejected or being wasted. And which means that regeneration in some sense decreases the heat input and therefore, fuel requirements for the same net work output because for generating the same net work output you are using lesser fuel because you are using part of the heat which is getting exhausted and using that energy in some part of the cycle. So, in the case of Brayton cycle regeneration is primarily carried out by using part of the compressor of part of the turbine exhaust which is energy which is primarily getting wasted in some sense that is heat rejected and it is used reused in another part of the cycle that is to heat up the compressor exit flow. And so, regeneration basically the understanding which we had from the Sterling and Ericsson cycle was that regeneration increases or improves efficiency. And so, in the case of a Brayton cycle heat rejected from the cycle is being reused and therefore, regeneration decreases the heat input and therefore, the fuel requirements for the same net work output. So, let us now look at regeneration cycle Brayton cycle with regeneration. Now, this was the basic Brayton cycle we had discussed on TS coordinates process starting from state 1, process 1 2 is an isentropic compression process, process 2 3 is constant pressure heat addition, process 3 4 is isentropic expansion and process 4 is constant pressure heat rejection. So, what happens in regeneration is that as we can see the temperature at state 4 T 4 is much higher than temperature at state 2. So, regeneration is a process wherein you would like to transfer some part of this heat which is being rejected back to the cycle. So, that is from state 4 the ideal scenario would be that we have an increase in temperature which is equal to that of T 4 itself which is obviously, not possible in actual practice. So, you would transfer some heat which is indicated here as Q region which is the heat which has been regenerated or transferred to the cycle. So, Q region which is equal to the Q saved which has been transferred back to the cycle that is at the end of the compression process. So, the heat rejected that is Q out is now equal to what is shown here between 1 and 6. So, heat rejected in the simple Brayton cycle would have been equal to Q 4 minus Q 1 this has now been reduced to Q 6 minus Q 1. And since that much heat has been transferred to the cycle now the heat input required will now be instead of Q 3 minus Q 2 it is now equal to Q 3 minus Q 5. So, we can see that there is a tremendous saving in terms of the heat required that is Q in as well as the decrease in the amount of heat that has been rejected because of heat regeneration on the simple Brayton cycle. So, regeneration primarily involves transfer of energy which in this case has been shown as this part of heat which has been transferred back to the cycle. And so, Q 4 minus Q 6 has been transferred back and therefore, heat input has also reduced significantly because of this. So, which means that ideally you should be able to get higher efficiencies because Q in has now reduced. Now, in an ideal case you could have a scenario where the air actually leaves the regenerator exits the regenerator at the temperature of the exhaust gas which was G 4. If that was the case the maximum regeneration that you could get is H 5 dash minus H 2 which is H 4 minus H 2. But actual regeneration takes place to a temperature less than that which is why the actual regeneration is H 5 minus H 2 that is on the T S diagram ideally you can actually you should be able to get regeneration heat equal to H 5 prime minus H 2 which is equal to H 4 minus H 2 the actual regeneration is less than that being equal to H 5 minus H 2. And so, we can define what is known as effectiveness of a regeneration process or regenerator. So, effectiveness is basically the ratio of the actual regeneration to the maximum regeneration which is H 5 minus H 2 divided by H 4 minus H 2. So, we have already derived an expression for the cycle efficiency thermal efficiency of an ideal Brayton cycle the simple Brayton cycle. Now, with regeneration if we were to derive an expression we would get thermal efficiency as equal to 1 minus T 1 by T 3 times r p raise to gamma minus 1 by gamma. So, here we have thermal efficiency which is not just a function of the pressure ratio it is also a function of the temperature ratio. For the simple Brayton cycle we have seen that the thermal efficiency is only a function of the pressure ratio. And as well as the ratio of specific heats with regeneration now the efficiency is also a function of the temperature ratio which means that as you increase or decrease the temperature ratio it can affect the efficiency of the Brayton cycle with which has a regeneration process or regeneration mechanism. So, efficiency of a Brayton cycle basically can be improved using regeneration because you are reducing the heat input to the cycle. Primarily because you are transferring some amount of heat which has been which is normally being rejected part back to the cycle. Therefore, the amount of heat that is required for generating the same work output will be reduced and which means that there has to be an increase in efficiency of the cycle. And so Brayton cycle with regeneration is one regeneration primarily is one of the mechanisms of improving the network output or the efficiency of a Brayton cycle. Now, there are other methods of increasing the efficiency or work output of a cycle as well. One such method is what is known as intercooling and the other method we shall discuss is reheating. Now, intercooling is something which concerns the compression process of a cycle reheating is something which concerns the expansion or the expansion process part of the cycle. So, the network output of a gas turbine cycle is basically the difference between the turbine work and the compressor work that is turbine work output and the compressor work input which means that the network output can be increased either by decreasing the compressor work