 So I am back online. Now let us look at the meanings of the various terms here. Let us now write down words. First this w u max is known as and is the maximum useful work that can be obtained during the process of the system from 1 to 2. What is this? This is maximum useful work that can be obtained from heat absorbed Q1 at temperature T1 and what is this? This is along with the negative sign, maximum useful work that can be obtained when system changes state from 1 to 2. And again let me make a correction here. This is the term T naught SP. So I should not have written T naught delta S T naught SP. What is this? I should put it in red. This T naught SP is the so called lost that means the maximum work which we could have obtained is this w u max. The actual work is w u. So w u minus w u max will be T naught SP which is the lost work. So if you want to define lost work, lost work or w lost is defined as w u max minus w u and we have shown it that it is T naught SP. Now when you use this term max, max it is not unconditional. All this maximum work and this maximum useful work that can be obtained from Q1 which is absorbed at T1 and the maximum useful work that can be obtained when the system changes state from 1 to 2. And this formula for u max and this idea of lost work is true only when we are considering a process where there is no change in the state 1 and 2 and no change in the interaction Q1 T1 Q1 and also in the temperatures T1 and T naught and the pressure P naught. So the only change allowed is the thermal interaction with the environment Q naught. If some other constraints are there then this standard classical exergy analysis derivation is not applicable and we will have to modify it to consider the constraints specified. Now a few things out of this. First thing is what we have gained number one to understand is that SP the entropy produced represents lost work and W lost that means which we could have obtained but did not obtain is shown to be equal to T naught SP. So this gives some importance to the term SP and in the morning I posed the question that a process in which SP is 3 kilo joule per Kelvin and the process in which SP is 10 kilo joule per Kelvin. We knew that the 3 kilo joule per Kelvin process is better than the 10 kilo joule per Kelvin process and we also know we did not know what was the difference between an ideal process and the 3 kilo joule per Kelvin process. Now we know that if our process produces 3 kilo joule per Kelvin of entropy our SP is 3 kilo joule per Kelvin and if our environment is say T naught is 300 K then the entropy production of 3 kilo joule per Kelvin means that T naught into SP which turns out to be 300 into 3 which is 900 kilo joule is the work which we could have extracted by executing the process in a reversible way but we could not because the actual implementation of the process required or led to an entropy production of 3 kilo joule per Kelvin. So the last work by the 3 kilo joule per Kelvin process was 900 kilo joule. Similarly the other process which was comparatively not so good a process which produced 10 kilo joule per Kelvin of entropy the last work for that process is equivalent to 300 into 10 that is 3000 kilo joule. So naturally losing the opportunity to extract 900 kilo joule is better than losing the opportunity to extract 3000 kilo joule. So if in a process you reduce your entropy production from 10 kilo joule per Kelvin to 3 kilo joule per Kelvin that means you have now increased your ability to extract work and you have extracted an additional work which is 10 minus 3 that is 7 into 300, 2100 kilo joule per Kelvin this is the idea. Now the second one is definition if you go back now let us look at this term and this term. This term does not have a specific name to it but we can understand that look at this term q naught into 1 minus t naught by t 1. What is this equal to? This is equal to q naught into eta of a Carnot or a I would simply say of a 2 t reversible heat engine working between t 1 as the higher temperature or t 1 as the source temperature and t naught as the sink temperature. Actually there is some question mark on this because I can consider this to be an engine only if t 1 is higher than t naught. If t 1 is lower than t naught it will not be working as an engine and this will turn out to be a negative number it does not matter but this has something to do with a 2 t reversible machine working between t 1 and again I am making mistake it should be q 1 I think I have yes it should be q 1. So, this is also q 1. So, this sometimes is referred to and here where we get into the confusion of terms this is sometimes referred to as exergy of heat q 1 from a source at t 1 and again I would not be surprised if you find me in a few hours say where instead of exergy the word availability is used. So, this is a confusing definition because this term will be as it is up to this point it is ok, but beyond that the naming given is done independently by independent authors, but for us to understand is that if you absorb heat q 1 from a source at t 1 and if the environment is at t naught naturally the maximum useful work you can extract is q 1 into 1 minus t naught by t 1 what it is called is immaterial. Now, here we have seen the meaning of lost work which gives some substance to the magnitude of s p and the meaning of this term and it is understanding that the maximum work can be obtained only when you have a reversible 2 t machine and by machine I include both engines and refrigerators working between t naught and t 1. Now, let us look at this term this term