 So, we have been saying in the previous lecture or previous couple of lectures that the reversible isothermal and the reversible adiabatic process both have played a crucial role in the development of fundamental ideas such as the coronocycle absolute temperature scale and so on. So, these are very special processes plus we also showed how any arbitrary reversible cycle may be decomposed into a series of infinite number of into an infinite number of infinitesimally small coronocycles. This was made possible because we were able to take an arbitrary internally reversible process and replace it with a sequence of reversible adiabatic and reversible isothermal processes. So, what this suggests is that when we are looking at issues related to second law irreversibilities and so on, depicting processes using TS coordinates will actually be better because these two processes namely the reversible isothermal and the reversible adiabatic will then fall along coordinate lines. So, for instance if I use TS coordinates then reversible adiabatic process which we said just in the previous lecture to be. So, reversible adiabatic process is one for which S2 is equal to S1 but DS is also equal to 0. So, this would be a reversible adiabatic process and similarly a reversible isothermal process would fall like this. So, this would be a reversible isothermal process. So, these two processes which are so special would then fall along coordinate lines and if there is any irreversibility then this can be illustrated very easily and very nicely using the TS coordinates. So, now we were using PV coordinates so far PV coordinates have their own merits lot of information can be drawn from them. Now we will also start using TS coordinates. So, there is no preference of one over the other you know students who are going through the course must be able to illustrate processes both on PV as well as TS diagrams and must learn how to interpret diagrams drawn using PV or TS coordinates. So, each one has its own advantages and so we will use them appropriately whenever we want to illustrate something and we will illustrate effectively using the most appropriate coordinates for the purpose. Now the other advantage of using TS coordinate lies here you may recall that we wrote this expression. So, we wrote this expression for entropy change of a system in the most general form. Now if the process were to be internally reversible then I can write for an incremental change in entropy of the system I can write delta S equal to delta Q over T plus delta sigma i n T. Now if the process were internally reversible then this would become equal to 0 because there is no internal reversibility. So, now I can write delta Q equal to T times dS that S is the entropy of the system or I may also write it in a specific form as delta Q equal to T times dS where I have divided both sides by the mass of the system or mass contained in the system. Now if I integrate this then I get Q to be equal to integral 1 to 2 T dS or capital Q equal to 1 to 2 T dS both are okay. So, what this says is if I plot a process in TS coordinates then the area under the process curve is nothing but if I plot a reversible process in TS coordinates then the area under the process curve is nothing but the heat interaction that the system has with the surroundings. Just like the area under the curve for a fully resisted and hence reversible process in a PV diagram is the work interaction here it is the heat interaction. So, this we illustrated earlier in this diagram. So, here A is a reversible process as we said before. So, A is a reversible process. So, the area under the process curve for A is nothing but Q 1 to 2. So, that is the additional advantage in plotting processes using TS coordinates. The area under the curve for reversible processes may directly be taken as the heat interaction that the system has with the surroundings but not so far irreversible processes obviously. In the just as what we mentioned in connection with PV diagram if the process is not a fully resisted process then we cannot really show the entire process and hence we cannot calculate integral PV. So, even if you manage to draw a process line between let us say states which are far apart it will still not be displacement work okay. The same way here area under the curve is heat interaction only for a reversible process not for irreversible processes. So, area under the process curve for a reversible process in this coordinate system is the heat transfer whether it is heat supplied to the system or heat rejected to the system that we can easily see. Again if you look at this expression notice that in the absence of any internal irreversibility if heat is supplied to the system then its entropy increases. So, based on this we can determine whether heat is supplied during the process or heat is being rejected during the process. So, if you see for instance for process A notice that entropy increases from the initial state 1 to the final state 2 which means that heat is supplied. So, Q is greater than 0 in this case heat is supplied to the system. So, by knowing the directionality of the process we can determine whether heat is being supplied or rejected okay and that comes from this expression here. So, if delta Q is positive then delta S is positive in the absence of sigma int okay. So, here we have illustrated the polytropic processes you may recall that we wrote down or we actually showed polytropic processes on a PV diagram earlier. Now we are showing the same thing on TS diagram okay. Notice that N equal to 0 corresponds to a constant pressure process P equal to constant isobaric N equal to infinity corresponds to isochoric process V equal to constant N equal to 1 of course is isothermal for an ideal gas. All these are for ideal gases okay isothermal and N equal to gamma as we mentioned earlier is isentropic for an ideal gas. We will show this in a minute we will show that this is isentropic for an ideal gas in a second. Notice that isochores or steeper than isobars on a TS diagram and that is actually an important point to keep in mind when we go and look at Brayton cycles or other air standard cycles that you may encounter later okay that isochores or steeper than isobars on a TS diagram. This information is important in gas dynamics also when you study gas dynamics in the next level course. So, how do we illustrate the phase change of water in TS coordinates that is shown here. So, you can see that you still see the dome shaped region which encloses a two phase mixture region. So, this side is