 So, in this module, we will discuss thermodynamic cycles and as I mentioned in my previous lecture, we will look at three types of cycles, two of them power producing, one power absorbing. The first one is a steam power cycle, which uses water as the working substance and the second one, second type of cycle that we will look at is the so called air standard cycle and the third one, which is a power absorbing cycle is actually the vapor compression refrigeration cycle. Now, in the category of air standard cycle, we will look at Brayton cycle, we will look at auto cycle, air standard auto cycle and air standard diesel cycle. So, let us begin with the steam power cycle or Rankine cycle. Now, most thermal power plants which employ for example, coal as the fuel utilize the Rankine cycle, modified form of Rankine cycle, but here a fixed quantity of water executes a cyclic process and this qualifies as a heat engine as we discussed in the previous course. Now, what the strategy that we will use while discussing this is as follows, we will start with the most basic form of the cycle as shown here and we will evaluate three performance metrics for this cycle. First one being the specific power, specific power is nothing but WX dot over M dot that is the amount of power developed per unit mass flow rate of the working substance. This is a very important performance metric because it decides or determines the size of the equipment. For instance, let us say we have two designs and both produce the same power, but one requires let us say two times the mass flow rate of the other. So, which means that the size of the equipment should also be bigger in the latter case to handle twice the amount of mass flow rate. So, specific power is very important when it comes to sizing of the power plant. So, we will look at that metric. The second metric that we will look at is the thermal efficiency of the cycle. This was evaluated in the previous course itself. So, this is an energy based efficiency, it is nothing but the ratio of the power produced divided by the rate at which heat is supplied. So, that is the second metric that we will look at. So, this is WX dot over QH dot, that QH dot is the rate at which heat is supplied in the high temperature reservoir or supplied from the high temperature reservoir. The third performance metric that we will look at is the second law efficiency of the cycle. So, after evaluating these three metrics, what we will do is look at the rate of exergy destruction in the individual components in the cycle. This will tell us where the exergy destruction is a maximum and that also indicates where we should actually focus our attention and make improvements in the cycle. So, if you want the performance of the cycle to improve, we should focus on these components where the exergy destruction rate is a maximum and then see how we can overcome this. So, we will propose means by which this can be done. So, we will evaluate rate of exergy destruction in individual components and we will propose means by which this is usually addressed and re-evaluate these performance metrics to see whether improvement has taken place or not. So, we will proceed in a systematic manner and address each one of the deficiencies and then come up with the final form of the cycle which is probably somewhat close to the actual cycle that is employed in practical power plants. So, we will close this module by looking at the layout of a typical 1 gigawatt power station and then identify components there which are similar or same as what we have discussed in this cycle. So, at the end of the module by looking at the actual or the cycle of the actual power plant, you should be able to realize that whatever we have discussed carries over completely to actual applications and you may recall that this was the objective of the course. So, we look at the actual application then you should be able to relate each one of the components there to what we have discussed here and be able to do the analysis of these types of cycles should you encounter them at any time in the future. Now Rankine cycle is not only used in coal based thermal power plants, even nuclear power plants also in the secondary loop use Rankine cycle because that uses again water as a working substance. So, this is very general and let us start with the basic cycle here. So, the basic cycle let us say we begin looking at the cycle from state 1. So, state 1 is entry to the turbine. So, basically this has 4 components a boiler where heat is added turbine where power is generated condenser where heat is rejected to the ambient pump where the liquid water is pumped to the boiler pressure. So, let us look at the cycle on TS diagram and then go back to the block diagram. So, state 1 as you can see is at the boiler pressure. So, the cycle basically operates between 2 isobars one is the boiler pressure and the other one is the condenser pressure. So, at the entry to the turbine we have in the basic Rankine cycle we have saturated vapor at the boiler pressure it expands in the turbine. If the expansion is isentropic then we reach state 2S. On the other hand in a realistic situation where the expansion is not isentropic there is going to be some internal irreversibilities. So, in that case we will reach state 2 either way we reach the condenser pressure and if the water is a saturated vapor at entry to the turbine then 2S and 