 So, before we proceed with the work example, let me just illustrate alternative method for using the air table and that looks like this. So, we mentioned that for an isotropic process, we can calculate the reduced pressure using the actual pressure ratio. So, by using the fact that PR2S over PR1 equal to P2 over P1, we computed the value for PR2S and then with this value for PR2S, we went into the table to retrieve the value for H2S. This can be done in a slightly different manner without using the reduced pressure itself as follows. So, for any process, for example, for a process 1 to 2S, I can write the entropy at state 2S as S0 at T2S minus r natural log P2S over PRF from the definition of specific entropy itself. And this is equal to S1 because we are saying 1 to 2S is an isotropic process. Therefore, I can write S0 of T2S is equal to S2S plus r natural log P2S over PRF. S2S is equal to S1. So, the value is known. We looked up the value from the table that is 1.70203. So, we can actually evaluate S0 of T2S using this expression. So, with this value for S0, we can now go to the table and retrieve the enthalpy value that we are looking for. So, S0 of T2S is 2.36287. So, if I go to this table, 2.36287 would fall right here. So, that is the value that we are looking for 2.36287. I am sorry 2.36287 would actually fall here. 2.36287 will fall here. And the previous procedure where we used PR2S, PR2S was 13.86. So, let us see where that comes. 13.86, notice that this also falls in the same in the same between the same two entries in the table. So, both the procedures will give you identical values. So, depending upon which one you prefer, you can either use S0 of T2S to look up enthalpy or calculate reduced pressure and then do it that way. Both are okay. Now, let us go to the worked example. So, we have an ideal Brayton cycle, which means isentropic efficiency of the components is 100 percent. It operates with the pressure ratio of 30, peak temperature 1300 Kelvin. So, air is the working substance. It enters the compressor at 100 kPa, 300 Kelvin. And we are asked to calculate heat supplied, power output, thermal efficiency and the second law efficiency and calculate the rate of exergy destruction in the components. Ambient temperature is given to be 300 Kelvin. So, we proceed in a methodical manner. So, what we do is the following. So, we start with state 1, look up the values that are required for state 1 and then walk through the cycle. So, here we have listed the properties at each state. The quantity that is shown in italics is the quantity that is used to enter the table. And the quantities that are retrieved from the table are shown in bold. So, the quantity used to enter the table is shown in italics and the quantities that are retrieved from the table are shown in bold. So, let us see how this proceeds. State 1, temperature is given to be 300 Kelvin. So, we go to the table corresponding to 300 Kelvin. So, corresponding to 300 Kelvin, we retrieve the specific enthalpy S0 at 300 Kelvin and pr. So, these three values are retrieved and S itself the specific entropy at this state is evaluated using the familiar expression S of t comma p equal to S0 of t minus r natural log p over p ref. Because p it is at this state the pressure is the same as the reference pressure S comes out to be just equal to S0 at that temperature. Now, we move on to state 2 S and the pressure ratio is given to be 30. So, that means the pressure at state 2 S is 3000 kilopascals. And since p 2 S over p 1 is equal to so remember 1 2 S is an isentropic process. So, p 2 S over p 1 is equal to pr 2 S over pr 1. So, that means I can evaluate pr 2 S to be 41.58. So, we go to the table with this value 41.58 for pr 41.58 would fall between these two entries. So, we interpolate for h and we interpolate for S0 from these two entries. So, let us see I am sorry S0 is actually not required for this case because it is an isentropic process we know the value of S itself. So, the h falls between these two entries here. So, let us go here. So, we look up the value for h like this. And since S 2 S is equal to S1 entropy value is known. Now, we move on to state 3 remember 2 S to 3 is an isobaric process. So, the pressure is known and the temperature is given to be 1300 Kelvin. So, we use this value to enter the table and retrieve quantities from there. So, 1300 Kelvin. So, we pick up the value for h from here and S0 from here and pr from here. So, h3. So, this is h3 this is S0 at 3 I am sorry S0 at state 3 and this is pr at state 3. So, h3 S0 at state 3 and pr at state 3 are all known. And we can evaluate S at state 3 using this formula S0 is known pressure p is 3000 kilo Pascal, p ref is of course, 100 kilo Pascal. So, we can evaluate S at state 3 using this expression. Now, we move on to state 4 S 3 to 4 S is an isentropic process. So, that means S 4 S is equal to S 3 that is known and pr 4 S over pr 3 is equal to p 3 over p 1 I am sorry p 3 I am sorry p 4 over p 4 over p 3. So, that means I can evaluate pr 4 S to be 11.03. So, this is the value which we use to enter the table pr equal to 11.03. So, let us see. So, this is where it falls. So, we can retrieve h at state 4 S0 is not required because S is already known. So, we retrieve the value for h at state 4 S from the table. So, that is given like this. So, now we have all the properties that we want for completing the calculations. So, at each for each state you need to identify the value that is going to be used to enter the table retrieve the values that are required and then proceed to the next state and then just walk through the cycle that is the procedure. So, once again like with the other worked examples my suggestion