 So, now the T s diagram for the air centered Brayden cycle with the two stage compression process and intercooling looks like this. So, rather than going from one all the way to some state like this, which is what we would have done with no intercooling, we go from 1 to 2 s, the air is then cooled to the same temperature as 1. Although this illustration does not show that remember T 3 is equal to T 1. And then again there is an isentropic compression like this and then heat addition expansion in the turbine and then heat rejection. So, as I said since the power required with multi stage compression is less than a single stage compression, we expect the overall efficiency of the cycle to improve. And this intermediate pressure is the square root of for two stage compression is a square root of the overall pressure ratio. So, let us redo the previous example with two stage compression. So, overall pressure ratio was 30. So, that means the pressure ratio across each stage is square root of 30. So, this is the same as before we need not do this again, but for the for state 2 s, the pressure is now 100 times square root of 30. Correspondingly, the value for P r 2 s is 7.59. So, we enter the table with this value for P r and we retrieve h like this. S need not be computed again because s 2 s is equal to s 1. So, for 7.59, so if we go to 7.59, 7.59 for P r 2 s. So, 7.59 falls between these two. So, we interpolate between these two values and 482 to 492 and we get h to be 488.64. Now, P 3 over P 2 s is also square root of 30 because with optimal intermediate pressure, the pressure ratio across each stage is the same. But remember T 3 is equal to 2 e t 1. So, we use this value of a temperature to go into the table and we retrieve P r just like before h just like before this also same as before. But s now has to be calculated using the expression that we derived that we wrote down earlier. Remember s has to be calculated using this expression here and if you do that you get the specific entropy at state 3 to be equal to 1.21386. At state 4 s, the pressure is 3000 and P r 4 s is nothing but P r 3 times square root of 30. So, we go into the table with this value for P r and we retrieve this value for the specific enthalpy. Entropy, specific entropy need not be evaluated because s 4 s is the same as s 3. State 5 at the end of the combustion process, we assume that there is no pressure loss during heat addition, temperature is given to be 1300 Kelvin. So, we go to the table with this value and retrieve h and let me just write it like this. So, we retrieve h and s 0 and specific entropy at s may be evaluated using the expression that we wrote down before. And that comes out to be 2.29715 5 to 6 s expansion in the turbine is an isentropic process. So, P r 6 s works out to be 330.9 divided by 30. Remember there is no reheat there is a single stage expansion. So, it is 330.9 divided by 30. So, we go to the table with this value for P r and retrieve this value for h. Yes, need not be calculated again because 5 to 6 s is an isentropic process. So, the specific power required for the compression process comes out to be 376.9. Notice that this actually this is equal to this because the power required in each stage is the same. So, we need not have done it like this, but it does not matter it comes out to be 376.9. Let us compare this with what we had before 376.9 against 488.92. So, you can see that there is a substantial reduction in the power required for compression as we expected power produced in the turbine remains the same. So, power required for the compressor decreases by 22.9 percent power produced in the turbine remains the same because there is no change on the expansion side. Now, q dot h comes out to be 907 against 606. So, you can see that there is a substantial increase in the rate of heat addition in the cycle heat rejected same as before because state 1 and state 6 s are the same as before. So, heat rejected is the same efficiency comes out to be 52.3 percent. So, this is actually a reduction when compared to the basic cycle primarily because of the substantial increase in the heat addition. Although the compressor power has reduced the amount of heat that is added to maintain the same peak temperature in the cycle has gone up. So, if we did not have intercooling then state 4 s would have been here and the amount of heat added from here to here was calculated before which would be less. But now state 4 s is over here and the amount of heat that has to be added to take it all the way to the peak temperature of 1300 Kelvin but this is. So, that is substantially higher as a result of which the overall thermal efficiency despite the reduction in the compressor power required goes down. The overall thermal efficiency still decreases. So, we need to find out how to address this. So, we like the fact that the compressor power comes down but this seems to be a problem. So, we need to address this which we will as we go along. So, rate at which exergy is supplied is 1074 compressor plus heat added in the combustor and rate at which exergy is recovered is 851.6. So, this is slightly more I am sorry the turbine work turbine work is the same yeah turbine work is the same net power is slightly more. So, the second law efficiency comes out to be 79.2 percent which is also less than the 89 percent that we had before. So, addition of intercooling two stage compression with the intercooling has reduced the compressor power and improve the