 So, that concludes our discussion on the principle of increase of entropy. We now come to the last module in the course which is thermodynamic cycles. So, what we will do here is look at the three cycle cycles that we have already considered that is during the definition of heat engines we looked at three different cyclic processes executed by each one of those heat engines. Number one, the Rankine cycle, the basic Rankine cycle, number two, the basic Brayton cycle and number three which is the vapor compression refrigeration cycle executed by the reverse engine. So, we look at two direct heat engines and two I am sorry one reverse heat engine and examine the cyclic process executed by each one of these engines. So, our basic idea would be to look at heat supplied, heat rejected, thermal efficiency and entropy generated as a result of operation of each one of these heat engines. Let us start with the Rankine cycle you may recall this illustration from before. So, basically water is the working substance and the system is what is in the in the loop. So, we have seen this already before this is a heat engine, direct heat engine. Now, let us look at the process undergone by the system in TS coordinates. So, state one as you can see here is at entry to the turbine. So, in the basic Rankine cycle, water is a saturated vapor at the entry to the turbine and if it is an ideal cycle then we expected to execute an isentropic process from 1 to 2 S like this. In the actual case because of internal irreversibilities the state point at the end of the process in the turbine will lie to the right of 2 S. So, the actual process would look something like this. So, this is the turbine 1 to 2 or 1 to 2 S. Now, we go to the condenser where the working substance loses heat and as it loses heat because it is undergoing a change of phase its temperature remains constant. So, the heat rejection process continues until the water is become saturated liquid at the condenser pressure and condenser temperature. Now, 3 to 4 S is the pumping process as we can see here. So, 3 to 4 S is the pumping process if the pump is ideal then we will reach state 4 S from 3. In the actual case due to internal irreversibility we will reach state 4. After this we have the heat addition process to reach state 1. So, this is the heat addition process in the boiler 4 S to 1 or 4 to 1 is the heat addition process in the boiler where heat is added. Now, in the case of the ideal cycle where there are no irreversibilities the amount of heat that is supplied during the heat addition in the boiler looks like this. So, this would be the we first erase this. So, the amount of heat that is added during the heat addition process in the boiler for an ideal cycle would be the area under the process curve. So, that would look something like this. So, that is the, so this is the heat that is added. Now, the heat that is rejected during the heat rejection process in the condenser looks like this. Let us use a different color for that. So, again it is the area under the process curve. So, the net heat that is added during the cycle is the difference between these two and that is this inner area here. Let us denote that either and since the system executes a cyclic process we know that cyclic integral delta Q equal to cyclic integral delta W. Since this area that we have shown in green is the net heat added in the cycle that is also equal to the net work that is generated during the cycle. So, the area enclosed in this in green or shaded in green is actually the net work that is generated by the cycle. Now, in the basic cycle as I said before the working substance leaves the boiler as a saturated vapor. But in the actual cycle it will always leave with a certain amount of super heat and the TS diagram for the super heat cycle, American cycle with super heat looks like this. So, the pressure remains the same. Now, state point one has moved further up on the isobar corresponding to the boiler pressure. So, basically this pressure this corresponds to the boiler pressure and this corresponds to the condenser pressure. So, the cycle operates between these two isobars. Notice that by allowing I said the amount of super heat in the cycle we actually improve the efficiency of the cycle. How do we accomplish that? Now, if you look at the additional amount of heat that has been added in the cycle assuming reversible process you can see that this would be the additional amount of since we used the red to indicate that that is. So, this is the additional amount of heat that we have been able to add as a result of super heating this steam. Now, the additional amount of heat that has been rejected looks like this. Now, notice that the increase in the amount of heat rejected is not commensurate with the increase in the amount of heat that has been supplied. Because the isobar is steep in the superheated region notice that the isobar is steep in the superheated region and horizontal in the phase change region. Remember the pressure remains constant in the phase change region because it is steep as we move along this isobar in the superheated region the amount of heat that we can add increases quite a lot and this usually results in higher efficiency because the amount of heat rejected still is in the two phase region. So, it does not increase that much. Since the heat rejection process remains in the two phase region the amount of heat rejected still does not increase that much because the isobar is horizontal here and the isobar is steep in the superheated region. So, which is the reason why the Rankine cycle is almost always operated with a considerable amount of superheat and in fact, as I had mentioned earlier today's thermal power plants operate in a ultra supercritical cycle where