would be for you to pause the lecture at this point work out the problem yourself and then come back and compare the values with what is what is shown here. It would it would be much more beneficial to you than just going through these examples just by listening to the lecture. You should actually work out the problem on your own. So, application of steady flow energy equation to the compressor gives us this again steady flow energy equation to the turbine on a per unit mass basis mass flow rate basis gives us this rate at which heat is added rate at which heat is rejected and the first law efficiency is 59.8 percent for the basic cycle. Now, let us calculate the rate at which exergy is supplied remember in this cycle. So, exergy is supplied here to the compressor and exergy is supplied here in the form of heat to the combustor and exergy is recovered here. Notice that there is no exergy recovery associated with this because this is rejecting heat to the ambient. So, there is no exergy recovery corresponding to Q c dot unlike what we saw in the Rankine cycle where the condenser was operating at a temperature of 45 degree Celsius not the ambient temperature here heat is being rejected to the ambient. So, there is no exergy recovery at this state. So, let us go back and do the calculations. So, exergy supplied compressor plus Q h dot we get 955.73 exergy recovered as I mentioned is only from the turbine. So, that gives me 851.62. So, the second law efficiency comes out to be 89.1 percent which is pretty high. Rate of exergy destruction during the heat addition process may be evaluated using this expression and it comes out to be 38.49 and rate of exergy destruction during heat rejection process may be evaluated using this expression and it comes out to be higher than the rate of exergy destruction during the heat addition process. So, this is actually 65. So, the scope for improvement it would seem in the case of the Brayden cycle is actually on the condenser side and not on the not on the combustor side in contrast to the Rankine cycle where the scope for improvement was on the boiler side rather than on the condenser side. In fact, practically no change was made on the condenser side at all in the case of the Rankine cycle. So, this one this is the reason why we compute these values particularly exergy the concept of exergy is very useful because it alone can offer insights like this which is not something that we would have expected when we start it out. Now, the first improvement that we will try to do to this cycle is try to reduce the compressor work. You may recall that earlier we actually discussed this or contraction of internally reversible steady flow processes and we said that the work that is required for an isothermal process which is this one is less than the work that is required for an isentropic process. So, we have assumed the compression work to be isentropic. So, it is there is scope for improvement here number 1, number 2 we also mentioned earlier that the work power that is produced in the case of an isothermal process is more than the power that is produced in the case of a isentropic process. So, this is s equal to constant. So, you get more expansion work in the case of a reversible isothermal process than an isentropic process. So, this is something that we will keep in mind, but for now let us focus on the compression process. So, this is the compression process and this is the expansion process. So, what we are about to discuss is applicable equally to expansion process as well. So, the idea here is to do the following. Now, the reversible isothermal compression process is actually not practicable because it requires excessive cooling and that is not very easy to implement. However, we can try to do a constructor process which sort of mimics a reversible isothermal process in spirit. Let us see how we do that. So, basically what we do is we start then we compress along the along the s equal to constant or isentropic isentrope and then we cool the gas like this. So, we cool it to the same temperature as state 1. Notice that this isotherm runs from state 1 to state 2. So, they are all at the same temperature. So, if I take air in at a temperature of 300 Kelvin after this compression process I cool it to 300. Then I run it along another isentropic process like this and then we cool it again to 300 Kelvin and then I again compress along another isentropic process like this and then cool it again to 300 Kelvin and then again compress along an isentropic process and then finally reach state 2. Notice that the cooling here in this case is not as excessive as it would have been had we carried out the entire process as a reversible isothermal process. Here we are only compressing partially and then cooling it to the initial temperature which is much more practicable from an engineering perspective. The more number of steps we have here the more I mean the easier the cooling processes. But again it is also impractical to have too many steps. In fact, if you have infinite number of steps obviously we approach the reversible isothermal process. So, we cannot have too many steps but at the same time we do not want to have too few steps which would actually increase the cooling requirements. That is the general idea and we do the same thing here also. So, in the case of an expansion process we go like this. So, here at each horizontal leg heat is removed. So, this is a cooling process whereas each horizontal leg here is a heating process. We add heat to reach the same temperature as state 1. So, we will come back to the expansion process later on but for now we see how this is done. Now, this sort of compression