 So, this is for an ideal cycle where we have taken the isentropic efficiency of the turbine and the compressor to be 100 percent. Now, if we relax that assumption and assume that it is not equal to 1 or 100 percent, then the expressions that we derived before look like this. The specific power now as before specific power depends on T3 over T1 and Rp, but the efficiency now depends on T3 over T1 as well in addition to Rp. So, the insights, notice that these insights that we have obtained here are possible only with an air standard cycle. Now, notice that this derivation allowed us to identify T3 over T1 and Rp as the parameters that control the performance of the cycle. Now, by looking at this expression and this expression like this, we are now able to understand the effect that these parameters have on the cycle. The effect of pressure ratio is generally to increase the efficiency. The effect of pressure ratio on the specific work is to increase it in the beginning, but beyond a certain value depending on T3 over T1, the specific output begins to decrease. So, that is the pressure ratio at which we would ideally like to operate where the efficiency is high and the specific work is the highest possible for that given value. So, these are insights that we are getting from the air standard analysis. That is very important. That was the objective that we started out with when we started discussing air standard cycle. Now, in the case of the actual cycle, obviously, both the isentropic efficiency of the turbine and the compressor are involved and if you look at this, these are shown using the dashed line here corresponding to eta T equal to 0.95 and eta C equal to 0.9. You can see that the general trends are the same. Of course, efficiency now depends on T3 over T1, but efficiency generally increases with Rp. These curves also have Rp going from 5 to 40. So, efficiency generally increases with Rp, but at higher values depending on the value of T3 over T1, efficiency also begins to decrease even though we are increasing Rp. So, here you see increasing, but beyond a certain point efficiency begins to decrease. Efficiency reaches a maximum here for 40, it does not decrease. Here also, it seems to reach a maximum, does not decrease, but efficiency certainly begins to decrease for T3 over T1 equal to 4 at higher values of Rp. If you look at specific work, again, you can see that it reaches a maximum and then begins to decrease. So, if you look at specific work, you can see that it reaches a maximum, then begins to decrease. Here also, you see the same, here is more or less flat, but yeah, you can see that it reaches a maximum, then begins to decrease with Rp. So, whatever trends we saw with the ideal cycle, we are seeing in the actual cycle also. Of course, the assumption that we have still not relaxed is that we are assuming the air to be calorically perfect. That is actually not realistic in the present application because the temperature ranges over such a, I mean ranges from 300 Kelvin to 1600 Kelvin, that is a wide range. And assuming Cp to be constant, Cv to be constant in this range is actually somewhat unrealistic, so we need to actually relax that assumption also, which we will do next. But the insights that we have obtained from the ideal cycles, more or less hold for the real cycle also. So, the approach that we will take with Brayden cycle is similar to or same as what we did for the Rankine cycle also. We will start with the basic cycle, look at the performance matrix, specific work, first law efficiency, second law efficiency, exergy destruction in the components and then identify scope for improvement and then improve in a systematic manner and after each improvement evaluate the performance see how it has changed, has it improved or has it become worse, then and then identify until we end up with the best possible cycle. Let us now talk about using air table, so I just mentioned that the temperature range encountered in the gas turbine application and spark ignition engine or auto cycle and compression ignition engine or air standard diesel cycle is actually far too wide for the calorically perfect assumption to hold. Remember, when we looked at internal energy of ideal gases, we said that internal energy U is a function of temperature and pressure and we argued that for gases the dependence on P is very weak, so we may essentially assume this to be a function of a temperature alone. Now, this would be called a thermally perfect gas, now if you assume that U is actually a linear function in T, that means U is equal to say Cv times T, then we have the so called calorically perfect assumption, which is what we have been using so far and as I said this is not realistic for the applications that we are looking at now, so we will relax this but still assume the gas to be thermally perfect, which is actually quite acceptable, there is no problem there for gases, so assuming gas to be thermally perfect is quite acceptable and that is what we are going to do. Of course, ideal gas equation of state may also be assumed to hold for this range of temperatures and that is also again realistic, there are no issues there and remember H is equal to U minus I am sorry U plus Pv or U plus RT. Now, the air table looks like this, so this table lists the specific enthalpy and specific internal energy and specific entropy at standard condition, so this is nothing but S at the given temperature and the reference pressure, so it lists H, U and S0 as functions of temperature, so this entropy is nothing but entropy corresponding to this temperature but at the reference pressure, but you must bear in mind that at any temperature here H is always equal to U plus RT, so if you want to evaluate the specific gas constant that has been used while compiling this table, you only have to take H and U at any entry, H minus U divided by T gives you the specific gas constant that has been used while compiling this table but that is actually good practice also. Notice that in keeping with our calorically perfect assumption, H and U are both functions only of temperature, so as I said the air table lists values for H and U for air at different temperatures and values of specific entropy at the reference pressure are also provided. Now, the value for specific entropy at any other pressure, same temperature but any other pressure may be evaluated from TDS relationship like this, so delta S equal to, so we have from the TDS relationships delta S may write like this, I am sorry TDS equal to VDP, so if I divide through by T, I