the peak temperature exceeds the critical temperature of this is the critical temperature of water. So, it exceeds even the critical temperature. So, the higher the temperature the higher the efficiency. This was something that we saw when we were discussing second law of thermodynamics when we looked at the best way to increase the efficiency of a Carnot cycle. Although lowering the temperature of the cold reservoir was the best way in reality we mentioned that efficiency gains have been achieved by increasing the temperature of the hot reservoir which means we keep increasing this temperature as much as possible both the pressure as well as temperature in the boiler. Let us look at a simple example involving the Rankine cycle with superheat. So, we have given the boiler pressure 160 bar and condenser temperature is given we can easily look up the the saturation pressure corresponding to this. Steam leaves the boiler with the 212.6 degree Celsius of superheat that means that the temperature above the saturation temperature is 212.6 degree Celsius. Determined heat supplied net power produced thermal efficiency in entropy generator we assume steady state operation and neglect KE and 20 changes. So, we can easily evaluate or retrieve or calculate property values corresponding to each one of the state point in the cycle. State 1 is a superheated state. State 2s most likely will be in the saturated mixture region. State 3 is saturated liquid at 45 degree Celsius. State 4s is usually a compressed liquid. Remember for state 4s because it is a compressed liquid you may recall that for this state S of T comma P is approximated as S f of T the specific entropy the saturated liquid at the same temperature and this has to be evaluated once again by using the factor this is a compressed liquid state. So, H at 4s you may recall that we had dH equal to dH equal to P dP. So, if I apply this to process 3 to 4s. So, if we apply this to process 3 to 4s we get H 4s minus H 3 equal to V 3. We assume that the specific volume remains a constant because it is a liquid and this may be written as P 4s minus V 3. Now P 4s is nothing but the boiler pressure P 3 is the condenser pressure. So, we may evaluate H 4s from this expression. We have to use this expression to calculate H 4s because the temperature at the end of the pumping process 4s is not given. If the temperature at the end of 4s were known then we would have evaluated H 4 H 4s if T and P are both known then we would have evaluated this as UF of T plus V times VF times P. Because the temperature is not given we actually use this expression to evaluate the specific enthalpy at state 4s. So, all these once these values are available it is relatively straight forward to calculate the quantities that are asked. So, heat added in the boiler per unit mass flow rate of steam may be evaluated as H 1 minus H 4s because the boiler is operating at steady state we apply SFEE and we can get this. Heat rejected in the condenser again per unit mass flow rate may be evaluated like this. Turbine work is nothing but H 1 minus H 2s. So, that comes out to be 1407.84 and pump work again may be evaluated as 16.15. You can notice how small the pump work is compared to the work that is generated by the turbine. So, the net power is the difference between the turbine work and the pump work and that comes out to be 1391.69 per unit mass flow rate which is why we have per kilogram here. And thermal efficiency of the cycle is nothing but net work divided by heat supplied and it is 42.68 percent. Now entropy generated in the inverse may be evaluated as delta S system plus delta S surroundings. The system here as I already mentioned is the water that executes the cyclic process in this loop. Since the water executes a cyclic process the entropy change of the system is 0. Entropy change of surroundings is result of heat that is supplied in the boiler and heat that is injected in the condenser to the low temperature reservoir. So, here we assume the high temperature reservoir to be at the highest temperature in the cycle. So, here this is the highest temperature in the cycle. So, we take this to be T H and again we take this to be T C. So, if I actually extend this, so this is my T H. The highest temperature in the cycle is T H and the temperature in the condenser is T C. So, we may now evaluate the entropy change of the surroundings. Since heat is supplied by the boiler there is a reduction in the entropy in the surroundings during this process. Since heat is rejected in the condenser there is an increase in entropy of the surroundings in this process, but the net sum comes out to be plus 1.9632 kilojoule per kg Kelvin. So, there is a net increase in the entropy of the surroundings as a result of entropy generation due to internal. In this case because the system is executing a cyclic process there is no contribution from there. So, this is largely this is all due to external irreversibilities in the heat transfer process. What we will do next is look at an analysis of the Brayton cycle which is also a power producing cycle and also look at the vapor compression refrigeration cycle which is a power absorbing cycle. So, the next cycle that we will look at is the Brayton cycle and you may recall that we discussed this also earlier when we discussed the notion of a heat engine. So, here you may recall that air is the working substance and it undergoes a cyclic process as illustrated here. The cutaway view of a gas turbine engine is shown here. So, you can see the compressor section which is this one and the circumferential combustor may also be seen. So, this has several units placed around the circumference and then you can see the turbine section. So, normally there may be a few more turbine sections or