 We are back after lunch. K. K. Wagnashik, over to you. Any questions? Good afternoon sir. This is regarding the work availability function. Just we have discussed before the lunch. What is the significance of the negative sign? Why there is a negative sign there? Over to you sir. See the origin of this is our original definition of delta E is minus W adiabatic. It is this minus sign which continues with this. If you write first law, we usually write first law as delta E equals Q minus W or Q equals delta E plus W. But you transpose W on to the other side and you get W equals minus delta E plus Q. So, it is this minus sign that we see in the availability formulation. Apart from that, there is no other significance to it. It is getting carried over from there. Over to you. Walchen call it sangly. Any questions? Over to you. Hello sir. My question is when they said power plants, is this exert analysis will be done for turbines and all. And if it is done to what extent it is possible to achieve WS maximum? Over to you sir. In actual power plants or for actual industrial equipment, exergy analysis, the textbook is exergy analysis is hardly ever done. Exergy analysis is used to write chapters in books, papers in journals and is also used to occupy a few lectures in a thermodynamics course. Just the way we are occupying one day of our course using combined first and second law. Well, everybody wants a process as reversible as possible because we know that a reversible process means no wastage. Perhaps wastage is the best word out there. But we have also seen in the morning, I think I sketched a small figure that if you plot cost against entropy production or lost work, exergy analysis makes us conscious of this lost work. And we can then guess the causes of that lost work and try to reduce those causes, trying to reduce the lost work or trying to reduce entropy production. But as we try to reduce the entropy production, the cost of number one manufacturing that equipment and maintaining that equipment goes up. And hence, at some stage economics will tell you that it is not worth entertaining or incurring that much of cost to do a particular task, whether it is producing something or doing something. Because a higher cost means it will soon become economically unviable. If you are producing a product, you will have to sell it at such a price that nobody will buy it. And if you are doing something for your own like air conditioning, then you will say that the cost is too much and I need not worry about entropy production. Let entropy be produced. Remember, producing entropy is as natural a thing as anything else. So, long as you are within a reasonable cost, reduce it. But beyond that, reducing it any further is just impossible. Over to you. Thank you, sir. Over to you. So, Omayam Mumbai, over to you. Sir, I want to ask you one question. For compressor or for turbine, we show the real process on H s diagram as a curve. So, what should be the nature of curve? The question asked is on a H s diagram, nothing special about the H s diagram except its utility. You might as well draw it on the T s diagram also. The nature of the curve, how should it be? If you take the H s diagram and let us take a turbine, a real life turbine. The inlet state will be somewhere else, exit state will be somewhere else. And I am assuming that the turbine is adiabatic. If it is isentropic, then the process is reversible. Hence, all the internal working will be known. It will also be quasi-static. That will be a straight line on the entropy axis, perpendicular to the entropy axis. The actual curve will be irreversible and unless we know the details of the working inside, we do not know what the curve is. The best thing is to draw a reasonable line. Can be curved, can be crooked, a bit does not matter from the inlet to exit. So, long as we look at the thermodynamics of the overall turbine, we cannot do anything better than this. However, if you start modeling the turbine in detail, including its internal thermodynamics, hydrodynamics as well as the mechanical dynamics, the relative motion between the moving parts and the stationary parts, you will not be able to do any obtain any intermediate state. But if you do stage analysis, it may be possible to obtain some intermediate states like this. May be at the end of every stage or a group of stages. In that case, you are free to again connect i to the point after stage 1, after stage 2, after stage 3, after stage 4. But even then, you will be doing justice to your background in thermodynamics if you join i to 1 by a dotted line, 1 to 2 by a dotted line indicating that I am sure of the states i 1, 2, 3, but I am not exactly sure of what happens in between. So, in this course, since we are not looking inside a turbine, we will be looking only at the inlet state and the exit state just connected by a reasonable looking dotted line. That is all I will say over to you. I am going to try JNTU, then College of Engineering Pune, then we will come to some problem solving. I will assign some problems, do a problem or two and then the rest you can do and the interaction will continue. JNTU, over to you. Good afternoon, sir. We have two questions. The