 We saw in the previous lecture that addition of reheat to the basic Brayton cycle increases the specific work output, but the efficiency both the first law and second law decreases. Since we supply additional heat, but the additional heat more than offsets the work that is developed. Number one, number two, the rate of exergy destruction in the reheat portion also is non-zero now, so that increases the overall rate of exergy destruction during heat addition in the cycle. And the rate of exergy destruction in the heat rejection portion also increases because as you can see the temperature at the exit of turbine 2 which is the low pressure turbine is actually more now. So just like before this is usually called the high pressure turbine and this is usually called the low pressure turbine just like what we said for the Rankine cycle. So the temperature at the end of expansion in the low pressure turbine is higher now. So that means that the rate of exergy destruction during the heat transfer to the ambient is also more and we could you could see that in our calculation. So if you look at rate of exergy destruction, I am sorry we have not calculated that, but if you do then it would be higher. So the rate of exergy destruction during the heat addition process is higher because of the additional heat addition in the reheat part of the cycle and the rate of exergy destruction during the heat rejection process is also more because the air enters the condenser or the cooler at a much higher temperature than before. And the exergy destruction in both the heat addition part and the heat rejection part are higher due to two reasons. Number one, the amount of heat itself increases and number two, the temperature difference across the source and the air when it enters the heat addition or heat rejection phase is also higher. So there are two factors which are contributing to increased exergy destruction in the heat addition and heat rejection part of the cycle and this is what we would like to address now. So reheat increases the specific power but both the first law and second law efficiency go down. Intercooling as we wanted reduces the compressor power. We wanted the compressor power to be as small as possible. So intercooling we saw does reduce the compressor power. However, for a given turbine entry temperature, the amount of heat that has to be supplied increases because the air now leaves the second stage compressor at a lower temperature than a single stage compressor. Now this has to be viewed as a penalty because the objective of intercooling is to reduce the compressor power. That has been met but that has been met at the cost of increased heat addition. So there is a downside to intercooling. So we would like to address these issues next by improving the performance of the cycle. Let us see what is possible. Now if I look at the, if I start with the basic Brayton cycle, let us start with the basic Brayton cycle. So in the basic Brayton cycle, you can see that the air enters the condenser or leaves the turbine at a temperature of 540 Kelvin whereas the temperature of the air as it enters the combustor is 770 Kelvin. One idea or one strategy that we wish to pursue is to see whether we can take the heat from the stream after expansion in the turbine and transfer it or use it to heat the air before it enters the combustor. Now if we are successful in doing this, it will address both the deficiencies simultaneously. By transferring this heat, the amount of heat from one stream to the other, the amount of heat that is rejected will go down and the amount of heat that has to be supplied will also go down. It will address that number one. Number two, because we are heating the air before it enters the combustor, the temperature difference between the source from which heat is supplied and the air when it enters the combustor becomes smaller. So the rate of exergy destruction due to the external irreversibility goes down simultaneously because the air after expansion in the turbine transfers part of its enthalpy to the incoming air. When this air enters the condenser or the cooler, its temperature is actually less than before. So the temperature difference of the air when it enters the condenser compared to the ambient temperature is less now. So the rate of exergy destruction will be less in the condenser. So this if it is possible to transfer heat from the stream after it leaves the turbine before it enters the condenser to the air after it leaves the compressor before it enters the combustor. If that is possible, then we would reduce the amount of heat that is supplied or that is rejected. Number one, we also reduce the temperature difference between the source and the air stream. Now if you look at the basic cycle, it is clear that the strategy is not possible. It is not viable because the temperature at the end of expansion is less than the temperature of the air when it enters the combustor. So heat cannot be transferred from the air at state forest to air at state 2S. So this is clearly not viable. Now if we look at intercooling, you can see that the temperature at the end of expansion remains the same 540 Kelvin. However, the temperature now at the temperature at entry to the combustor is 486. So definitely heat can be transferred from here to here. Now let us look at reheat. So the temperature at the end of expansion is even higher now. It was 540, now it is 850 and the temperature at the end of compression as we can see is still is 770 because this is single stage compression. So there is definitely a favorable temperature difference. So heat can still be transferred from the air after expansion of the turbine to the air before it enters the combustor. So this temperature difference is still appreciable. Some heat can be transferred. However, the maximum benefit will be realized if we have a cycle which incorporates both intercooling as well as reheat because in that case the temperature at the end of expansion will be 850 and the temperature at the end of the compression process is going to be 486. So this temperature difference is the maximum and so maximum amount of heat transfer is possible from the air after expansion to the air before it enters the combustor. So that is actually the strategy that we are going to pursue. So this strategy of transferring enthalpy from the stream after expansion to the air before it enters the combustor is called regeneration. So this we have already mentioned the temperature at the end of the second stage turbine is higher when reheat is employed. So there is scope for heating the air as we just saw before it enters the combustion chamber by utilizing the exhaust heat from the either the second stage turbine or even in the case of single stage turbine if intercooling is employed we saw that there was still about 100 plus Kelvin temperature difference. So there is still scope for regeneration there. So this strategy is very similar to its counterpart the Rankine cycle and so this is called the regeneration. Now in the case of Brayden cycle there is no penalty that accompanies regeneration unlike in the case of Rankine cycle. Rankine cycle the steam was extracted from the turbine partially. So there was a penalty on the specific power but in the case of the Brayden cycle no such extraction takes place so there is no accompanying or competent penalty in the specific power. So basically the regenerator is also a heat exchanger. So we take the air at the end of compression process