which I am this is the maximum useful work that can be obtained when the system changes its state from 1 to 2 and of course when q naught is 0 that means this term does not exist and no change of end states is allowed keeping delta e delta v and delta s same no change in the environment parameters is allowed t naught and t naught and the only change allowed is the interaction with the surrounding q naught and that we write as this is now the term minus delta e plus t naught delta v minus t naught delta s. Noting that t naught and t naught are constant this is written down as minus of minus of change of e plus t naught v minus t naught s while doing all this we should remember t naught and t naught are fixed while expanding this do not ever take a differential of t naught and t naught and that means this is written down as minus e plus t naught v minus t naught s final state minus t plus t naught v minus t naught s as the initial state. Now, notice that this is something like a pseudo property it is sometimes given the symbol phi. So, this can be written down as phi final minus phi initial. Now, notice that phi is not really a property because of the presence of p naught and t naught if it were a property of the system then it would not have depended on p naught and t naught the property of a system should not depend on what are the environmental parameters are. Properties of water at 1 bar 25 degree C do not depend on whether the environment is at 1 bar and 30 degree C or whether the environment is at 0.9 bar and 45 degree C that should not depend, but here we have this phi depending not only on the state of the system e v s but also on the state on the environment p naught and t naught. Now, notice that what does phi represent? Now, this phi represents or phi it is something like a potential. Now, our idea of potential from other branches of physics like mechanics is that a potential is an entity the decrease in which tells us how much work that can be obtained. So, we know that if the gravitational potential is high if you reduce it by bringing our mass to a lower height in the gravitational field Mgh is the amount of work that you can extract out of it. Similarly, the kinetic energy is also potential if you have initially high kinetic energy you can reduce it to a low kinetic energy and the reduction in kinetic energy can also be obtained as or can be extracted as work. So, a potential is an entity the decrease in which tells you the capacity of to do work and we have come across this when we studied the Helmholtz function or Helmholtz potential and the Gibbs function which is the Gibbs potential. So, is a potential the decrease in which the decrease because final minus initial would be increased, but there is a negative sign here notice that. So, the decrease in which represents the maximum useful work that can be obtained because of the change of state and again the change of state must be for a fixed environment and when there is no q 1 from t 1 and the only thermal interaction allowed is with the environment q naught which is a variable which is allowed to variable. Now, the nomenclature again let me tell you sometimes some authors define phi which is called as the e plus p naught v minus t naught s is called the availability of the system given p naught t naught. Remember this is some authors define and if you go from across the Atlantic to this side of the Atlantic and start moving towards Siberia it is likely that at some places instead of availability you will hear the word or see the word exergy. Now, there are some authors or some books which define what is known as a dead state. The idea of a dead state is if you leave a system to itself let it interact any way it feels like with the environment. What will happen? There is likely to be some heat exchange if the temperature of the system is not equal to the temperature of the environment if you allow heat transfer. Similarly, if you allow work transfer by expansion and contraction slowly the system you know the pressure of the system will become equal to the pressure of the environment. For example, if you loosen the elastic property of the skin of a balloon what will happen? Instead of pressurized gas inside remaining the balloon will relax itself slowly so that the pressure of the gas inside equals the pressure of the gas outside and similarly if the skin of the balloon is not an insulation and it is not expected to be an insulation slowly heat will be transferred between the gas inside the balloon and the air outside environment outside so that the temperature of the system also becomes T0. So, dead state of the system is system at T0, T0. Environment is anyway at T0, T0. So, what some authors define is they define alternative definition or I should not say alternative I said. Another definition phi is defined to be equal to not just E plus E plus P0 V minus T0 S which we have seen that that is the definition of availability in some places. Another definition is this minus E0 plus P0 V0 minus T0 S0 where if the system is at T0 P0 its state would be represented by energy E0 volume V0 and entropy S0. So, for example, if I have water at 2 bar and say 50 degree C say 2 bar and 350 Kelvin I hope whatever water or steam and environment is at 1 bar 300 Kelvin then the dead state for this water will be 1 bar 300 Kelvin. You calculate E B S for water at the given state 2 bar 350 Kelvin and substitute here in the first bracket then calculate the E B S for water at 1 bar 300 Kelvin which is the environment system then E0 V0 S0. So, remember here that E0 V0 S0 are properties