depending upon the power rating you may have only this many. Note that the number of compressor blade rows is actually much more than the number of turbine blade rows because as we said earlier the process that the fluid undergoes in the turbine section is actually an expansion process. So, the pressure gradient is favorable from a fluid mechanics perspective whereas, the process that is undergone by a fluid in a compressor section is actually a compression process which means the flow must actually go against the pressure gradient is adverse and the flow must go against that. So, that is always somewhat more challenging from a fluid mechanics perspective which is why we try to compress in many stages with a small amount of compression in each stage. So, that fluid mechanics of the design can be made much better. So, here only the rotor blades are shown. So, here you see the, so this is the shaft of the rotor and you can see the blades are mounted on the rotor. The stator blades which actually in between these rotor blades are not shown in this exposed view and you get an idea about the size of the blades in the rotor and so on from this figure. So, in the simplest form of the cycle as we mentioned earlier air is taken into a compressor. So, the air actually executes cyclic process in a closed loop. So, the air at entry to the compressor usually is at atmospheric pressure and temperature and then it is compressed to high pressure and relatively high temperature due to the compression process at state 2. Heat is then added in the combustor, fuel is burnt and heat is added in the combustor. Note that when we actually treat this as a heat engine, we treat the combustor as a heat exchanger. So, the air is in an enclosed circuit and heat is added to the air. So, in this manner only we will be able to take the air as a system. In a realistic gas turbine engine as shown here in this combustor normally fuel is sprayed into the air stream. So, the air is actually mixed with the fuel, the fuel is then burned and the air stream then undergoes further expansion in the turbine. So, the actual process in the case of the gas turbine engine is actually not a closed source cyclic process. Air is drawn in from the ambient, compressed, fuel is then injected into the air stream, burned and the combustion gases are expanded through the turbine and expelled into the atmosphere. So, we cannot really treat that as a heat engine. So, what we are looking at here is an idealized concept of a gas turbine engine where the air circulates in a closed loop. So, heat is assumed to be added in the combustor which is modeled as a sort of heat exchanger. So, after addition of heat the air is at high pressure and high temperature before entry to the turbine. It is expanded in the turbine and then it comes out at low pressure and low temperature. So, here instead of exhausting this into the ambient in the case of a heat engine we take this to a cooler or a condenser where heat is rejected to the ambient and the air is now at almost the same temperature as the ambient temperature and at the end of the expansion process here it is almost at the same pressure as the ambient pressure. It can, the air can now repeat the cycle. So, if we illustrate the cyclic process undergone by the air on a TS diagram it looks like this. So, if the cycle is ideal then the compression process in the compressor is ideal and goes from 1 to 2 s, it is isentropic goes from 1 to 2 s. So, heat addition takes place at constant pressure in the combustor. So, we go from state 2 s to state 3. If the expansion process in the turbine is also ideal meaning isentropic then we go from 3 to 4 s and then the heat rejection takes place in the in the cooler along 4 s to 1 that is the ideal cycle. In the actual cycle the compression process tends to be adiabatic but with internal irreversibilities. So, the process actually goes from 1 to 2 in this case and in the case of the turbine also the expansion process can be assumed to be adiabatic but with internal irreversibilities. So, the entropy increases as a result of the irreversibilities and the actual process goes from 3 to 4. Now, in the case of the ideal cycle as we saw before with the Rankine cycle the heat added during the cycle assuming that there is no I am sorry, assuming the process to be ideal the heat added during the cycle is given by the area under this curve. So, we may show this to be like this that is the heat added. Now, heat rejected may be shown on this diagram like this. So, that is the heat rejected and the difference between the two is the net heat and that would be this area here. So, that is the net heat that is supplied during the cycle and from first law for a cyclic process net heat is also equal to net work in the case of the ideal process. So, the area that is shown shaded in pink is actually the net work that is developed during the cycle assuming all the processes to be ideal. So, we are looking at the simplest possible cycles many additional enhancements to the cycle are always carried out to improve the efficiency of the cycle. You will probably learn those things in the next level thermodynamics course the effect of pressure ratio on the performance of the cycle and the effect of the peak temperature in the cycle which occurs at state 3 before entry to turbine. Effect of that on the efficiency of the cycle and the specific work output of the cycle these are very very important performance parameters of the cycle. So, generally for these cycles the performance parameters or efficiency specific work output. So, this is the specific work output that is work output net work output per unit mass of working substance. This allows us to draw inferences on scaling aspects of the cycle. So, if I can generate say