first one is regarding the exercise on open systems and it is problem number OS10. It is about the pump work where the process of compression is given as isentropic. The final state is given as 25 bar and the temperature there is going beyond 700 degrees centigrade for which the property cannot be obtained or the enthalpy could not be obtained from either Mollier chart or from the steam tables. Do we have to treat steam as ideal gas and evaluate enthalpy at that state at the final state? Over to you. I can only tell you, maybe I will show it at the end of the day. If you are getting something exit state, something like 70 bar or something like that, then there is something wrong or 700 degrees C, then there is something wrong in your calculations. Let me see. Maybe there is some conversion factor or something that you have mixed because I have solved this problem and I will solve it again in front of you today. At least the procedure and the order of magnitude tells us that the temperature will not be very far from the temperature of the saturation at 0.1 bar. So it is sub cooled liquid at 25 bar. You are going nowhere near saturation at 25 bar. Over to you. We will write again and we will come back to you, sir. There is another question regarding entropy. Can we define entropy in terms of availability, sir? Over to you. See, we have a change in availability related to change in entropy but there are other properties involved. And anyway, availability, etcetera is just an algebraic thing which has come out of the combination of first and second law. Entropy is defined properly through the second law of thermodynamics and let us stick to that definition. Other things, we can derive relations between change in entropy and any other thing we wish to define or derive. But let us not say that entropy can be defined using availability because through second law entropy is a primary property. Entropy changes properly defined. Availability, etcetera is absolutely secondary of no consequence. Entropy is far more important than anything else. Entropy change is the most important thing from the second law point of view and let us leave it like that. Over. Sir, the problem which I just discussed about is, okay, sir. There is no question regarding that. I got the answer for that. Thank you, sir. So, that is it with JNTU. I will try, last time, College of Engineering, Pune, there are three others but let me do some problems and then I will come back to the Trichy, Bharamati and KJ Somaya. First, I will try College of Engineering, Pune. Over to you, Pune. Hello, sir. We have applied first and second law to closed system and at that time we have written one equation that is delta S is equal to Q by T plus Q naught by T naught plus SP and afterwards we have multiplied by T 0 to both sides of the equation. So, Q by T, what type of temperature, what temperature we have to take if it is non-isothermal process? My question is over. I will come to this question when I solve problems because the explanation of that Q by T term is necessary in a few cases. I will take an illustration and we will see that this is explained properly. Over to you. Any other question? Over. Now, we are at a stage where I think we should start looking at exercises. So, now, if you look at exercises CL. So, in this, you know a question which is quite often asked, is there an energy associated with vacuum? If it is, it will say that look, an energy is associated with vacuum because it may have radiation in it. But rather than consider radiation and any such stuff, let us look at it purely from a simple thermodynamics point of view that we have a perfect vacuum in an environment which is at P naught T naught. Notice that if you come to page 13 of your exercises sheet, under combined first and second laws, it is very clearly written that assume that the environment is at P naught equal to 1 bar, T naught equal to 300 K, unless specified otherwise. So, that means for every problem it is assumed that we have an environment which is at this given P naught and given T naught. So, in this environment, we have a vacuum of volume V vac. What is the maximum work that can be obtained from a perfectly evacuated space of volume V vac? First let us just use the formula. The question now we are asking is, while collapsing this volume to 0, because if we allow this to come, assume that this is a chamber and it has nothing inside it. So, pressure of the atmosphere acts on it and suppose we reversibly slowly slowly allow that pressure to act on it reducing the volume to 0, what is the maximum useful work that we can obtain. Now, since the final state would be absolutely nothing the evacuated space will disappear. So, let us look at what will be the W useful maximum. This is going to be minus the availability of the current state that is availability of a volume evacuated minus the availability of the dead state. So, let me expand this E plus let me write the full formula E plus P naught V minus T naught V minus T naught V naught S the sorry we must this is with a negative sign this should be the final state that is the dead state I think that is our definition. So, this will turn out to be minus E of the final state