2 both actually will be in the saturated mixture region. So, from here heat rejection takes place in the condenser as you can see here 2 to 3 is heat rejection in the condenser. So, 1 to 2 or 1 to 2S is expansion in the turbine and 2 to 3 is heat rejection in the condenser and then the water which is now a saturated liquid as it leaves the condenser is now pumped to the boiler pressure by means of a pump as you can see here. So, this is saturated liquid and state 2 as we have discussed is a saturated mixture this one in the case of a basic cycle this is saturated vapor. So, the saturated liquid is pumped in the pump to a boiler pressure again if the pump is isentropic we reach state 4S otherwise we reach state 4 as a result of internal irreversibilities. So, the compressed liquid state so, when the water enters the boiler it is in a compressed liquid state so, heat is added here until it becomes so, heat is added at constant pressure until we reach state 1. Now, for the sake of simplicity and without any loss of generality we will assume the isentropic efficiency of the turbine and the pump or indeed any other turbines are out pumps to be 100 percent there is no loss of generality in doing so, you can always repeat the examples that we will work out using realistic values for the isentropic efficiency. What we wish to bring out through this work examples is the effect of the improvements that we are suggesting on the overall performance of the cycle. So, isentropic efficiency being 100 percent does not affect any of these conclusions in any manner so, there is no loss of generality. And as you can see because these processes are reversible you can see that the let me just erase these things so, the heat added in the cycle is nothing but the area under the heat addition process curve so, let us just show it like this so, the heat addition process curve looks like this so, this is the heat added during the cycle. Now, the heat rejection process is 2 S to 3 and again it is a reversible process so, the heat rejected during the cycle may be illustrated like this. So, this area that you are seeing here this area that you are seeing here is Q H minus Q C so, this area is Q H minus Q C the area shown in pink and since the working substance executes a cyclic process you know from first law that this is the net work that is produced during the cycle. Of course, when we illustrate processes on a TS diagram we are showing the cycle so, everything is on a per cycle basis when we actually do the calculation since it is a steady flow process everything will be on a per unit mass flow rate basis. So, let us look at an example involving basic Rankine cycle so, an ideal basic Rankine cycle operates between a boiler pressure of 160 bar and a condensate temperature of 45 degree Celsius. Determine heat supplied net power produced thermal efficiency and second law efficiency also calculate the rate of exergy destruction in each of the components assumes steady operation neglect KE and PE changes and also assume that heat is supplied from a reservoir maintained at the highest temperature in the cycle and heat is rejected to a reservoir maintained at the condenser temperature these are important for us to evaluate rate of exergy destruction in the individual components. So, basically we can look up these values based on the states that we have mentioned so, at the entry as I said before it is a saturated vapor and at state 2S remember this is located by showing by using the fact that S 2S is equal to S 1. So, this is a saturated mixture and state 3 is a saturated liquid and this is a compressed liquid state. So, heat supplied may be evaluated by applying the steady flow energy equation to the boiler and on a per unit mass basis it comes out to be 2375.63 kilo joule per kilogram heat rejected again on a per unit mass flow rate basis comes out to be 1465 what produced by the turbine is 926.25 kilo joule per kilogram what supplied to the pump is very small compared to work developed by the turbine. So, this is only 16.15 kilo joule per kilogram. So, net power generator which is W X dot turbine minus W X dot pump is 910.1 kilowatt per unit mass flow rate of steam. So, thermal efficiency of the cycle use this definition we get it to be 38.31 percent and as I mentioned pump work is only about 2 percent of the work produced by the turbines negligibly small. Now, one important performance parameter that is of relevance in in in Dragon cycle is the average temperature of heat addition and the average temperature of heat addition is evaluated using this expression basically what we are saying is the following let us show the cycle on a T S diagram. So, if this is the heat addition process 4 S to 1 then this entire area as we showed earlier is the Q H. So, we can actually evaluate the average temperature by taking this area dividing it by the entropy change S 1 minus S 4 S. So, what that will give us is an equivalent heat addition process at a constant temperature for instance. So, when we do this we may get an equivalent process like this between the same entropy limits and it may look something like this. So, it is between the same entropy limits. So, we will probably have something like this. So, this is the equivalent process. So, this temperature here is the equivalent high temperature T H prime. So, T H prime is Q H dot divided by M dot divided by S 1 minus S 4 S and that comes