get DS equal to DH over T minus V over T and if I use PV equal to RT, so remember PV equal to RT, so V over T becomes equal to R over P, so minus RDP over P. So, if I want to evaluate entropy at any other pressure but the same temperature, then since H is dependent only on temperature, this term will go away, we can integrate and end up with this expression, S of T, P equal to S at T, P ref minus R natural log P over P ref. Now, in case the process is an isentropic process, then of course, there is no change in entropy, so S0 of T2 minus S0 of T1 minus R natural log P2 over P1 is equal to 0. In fact, it would have been better if we had written instead of 2, if we had written this as 2S, but anyway that is understood because we have said that this is an isentropic process. Remember, we are very much interested in isentropic process because we take the isentropic efficiency to be 100 percent, then we have only isentropic processes in the cycle. If you take the isentropic process to be less than 100 percent, then we still need the isentropic state to work out the either the actual state at the exit of the compressor or turbine or calculate the power. So, either way we need to be able to handle isentropic processes and with varying Cp, how do we do that is what we are looking at right now. So, basically what we have done here is we have taken two states S2 of T2 comma P2 and S1 of T1 comma P1 and we have said that going from S1 to S2 is an isentropic process which allowed us to write this expression. So, if you rearrange, you can actually write this expression like this P2 over P1 equal to e raised to this quantity divided by e raised to this quantity. Notice that this depends only on temperature, this also depends only on temperature. So, we can in fact write this as the ratio of two pressures PR the reduced pressure PR2 over PR1. So, P2 over P1 is equal to PR2 over PR1. Now, since PV equal to RT, we may write the ratio of specific volumes like this V2 over V1 is equal to this and for an isentropic process I can replace P1 over P2 with PR1 over PR2. Now, if I regroup the term slightly differently, I can actually write this as a quantity in the numerator which depends only on temperature. Remember PR2 depends only on temperature. So, and you have T which is of course temperature, same thing here. So, I may actually write V2 over V1 also as the ratio of VR2 over VR1 which are both dependent only on temperature. So, dependent only on and that is what the air table shows in the remaining two columns. So, we can see that PR and VR are listed in the air table, they depend only on the temperature. So, they are shown as a function of temperature. Of course, these values are not exactly the same as what we have shown they are scaled suitably for use in the table. So, that the numbers look decent. So, these are the values that are shown here. These are applicable only for an isentropic process. They are not meaningful for any other process they are meaningful only for an isentropic process. So, you must bear that in mind. Now, let us see how we use the air table by means of an example. So, before we do that let us just recapitulate PR and VR depend only on T, they are listed in the table after appropriate scaling. So, please do not use these expressions to evaluate them they have been scaled further and they are meaningful only for an isentropic process PR and VR. So, if state 1 and state 2 or isentropic states then P2 over P1 equal to PR2 over PR1 if and only if 1 and 2 are isentropic. V2 over V1 is equal to VR2 over VR1 if and only if states 1 and 2 are isentropic. So, we have a compressor where air is compressed from 1 bar 300 Kelvin to 10 bar. The isentropic efficiency is 92 percent determine the power required and the exit temperature. So, the temperature at state 1 is given to be 300 Kelvin which means we can retrieve which means we go into the table with 300 Kelvin and retrieve H to be 300.19 and PR to be 1.386. So, we retrieve H1 and PR1 from the table. So, P2 over P1 is equal to 10. So, in case the compression process is isentropic that means PR2S would have been equal to 10 times PR1. So, that would be 13.86. So, what we are doing is we are saying this P2S over P1 is equal to PR2S over PR1 that is equal to 10 which means that PR2S is 10 times PR1. So, it comes out to be 13.86 for state 2S not state 2 state 2S. Now, we go to the table again and corresponding to a PR value of 13.86. So, value of 13.86 falls between these two. So, by means of interpolation, we can actually retrieve the value for H and if required the value for temperature at the exit of the compressor also. So, we use this value and then interpolate for H interpolate for T if required. So, interpolate for H and T if required. So, if you do that we can get H2S to be 579.865. Remember this is H2S PR is applicable only for an isentropic process. So, from the definition of isentropic efficiency, H2S minus H1 divided by H2 minus H1 is equal to eta C. Eta C is given to be 92 percent except we know all quantities except for H2. So, we may evaluate H2 to be 604.185 kilo joule per kilogram and the power required actual power required may be evaluated as H2 minus H1 specific power required may be evaluated as H2 minus H1 304 Kelvin. We are also as to calculate the exit temperature. So, we go to the air table with this value of H 604.185 what is that 604.185 we just change color here. So, 604.185 falls between these two entries in the table. So, we may interpolate and get the exit temperature to be 597 Kelvin. So, we can see that although PR and VR are applicable only for isentropic processes and the actual process need not be isentropic. Remember here we have taken 1, 2 to be a non-isentropic process because the isentropic efficiency is given. It is adiabatic but not reversible. So, you can make use of the values of PR and VR from the table and then evaluate the actual quantities that are required for the process that is being executed. So, that is the usefulness of the air table. It accounts for variation in CP with temperature and so it gives us much more accurate values for the quantities that we are calculating specific power, efficiency and so on. So, in the next lecture we will work out an example involving the basic Brayton cycle and we will allow for property variation. So, we will not assume CP to be constant. So, we will do the calculation with the air table. That would be for the basic cycle. Then we will analyze the performance parameters of the cycle like we did before in the case of the Brayton cycle and then keep going like that.