of the system when it comes to dead state whereas P0 T0 and P0 T0 here are the environment and temperature and this is this is alternative definition of availability. This multiple definitions cause confusion and I am sure this will cause confusion even during examinations because the paper setter may have one idea and the student may have another idea for all you know the evaluator may have a third idea. But whether you use this definition the A definition or whether you use the B definition this expression that maximum useful work or that can be obtained because of the change of state would be phi final minus phi initial with a negative sign because you would notice that this term will get cancelled out from either case. So, in either case w u max due to change of state would turn out to be equal to minus phi final initial remember this. So, this is true. So, that brings us to the end of the first part of this topic classical analysis of availability exergy for clothes system now let us do some exercises before we go to open system. The exercises here are actually SL 10 and then we come to page 18 to solve problems. Now remember that because all that we have done is combined the first and second loss and since first and second loss can always be applied in a basic ab initio format many of these exercises can be done without really using these formulae and just applying first law, applying second law and our thermodynamic understanding and thermodynamic commonset. So, let us look at a very simple situation which is CL 1 what is the maximum work that can be obtained from a perfectly evacuated space of volume V vac that means you have perfect vacuum in a chamber of volume V vac and just above this you will notice that you are given that the environment is at P 0 equals 1 bar P 0 is 300 K unless specified otherwise. So that means for CL 1 we have environment at 300 K and 1 bar and in that we have a chamber which is perfectly evacuated the volume of the chamber is V vac. Let us solve this problem in two different ways one is we have a chamber of volume V vac contains nothing and the environment obviously is at P 0, T 0. Now let us use the idea that we are not there is no mention of a heat source anywhere. So we are not allowed to absorb any heat from any source but of course for extracting this work Q 0 with respect to T 0 is always available with the environment we are allowed to have any amount of heat interaction. Now let us use try to use the formula W U max for the change of state. Now let us look at this I have a vacuum and what is the equilibrium state of the vacuum when it is pressure and temperature are brought equal to the environment. So what is the dead state of a vacuum? Now that means if I allow the vacuum to interact with the environment of course there is no need for any heat interaction because there is nothing inside for the heat to go into. But if I leave it in a balloon with inelastic boundaries the balloon will simply collapse and the vacuum will vanish. So the dead state is nothing I do not even need the boundaries of the chamber. So if I have a bottle with perfect vacuum if I allow it to interact the vacuum will simply vanish and while making the vacuum vanish the bottle to come to pressure of 1 bar what is the maximum work that I can obtain that is actually the question. So let us say since system is vacuum that means contains nothing we can say that it does not have any energy E let us not complicate it by assuming that it has a radiation field and all that. The systems energy E is 0 systems entropy also is 0 only the volume of the system has a value equal to V vac the dead state what is the dead state? System itself vanishes so the dead state will be E equals 0 S equals 0 V equals 0. So if this is understood then we can now blindly use the formula WU max I should say change of state would equal minus phi final minus phi initial or better way of writing E plus E final P naught V let me write this is initial is I and final is F P naught V F minus T naught SF minus E initial plus P naught V initial minus T naught S initial. Now E final is 0 I am just cancelling it out E initial is 0 I am cancelling it out S initial is 0 S final is 0 V final is 0 the only term remains is V initial which is non-zero. So this V initial and of course P naught is non-zero this is not equal to 0. So all that happens is this negative sign and this negative sign gives me a positive sign. So this turns out to be P naught V vacuum. So all that you have to do is multiply the volume of the space which is perfectly evacuated multiplied by the ambient pressure and you will get the maximum useful work that one can obtain. For example if you were to suppose this is the chamber which has perfect vacuum inside I do not have evacuated bottle here but you can imagine then how will I do this work may be I will using thermodynamic magic drill a small hole but prevent any air from going in then in that small hole I will design a thermodynamically ideal reversible gas turbine just a turbine few stages which job will be as the air rushes in extract the thermodynamically maximum amount of power and as the air rushes in finally the pressure is equalize the turbine will start working and when the flow stop the turbine will stop. Without going into the details of that turbine and how it will work and naturally you will say initially the pressure difference is large finally the pressure difference is small. So flow rate will be different all those complications we do not have to worry about and say that it will be P0 back initially when I got this