or by increasing the turbine work or doing both that is either you reduce the compressor work input or increase the turbine work output or you could do a combination of these two which will eventually lead to an increase in the network output. So, these are two options that we have for increasing the network output of a cycle which means that W net if we increase we can increase the efficiency as well. And so in the first case that is if you wish to reduce the net work input for the compression process we can do that by what is known as intercooling that is we split the compression process into different stages and then between these different stages we apply a certain cooling. So, that was to bring down the net work required for carrying out the certain amount of work compression process. So, the work required for a compressor to compress a gas between two specified pressures can basically be decreased by carrying out the compression process in stages and cooling the gas in between. This is basically known as multi stage compression with cooling that is between do a different compression processes or compression stages we apply cooling of the gas in between them that is known as multi stage compression with intercooling. Similarly, the work output of a cycle of a turbine can be increased by what is known as multi stage expansion with reheating that is we split the expansion process into different stages and apply reheating in between them and so that the net work output of the turbine can be increased. So, in a limiting case as you increase the number of stages of compression and expansion the process basically approaches an isothermal process. So, as you increase the number of stages of compression and expansion the process basically approaches an isothermal process and a combination of intercooling and reheating will basically increase the net work output of a Brayton cycle that is because you are increasing the net work output by using intercooling and reheating. What we shall see a little later on that using either intercooling alone or reheating alone is not sufficient it is also necessary that we use some amount of regeneration as well so that the overall cycle efficiency can be improved. So, intercooling plus reheating with regeneration is one of the methods or mechanisms by which one can increase the efficiency of a Brayton cycle significantly. So, let us look at how intercooling can be carried out at least thermodynamically as we as we look at a P V diagram of a compression process. So, process 1 A C is the compression process without intercooling or in single stage compression. Now, if you were to split this compression process into two stage compression with intercooling then this process becomes 1 A B D that is at the end of process 1 to A we apply a cooling of the gas and ideally we assume that cooling to takes take place at constant pressure which means after the gas reaches state A which is end of the first compressor as we apply cooling of the gas then it reaches state B at constant pressure and from B to D we again apply the compression process which takes it from B to D. Therefore, the amount of work that has been saved as a result of intercooling is given by this shaded area which is given here. So, this is the amount of work that is saved as a result of intercooling. So, process 1 A C can be replaced which is a single stage compression process can be replaced by process 1 A B D which is splitting the compression process into two stage compression process with intercooling. Similarly, we can also look at the reheating we will also be able to see the benefits as we get higher work output by reheating at the end of the expansion process through a turbine. So, let us look at one example of an ideal cycle gas turbine cycle basically with intercooling reheating and regeneration. So, what we have shown here is a two stage compression process and a two stage expansion process. So, the cycle begins with at state 1 and there is an isentropic compression which results in state 2. At the end of state 2 we carry out intercooling which means that the process will now move along this path process 2 3 and then it moves from state 3 to state 4. And the end of process that after it reaches state 4 then it moves along the process path which is a constant pressure process from process 4 to from state 4 to state 6. And what is shown here are the heat inputs and the heat rejection taking place. There is a certain amount of heat rejection taking place here because the gas has been cooled from state 2 to state 3. So, there is a heat rejection which is also taking place during process 2 3. Now, after the gas reaches state 6 which is the beginning of the expansion process 6 to 7 is expansion and 7 to 8 is reheating which means there is heat input here as well. So, Q in takes place during constant pressure process 4 6 as well as during 7 8. 8 to 9 is another expansion process the second expansion process and 9 to 1 is the heat rejection process. Now, this is only with intercooling and reheating. So, intercooling of the compression process reheating during the expansion process. Now, if you also apply regeneration during this process then ideally we should be able to get a temperature which is equal to that of temperatures at 7 and 9. And therefore, the amount of heat that is saved is equal to the difference between 9 and 10 which is the heat which has been regenerated. And so, this much amount of heat has been saved or regenerated during this process. Heat input is during process 5 6 and 7 8 and heat rejection is during process 2 3 and process 10 1. So, this is one example of a Brayton cycle which has intercooling reheating as well as regeneration. Intercooling is during process 2 3 reheating during process 7 8 and regeneration during process 9 10 or 4 5. So, Brayton cycle with all the three intercooling reheating and regeneration is likely to have higher efficiencies than that of the simple Brayton cycle which has no intercooling reheating or regeneration. Now, if