superheated and this side is compressed liquid or subcooled liquid. The most important thing that you should take away from this illustration is the shape of the isobar. So, an isobar in the TS coordinate diagram looks like this okay. This is important because when we look at thermodynamic cycles later on for example Rankine cycle usually Rankine cycle operates between two pressures the condenser pressure and the boiler pressure. So, it operates between these two isobars. So, knowing the shape of isobar is important when we depict cycles on TS coordinates okay. So, this is what the isobar looks like and similarly when we look at Brayton cycle later on once again the Brayton cycle operates between turbine pressure and intercooler or which is the equivalent of a condenser. So, that pressure and so we need to be able to show isobars on a TS diagram for an ideal gas and then illustrate processes. So, you know how to depict isobars on a TS diagram. So, this is what an isobar looks like. So, we should be able to show isobars on a TS diagram for an ideal gas as well as for a two phase mixture. Although we have shown a TS diagram for liquid water, water vapor mixture here and the one corresponding to R134A or any other refrigerant will look identical except for the numbers. And what is also shown in this illustration is a process 1, 2 which as you can obviously see is an isentropic process okay. Now, let us look at a worked example. So, we are given two internally reversible cycles 1, 1 and 2 between the same temperature and entropy limits. So, both of them operate between the same temperature and same temperature limits and same entropy limits. Delta S is the same for both cycles and temperature limits are also the same for both cycles. We are asked to determine if they are power absorbing or power producing and also determine which one is more efficient okay. So, as you can see since these are internally reversible cycles all the processes are internally reversible which means the area under the process curve on a TS diagram is the heat interaction that the system has with the surroundings. So, you can see that based on the change of entropy you can see that Q is positive for this process that means heat is supplied to the system Q is less than 0 for this it is rejected and Q is less than 0 for this which means heat is rejected. And similarly here Q is positive, Q is positive, Q is negative for this heat is rejected in process 3, 1. So, heat is added in process 1, 2 here and rejected in 2, 3 and 3, 1 here heat is added in process 1, 2 and process 2, 3 and rejected in process 3, 1 okay. Now, let us look at the look at the net heat that is supplied during the cycle notice that the area under this curve which would be the total. So, if I look at the total area under process 1, 2 where heat is supplied it would look something like this. What is that? That is more than the total heat that is rejected during process 2, 3 which would look something like this okay. So, the heat rejected under process 2, 3 curve will look something like this and under process 3, 1 would look something like this. So, the total heat supplied is more than this. So, the net area enclosed in the cycle is actually so the net area enclosed in the process curve 1, 2, 3 is actually Q net and that is positive because heat supplied is more than the heat rejected. So, Q net is greater than 0 for this cycle in the same manner. So, you can see heat rejected in process 3, 1 here looks like this and heat supplied in process 1, 2 and process 2, 3 look like this okay. So, the net heat supplied as you can see is greater than 0 because the total heat supplied is greater than the heat reject. Since net heat supplied is greater than 0 from first law of thermodynamics remember in a cyclic process delta W equal to delta Q or W net equal to Q net. Since Q net is positive, W net is also positive which means that both the cycles are power producing cycles. They are producing a positive amount of net, positive amount of work okay. So, the net heat interaction is positive and so the net work interaction is also positive for the cycle. Therefore, both the cycles are power producing cycles okay and this analysis was made possible because the processes are internally reversible. All the processes in both the cycles are internally reversible cycles. Now, we are also asked to determine which processes is more efficient okay. Now, you can see from this illustration that heat supplied in cycle 1 which is the total heat supplied in cycle 1 is this. So, let me just erase all this. So, total heat supplied in cycle 1 is this much and total heat supplied in cycle 2 is this much. What is that? The net work produced by both the cycles is the same because both of them have the same shape and they are operating between the same temperature limits and same entropy limits. So, area of the triangle, this one is an inverter triangle, this one is an upright triangle but the net work is the same for both the cycles okay. So, this cycle which is cycle 1 is given more heat and it produces a certain amount of work whereas cycle 2 receives lesser amount of heat but produces the same amount of work which means that cycle 2 is more, cycle 2 is more efficient. So, heat supplied in cycle 1 is more since the net work produced in both the cycles is the same, cycle 2 is more efficient. Now, you may wonder for instance since both are reversible cycles and they operate between the same temperature limits and the same entropy limits should they not have the same efficiency okay. Now, bear in mind that we are supplying different amounts of heat to each one on the cycle which is why their efficiencies may be different. All the statements that we made earlier had certain very important constraints. They should operate between the same reservoirs and they should be supplied with the same heat. So, only then the comparison becomes fair. So, here we have supplied different amounts of heat which is why they have different efficiencies. We are also asked to determine what would be the temperature limits of an internal reversible corno cycle between the same entropy limits okay. So, if I have a corno cycle which operates between the same entropy limits, what is that a corno cycle on a TS diagram would look like this okay. Let us let me see okay. So, a corno cycle on a TS diagram operating between two reservoirs TH and TC and two isotherms I am sorry two isotropes would simply look like this. 1, 2 is reversible isothermal which means on a TS diagram this will become a horizontal line. 2, 3 is reversible