 look at 7.2 before we go to open. And in 7.2 we will look at B because A is only formula substitution. Here it is given that actually 7.2 should be made separately as a problem. You have two identical systems of the same mass M and specific heat capacity Cp. Temperatures T A and T B, pressure P naught. Final state equilibrium with each other that means some common temperature. Pressure is already common. Process no change of phase, adiabatic and constant temperature, constant pressure. So let us say that we have one system and you have one other system. This is at T A, this is at T B. They work at constant pressure. So this remains at T naught, this remains at T naught. The constant heat capacity so Cp also is common. And all that we will do is we will combine them together. So let them interact and let us extract a work W. What is this W mass? How will you calculate? That is one way, more engineering way of doing it. One you can substitute formula. But if you do change in availability, remember that they are not interacting with the environment at all. Which dead state? They come into equilibrium with each other. So there is no mention of an environment. Environment does not come into picture. They will come to some common temperature. By if we do not do anything, just put them in an isolated environment. All that will happen is A will lose it to B or B will lose it to A till they come to a common temperature T naught. And that is our simple energy balance we can do. Apply first law and do. But then we will find that there is some entropy produced. Entropy produced means you could have obtained some work but which you have not obtained. Suppose what we could have done? We could have imagined a reversible engine working between the two. Assume for example that T a is higher than T b. Then that reversible engine could take a small amount of heat dQ a from a dQ b from reject dQ b to b and do a small amount of work dW. DQ, ratio of dQ a and dQ b will be related to T a and T b. But because these are finite bodies of mass m each, same mass m. Moment dQ is extracted, T a will reduce a bit. You can relate it to dQ and m and C b. dQ b is made available to T b. So you can relate delta T b to dQ b and m and C b. So after the first cycle or the first interaction, T a will reduce, T b will increase not by the same amount because higher energy is extracted, lower heat is provided. So reduction in T a will be higher, increase in T b will be lower. But there will still be some difference and you can continue running the engine till T a becomes equal to T b. But then it will not be T a plus T b by 2 although they are of the same mass and same heat capacity. In fact, the resulting temperature will be nearer T b than to T a because higher amount of heat is extracted from T a, lower amount of heat is made available to T b. The difference is the work done. And you can write down this equation and integrate it and spend two pages of algebra. There is another way of doing it. Apply first law, apply second law. I am not saying apply availability or exergy analysis. Apply first law, apply second law. Let us say that final initial state, one system at T a, another system at T b. Final state both systems at say some temperature T c. A as well as b. What would be the process? For a and b together it should be an adiabatic process. There will be a w and we want our h p to be equal to 0. That means in the process for the whole system delta s should be 0. Now apply first law to the system q equals delta e plus w. What is q? 0. How do you treat delta e plus w? Both are executing constant pressure processes. So that means after some algebra you should be able to show that this is delta h plus w non PDG because they are executing constant pressure processes without a change of phase. So there could be some change in volume. So let us neglect that work saying it is done. So this is w non PDG and this is what we are going to maximize. Is that right? Because when only PDG work is done q is delta h. So this is our first law. What about second law? It is adiabatic. So second law in principle says that q by T whatever plus s p is delta e. This is going to be 0. We want it reversible. So this also is 0. This is given. This is needed to give us maximum work or maximum w non PDG. Now that gives us the following. Delta s is 0 and w non PDG will be equal to minus delta h. Now delta s equals 0 expand in terms of the two systems. They are going to work at constant pressure. There is no change of phase. So system A goes from T a to T c. System B goes from T a to T c. So delta s is going to be equal to m a P p log of T p by T a that is the delta s of the a part plus m b P p log of T c by since this has to be equal to 0 and m a equals m b C p equals C p. This means that T c T c divided by T a T b will be 1. So that means T c equals square root of T a T b. Geometric mean of T a T b. Not arithmetic mean. If we were to just leave it to themselves not extract any work we would have got arithmetic mean. And we know that the geometric mean of two positive numbers is lower than the