and send it to the regenerator. So the air goes from state 2 to state x where its temperature increases. So Tx is greater than T2 or more importantly Hx is greater than H2. So the air at state x is then sent to the combustor where heat addition takes place. Now the air that comes out of the turbine at state 4 is sent to the regenerator and it transfers its enthalpy to the air before it enters the combustor. So enthalpy transfer takes place and this air then leaves at state y and as you can see Hy is less than H4 and then it enters the condenser or cooler where it loses heat to the ambient. So that is a general idea and performance metric for the regenerator is the effectiveness epsilon. So this is the effectiveness epsilon and that is defined like this Hx minus H2 the actual enthalpy transfer for the air stream before it enters the combustor divided by the maximum enthalpy transfer that is possible. So the maximum enthalpy transfer is H4 minus H2. So that is how effectiveness is defined. So let us now incorporate regeneration in a cycle which has intercooling, reheat, intercooling and reheat. So in such a case as you can see the air undergoes two stage compression process so the temperature or state at the end of compression is 4s and then it is heated from state 4s to 4x in the regenerator. So this is the regenerator and this is the combustor. So from state x to state 5 heat is added in the combustor, it undergoes expansion in the high pressure turbine then this is the reheat portion and this is the low pressure turbine it undergoes expansion in the low pressure turbine and this air exchanges its heat with the incoming air. So this is also in the regenerator then finally heat is lost to the ambient in the cooler. So let us look at the performance of this cycle. Let us first start with the table then we will go back to this. So state 1 so if you look at this cycle state 1 2s 3 4s up to 4s is the same as what we had in the 2 stage compression with intercooling. So these values may be directly taken from there 2 stage compression with intercooling up to 4s. State 5 is the same for all the cases it is the temperature it is a state at the exit of the combustor so we already have the values for this. What is that state 6s 7 and 8s are the same as in the reheat cycle. So these values may be taken from the Brayden cycle with reheat. Now all we need to do is determine Hx and Sy corresponding to state x and y. So the effectiveness of the heat exchanger is given to be 0.9. Notice that all the other values in this definition are known so we may evaluate Hx to be 838.569. So we may go to the table with this value for the so we may go to the table with this value for Hx and retrieve specific entropy to be 1.76205 I am sorry we get S0 to be this and then from S0 by using the expression for S we may calculate S to be equal to this. Remember the pressure at X is the same as the pressure at the end of compression. So we get S0 from the table then we can use the expression given earlier to calculate S. Now if we apply steady flow energy equation to the to the regenerator we get this expression from which we can evaluate Hy. We can go to the table and retrieve S0 and then calculate S. Notice that since the pressure at the end of expansion is 100 kilopascal the S0 that we retrieve here is the S that we calculate is the same as the S0 that we retrieve here. So now we have the property values that we need at all the state points. Notice that once we have the effectiveness for the regenerator Hx may be calculated and then by applying steady flow energy equation simple energy balance we can get Hy also. So this is a two stage compression process so we may evaluate the compressor work like this which is the same as what we had earlier with two stage with intercooling two stage compression with intercooling. Again the power produced by HPT and LPT together come out to be 1037. Heat added now as we have shown here notice that the heat added in the combustor is from state X to state 5 and the heat added in the reheat portion of the cycle is from state 6s to state 7. So heat added is H5 minus Hx plus 7 minus 6s which works out to 1075.92. Now heat rejected in the cycle Qc is H2S minus H3 so H2S minus H3 is the heat rejected here and heat rejected in the cooler is over here. So that is Hy minus H1 which comes out to 417.78. So the efficiency of the cycle now comes out to be 61.36% which is definitely higher than what whatever we had earlier. So the rate at which exergy is supplied comes out to be 1204. The rate at which exergy is recovered is 1037. So the second law efficiency is 86.1 which again is approaching the value that we had for the basic cycle. The rate of exergy destruction in the heat addition process now remember the peak temperature in the cycle remains the same. So the rate of exergy destruction now comes out to be 58.686 and this should be compared we can compare this with the value corresponding value for the intercooling cycle let us just do that. So it is 115.6 so rate of exergy destruction during the heat addition process certainly has come down and rate of exergy destruction during the heat rejection process has also decreased. But now we have exergy destruction in the regenerator which has to be added to the process. But still that is not a very large amount that is a small amount. So overall we see an increase in specific power, we see an increase in first law efficiency and we see an increase in second law efficiency which is summarized in this table here. So for the basic cycle we started out with 59.8 for first law efficiency 89.1 for second law efficiency and specific work was 362.7. Now intercooling as a result of intercooling thermal efficiency reduced, second law efficiency also reduced but the specific power improved so this went up. Now with addition of reheat to the basic cycle the specific power improves even more. However the first law and the second law efficiency both decrease further. Now with the addition of regeneration to this cycle with intercooling with the two stage compression with intercooling as well as reheat the first law efficiency is higher now even higher than the basic cycle. Second law efficiency is also quite high very close to what we had for the basic cycle and the specific power is the highest. So all three performance parameters in the cycle show extremely good values when we employ regeneration so that is the key. So reheat and two stage compression or multi stage compression they affect the specific power output in a positive sense. Both of them increases specific power output but at the cost of efficiency both first law and second law. The addition of regenerator to the cycle retains the advantage that we had as a result of reheat and multi stage compression but it improves the first law and the second law efficiency to values which are comparable to the basic cycle if not even slightly higher than that of the basic cycle. So any practical bread and cycle of course for land based power generation utilizes this cycle. But regeneration is difficult to implement in the case of aircraft engines because it adds to the weight of the engine and for aerospace applications propulsion applications weight is a very critical factor first two weight of the engine is a very important performance parameter so the regenerator it is not practical to add regenerator to aircraft engines. So we will start discussing air standard cycles corresponding to internal combustion engines in the next lecture.