naught plus P naught V naught minus T naught S naught, where naught is the final state which essentially means it does not have anything minus the initial state E plus P naught V minus T naught S. Now, look at it this is important our vacuum contains no mass. So, from our thermodynamic definition mass cannot have any energy mass cannot have any entropy. So, our E 0 E is 0 E 0 is also 0 our S is 0 S 0 is also 0. Now, what is our final volume the final volume of the dead state is also 0 because the vacuum is simply collapse nothing. So, this is also 0, but of the initial thing what is the initial volume initial volume is V vac. So, that means the maximum useful work is going to be P naught V vac a positive amount. Now, this has come out of formula and this does not really give us any physical inside. Let us demonstrate that it is possible to really do work which in thermodynamics means raise of a weight and we will do it like this. We will rearrange the vacuum in such a way that the vacuum is inside a cylinder piston arrangement and although the volume is V vac let the piston have an area A and A into whatever is the distance from the cylinder head L will give you V vac. Now, since there is no pressure acting on this side on the outside there is a pressure acting which is P naught the ambient pressure because this pressure is acting the piston will try to go into the vacuum. So, we must connect something to it let us say we have a mass less strong inflexible unstretchable string put it on a pulley and put a mass here such that M into G the force which acts on the this thing because of gravity F of gravity because of this pulley it will be transmitted here and F into F that is M G should balance P naught A which is the force on the piston because of the atmosphere. Now, this is the balance situation this way I can maintain the vacuum as long as I feel like. Now, all that you do is this is our ideal leak proof, but friction less piston. So, I just tap the piston giving it a very small velocity imperceptible, but non-zero velocity what will happen? Piston will slowly start moving and we make it stop here and not bounce by putting it a tap exactly equal and opposite tap. So, that it does not bounce back and what has happened? Piston has gone in now we have 0 volume of the vacuum remaining. So, the vacuum has vanished and what has happened to the force F? The force F has gone up by a distance L and this weight rise you can show by simple algebra which we have already done is the force on the piston that is P naught into A into the distance travelled, but A into the distance travelled is the V vac. So, the work done here while raising this piston is W sorry P naught into V vac and what we have shown is the basic definition in thermodynamics of work being done raise of weight in a gravitational field. So, this is an absolute clear clarification of C L 1. So, we know that there is some availability associated with if you say so, availability associated with even a perfect vacuum, but without using even any laws of thermodynamics we can show that the work obtainable is P naught into V vac and if you go by the crooked long winded formula method then also you get the same answer. I would prefer that for such situation you look at the basic physics rather than the formula answer that was about C L 1. Now, we will take C L 4 and I will spend some time on it, C L 4 and other exercises are for you to do. C L 3 is very similar to C L 4, C L 5 is on a closed system, C L 2 is also on closed systems. I am taking C L 4 because it is with air we do not have to look at steam tables and we can do the calculations using calculator quickly. C L 6 is fun I leave it to you as your homework, but let us say C L 4 let us read it. We have air at a high temperature and high pressure 1200 k and 8 bar expands in an adiabatic turbine to 1 bar with an isentropic efficiency of 0.85. The air flow rate is 2 kg per second. Determine a power output to maximum possible power output for the same inlet and exit conditions and maximum possible power output if the turbine were to remain adiabatic. What is the exit state in this case? So, I will sketch it and then I will do the calculation here, but not online. I would also like you to do the calculation and send to me by a chat what is your answer. First I will sketch the diagram so all of us are on the same coordinates. I will sketch a TS diagram. For air a TS diagram and an HS diagram would essentially be the same just a change of scale and the shift of origin if we assume air to be an ideal gas with constant specific heats which we will assume. So, we will assume air to be an ideal gas with constant C pressure. Cp Cv and in this case we will assume Cp is 1 kilo joule per kilogram Kelvin. Fortunately, for us the Cp of air around the room temperature and low temperatures is 1.005 or something simple calculations 1 is good enough and gamma is 1.4 that also is a good even good enough assumption. So, we have a 12 bar line and we have sorry 8 bar line and this is 1 bar line. Let us say this is state I. Isentropically we would go to state E star even these lines are not seen what happened 8 bar line is now seen 1 bar line is also now seen. So, this is E