arithmetic mean. So the mean temperature of the system will now be lower. So whatever is the lower reduction in energy that will be made available at work. And now you substitute this in delta h and you will get w non PDG and that too this will be the maximum will turn out to be minus m P p P a minus T c minus m P p for the second part. Why cannot we apply our standard availability analysis to this? Because there is no heat transfer with the environment that is enforced to be 0. So that is not a variable that we have. But what is it that we have changed? We have changed the end state. See if we were to leave the two systems to themselves they would have reached a temperature T a plus T b by 2. If one were at say 300 Kelvin another way were to be at 600 Kelvin they would have reached 450 Kelvin. But now they are not reaching 450 Kelvin. They are reaching square root of 600 plus 600 into 400. How much is that? 600 multiplied by 300 sorry. They reached 424.3 Kelvin. So that difference between 450 and 424.3 that difference in energy corresponding energy is what is available to us at work. Energy reduces so there is a work out. Any questions on this so far? I am sure you are a bit confused because if you have read something about availability and exergy there is bound to be some difference in nomenclature some difference in terminology. So I think if you take care of that you should appreciate in the end that it is nothing but application of first law and second law. And although jazzy things are only definitions and terminology and any problem can always be solved by routine application of first and second law. Yes sir. Does it mean that if we are not extracting any work then the system is reaching 450 the temperature for example and in this case it is reaching 346. So 424. So what. So 450 it is reaching 424. So whatever the difference between those two energies that is unavailability for this particular system. Yes. See if you do not extract any work this is the lost work. Yes. Lost work you can say is unavailable work but actually it is not unavailable work. It is work which you could have obtained. So I think lost work or unextracted work is a better term rather than unavailable work. Because in Mumbai New York City syllabus they are in every exam they are specifically asking about unavailability. So every time they are asking 4 definitions effectiveness of the industry, availability, reversibility and unavailability. Sorry dead state. I would like to see what those question papers are and more than that I would like to meet the papers later. Sir if the time permits can you explain this unavailability with some T.S diagram. Yes we will come to that when it comes to open systems it will be it will be clear to some extent. So shall we move on to open systems. Now let us look at the same standard analysis for open system. This is the classical exergy analysis and just for simplicity we will assume that a steady state exists and there is one inlet and one exit and we will do what is known as the standard analysis which means end states inlet and particularly exit. Remember that in an open system you should emphasize this to the students that the inlet state is dictated by some other system. So whenever we change an inlet system inlet state is something we just cannot change unless somebody else tells us ok now work with a different inlet system. Whereas our exit system exit state we will provide this is my exit state and now whatever is the next piece of equipment you design and of course the interface pressure is pre decided by mutual discussion as between a boiler and a turbine turbine and a condenser and so on. So our standard analysis we consider an open system single inlet I single exit E and this is our control volume steady state and let us say that we have Q at E and we have from T this is Q naught from T naught and of course there will be some power output W dot S. The so called shaft work we will take care of flow work by using enthalpy first law Q dot minus W dot is m dot into all that. Since it is steady state the same m dot comes in the same m dot goes out. Now Q dot is made up of Q component Q dot and Q dot 0. So Q dot plus Q dot 0 minus W dot S is m dot. Now I will write here delta H plus delta EK plus delta EP. But henceforth I am going to not write this I am not going to neglect it but I am not writing it just for simplification wherever you see delta H you can add delta EK delta EP if they are not significant wherever you see HR you can add VI squared by 2 plus GZI if you feel like and if needed that will just save some time space of course there is no question of saving paper ending here but if you are writing down you will save paper ending. So this we will write as m dot into HE minus second law Q dot by T naught Q dot by T sigma or integral will become Q dot by T plus Q dot naught by T naught. W dot S has nothing to do with second law plus S dot P is m dot into HE minus S. Multiply this by T naught and you will get Q dot into T naught by T plus Q dot 0 plus T naught S dot P is m dot T naught by T minus S. Now we