out to be 243 degree Celsius. Notice that the saturation temperature corresponding to 160 bar boiler pressure is 347.4 degree Celsius. So, the average temperature as we have indicated here is considerably less than the saturation temperature corresponding to the boiler pressure. Of course, in the basic Rankine cycle this is also the maximum temperature. So, which means that in the boiler when we add heat we assume that the boiler is maintained at this temperature as heat is added to the working substance. So, now we go on to evaluate the second law efficiency. So, rate at which exergy is supplied. So, if you look at the cycle you can see that exergy is supplied in the boiler and exergy is recovered in the I am sorry exergy is supplied in the boiler and the pump and exergy is recovered in the exergy is recovered in the turbine and condenser. So, rate at which exergy is supplied as I said is the pump and heat supplied in the boiler and this may be notice that we assume that the source is maintained at a temperature which is equal to T H which is nothing but. So, this is T H. So, this is the highest temperature in the cycle. So, that comes out to be 1250 rate at which exergy is recovered in the turbine and in the condenser that comes out to be 101 H. So, the second law efficiency is 81.43 percent which is reasonably high. Now, let us look at the rate of exergy destruction in the individual components. Now, we have assumed both the turbine and the pump to be isentropic which means that there is no internal or external reversibility. So, the rate of exergy destruction in the turbine and pump power 0. Again as I said there is no loss of generality in assuming these to be isentropic. Now, rate of exergy destruction in the boiler may be worked out in using our familiar expression in this manner. So, that comes out to be 231.64. Now, in the condenser because the process is reversible isothermal. The exergy destruction in the condenser is 0. So, exergy destruction in this case occurs entirely in the boiler. And so that is where the scope for further improvement is. And this exergy destruction in the boiler as we discussed earlier is entirely because of the large temperature difference across which heat is transferred to the fluid. So, if you look at the cycle, you can see that the fluid is at a temperature which is considerably less than the peak temperature at which the reservoir is maintained. So, there is a large temperature difference between the reservoir and the fluid when it enters the boiler. And this difference diminishes as the fluid approaches the saturation temperature here, but the difference is very large nonetheless and that is what causes this large exergy destruction. So, this actually demonstrates the usefulness of the notion of exergy. You may recall that this was primarily the reason why we developed the notion of exergy so that we can look at individual components, identify those with poor performance for improvement. Now, a few points are worth noting about the basic Rankine cycle. The basic Rankine cycle is discussed here for academic purposes only. It is not practically useful because the expansion in the turbine takes place entirely in the two phase region. And although it is possible to construct very specialized turbines which can handle saturated mixtures, it is generally not preferred because the complexity is there. Number one, number two, the erosion of the turbine blades as a result of handling liquid droplets and vapor is also high. So, life is also not very long for such turbines. So, it is preferable to have the expansion occur in the superheated region. And at the end of the expansion, we would ideally like to have a dryness fraction around 0.9 or so or higher than that, that would be even better. So, that is one very important point. So, we would like to move these points further closer to the saturated line better even here. So, as you can see, the only way we can do this is to actually allow the fluid to become superheated at the exit to the boiler. By allowing some amount of superheat, we can shift these two data points, I am sorry these two state points towards the saturated vapor line. So, that is a strategy that we are going to use next. The efficiency of the cycle can be improved by increasing the average temperature at which the heat is added. Now, average temperature at which heat is added can be accomplished in two different ways. One is to simply increase the boiler pressure, at least for the basic Rankine cycle. So, if you consider the basic Rankine cycle, so if you consider the basic Rankine cycle, the average temperature at which heat is added may be increased. Number one, by increasing the boiler pressure, which means this will operate at a higher boiler pressure. So, forest will now move to this location and if you want it to be a saturated vapor as in the basic Rankine cycle, this will be the state at entry to the turbine. Usually, what is done is we would prefer one to move like this and this should become like this. So, this is operating at a higher boiler pressure. So, that state one is superheated at entry to the turbine. That is one way of increasing the average temperature in the cycle. The other way of increasing the average temperature in the cycle is to increase the degree of superheat. So, we keep the boiler pressure the same, but we increase the degree of superheat. So, we move this along this