star this is I the turbine as an isentropic efficiency. So, let us say this is E ideal process I can show by a continuous line, but the actual process I will just show by a dotted line whether it is state or a bit crooked. So, long as you do not show it going all over the place it is now this is 1200 K. So, calculate in order the data given is P I is 8 bar P E also P E star is 1 bar. So, first calculate T E star then using the isentropic efficiency which is given to us calculate T E and from that calculate W dot s for the turbine, but this for the turbine is very clear I will not use that subscript anymore this in K and do this in say kilowatt do it and send the answer to me by chat. I am now old enough and I am slow enough on the calculator young people out there should be very fast on the draw. The first correct answer the first answer almost correct is from K K Wach that is T E star is given at 662.4 my recommendation is temperatures let us write up to two places of decimals in Kelvin the two reasons number one these days it is very easy not much difficult to monitor temperatures up to 0.01 Kelvin unless they are very high or very low and the second thing is we have 273.15 as the addition or subtraction from and two Celsius scale. So, then the number of decimals remain consistent now N i T T tells me that T E is 742 K my calculation show it as 743.08 K that is ok maybe they are using a slide rule or a quick method K K Wach says 743 go ahead maybe I could hand out chocolates for the first correct answer 913.84 kilowatt my 835 or something so 913.84 kilowatt that is the first answer from M K SSP so Cummins. Now do not send me any more answers because I have enough of them I have from K K Wach 662.4 for T E then N i T Trichy gave me 743 K. 742 K for the temperature T E then K K Wach gave me just 743 and T P G I T also gave me 743.08 which units do not ever forget the units and then there is a small mistake according to me in M K SSP in kilowatt K is small that is right, but your watt has to be capital K as written by N i T Trichy. So now let us look at let me put in some paper here. Now this is the actual situation and now let us see and with this actual situation we have taken care of power output. So this is part A of the problem actually while doing this we have half way through solving part C what is the maximum power output if the turbine where to remain adiabatic and what is the exit state in this case. Now we know that if the turbine where to remain adiabatic then the ideal situation is remember an ideal situation is always reversible and if the turbine where to remain adiabatic then the ideal situation is adiabatic as well as reversible and that means the ideal situation is isentropic and that means the for part C of the problem adiabatic plus ideal that is reversible we know from our thermodynamics background that this would mean that the turbine should be isentropic. That means the in this case C ideal exit state is E star we have already determined that and that means T E star is what we have determined 662.45 k and since and that means W dot S star would be M dot C p T i minus T E star which would be M dot C p T i minus T E star which be actually your W dot S divided by eta turbine. Now I am not going to ask you for answers because this is only one this thing our turbine output was 913.84 kilowatt divided by isentropic efficiency which was 0.85. So divide by 0.85 and you get 1075.1 kilowatt. So this is the answer to the maximum power output if the turbine where to remain adiabatic and the exit state in this case is pressure at exit is 1 bar temperature at exit is 662.45 Kelvin. This is something we have not worried about availability, exergy nothing nothing nothing. Now let us look at B what does B say? B says maximum power output for the same inlet and exit conditions W dot S max same inlet and exit conditions. Let me go to the next page but I would essentially be sketching the same diagram as earlier. So long as they understand that I am trying to look at that temperature and this temperature it is okay. This is 1200 this is 743.08 and this is 662.45 I am not muted 662.45 the actual process is something like this. Now we want the inlet to remain at I we want the exit to remain at E but since now our process is not isentropic. The non isentropic process which is reversible which will now definitely be non adiabatic and our I will write the formula here and may be the computation I will do on the next page. W dot S max will now be equal to m dot into H i minus T naught S i minus H e minus T naught S e. The neat way to remember this in a simple form is to write m dot into H i minus H e which is the first law without max and if you add max here all that I think I should use write this as m dot into H i minus H e without the max and then add max here and subtract from H i and H e T naught S i and T naught S e respectively. So if you want to write this this is m dot into one term H i minus H e and the second term notice that this is minus T naught S i plus T naught S e and since S e is greater than S i this I might as well keep it as T naught S e minus S i which one? Oh inlet condition is 8 bar and exit condition is why did I write 12 and 