will eliminate Q dot 0 from this equation and this equation by subtracting this equation from this equation and you should get Q dot 1 minus Q naught by T S right that is what we will get from here. This will get eliminated W dot S minus T naught S dot P minus W dot S minus T naught S dot P equal to m dot HE minus HI minus m dot T naught SE minus HI. Now take W dot S on one side and put everything else on the other side and what you will get is an expression quite similar to the earlier one P S. Check that the algebra is alright. Is that right? Now notice that S dot P can only be positive best situation 0. So this T naught S dot P is only going to reduce whatever is the remaining expression. So this expression is maximum power that can be obtained. Again remember standard analysis by keeping the inlet and exit or the end states fixed. The interaction Q at Q dot at T fixed, the only change allowed is Q naught. You are allowed to vary with T naught, vary at T naught. So the only modification allowed is exchange of heat with the environment. What is this? Without the negative sign? We called it lost work there. We will call it lost power. This is maximum power available because of Q naught at T naught. So you can say this is the max power Q naught at T and this you can say because there is a power exergy of Q dot. When generally they talk of exergy, they say exergy of Q dot and they say it depends on the temperature on at which it is available. We should be clear. Thermodynamics says heat is always associated with the temperature level at which it is available. So when you say Q dot, Q dot at T. Only in first law we talk of Q dot. In the second law, the moment it is second law, every Q dot is associated with a temperature. What is this? With the negative sign is the maximum power that can be obtained when the stream goes in at I and comes out at E. This is the maximum power. This is quite often known as the exergetic power. But mind you bring 10 books and 10 papers on availability or exergy, you will have 10 different nomenclatures and 10 different terminology. This is maximum power due to flow of M dot from inlet to exit. And notice that not only does it depend on inlet and exit enthalpies, it also depends on inlet and exit entropies as is expected because second law has been used and the environment temperature T naught. And of course this is loss power. Now nomenclature, this term is quite often written as equal to minus M dot BE minus DI, where B is defined as H minus P naught S and is exergy of flow. So B also is something like a potential, the decrease in which multiplied by M dot gives you the maximum power obtainable because of that flow through an open system. This is perfectly analogous to what we wrote for the closed systems. Now the restriction here is we have considered one inlet one exit. If there are more than one inlet exit, we know how to take care of those terms. Similarly, if it is not a steady state, you will have that D by dt of that phi, that term will also be coming in. You can derive an expression. I derived it like this without taking into account that term just to reduce the clutter. And of course I have reduced clutter. Remember when I say H e here, it is actually H e plus V square by 2 plus G z. Similarly, H i is H i plus V i square by 2 plus G z. Put those terms whenever necessary or whenever needed. Now shall we start looking at exercises or is there a question? Realize that go back if you have written things down, you will find that this expression I will call this y. It is very similar to this earlier expression. Look at the similarity and look at the differences. The similarity becomes striking because this gets replaced by that M dot delta B. Here it is because of change of state of the system. There it is because of change of state of the flow from inlet to exit. And this term also exists. Actually D by dt of this term will exist there. But we have considered a steady state. So, that term does not exist. Otherwise this will also exist there. But even otherwise look at the similarity between the first term, second term, third term here in the expression x, the first term, second term and third term here in the expression y. And you will the similarity is more striking if you replace H e by U e or E e plus P e e. Then there also a E term, a P v term and a T s term. Here also there is a E or U term, a P v term and a T s term. So, as Mandar, Tendulkar says they have reached nano Kelvin. The question still arises is are you sure it is nano Kelvin? There are issues of measurement, not only issues of reaching that, but how do you measure the temperature? That is only deduced. Sir, but this is really where the sum of time we have got. It is, if you look at it, it turns out that it is always better to work at a lower temperature. But T naught, that one conclusion one can always draw is look, why consider T naught as fixed? Consider as variable or adjustable. Adjustable is not easily variable, but you can consider another situation where it is different. But that means for example, Delhi in summer, most of