isobar. So, we keep the boiler pressure the same and at exit to the boiler or at entry to the turbine, we have a superheated vapor. So, this also increases the average temperature at which heat is added. So, that is th prime. So, in this sequence of examples, progressive examples that we are going to look at, we will keep the boiler pressure and condenser pressure the same and evaluate what superheat does to the cycle. Now, in reality today, both these strategies are widely used namely increasing the boiler pressure, increasing the degree of superheat or widely used. In fact, the boiler pressure of most modern installations today actually are higher than the critical pressure of water. So, these are supercritical and ultra supercritical steam power units. So, the critical pressure for water is 221 bar. So, most modern plans today operate at pressures of 300 bars or even above that 300 would be supercritical. If it is even above 300, then it would be ultra supercritical. So, the efficiency is can be quite high or efficiency of these units can be quite high. So, as I demonstrated earlier, adding superheat while keeping boiler pressure the same has this effect on the cycle. So, state point 1 is moved from here to here and immediately it becomes apparent that the heat rejected has increased by this amount, but the heat added has increased by a larger amount. So, the heat added has increased by, so this is the increase in the heat rejected and this is the increase in the amount of heat added because the isobar is steeper in the superheated region. The increase in the heat added is much more and consequently, you can see that increase in the work produced by this in this cycle is also much higher. So, this is also the increase in work produced and the increase in heat rejected is smaller because we are still operating within the two phase region where the isobar is horizontal. Whereas, the increase in heat added is much higher because the isobar is steeper in the superheated vapor region. So, consequently W net is more and increase in heat rejected is less. So, we expect this cycle to have higher thermal efficiency and let us see if that comes out to be the case. So, state 1 is now superheated as you can see here. So, we have added 212.6 degrees of superheat. So, previously as if you recall T H was 347.4 degree Celsius that was the saturation temperature. So, this temperature is 347.4 degree Celsius. This temperature now we have added another 212.6 degrees of degree Celsius of superheat. So, this state is a superheated state 2S still is a saturated mixture and 3 is saturated liquid same as before and 4S is compressed liquid same as before. So, these two states remain same as before 1 and 2S have changed now. So, we repeat the calculations as we did before. So, rated which heat is supplied is now higher as you can see 3, 2, 6, 0 compared with 2, 3, 7, 5. Heat rejected is 1869.14 now compared to 1465. So, you can see that heat rejected has not increased by that much. So, what produced in the turbine is equal to 1407 kilojoule per kilogram power required by the pump is the same. So, the net power produced in the cycle is 1391.69. So, 1391.69 compared to 910. So, you can see that there is a considerable increase in the specific power in the cycle as we anticipated. So, the thermal efficiency of the cycle works out to be 42.68, but only an increase of 4 percent which is actually quite high in practical terms, but it seems small, but it is actually quite considerable. Now, the dryness fraction at the exit of the turbine has increased from 0.61 to 0.78, but still it is much less than the value of 0.9. 0.9 is the lowest value that is generally acceptable. So, the specific power has increased as a result of super heat efficiency has increased somewhat. Now, let us look at second law efficiency. Average temperature at which heat is added has increased now, because understandably, because we have super heat also now. So, rate at which XIG is supplied is 2110.44 kilojoule. Notice that we have assumed the reservoir to be at a temperature of 560 degree Celsius. So, as we indicated here 347.4 plus 212.6. So, that is 560 degree Celsius. So, the reservoir is maintained at 560 degree Celsius and heat is supplied from that reservoir. So, rate at which XIG is recovered comes out to be like this and second law efficiency is 72.3 percent. So, you can see that thermal efficiency has increased slightly, but second law efficiency was 81.43 that has now decreased to 72.3. So, that has decreased considerably and let us see why that is. So, rate of XIG destruction in the turbine and pump are 0 as we discussed earlier and rate of XIG destruction in the condenser is also 0, because it is still operating in the saturated mixture region. Now, rate of XIG destruction in the boiler has now come out to be 584.15 compared to 231. So, you can see that it has doubled in the rate of XIG destruction in the boiler has doubled as a result of super heat, because the temperature difference across which heat transfer takes place has become even higher now, because the reservoir is being maintained at a higher temperature. So, how do we actually reduce the rate of XIG destruction in the boiler? So, this is now more than twice the value that we calculated earlier. So, the XIG destruction here as I said is primarily because of the large temperature difference between the source and the working substance.