8? Anyway no harm done except that you will have to erase and write it. So notice that the first part here m dot H i minus H e minus H e represents the actual power output that is 913.84 kilowatt. This is a positive number so m dot T naught into S e minus S i is the additional power you get or in the other words this is our lost power. So let us now calculate this but rather than calculate the whole thing we will write this as W dot S plus T naught S e minus S i plus T naught S minus S i into m dot. I am going very slow and step by step just to show that just to show how neatly things can be done and so that everyone understands this perfectly. Now S e minus S i equal to you notice that first thing is you notice that S e star and S i the same. So S e minus S i is S e minus S i star since S i is S e star. These are isentropic states by definition or by the requirement of the earlier adiabatic isentropic term. Now we notice that e and e star are at the same pressure and since e and e star are at the same pressure I can write this as C p logarithm of T e by T e minus S i minus S i minus S i T e star and if you just calculate this, this is 1.0 kilo joule per kilogram Kelvin into logarithm of what was our T e? 743.08 yes 743.08 divided by 662.45 we calculate this 743.08 divided by 663.45 equal to this anyway is 1. So this comes to 1.120 kilo joule per kilogram Kelvin. So the second term I will write it as now power lost is m dot T naught is e minus S i minus S i that will be 2 kg per second multiplied by 300 multiplied by this is 300 is in Kelvin's multiplied by 1.120 kilo joule per kilogram Kelvin that gives this as 1.12 multiplied by 300 multiplied by 2. This is 670 in the chat on that 12 bar 8 bar. Now only thing I want to know is whether this 1.120 is all right. I have some 743 divided by 663 kilo joule per kilogram Kelvin divided by 662.45. Oh that itself is 1.12. So I think I forgot to take the logarithm or the logarithm key did not work. How come nobody shouted at me divided by 662.45 equal to logarithm? Oh sorry mistake here because suddenly I notice that this value is too large. This is 0.1148. So multiply that by 300 multiply that by 2. Yes now it is more like it. This is 68.92 kilowatt. I should not or may be I should blame this old calculator of mine. It is 1983 vintage but since it is a you know what you say childhood sweetheart I do not want to give it up. But now you see that our W dot S max by our exergy method is the original power which is 913.84 kilowatt. Plus 68.92 kilowatt which turns out to be 913.84 plus 68.92 982.76 kilowatt. Notice that this is different from the maximum power which we obtained what was that 1075.1 kilowatt notice that this one 982.76 kilowatt is the maximum power that we could have obtained when it were purely isentropic adiabatic 1075.1 kilowatt. And if that power was based on isentropic efficiency we can you know this is a question mark quite often an exergetic efficiency is defined. The exergetic efficiency for a turbine is defined as W dot S actual divided by W dot S maximum by the common exergy analysis. So, in our case this turns out to be the actual power of 913.84 divided by the maximum power is 982.76 kilowatt I do not want to write do not have to write kilowatt because I am know the numerator and denominator have the same units. So, 913.84 divided by 982.76 equal to this comes out to be almost 93 percent. So, that was 85 percent this is 93 percent. What is the meaning of this well it is another ratio just like the isentropic efficiency and there are many people who say that look those who believe more in the exergy analysis than the standard isentropic behavior of a turbine say that it is this efficiency that one should look at and not the isentropic efficiency and they also claim that if you do enough effort you should be able to you know go up to 0.872.76 kilowatt, but I do not think there is anything special of sorry 982.76 kilowatt W dot S max, but I do not think there is anything special about it. We would like our turbines to be adiabatic and hence we would like it to be as isentropic as possible. Hence for us the actual attraction should be that 1015 kilowatt our aim should really be that, but we also know that anything which leads to reduction in the entropy of E and tries to bring it down to E star like this is going to be a pretty costly affair. So to the extent possible we will do and that is why whenever we buy a turbine when you compare two turbines apart from cost and other technical and commercial details we also look at the isentropic efficiency. A turbine which has demonstrated an isentropic efficiency of 85 percent is definitely better than a turbine which has demonstrated an isentropic efficiency of just 80 percent. So higher the isentropic efficiency is definitely better. So, when we have such a nice parameter to work with there is no real reason to define other other efficiencies like exergetic efficiencies. So that brings us to the end of CL 4. I would recommend that you do first come to CL 2. In CL 2 the first part the A part you are given water at three different states 10 bar saturated liquid, 10 bar dry saturated vapor, 10 bar 600 degree C and the final state is P naught T naught that