specific power output from the engine. The power output increases because turbine work remains the same compressor work has gone down. So, the net specific power increases. So, it improves the specific power but both the thermal efficiency and the second law efficiency decrease because of the first one increases because of the decreases in the amount of heat that is added that is this one. And the second law efficiency decreases because so this is the rate of exergy destruction in the heat addition process. So, this is 115.6 compared to compared to 38.49. So, you can see that there is a substantial increase in the exergy destruction in the combustor. Whereas, the exergy destruction in the in the condenser and the cooler remains the same. So, this remains the same this has increased substantially and this has also increased substantially. So, as a result of this the first law efficiency has come down and as a result of this increase the second law efficiency has come down. So, we now will take a look at the effect of reheat. Remember reheat is the same as multi-stage compression with intercooling just on the expansion side. So, we expand up to a certain intermediate. So, this is an intermediate pressure again the optimal pressure is already known. So, that would be such that the pressure ratio between 3 to 4 S and 5 to 6 S is the same. So, we expand to this optimal pressure from here to here and then it is taken to the combustor where heat is added in such a way that T 5 is equal to T 3 and then it undergoes further expansion here. So, for optimal operation T 5 is equal to T 3 and P 4 S over P 3 is equal to P 6 S over P 5 is equal to square root of P 3 I am sorry P 4 S over P 5. So, this should be P 5 over P 6 S and this is equal to P 3 over P 6 S equal to square root of R P in the cycle. So, let us see how this impacts the performance of the cycle. We start in the same manner as before state 1 is the same as the basic cycle. State 2 is again is the same as in the basic cycle not in the operating cycle with intercooling. State 3 also same as the basic cycle. So, we do not need to repeat the calculation. Now state 4 S is at an intermediate pressure 100 root 30 which means P R 4 S is 60.25. So, we go to the table with this value for P R and retrieve H 3 to 4 S because it is an isentropic process we need not evaluate S again for state 4 S. State 5 again state 5 pressure is 100 root 30 because the heat addition reheat process is a constant pressure process, but it is reheated to a temperature of 1300 Kelvin. So, we go into a table with this value and retrieve P R and retrieve H and of course, S 0 also S for 5 may be evaluated using the expression given earlier. So, S of state 5 is equal to S 0 of S 0 corresponding to 1300 Kelvin minus or natural log P 5 over P ref. So, if you do that you get S to be equal to 2.7853 5 to 6 S is an isentropic process. So, we go to the table with this value for P R which is nothing but 330.9 divided by square root of 30 square root of RP and we retrieve this value for H and the specific entropy is the same as 5 because S 6 is equal to S 5. So, we have all the values that we require now to proceed with the calculation. Specific compressor power same as what we saw in the basic cycle. Specific power output from the turbine we now have two turbines each producing the same amount of work. So, the specific power output from the turbine has increased now as we can see. Heat added has also increased when compared to the basic cycle because we now have a reheat stage also. Heat rejected has also increased because state points this state point would earlier have been over here. So, earlier would have been over here. So, the heat rejected earlier was equal to this. So, now because the state point 6S has moved over here there is an additional heat rejection that is taking place. So, this is the additional amount of heat rejection that is taking place. So, heat rejected has also increased. So, the thermal efficiency or first law efficiency comes out to be 48.7 percent which is still much less than that of the basic cycle. But the specific power has increased by almost 22 percent. So, specific power has increased but the efficiency has decreased due to the increased heat addition. And again exergy supplied in the compressor and in the combustor exergy recovered is in the turbine only. So, the second law efficiency is 76.56 still not close to the 89 percent that we had for the basic cycle. And as we said the primary reason for this is the fact that the exergy destroyed in the during the heat addition process is now higher because we have additional heat addition also. So, both the heat addition as well as heat addition in the combustor as well as heat addition during reheat has actually increased the overall heat addition resulting in reduction in thermal efficiency. And heat rejected has also increased you have to bear that in mind as well. So, reheat also just like multi stage compression with intercooling reheat also increases the specific power but results in a reduction of thermal efficiency, first law efficiency as well as second law efficiency. So, what we need to address is how to actually combine intercooling. So, intercooling is beneficial because it reduces the compressor power but there is a penalty that is associated with that. And so, we will try to address this in the next lecture how to actually not only improve the specific power but also improve the first law and the second law efficiency.