result I was also sceptic saying how can we have such a simple result but then few hours later actually it dot on me that I can demonstrate that this is so the demonstration is simply like this this is CL1 this is the if I call this is the formula method now let us see the first principle again let us go back to our thermodynamic drawing board let us say that we have a cylinder piston arrangement and let us say this is the chamber which is perfectly evacuated of course the ambient pressure is P0 let me say that this is a right circular cylinder or right cylinder of some kind for simplicity and let us say the length is LVAC and let us say that the area of the piston is AP then the P0 into AP will be the force by which the atmosphere is pushing against the piston if we leave it like that the piston will move in and make the vacuum vanish but if we leave it like that we will not extract any work so first I should see to it that I must have here a non stretchable string and I will put it over a frictionless lightweight pulley and I will put a mass here so that the force which the if I show the force now this will be P0 AP would be the force acting on the piston because of the atmosphere and I will put a mass M here so Mg would be the gravitational force and because of the non stretchable string and frictionless lightweight pulley this will be Mg and the two forces will balance each other if I want to have an equilibrium that means the state of the system as represented by VVAC is maintained so I must have Mg equal to P0 AP of course I must have a piston which is leak proof as well as frictionless which is again very very difficult or almost impossible in practice but well we are discussing thermodynamic ideal and well if you go to your library physics popular physics there will be old books by a physicist cosmologist called George Gamow and in one of his books he uses the term thermotopia indicating a world where everything is thermodynamically ideal or thermodynamically utopian so he called it that word thermal utopia and called it thermotopia so let us assume that we are in such a thermotopia in fact that book has various chapters the hero of that book is one Mr. Tompkins and I think the title of the book is Mr. Tompkins in Wonderland and he explains various aspects of physics by taking Mr. Tompkins to various worlds one word is this thermotopian world another word is one in which the velocity of light instead of 3 into 10 raise to 8 meters per second it is of the order of say 10 or 20 kilometers per hour so the you know the relativistic this time dilation and length contraction things are obvious even when you ride a bicycle or drive through a driven in a tram then another one in which the Heisenberg constant instead of some 10 raise to minus 27 arc second is one arc second so what happens when you try to aim at someone if somebody is moving you know that he or she is moving but you do not know the position and so on but anyway let us come back now this is the situation an equilibrium maintained by providing a gravitational force to counteract the pressure of the atmosphere on the chamber so now this is the initial condition now if I want to extract work that means I have to raise this weight and for that all I have to do is do a reversible quasi static process by which the vacuum vanishes all that I have to do is tap this by a small tap a very small tap so providing an impulse to the piston so that the piston slowly but surely without any jerky motion starts moving and stops when it comes to the other dead center or the inner dead center and of course in the ideal case everything will be elastic so it will bounce against the inner surface of the head of the cylinder and bounce back so all that I have to do is when I go there just before it hits it provide a small tap so that it stops there and what happens while doing that the piston has moved from here to here a distance of L-vac because this is a non stretchable string this thing has moved up by V-vac so the amount of work which has been extracted and by first principles of thermodynamics the work is equivalent to the raise in weight in a gravitational field so the work which would be because it is ideal work and w u max would be L-vac into mg and we know that because it is just a small tap at every location this equilibrium relation would be applicable so I replace this by p0ap and since we know that ap into L-vac is V-vac we end up with p0vac this is the solution from absolute first principle I have just argued out things now let us take some other example another example is cl2 in cl2 there are two parts part a and part b part b part a is essentially formula substitution 2 kg of ordinary water substance that either 10 bar saturated liquid 10 bar dry saturated vapor or 10 bar 600 degree C 3 sub cases and we have to determine the maximum useful work that can be obtained when the following systems undergo the specified change here the initial state is specified the final state is p0 t0 so 1 bar 300 k so this is the situation where the system is brought to its dead state and we are asked what is the maximum useful work that can be obtained and remember that from an absolute thermodynamic point of view the entropy production or entropy produced sp comes properly out of the first law and second law actually out of the second law first law to be used as needed and that is something which is invariant so long as we understand what is entropy production and that is the difference between the two sides of the second law delta s minus integral dq by t they should be absolutely