is 1 bar 300 K. Find out the maximum useful work that pen be obtained when the system goes from the specified initial state to the final state not specified here, but naturally the condition is no heat transfer from any source, but heat transfer to or from the environment is allowed. The B part is more interesting you have two identical systems of the same mass same specific heat capacity and same temperatures T A and T B and they remain at the pressure P naught. The initial state is that they are at two different temperatures T A and T B. The final state is they come to equilibrium with each other that means they come to the same temperature and same pressure and the process during this for each one is no change of phase they remain adiabatic except that adiabatic with respect to any other systems they may exchange heat with respect to each other and they remain at constant pressure. This just indicates that for every each system you can write DQ is C P into DT of that system you will have to make the assumption that there is no stirrer work and you can part B of C L 2 can be done by two methods one is apply our first law second law just check the check whether because there is no environment here the common method of availability analysis I do not think will be applicable. The other method is more elaborate method and then that you say that if you have two bodies at temperatures T A and T B I will extract work by running a reversible two T heat engine between them. But then there is a small modification to this problem or to this situation of a reversible two T heat engine that has to be a continuously adjustable reversible two T heat engine. So, what is going to happen is if you I leave it to you as a homework we may not look at it in any detail C L 2 part B. You have a body constant temperature no constant pressure but initially at T A you have another body at the same pressure T B. Let us assume for the sake of computation that T A is greater than T B it does not matter you are just given T A and T B the problem is otherwise symmetric. Let us run a reversible engine between the two. So, if T A is higher it will take a D Q from here but what it will reject is not D Q some work will be done D W and what will be rejected is D Q into T B by T A and D W will be now D Q into 1 minus T B by T A. But because of this D Q the temperature of T A is going to go down because T A loses heat D Q. So, minus D Q would be M the same mass M I am told M same C P into D T A. This comes up as the D H of A and that comes up because it is a constant pressure process and we assume no stirrer work and no change of phase. So, this is one equation this is one equation the third equation is what happens here this T B absorbs heat equal to D Q into T B by T A and that would be by first law like this equal to M mass of B C P the specific heat of B into D T B. So, as the engine executes its first cycle as the engine executes its first cycle absorbing D Q and providing this smaller D Q to T B T A goes down a bit T B goes up a bit because notice if we take D Q to be positive D T A will be negative. So, T A will go down T B will go up and finally it will continue till while going down of T A and rising of T B they reach the same temperature each other and I will leave it to you, but I get a feeling that just check the T final for both T A and T B both A and B I think turns out to be square root of T A and T B the initial temperatures of T A and T B and then you may attempt C L 3 which is similar to a slight modification of C L 4, but it is with steam. So, it is a bit tedious because we will have to do the detailed calculations using steam tapers at this stage there are three hands raised I hope they are still around Trichy, Somaya and Amal Jyoti just check with them in order over to you Trichy. Sir, this is regarding the flow work boundary work and steady flow work. So, we can write the flow work in differential form D of P V is equal to D P V plus P D V. So, it seems that it is a there is boundary work minus steady flow work is the flow work. So, mathematically we can arrive this one is there any physical meaning is there any relation between flow work non flow work and steady flow work over to you. I think what you said is something like this you wrote flow work as D P V as P D V as P D V plus V D P I think you did that then you said this is what is known as boundary work and this I do not know what you said and this you claim to be flow work. So, am I right over to you. Sir, what about this V D P V D V we can call it as steady flow work over to you then is there any between this work physically. The basic issue here is our flow work is not D P V, D P V is a small change remember that our flow work is P E V E into m dot at the exit and minus m dot P I V I there is nothing differential about it it is not a small change this can be at somewhere else this can be say 20 bar V E some value this can be 0.1 bar V I some value. So, I do not think this writing it as a D and expanding this as a is a proper thing to do we do not do this thing at all and anyway finally, this gets embedded into that H I minus H E or H E minus H I and we leave it at that do not try to do this and assign