we look at the benefits of intercooling or reheating as a standalone component towards the overall efficiency then it is not always the case that you have an improved efficiency. That is intercooling alone or reheating alone normally does not mean that you would have an increase in efficiency if it is not used in conjunction with regeneration that is intercooling and reheating with regeneration can result in increased efficiency, but not just intercooling alone or reheating alone would necessarily lead to increased efficiency of the Brayton cycle. So, network output of a gas turbine cycle will increase which is already known as a result of intercooling and reheating, but for thermal efficiency to improve it is required that there is regeneration taking place because during reheating there is heat input additional heat input that is required. So, it is possible that you may not necessarily get increased thermal efficiency, but if regeneration is used it is possible to basically have increased efficiency of the system. Now, primarily what is happening is that during intercooling process the average temperature at which heat is added decreases and reheating basically tends to increase the temperature at which heat is being rejected. And so basically that can lead to that necessarily does not mean that you get an increased thermal efficiency unless you have regeneration as well applied in the process. Now, if you as I mentioned if you keep increasing the number of stages of intercooling and reheating you can ensure that you have isothermal it is possible that you have compression processes which are if you extend the number of processes at which intercooling and reheating are taking place. Then it is possible for us in an ideal scenario that if I if you have infinite number of compression stages and expansion stages which means Brayton cycle with intercooling reheating and regeneration then Brayton cycle will kind of approach an Ericsson cycle that is you have isothermal heat rejection as well as isothermal heat addition. And if you apply regeneration as well then this is like an Ericsson cycle. Ericsson cycle consists of two isothermal processes and two constant pressure processes. And so Brayton cycle with infinite number of intercooling reheating and regeneration can approach an Ericsson cycle which means that this will have efficiencies which are equal to that of the Carnot efficiency. So, Brayton cycle with several infinite number of intercooling reheating and regeneration stages can cause efficiencies which are equal to that of Carnot cycle efficiencies. So, we have discussed now about the Brayton cycle and which forms the basic cycle for gas turbine engines. And we will now have a very quick look at one of the vapor cycles most commonly used vapor power cycle that is known as a Rankine cycle. Rankine cycle is the ideal cycle for vapor power plants like the steam power plant for example. We will have a very quick look at the Rankine cycle and basically Rankine cycle like the other ideal cycles we have discussed does not involve any internal irreversibilities. And an ideal Rankine cycle consists of four processes an isentropic compression which takes place in the case of steam power plant in a pump. The second process that is process two three it is a constant pressure heat addition which could be in a boiler in a steam power plant. Process three four is an isentropic expansion which is in a in the case of steam power plant in a turbine. Process four one is constant pressure heat rejection which could be in a condenser. So, these are the four different processes which constitute a Rankine cycle an isentropic compression constant pressure heat addition, isentropic expansion and constant pressure heat rejection. So, let us look at a Rankine cycle on one of the coordinates that we have been plotting these cycles with on a T S diagram for example. So, an ideal Rankine cycle on a T S diagram looks like this what is shown by the black line here the black curve here is the saturation line for water that is basically indicating that outside the saturation line water exists as steam that is there is no content of water in steam in this region is basically known as super heated steam. And within this curve you may have water in both these phases water as well as in steam. Now, the process begins at state one state one is an isentropic compression in the case of steam power plant it is through a pump and process two to three is a constant pressure heat addition process. So, constant pressure lines on a saturation curve are shown here. So, this line here corresponds to a constant pressure line. So, process two to three all the way up to this is a constant pressure heat addition process where in as you add heat water moves from its saturation states and it becomes a super heated steam at state three. And process three four is the isentropic expansion process during process three to four there is a network output from the turbine. And process at the end of process the third process that is expansion process we have a constant pressure heat rejection. So, process four to one is a constant pressure heat rejection process. So, these are the four different processes which constitute the steam power cycle that is process one to two which is the work input in the case of compression in the case of steam power plant it is through a pump. So, it is an isentropic compression process two three is the constant pressure process during which heat is added to the cycle process three four is the isentropic expansion process during which there is a network output from the turbine. And process four one is constant pressure heat rejection from the cycle. So, the network output here is the difference between the turbine work output and the pump input. There is a power required for the pump and the difference between turbine work output and the pump input is the net work output. Heat input and difference between Q in and Q out is the net heat transfer during this cycle. So, if