the engineering systems are assumed, assuming T naught of the order of 35, 40. Some days it will even be 45. You go to lay nowadays, lay in December, minus 20, minus 30 Celsius. So, why not do everything at the North Pole or everything in the Antarctic? Then we will be able to extract more power out of everything. Then there will be other problems, other issues. So, what you are thinking of is 0 Kelvin, but then apparently the interstellar space temperature, you know the background radiation, that is about 4 Kelvin. So, naturally perhaps that is the lowest temperature you can expect to reach. But that means put all our power plants on a huge satellite, let it go far away. Then how do you transmit the power from there to here? How do you make fuel available to that? So, we can think about it. You know such flights of fancies are useful because that leads to science fiction, entertainment. But then again that now that you have brought it up, that also you know reminds me of Arthur Clarke's statement that any sufficiently advanced technology is indistinguishable from magic. So, today we may consider it magic if something like that can happen. But maybe another few decades, say most of the things which have been presented in science fiction novels, whether that of Jules Verne or even Arthur Clarke, who expired only few years ago. They are coming true. So, whatever was science fiction 50 years ago is almost a fact today. So, nothing wrong in imagining, but we being engineers we should realize the current difficulties in achieving that. Maybe our grandkids will achieve it. It would not be that much difficult for them. Okay, so now let us start looking at exercises. Let me look at 7.2 we have an adiabatic turbine expanding air inlet 8 bar 1200 Kelvin exit 1 bar isentropic efficiency 0.85 air flow rate 2 kg per second inlet 12 bar sorry 8 bar 8 bar 1200 Kelvin exit 1 bar. This is ad-specified adiabatic turbine and of course eta T 0.85. Let us first work our standard problem power output. That is what first thing is on. So, we sketch the T s diagram because it is an air the T s diagram also represents the H s diagram. So, you will have an 1 bar line and we will have a 12 bar line 8 bar line. Let us say this is the inlet state I this is the isentropic exit state E star actual exit state E ideally exit state E star this is 1200 Kelvin. How do you proceed? First we calculate T e star using the isentropic relation T e star by T i is T e by T i raise to gamma minus 1 by gamma. We will assume air to be an ideal gas constant C p. So, here T e is given T i is given T i is known gamma is been assumed. So, that gives us T e star. Now, the isentropic efficiency eta s is given to be 0.85 which turns out to be W dot s divided by W dot s T i star max let us not use the word here which is m dot C p T i minus T e divided by m dot C p T i minus this equals this this equals this. So, the idea is from we have T e star T i m dot and C p we get W s star from W s star we get W s W dot s from W dot s we get T e star divided. So, this gives us this expression gives us both T e as well as W dot s and that solves the problem part A. Now, part B maximum power output for the same inlet and exit solutions. Notice that it is a otherwise adiabatic turbine. So, our Q does not exist Q at T does not exist. So, W dot s e fixed would be by our formula minus m dot B e minus B i or m dot into B i minus B which is equal to m dot into h i minus h e minus T naught s i minus out of which m dot h i minus h e is here is that right W dot s this is equal to W dot s from A first part this part what is second part there is a negative sign here there is a negative sign here. So, I will write plus m dot T naught h e minus h i I have just flipped it around by taking care of this negative sign and I also will write this as instead of s i I will write s e star since s e star is s i look at this figure s e star is s i that is used for definition of isentropic efficient. Now, our W dot now it is not no new principle W dot s max i e fixed will be W dot s plus m dot right m dot T naught now s i h i minus h e star are at the same pressure level. So, there will be C p l n T e by T e star C p l n now in this expression you have already W dot s has been computed in A T naught is given assume 300 k unless otherwise stated m dot is specified C p has been assumed T e has been calculated T e star has been calculated. So, this gives us the answer to this now the question is how will you obtain this maximum power this much power you are obtaining already how do you obtain this power this is the lost power is not it this is maximum this is actual this is the lost power and if you look at the expression it is nothing, but T naught into s dot p because it is an adiabatic turbine s dot p is nothing, but s e minus s i into m dot. So, how will you obtain this think you are given ideal I will give you any amount of ideal pieces of equipment tell me how will you obtain this once you consider that you will realize the difficulty in obtaining this do not talk of weight term use ideal pieces of equipment ideal pieces