anything as boundary work and all that because this is not a P D V in that sense it is not P D V this is there was if you see the derivation we had an exit plug there was a pressure acting on it and this thing was physically moving by exit velocity multiplied by delta t and from that using the mass flow rate we brought in m dot and V E. So, it is it is not directly comparable to our P D V type of work, but it is comparable to force acting on this pressure into area multiplied by the displacement because of that force that is how it comes up over to you. Over and over I think one of the problems we have is in many text books flow work for some reason is defined as minus V D P or open system work is defined as the problem quite often arises that many text and many books refer to some work component like minus V D P particularly pertaining to open systems how does that come about I have never been able to understand, but it may be some jugglery similar to what we see here as V D P and all that, but straight forward thermodynamics does not bring us to anything like that. Vidya Pratishthan, Bharamati. My query is supplemented with the query which is asked before the lunch it is regarding the compressor if we are considering the compressor then the limiting case is there that inlet entropy and exert entropy must be same, but in practical cases we are seeing that the temperature of air is more. So, in that case the heat may be lost and entropy may be reduced. So, please be make me clear regarding this over to you sir. This is a question pertaining to compressors and unless in case of turbines see in case of turbines the inlet temperature is already at a higher temperature inlet temperature is already high and reduces as the flow progresses through the turbine and because of this the perhaps the most severe situation is at the inlet. We do not have to worry about behavior of materials structural fluid as well as lubricating materials as we proceed to lower and lower temperatures. For compressors it is the other way round. You take an air compressor on a T S diagram this will be the inlet pressure this will be the exit pressure P against exit P inlet and we have a compressor and remember W dot C compressor will be less than 0 that means it will be a power consumption although I am showing the conventional arrow. Now what happens is if like a turbine if you make the compressor insulated saying that look the ideal compressor should be adiabatic isentropic and hence sorry adiabatic and reversible isentropic then from the inlet this will be our exit state and the actual state will go something like this the actual process will go something like this and this is the actual exit state. The temperature here would be pretty high that is one thing and that may be detrimental to some components of the compressor or to the exit duct or even to the lubricants. Back to that there is another issue a compressor raise the main job of a compressor is to raise the temperature raise the pressure but because we invariably compress a gas the gas will raise its temperature as it is being compressed because by default a compressor is near adiabatic and our requirement of that compressed gas may not be at the compressed or anywhere near the adiabatic compression temperature. We usually need the gas at near room temperature for example compressed air for cleaning purposes or instrumental purposes is needed at room temperature and 8 bar or room temperature and 6 bar or 12 bar or whatever but when we compress it we compress it to 8 or 10 bar and it is hot. So we add a cooler sometimes in the compressor itself or sometimes an after cooler in the compressor. So if a compressor is cooled then we also have a q dot which is negative because of the cooling jacket and then our the moment you do this your compressor is no more isentropic and hence for our earthbound compressors which are used for providing compressed air for say cleaning purposes or instrumentation purposes or even for inflating tubes of various vehicles or various sports goods like footballs. We never talk of isentropic efficiency, we talk of isentropic efficiency of a compressor invariably when the compressor is part of a gas turbine power plant or a gas turbine engine because there after the compressor we heat up the gas further by the process of combustion. So there is no cooling requirement at all there is an further heating requirement and there we talk about isentropic compressors. So if you take earthbound compressors usually the situation is a compressor and a cooler or including a compressor with a cooling jacket and in that case we have a non isentropic system and then the first law will become q dot minus w dot c I am just indicating c meaning compressor actually it should be w dot s for compressor remember q dot is negative w dot c is also negative this will be m dot into h e minus h i. So consequently the amount of power required for driving the compressor minus w dot c will be m dot h e minus h i minus q dot for a given exit and inlet temperature. And remember q dot is less than 0. So since this is a negative quantity minus q dot minus will be a positive quantity this is also a positive quantity minus w c is a positive