you now carry out the energy balance for this particular cycle we again assume that all the components are steady flow and reversible systems internally reversible. The energy balance for each of these systems can be basically assumed expressed as Q in minus Q out plus W in minus W out is delta H. Let us look at each of the components let us say the pump where we have work required for the pump work input is H two minus H one which is basically V times P two minus P one. And for a boiler there is heat input Q in is equal to H three minus H two condenser is heat rejection Q out which is H four minus H one turbine is work output H three minus H four. So, basically the thermal efficiency of the ideal Rankine cycle can then be a determined using the cold air standard assumptions. And so the Rankine cycle efficiency is W net by Q in which is one minus Q out by Q in where W net is equal to Q in minus Q out which is in fact equal to W turbine output minus W pump input. So, these are the different components or constituents of the efficiency we can calculate the net work output and the heat input to determine the thermal efficiency of a Brighton cycle of a Rankine cycle. So, a Rankine cycle efficiency is primarily as we have seen we need to look at the different constituents of the Rankine cycle because you have a work input for the pump work output from the turbine heat input at the boiler end and heat output from the condenser. So, from this we can calculate the net work output and the heat input and this can help us in determining the thermal efficiency of a Rankine cycle. So, Rankine cycle efficiency again depends upon many of these parameters and unlike an auto cycle diesel cycle they are not really able to calculate or express the efficiency in terms of one parameter which could be a cycle pressure ratio or so on. Because it involves multiple phases in the case of vapor power cycles you have different phases of the working fluid that is water which can exist as water in the pump to steam superheated steam in the turbine to saturated steam in the condenser and so on. So, it is not possible for us to basically express the efficiencies in a single term. Now, Rankine cycle can also be operated with reheat and regeneration and also with in fact intercooling by multiple staging. So, Rankine cycle operated with reheat and regeneration obviously can increase the net work output and efficiency substantially. So, average temperature during reheat can be increased by multiple stages of reheat and you can also increase the number of expansion stages during reheat process. So, that you can increase the average temperature during which at which heat is added and therefore, Rankine cycle with reheat and regeneration will have efficiencies which are substantially higher as compared to the simple Rankine cycle. So, I will not be taking up discussion on reheat and regeneration during this particular lecture. I guess I leave it to you as an exercise to look at Rankine cycle with reheat and regeneration. We have already discussed that for the Brayton cycle. So, it should be possible for you to extend that same for a Rankine cycle. So, let us look at what we had discussed in today's lecture. It is we started our discussion in today's lecture by the Sterling and Erickson cycle which are basic ideal cycles which can have efficiencies as high as that of Carnot cycle because they are totally reversible cycles. Brayton cycle is the ideal cycle for gas turbine engines. We had discussed on the Brayton cycle and different variants of the Brayton cycle, methods of improving efficiency, Brayton cycle with regeneration, Brayton cycle with intercooling, reheating and regeneration. Then we also discussed about the Rankine cycle that is the ideal cycle for vapor power cycles. So, in today's lecture, we had discussed a lot about the various power cycles that are used in engineering applications. Some of the power cycles which we discussed today were the Otto cycle, the diesel cycle and the dual cycle as well as the Brayton cycle with a lot of modifications. And then we also discussed in short about the vapor power cycle. What we are going to discuss in the next lecture is a slightly different topic. It is a lot to do with power cycles in any way. It is basically something to do with basic thermodynamic relations. So, we will be discussing a lot about some of the important thermodynamic relations that are used very commonly in thermodynamic analysis. So, let us take a look at what we are going to discuss in the next lecture. In the next lecture, we will have some discussion on what are known as Helmholtz and Gibbs functions. We shall then talk about the Legendre transformations which are basically used on the Gibbs equation, which we will discuss in the next lecture. And then we will also discuss about thermodynamic potentials. We will then discuss a very important set of relations known as the Maxwell relations. Maxwell relations are very basic thermodynamic relations which are very commonly used in the analysis. And we will spend some time on discussing about the Maxwell relations. Then we will talk about the ideal gas equation of state. I guess you are already familiar with the equation of state. But what we should remember is that the equation of state which we know as p v is equal to r t is basically for an ideal gas. And as we have tried to apply it for real gases, there are certain effects that need to be considered. We will then talk about the compressibility factor which is a factor which is applied for real gases. And then we will also discuss about some of the other equations of state like the Van der Weyl's equation etcetera. We will discuss about those equations in little bit of detail. And towards the end of the next lecture, we will be talking about what is known as the Joule-Thompson effect which is basically to do with fluid passing through a throttling device. So, these are some of the topics that we shall be taking up during our next lecture which would be lecture 19.