a large amount of power with the small amount of working substance that means that the plant size can be small. So, this actually affects the specific work output affects the plant or equipment size. So, that is an important parameter. And of course, entropy generated during the process is also an important quantity and we would like to have an idea on how much entropy is generated as a result of the cyclic operation in the Brayton cycle. We will work out an example and show how these things may be evaluated. So, an ideal Brayton cycle operates with the pressure ratio of 30 and the peak temperature of 1300 Kelvin. Air is the working substance and it enters the compressor at 100 kPa and 300 Kelvin. So, we are asked to determine heat supplied power output thermal efficiency and entropy generated. So, we are given that P1 is 100 kPa, T1 is 300 Kelvin. The pressure ratio or P is given to be 30 and T3 is 1300 Kelvin. Notice that the Brayton cycle also just like the Brayton cycle operates between these two isobars. And the pressure ratio of the cycle is nothing but the higher pressure, compressor pressure or compressor exit pressure divided by compressor inlet pressure that is the pressure ratio of the cycle that is given to be 30. So, assuming compression process to be ideal, we may evaluate T2S as the temperature at the end of an isentropic compression process and we can get that to be 793 Kelvin. Notice that P2S over T1 is equal to 30. Similarly, T4S which is the temperature at the end of an isentropic expansion process in the turbine may be evaluated as 492 Kelvin and again the pressure ratio P4, I am sorry P3 over P4S. So, this is equal to 1 over 30 and the temperature at the end of the expansion process is 492 Kelvin. Now by applying steady flow energy equation to each one of these components in turn, we can evaluate the required quantities. So, compressor work, specific compressor work is nothing but H2S minus H1 and that comes out to be 498 kilojoule per kilogram. Heat added in the combustor per unit air mass flow rate is 512.263 again by application of SFE to the combustor. Specific turbine work output comes out to be 816.387. The biggest difference between the Rankine cycle and the Brayden cycle is the amount of power that is consumed by the compressor versus the amount of power that is consumed by the pump. So, here we can see that the compressor consumes more than half of the work that is or power that is generated by the turbine. In contrast for the Rankine cycle, we saw the pump work to be negligibly small compared to the turbine work. So, that is a big difference and the difference comes as a result of the working substance. Here we use air as the working substance and that is a highly compressible substance. So, the power required to compress air through a pressure ratio of 30 is quite high. On the other hand, in the case of the Rankine cycle, we are pumping water and since water is almost an incompressible liquid, the power required to increase the pressure of water is actually quite small. So, heat rejected may be evaluated as 194 kilojoule per Kelvin. So, the net power that is generated during the cycle comes out to be 318.27 that is WX dot turbine minus WX dot compressor 318.27 kilojoule per kilogram. So, the thermal efficiency of the cycle is nothing but network divided by heat supplied and that is 62.13 percent, which is much higher than the value that we saw for Rankine cycle and this is typical of gas turbine power plants. They have much higher efficiencies compared to Rankine cycles, but the amount of power that can be generated in a Brayton cycle is not as high as what may be generated in the case of a Rankine cycle. Because the Rankine cycle works with water and the enthalpy change is that we are looking at there where of the order of thousands of kilojoule per kilogram. So, the power generated consequently is also much higher in a Rankine cycle. Whereas, here for instance, we are working with air and the enthalpy values that we are looking at are only of the order of about few kilojoule per kilogram. In fact, all this may be evaluated as this quantity itself may be evaluated as Cp times T2S minus T1 because we are assuming the gas to be calorically perfect. So, you may recall that Cp is of the order of about 1 kilojoule per kg Kelvin. So, enthalpy changes that we are talking about are not very high. So, consequently the power that can be generated or specific power that can be generated by gas turbine engines are relatively smaller compared to the Rankine cycles. Now, the overall entropy that is generated during the cycle may be evaluated as delta S system plus delta S surrounding. Here the air that executes a cyclic process in the closed loop is a system. Since it executes a cyclic process delta S for the system is 0 and the change in entropy for the surrounding may be evaluated as minus Q H dot over T H where Q H dot is the amount of heat that is rejected. Amount of heat that is supplied in process 2, 3 and since the surroundings are supplying heat to the system the entropy of the surroundings decreases which is why we see a negative sign here. The entropy change of the surroundings during the heat rejection process is plus Q C dot over T C because heat is being rejected to the surroundings and the algebraic sum of the 2 comes out to be 0.2526. So, delta S system is 0 delta S surroundings the entire change or entropy generation is due to change in surroundings of the entropy in this case and that is positive as it should be. In the case of the Brayden cycle we took T H to be like just like the Rankine cycle the highest temperature in the cycle. So, this we took to be T H and the ambient temperature was taken to be the T C was taken to be equal to the ambient temperature in this case.