process is known as compression with intercooling. So, this is compression with intercooling and let us see. Let us go back and then look at a simple situation. So, let us summarize what we have said so far. So, for a given pressure ratio, steady flow reversible isothermal compression process requires less power than an isentropic compression process. So, if the work required in compression is less our expectation is that the efficiency of the Brayton cycle can be improved because the work that is required is less. We will see whether it comes out whether it turns out to be the case or not. But as I already mentioned, a reversible isothermal process is very difficult to execute in real life because of the excessive cooling requirement. So, what we try to do what is more practicable is to achieve the compression in multiple stages with the air being cool to its initial temperature at the end of each stage. But the compression process across each stage as we showed before is still isentropic. So, here we have illustrated two different versions of a single stage two stage compression process with single stage intercooling. So, the air is compressed from here to here and then intercooled cooled and then taken all the way there. Here the air is compressed from here to here cooled and then compressed up to this level. So, it is two stage compression with intercooling in between the stages. Now, so you can see that both these are two stage compression processes. So, the question is what should be the intermediate pressure? So, this is the intermediate pressure. So, what should the intermediate pressure be in order for the work to be or the power consumed for the process to be as small as possible or the least possible? So, you can see that the area shown here the gray area is the savings in work. So, this is I am sorry savings in power. So, in one case you can see that this area is long and skinny whereas in the other case you can see that it is actually more like a rectangle short and fat. So, as this pressure intermediate pressure at which we are cooling the gas as it is changed the power saving also changes the shape of the area that we have shown in gray also changes. So, there is an optimum pressure in between at which we expect the power savings to be a maximum. How do we determine this? So, we apply SFE to the overall compression process. So, the first stage compression process we go from 1 to X and in the second stage compression process we go from Y to F where F is the final temperature. Depending upon where the intermediate pressure is the final temperature of course will be different, but the pressure ratio is the same P2 over P1 is the same. Now also notice that Ty is the temperature at state Y and Y state Y and state 1 both lie on the same isobar. So, I am sorry isotherm. So, state Y and state 1 both lie on the same isotherm. So, Ty is equal to T1 and states X and Y lie on the same isobar. So, Py equal to Px and Ty equal to T1. Notice that here we have assumed the air to be calorically perfect for the sake of simplicity. Remember we are just trying to get an idea of what this optimal value of what the optimal value for the intermediate pressure is. So, without any loss of generality and with minimal loss of accuracy we can actually and for the sake of simplicity we may assume the gas to be or air to be calorically perfect. So, this is the first stage compression. This is the second stage compression. Now 1 X and YF for isentropic processes what is that 1 X and YF I am sorry Y to the final state wherever it may be in this case it is 2 prime in this case it is 2 double prime. So, wherever the final state is 1 X and YF are both isentropic processes. So, which means I can replace this ratio Tx over T1 using the familiar isentropic relation Pv raise to gamma equal to constant. So, we get an expression like this again remember Px equal to Py. So, we differentiate W this expression for W x dot compressor compression with respect to Px and set the derivative equal to 0 and quite surprisingly it turns out that Px equal to Py equal to P1 P2 to the power one half. In other words the optimal intermediate pressure is such that the pressure ratio in each stage of compression is the same. So, this means that the pressure ratio in the first stage of compression Px over P1 and the pressure ratio in the second stage of compression P2 over Py are both equal and not only that it is also equal to square root of the pressure ratio across the entire cycle. So, this is nothing but P2 over P1. The additional fact that happens in this is since the temperature once we have an optimum once the pressure ratio is the same in each stage. Since the temperature at the beginning of each stage of compression is always the same. So, we started state one we go up to the intermediate pressure we then cool the air to the same temperature as state one then we continue the compression. So, the power required in each stage is also the same. So, our calculations actually become somewhat simpler if you employ the intercooling or multi stage compression with intercooling. Now, in case we carry out the compression in n stages instead of two stage the pressure ratio across each stage simply becomes equal to rp raise to 1 over n the nth root of rp. If it is two stages then it is a second root of rp it is n stages then it is the nth root of rp. So, the same consideration applies to an isentropic expansion process also. So, the intermediate pressure again during the expansion should be the square root if it is a two stage process should be the square root of the overall pressure ratio. And we will make use of this fact when we do reheat in when we add reheat to the Brayton cycle.