no confusion but then when it comes to lost work you can say lost work is t0 sp so it depends on what is the environment temperature is and when you say what is the availability for this particular state there is still more confusion because you have to know what is t0 you have to know what is t0 and you have to select one of the two definitions of availability availability is either e plus p0 v minus t0 s of course you can replace e by u if the contribution of other components of energy is negligible or you will have to use the b definition where it is e plus p0 v minus p0 s minus e0 plus p0 v0 minus p0 s okay so the safest thing for us to ask is what is the entropy produced you are entering a bit of a danger zone when you are saying what is the lost work but when you ask what is the availability of this particular state it is like asking what is the entropy of this particular state well 10 different students can write 10 different values because they may be using 10 different reference okay let us look at cl2 part b we have two identical systems of the same mass m and specific heat capacity cp that is isobaric is capacity temperatures ta and tb and pressure p0 this is the initial state and final state is equilibrium with each other the process is given to be no change of phase adiabatic and constant pressure so let us say that we have two systems a and both are at ambient pressure p0 the advantage of ambient pressure is that if there is a change in volume you are not going to extract any work because p0 into delta v minus the ambient tax which also be p0 into delta v we will take care of that so we do not have to worry about the expansion work here now initially this is at a temperature ta and this is at a temperature tb and if the temperatures are the same then naturally there will be no interaction and no work to be extracted but if the temperatures are different suppose ta is higher than tb then if we allow them to interact some heat will flow from system a to system b because the pressure is remaining constant the volume of a will reduce volume of b may increase but there will be no useful work done but we know that if the temperatures are different actually I can run a adjustable reversible 2t heat engine between them by adjustable I mean suppose let it execute one cycle with a small interaction dqa from here then it will produce some dw and reject dqb now what is going to happen dw will be related to ta and tb through our Carnot relationship so will dqa and dqb to each other and the small amount of dqa if we take that as the controlling variable will give us a certain amount of dw which will be dqa into 1 minus tb by ta and dqb would be dqa minus dw and the dqa will reduce temperature from ta to ta minus delta t dqb will raise the temperature of tb to tb plus delta t and in the next cycle the engine will have to work between two new temperatures ta minus dta and tb minus ta tb plus dtb so that is why it needs to be adjustable and in principle you can do the necessary algebra calculus and all that but let us do something simpler we will say that if we bring this down reversibly to equal temperatures let tc be the final temperature of both a and b so what we will do is if the engine works reversibly then delta s of that r where r represents the engine would be 0 and the w max or w u max will be obtained when delta s a plus delta s b is 0 because this total system is an adiabatic system so all that you have to do is assuming tc constant pressure process so the final states are known calculate the change in entropy as entropy of a as a goes from ta to tc at constant pressure p0 so the change in a is from p0 ta to p0 tc and change in b is from p0 tb to p0 tc and you should be able to show using the fact that it is the initial and final states are at the same pressure the masses are equal of the same mass m the same heat capacity cp and no change of phase you should be able to show that tc is the geometric mean of ta and tb and then you can determine what is the maximum amount of work those who enjoy or do not mind doing detailed algebra and calculus are encouraged to use the first principle more or less dumb approach where you set up an adjustable reversible to t heat engine here and let us say that you obtain dqa work the detail out and you will get that the final temperature is this and the same amount of work as is indicated by this method is produced there are only three centers 1093 1176 and 1146 and I am going to look at only those three centers so 1093 LDRP institute of technology Gandhinagar over to you I have question regarding a optimization of the HRSG ok go ahead HRSG system I think there is some problem in your mind your question I suppose is on optimization of an HRSG is that right exactly yes ok by HRSG I think you mean a heat recovery steam generator yes ok and I suppose you are thinking of a heat recovery steam generator in the context of a combined cycle power plant because HRSG can be used in two three different situations so which one are you looking at over to you sorry it is connected to combined cycle power plant and gas turbine gas system ok so it is the it is a heat exchanger which uses the heat available from the exhaust of a gas turbine plant to raise steam that is the situation then actually the problem is a combination of thermodynamics heat transfer and economics so thermodynamics will give you the basic relations heat transfer will lead to a link between the heat transfer rate and the various areas for the economizer part and evaporator part or if you have a desuper heater part even that and