quantity. So the higher the amount of cooling the more power you will have to supply to the compressor and this is our first law for the compressor similarly you can write the second law for the compressor and determine the entropy production of the compressor. If you write the second law for the compressor and assuming that this q dot is taken from some t naught the second law for the compressor would be q dot by that t naught plus s dot p equal to m dot into s e minus s i that will be the second law. And thus that satisfy you over 2 times the power output. So, preciprocating compressor we have to cool the compressor or that cooling is provided for reducing the power output. As you have mentioned here that if the cooling losses are more then power required for compression is more, but in case of preciprocating compressor it is a reverse. So, can you explain that issue over to you sir. Actually what you are comparing are two different situations when I said that with cooling you have to provide more power I had underlined this I will show you that I said that if h e and h i are the same then cooling of that will require more power because if you are cooling your q dot is negative you are extracting q dot and that means this minus q dot is a positive sign minus w dot c is the power consumed by the compressor. But if you are think it just to reach that pressure then a reciprocating compressor without cooling will take you at shown in this figure to a higher exit temperature that means a higher enthalpy and that will require and that will definitely required higher power. I think one will be able to look at it if you really do calculations with a certain class of compressors, but if you keep your if you decide that my exit state is a particular state then I think this will hold, but if you just say that without cooling if I run a compressor then naturally your exit state will be at a much higher temperature and it will consume more power for the same pressure ratio over to you. Thank you sir over to you and now KJ Somaia this is the last interaction before t I am not going to encroach on the t time because my sort is throat I need that t cup desperately. So, over to you Somaia quick one question later on I will take more questions over. Good afternoon sir I am Parag Mule from KJ Somaia sir I have one simple doubt we have derived one expression dsp is equal to ds minus dq by t or a change in entropy delta s is equal to dsp plus dq by t during the second law analysis and in a steady flow process or steady state process entropy being the property the time differential of entropy will be equal to 0. When now delta s is equal to 0 there is some generation of the generation of the entropy so can you please highlight on such illustration that there is no change in entropy, but there is entropy generated I mean what is it how we can understand why a better example that there is no change in entropy, but entropy is generated. So, you want an illustration of a situation where there is no change in entropy sorry there is no change in entropy, but entropy is being generated I will just give you a very simple illustration which I have taken earlier maybe it was not noticed at that time that there is no change in entropy, but entropy is being generated I have a fluid at temperature T naught pressure P naught say it is open to ambient. So, the pressure is P naught and some temperature T naught and let us say the temperature is also ambient I am keeping it on a cold plate at large thermal capacity plate again at T naught. So, good thermal contact any amount of heat transfer can take place now here I have a stirrer I am doing stirrer work which is negative because I can do only work on this and the state of this is maintained steady because it is absorbing heat which is also negative that means it is rejecting heat to the surrounding. So, energy is being supplied to it through the stirrer W stirrer being negative and energy is being extracted by heat transfer the heat transfer is from the system to the surrounding. So, that also is negative and if I balance W stirrer equal to Q by appropriately providing the cooling here then my delta E is 0. So, delta T is 0 delta P is 0. So, this is a steady state state does not change state does not change means your delta S is 0 or in a differential form delta S by d S by d T is 0. But entropy is being produced it is a irreversible process because if you apply first law to this the first law says d E by d T equals Q dot minus W dot these two balance each other and hence this is 0. The second law says that d S by d T will be Q dot by T plus S dot P where S dot P has to be greater than or equal to 0. Remember when you write like this this equation becomes definition of S dot P and this becomes the requirement coming out of second law. So, now here your d S by d T is 0, but since Q dot Q or Q dot is negative this is a negative number since the left hand side is 0 this has to be a positive number. So, S dot P turns out to be minus Q dot by T which is a positive number. This is the simplest illustration that I can give you over to you. Thank you sir. So, we break take a break for T and we will continue our interaction after T. Thank you.