of equipment are isentropic turbines isentropic compressors heat exchangers with 0 delta T m 0 l m T d you understand what is mean temperature difference of heat exchanger right heat exchangers with 0 l m T d and that means heat transfer with 0 delta T should be possible and the pressure ducts with 0 pressure drop is possible how will you do this these are the ideal components available to you mind you you have to maintain the same inlet and exit condition let me draw this figure T s diagram we had a 1 bar 9 and we had an 8 bar and we expanded it from 8 bar to 1 bar, but like this now we have to keep this I fixed we have to keep this e fixed and we must extract this much amount of power you know how do we go about doing it now we are extracting this much power we must extract this additional power for that additional power additional interaction has to take place where is that interaction and what type that has to be a heat interaction with the ambient so it has to absorb that much energy from the ambient and how will it absorb if you look at it let us say this is the ambient temperature T 0 then the requirement is I will draw it in a different color so that it is clear which one only one term will come that is because delta s is tp ln Te by Te star actually it is minus r ln Te by Te star but this is the same P equals Pe star so this term drops out is that right otherwise I will explain to you in detail let me complete this notice that we have to go from here I to e reversibly so what I do is I use a reversible isentropic expander right up to the ambient temperature then I use an isentropic compressor from ambient temperature to the exit state but since the exit state is at a different entropy these two streams have to be joined together by having an isothermal heating process at the ambient temperature so I will need three ideal components an ideal isentropic turbine to bring it up to ambient temperature then at ambient temperature heat it up so that it reaches the entropy that is needed to go up to e fixed exit state and then at that entropy from ambient temperature whatever is the pressure compress it up to the required pressure of one bar just as an illustration I am assuming that this line happens to be below this but it could even be above this does not matter just for clarity I have shown it like this now imagine how difficult it will be to have an isentropic turbine an isentropic reversible compressor and a 0 delta T heat exchanger at ambient temperature that to at ambient temperature mind you this is the ideal process and the exergy expert tells us that no no no this is what you should do otherwise you are losing this much power as engineers what shall we say shall we try doing this or shall we live with losing this power no but there is something better we can do we can say that look as engineers turbine engineers we will look at it in a slightly different way and that brings us to part c we have taken care of part b here what is part c part c says what is the maximum power output if the turbine way to remain adiabatic mind you when you do this to obtain the maximum power the turbine does not remain adiabatic it has to exchange it with the environment if the turbine way to remain adiabatic that is problem c what we will do is we have one bar we have eight bar so from I we are going to ideal adiabatic we will say that for a turbine engineer turbine designer a better option would be keep adiabatic and do not worry about this this is your going to be your q naught making it non adiabatic keep adiabatic but make it a better and better turbine by reducing irreversibility between it try to bring e to e star if I were to keep it adiabatic the best one is adiabatic reversible which is isentropic so the maximum power is if it expands from I to e star and that answer we have already determined here that is this W s star that is the answer for c mind you those of you who are working with turbine the aim of a turbine designer will always to reach e star rather than to reach this some funny maximum work out so we look at exercise 7.3 we have an adiabatic steam turbine inlet 10 bar 350 degree c exit pressure 1 bar flow rate 5 kg per second calculate the maximum useful power output from an adiabatic steam turbine this is a now what is given to us is the following we have an adiabatic steam turbine and for a part diagram let us say on the T s diagram so this is our f line dry saturated vapor line this is the 1 bar isobar this is 10 bar isobar this is 350 degree c isotherm so this is the state I in part a it is asked what is the maximum useful power output from an adiabatic steam turbine if the inlet state is 10 bar 350 degree c exit pressure is 1 bar flow rate 5 kg per second now in the part a since it is adiabatic the best is reversible and adiabatic and reversible means isentropic yesterday when you have gone through that exercise that the 3 adjectives adiabatic reversible and isentropic are vaguely loosely connected to each other the process which is adiabatic need not be isentropic need not be reversible and anyway but a process