adiabatic which means this will be a vertical line. 3, 4 is reversible isothermal at T equal to TC. So, that is a horizontal line and this is a vertical line. So, that is what we have shown here. So, illustration of the corno cycle on a TS diagram also becomes very nice when you use I mean TS diagram becomes very nice. So, that is yet another reason why TS coordinates are preferred when we start discussing second law irreversibilities and so on okay. So, what we are asked to determine now is the temperature limits for a corno cycle that would operate between the same entropy limits. So, notice that for cycle one the upper limit would be upper temperature limit would be the same as TH and the lower temperature limit based on the heat that is rejected we can calculate the lower temperature limit to be. So, total heat rejected is equal to this okay. So, if I write it in terms of delta S so this is what it looks like. So, for a corno engine that operates between the same entropy limits the sink temperature will be TH plus TC over 2. So, the equivalent corno cycle for cycle one would be one that operates between TH and TH plus TC over 2. That is what is given here. You can show similarly that for cycle two the equivalent corno cycle will operate between a temperature of TH plus TC over 2 as the hot reservoir temperature and TC as the cold reservoir temperature okay. So, we have gone just about as far as we can go with this expression. As I said earlier this expression is very, very good for drawing inferences on entropy changes of processes, effect of internal irreversibilities and so on. But we really cannot calculate delta S using this expression. We can only draw inferences on delta S whether it is positive equal to 0 or negative and so on and so forth. So, what we are going to do next is calculate delta S for a reversible process. We will develop relations for calculating reversible process remember S is a property. So, as long as you know delta S for between two states any irreversible process operating between those two states will also have the same entropy change. So, that is the basic idea. So, we develop what are called TDS relationships. So, for a fully resisted process which means is a reversible process where there is no KE or PE change we can write first law in differential form like this. Since the process is internally reversible we may write delta Q as TDS because it is internally reversible and we can write this relationship TDS equal to du plus pdv. We may also eliminate du by using the fact that h is equal to u plus pv and then taking the differential and write TDS equal to dh minus vdp. For some situations this relationship is useful for other situations this is more useful. Generally for steady flow systems this is more useful because it involves the enthalpy. For systems with fixed mass this is more useful because there we are only seeing internal energy changes not enthalpy changes but both can be used. Notice that this is written for reversible process and as I said before when we calculate entropy change when we want entropy change between two states as we showed here. So, entropy change between state 2 and state 1 may be calculated in any way I wish. So, I may write I may use a sequence of reversible processes between state 1 and 2. For example, since the TDS diagram I may use isentropic process like this and then an isothermal reversible isothermal process like this calculate the entropy change for each one of this and then get s2 minus s1. Now an irreversible process like B which operates between 1 and 2 we will also have the same entropy change. So, that is the basis of TDS relationships and that is what we are doing here. So, just like what we did earlier when we wanted to calculate property change or when we wanted to calculate delta u for a pure substance we will now just look at how to calculate entropy change of different forms of pure substances. For solids and liquids which are incompressible we can actually use this relationship this is more meaningful dv is 0 for solids and liquids because they are incompressible. So, TDS equal to du and du itself may be written as mc dt. So, if you integrate delta s for this system which consists of a solid or liquid of mass m comes out to be m times c times natural log t2 over t1 where c is the specific heat capacity. Now the same relationship may be written for a perfect gas for example, if I write this for a perfect gas I may write this as cp times dt calorically perfect gas. So, v itself may be written as since pv equal to rt v itself may be written as r times t over p. So, I may write this as r times t over p times dp. So, if I bring t to the denominator here I can integrate and get one expression for delta s involving temperature and pressure. If I use this then I can get another expression for delta s involving t and the specific volume v. So, for ideal gases three such expressions may be obtained. So, this involves t and volume, this involves temperature and pressure, this involves pressure and volume. Notice that for a perfect gas that undergoes an isentropic process for a perfect gas that undergoes an isentropic process delta s equal to 0 and so we can show from this last relationship that p2, p2 raise to gamma is equal to p1 v1 raise to the power gamma. So, this becomes a polytropic process with index gamma which is what we had illustrated earlier that is why we said this was an isentropic process for perfect gas. For mixture of perfect gases the expression remains the same where the pressure we have to be careful about the pressure we are using the Dalton's model here. So, we use the partial pressure of that particular component. And we can do this on a mass basis or on a molar basis, but only thing is partial pressure of that particular component has to be used. So, one has to be a little bit careful when dealing with mixture of perfect gases. Now, in the case of water and R134a, we can get the entropy values from the table. So, for so, for superheated water or superheated R134a, we can directly retrieve the values from the table. For a two phase mixture, we use the same expression as before. So, s equal to sf plus sg minus sf times x for a two phase mixture. Just like we calculated specific internal energy or specific volume or specific enthalpy. For a compressed or subcooled liquid, we may approximate s as sf of t, the corresponding specific entropy of saturated liquid at the same temperature. So, what we will do next is work out a few examples which illustrate how to calculate entropy change using the tedious relations.