once you use the areas and the material which is required then economics will tell you what is the cost that is the analysis of the thermo economic analysis yes but sir but when we are doing the thermodynamic analysis of the system right at that time sir if we calculate the f p calculate the efficiency of according to the first law and the second law then why there is a major difference between the first law and second law analysis ok well you are not really talking about the HRSG you are actually your question is what is the difference between the first law efficiency and the so-called second law efficiency I am coming to that after the tea break when I come to open system ok so wait I will now go to see when you say optimization of HRSG that is an entirely different issue and then finally when you come to the second law analysis and first law analysis the first law analysis is purely an energy balance the second law analysis is simply determination of s p or s dot p that is where thermodynamic stops but finally the decisions are taken by considering the economic aspect ok. So from that point of view I will stop after the I will come back to this after the tea break but then I am not going to go into the economics part ok over and out 1176 University BDT college Dhawan Giri over to you. Good morning sir. Static and stagnation condition with reference to the system. Oh these are something to do with compressible fluid flow but I would say you may put this question again to Professor Puranik when he comes here is to be here in the afternoon sometime and we will announce when he will be here. Static is as measured by an instrument moving with the flow so the word static means that the instrument and the flow are moving together so the instantaneous thing is the static so no effect of velocity you need not really move with the flow but see to it that a lower or a higher velocity does not affect your condition whereas a stagnation condition means whatever you have the flow reduce it to V equal to 0 and reach a P0 T0 so that this is done by a reversible isentropic that means adiabatic also process. So if you that means reduce your flow to 0 velocity that is the stagnation through a reversible isentropic process in that case what you have are the stagnation condition so the stagnation pressure and stagnation temperature comes out of this in fact T and T0 are related to the first law and then P and P0 are related to T T0 and the fact that you have a isentropic situation. Over to you any other question right there is one more question yes can you define the practical example of the combined first and second law which we can give an example to the students the two examples which I have taken are the combined examples of first and second law there is nothing anything special in fact I would prefer that you look at the exercise SL10 again that is an excellent example of where first and second law are combined if you need after the T break during discussion I will bring that again over and out before T let me just talk to the third centre which was 1146 knowledge institute of technology tell them over to you. Good morning sir and from our centre there are two question yes one question kindly clarify about the dead centre with an practical cyclic process. I think there is some confusion about the dead state and the dead centre the dead centre comes in reciprocating engines and we will talk about it when we consider cycles tomorrow ok whereas the dead state is nothing but if you bring the system somehow to a pressure and temperature which are equal to the pressure and temperature of the atmosphere then the state of that system is known as the dead state this is just a definition so if you have in your pressure cooker suppose you have saturated dry saturated steam at 2 bar you want to know what is the maximum work that can be obtained then by bringing it to the dead state then in that case the dead state is executed process by which that 2 bar dry saturated steam is brought to a situation where you have 1 bar say 300 K assuming that is the environment. So the steam at 1 bar 300 K will be the dead state for that steam the dead state is nothing but the state of the system at P0 and T0 where P0, T0 are pressure temperature of the environment over to you thank you sir one more question from our centre regarding energy consideration have a higher temperature the work convert is lot of source is available but have a lower temperature if energy or work conversion just like a energy cascading just I clarify the energy cascading methods any different ideas if you look at our derivation here it is very clear look at we have not looked at that term in any detail but this first term this term Q1 into 1 minus T0 by T1 maximum useful work that can be obtained from when you receive Q1 at T1. Now assuming this is a heat source and T1 is the temperature it is obvious to you that given a fixed value of Q1 if T1 is higher then you can obtain more amount of work and that is very clear because 1 minus T0 by T1 is the maximum efficiency for a 2T engine or a Carnot engine working between T1 and T0. So just thermodynamics tells us that if you have to absorb it you should absorb it at high as high a temperature as possible that is about it and if you go back it has its origin in the definition of entropy which is dQ by T so that also says that a higher the temperature is better because it will lead to a smaller rise in entropy and hopefully that will lead to a smaller rise in entropy production. So thank you over and out it is time for T we will meet again at 1130.