which is adiabatic and reversible must be isentropic so any one of the 3 does not imply any of the remaining 2 but any 2 of the 3 automatically means the third one because the relation is like a plus b plus c is 0 if one of them is 0 the other 2 can have appropriate values positive negative 0 but if 2 are 0 a and b then c has to be 0 it has no choice so this means reversible and hence in the case a the process will be like this is e for e so the w.s or w.s max is obtained by grouping an isentropic process from i to e whatever is the inlet state i to the exit pressure e yes it is max because we are restraining it to adiabatic so under adiabatic condition the best thing one can do is make it adiabatic reversible which is isentropic in this case entropy produce would be 0 so this will be maximum possible power output if you constrain it to be adiabatic. Now let us look at the second part in the second part rework the problem if the steam loses 40 kilo joule per kg of heat to the atmosphere which is at 300 k what is the exit state in this case so now in case of b we are told that although we make an effort to keep it insulated it is not insulated so you have an inlet state i which is 10 bar 350 degree pressure you have an exit state which is 1 bar but the thing is not adiabatic there is a q dot which comes from the ambient which is at 300 k and q dot is given as m dot into 40 kilo joule per kg is the m dot specified flow rate is 5 kg per second so 5 into 40 200 kilowatt is the heat lost and this is with a negative sign minus 40 it is lost to the ambient and that means it is no more adiabatic I cannot write q dot equal to 0 so how do you proceed what is the best exit state not isentropic the best exit state is the state which produces no entropy okay so the ideal exit state in this case is the one which produces so with this case remember our first law we say let the exit state be the first law says q dot minus w dot s equals m dot into h i minus h i in this w dot s is to be determined m dot is given h i is known h e is not known what about q dot q dot is given to be minus 40 kilo joule per kilogram into m dot so q dot here turns out to be minus 200 kilowatt so we cannot determine w dot s unless we determine h e so how do you get h e we have to use second law what is second law q by t q dot by t so there is a q dot q dot by t not plus s dot p equals m dot into s e minus s i what is s dot p should be 0 because our basic idea is s dot p is the best for us utopia heavens so now here q dot is known t not is known m dot is known s i is known so you determine s e and now notice how does s e compare with s i s dot p is 0 q dot is negative so s e is lower than s i got it so from here obtain s e exit pressure is known to us so p e s e fixes the exit state which fixes h e from which we get w dot s this thing will be clear when we sketch the t s diagram this is i this is s i but our s e is less than s i since q dot by t not is less than 0 it is losing it so q dot is negative so our s e which is actually the s e star the ideal one already based on s dot p equal to 0 is below s i the actual process would be something like this is that clear now before we proceed one last parting shot that is for an adiabatic turbine w dot s is less than w dot s star and this is known as the isentropic efficient we are already familiar with this for any turbine for that matter any turbine or some similar thing we can also have this w dot s will always be less than or at most equal to w dot s max as determined by our combined first and second law but we know that w dot s max and w dot x star need not be the same that 7.4 as told us so this is another ratio w dot s by w dot s max would be defined this would be less than or equal to 1 this is quite often defined as and this is known as the exergetic efficient another term no new concept it is an exergetic efficient now which efficiency shall we really look at as exergy engineers we should always look at this because this is the maximum you could have obtained you have obtained only this so this must be the exergetic efficient but a turbine engineer will always look at it because a turbine engineer knows that look I have to have my turbine adiabatic with absolutely no point in extending it so maximum I can do is isentropic turbine so a turbine engineer will always look at isentropic efficient we will not look at exergetic efficient because as an engineer that is what one has to aim for he will well insulate the turbine he will not say let there be some heat loss I will run a heat engine between that and the atmosphere he will not worry about and finally one should draw a line and so something like this if this is 0 this is positive and this is negative on this side we have turbine produce power on this side we have compressor pump etc. with need to be driven they absorb power nw.t will be here nw.t star I am writing t because I am writing it for a turbine so this will be less than this or at most equal to these two depending on the condition may be higher or lower and hence the ratio of this this goes in the numerator this goes in the denominator whenever we define an efficiency when it comes to a compressor the relative location is the same thing but both are negated w.c w.c star w.c the so called max but it is an algebraic maximum it is not a numerical maximum numerically it may be a minimum the relation is still less than like this but here both are positive here both are negative and hence when it comes to defining efficiencies of any kind it is w.t by w.t star or w.t divided by w.t max because psychologically we are whenever we say something is an efficiency we want it to be a number less than or equal to 1 that is why for refrigerators we say COP we do not call it efficiency refrigerators but since both are negative in magnitude w.c will be higher than that of w.c star or w.c max so when it comes to efficiency here we define it as w.c star by w.t or by w.c max by w.c this diagram I think if you show then there is no confusion about why one efficiency definition it is sort of reciprocal of the other efficiency definition but the less than remains always less than but in an algebraic sense here it may be 1000 ideal 800 real but both are positive here it may be 1000 ideal but minus 1000 the real may be minus 1200 it is lower than minus 1000 magnitude wise it is higher hence for efficiency we put it the other way yes sir. Sometime a term used second law efficient. That is exergetic efficiency also it is sometime known as second law efficient. Excuse me sir is it possible to get exotic efficiency as one in principle yes if you for example if you are talking of isentropic efficiency make the turbine run as isentropic in that case you will get isentropic efficiency one or in case of a general process where you may be talking about exergetic efficiency make the whole process run as a reversible process in that case you will get the second law efficiency or the exergetic efficiency to be one. Notice that both a isentropic reversible process which means a zero entropy producing process or a zero entropy production process any other kind is a thermodynamically utopian process ideal it is used only as a reference engineers will achieve for example from an isentropic over the here the isentropic efficiencies of the turbine must have gone up from 0.7 to 0.8 may be slightly higher than 0.8 but as you start approaching the ideality the cost goes up so if you look at any process and plot it this is positive this is negative and here you put cost so a turbine a very simple turbine not much of technologically advanced one will produce loss of entropy but will not cost much but the same turbine if you try to make it jazzier less entropy producing more isentropic efficiency the cost will shoot up as you come to near reversibility will shoot up to such high values that the management says that look it is too costly people will prefer to have a slightly inefficient turbine at a much less price than a efficient turbine at may be 10 times the cost and this zone is impossibility whatever you invest this is impossible you turbine with negative entropy production is just not possible why turbine nozzle anything anything someone who does not understand the second law we try to design nozzles which work here but they do not work on paper you can do anything you can make it work which one entropy entropy production is 0 exit entropy is lower than inlet entropy so that is there just to indicate that one should not have a mental set that ideal case means exit entropy equal to inlet entropy that is only in case of adiabatic if you are losing it ideal exit state is like to be at a lower entropy than the inlet exit state is that ok now. Sir for the reversible adiabatic process isentropic efficiency and the exotic efficiency matches. For a reversible adiabatic process reversible adiabatic process means ideal process that means n s dot p is 0 and that means both the definition both the efficiencies will be one when they will match I think they will definitely match when they are one but I do not know whether they will match in some other cases also they may accidentally they may but naturally when one when an efficiency is 0 by any definition it will be 0 when efficiency is 1 I think by any reasonable definition any reasonable definition efficiency is 0 but this is where thermodynamics sort of diverges from engineering you know actually basic thermodynamics does not comment it says this is what is possible this is what is not possible it is the when you do exergy analysis you say that look this is the best possible and you should try to approach it that is where engineers are likely to differ a turbine engineer will look at an ideal turbine in a different way and exergy engineer will look at an ideal turbine in a different way and when it comes to real engineer real engineer will always be comfortable with economy so you will have a figure like this you will understand that it makes sense to produce lower entropy but it does cost more money to do it and in an environment of competition you have to always see what your competitors are doing and telling so I think we stop here