 Welcome back, in the morning session we looked at the availability and exergy analysis, actually I should only say combine first and second law analysis and now we go to the second part of the combined first and second law analysis and that is for open system. So combine first and second law, again the procedure is very similar but we will simplify the procedure and consider a rather special case. As you know we have reduced or we have derived the first law for open system considering a typical one inlet one exit type of a system. So we derived the general case of a transient for many many problems we looked at the steady state. So for open systems we will consider the special case one inlet one exit and steady state. Extending this to a situation where you have more than one inlet or exit is reasonably straight forward, you just have to replace some term by appropriate summation. Taking care of the general case of unsteady state requires you to derive in a fashion similar to the case for the closed system. So even that appropriate term you can add. So I would write here general case that is m inlets n exits and transient is left as m exercise. Now the typical situation which is considered is as follows. We have a system one inlet i one exit e power output w dot s of course this does not include the flow work at inlet to exit that is why that subscript s. Now the heat interactions again we will say as earlier that there be some heat interaction rate q dot one from a source at t one and another interaction rate q dot not to t not and it is this interaction rate which is the so called slack variable and our aim is to derive a relation between w dot s and the other parameters and then try to find out the situation where the entropy production rate s dot p can be reduced to 0 we would expect that to increase our w dot s to w dot s max. We will use first law we will use second law so as before we will be using the first law we will be using the second law we will be using the definition of s dot p and we will proceed from there this is going to be the situation and of course without saying so which is not written here we will be assuming steady state and at the inlet and exit notice that because it is steady state we do not have to worry about the change or rate of change in the state of the system. So rate of change of energy of the system will be 0 with respect to time rate of change of entropy of the system with respect to time will be 0 rate of change of volume of the system with respect to time will also be 0 and since the system is not changing there will be no difference between that we do not have to define as w dot s useful because that p not data v term will become a term which would be p not into dv of the system by dt rate of change of volume of the system with time and since we are considering steady state the rate of volume of the system with time is 0 because the volume of the system is not changing with time. We will also assume that at inlet and exit the energy is made up of u plus the kinetic energy v square by 2 plus g z at inlet and exit. So we do not have to worry about any electrical work etc. This is the assumption which we have made and which gives us the term h e plus h plus v square by 2 plus g z and what I am going to do is again to keep the algebra simpler I am not going to write these terms and say that I will just use h but wherever you see h if appropriate you will replace it by h plus v square by 2 plus g z that will just simplify my algebra our algebra. So let us apply now first law to this system the general expression because it is steady state and there is only one stream flowing one inlet one outlet q dot minus w dot s equal to m dot inlet h that is enthalpy rise from inlet to exit h e minus h i plus v square and all that. Here q dot is made up of two streams so we will have a q dot 1 plus q dot naught minus w dot s equal to m dot into h e minus h i. The right hand side turns out to be simply h e minus h i if you want to be particular you can write plus delta e k plus delta e p but these things you do not have to actually I think I should put it in some other color because I am not going to for example I will put it in green. So if you insist you can have plus delta e k plus delta e p but I am not going to carry over these terms you can add them as needed. So let us say this is my equation one representing first law. Now the second law and the definition of s dot p would bring us to the following equation q dot 1 by t 1 plus q dot naught by t naught plus s dot p equal to m dot into s e minus h i. This will be my equation two with the second law actually being represented by s dot p being greater than or equal to 0. Now again our aim is to reduce s dot p to 0 in the standard case q dot 1 t 1 the inlet i exit e and of course because it is steady state p naught does not have a presence here but sorry p naught does not have a presence all these things you keep fixed. Reduce s dot p to 0 and w dot s will then go to w dot s max because if you want to change q dot s dot p nothing is allowed to be changed here. So actually here you should put even m dot nothing is allowed to be changed here nothing is allowed to be changed in q dot 1 t 1 and t naught. So s dot p as it changes only q naught will change and as q naught changes w dot s will change and we want a direct relation between s dot p and w dot s. So we will eliminate q dot 0 from these two equations. So the second equation after multiplying it by t naught will become I hope I write it correctly now q dot 1 into t naught by t 1 plus q dot 0 that takes care of the first two terms then we will have t naught s dot p on the right hand side we will have m dot into s e minus s i into t naught yes. The first equation remains unchanged q dot 1 plus q dot naught minus w dot s equals m dot into h e minus h i. So let me call this equation 3 and equation 4 equation 3 is nothing but equation 2 multiplied by t naught equation 4 is actually nothing but equation 1 let me not confuse. So now all that you do is subtract equation 1 from equation 3 or equation 3 from equation 1 and you will get q dot 1 into 1 minus t naught by t 1 this will get eliminated q dot minus q dot you will have minus w dot s minus t naught s dot p equal to m dot into h e minus h i minus t naught into s e minus s. And we will now get to write w dot s on one side and we will put everything q dot 1 this term is taken care of now I will write here minus m dot into h e minus t naught s e minus h i minus t naught s i and now all that remains is minus t naught s dot p. Now this gives us a direct relation between s dot p on one side and w dot s on the other side and notice that none of these parameters are allowed to be changed inlet and exit states are the same should write this correctly does not look nice h e minus t naught into. So t naught is fixed t 1 is fixed q 1 is fixed exit state is fixed so h e and s e is fixed inlet state is fixed so h i s i is fixed m dot is also fixed. So that means now I can rewrite this as the maximum power output minus the so called lost power. So let me write this as w dot lost is t naught s dot p. Let us go back a few slides and compare this with this slide w dot lost is this let me write it in exactly this format. Now this w dot s max will be made up of two terms q dot term and m dot term. The q dot term will be now again let us write down the terms first this is the max power obtained from q dot 1 at t 1 a term we are familiar with from the earlier analysis. Now this one including the negative sign is max power that can be obtained or delivered same thing when the stream m dot goes from inlet i to exit without having any heat interaction with anything other than the surroundings that is only q naught is allowed this q dot 1 of any kind is not allowed and this one is lost work and we will now see lost power and w dot lost it is t naught s dot. So this one is very clear and now look at these terms there is a term here h e minus t naught s e there is a term here h i minus t naught s i. So I would say common definition but not the definition not a definition we define now the term here used sometimes is b sometimes it is used as small phi let me use the word b because small phi capital phi sometimes is confused b which is defined as h minus t naught s is often called the exergy of state h s state at h s when environment is at t naught just the way we had the definition of availability of a state given by e v s when the environment was at t naught t naught sometimes this is also known as flow exergy or exergy of flow that is because as I said sometimes even that e plus p naught v minus t naught s is also known as exergy so that is exergy of a non-flow system this is the exergy of a flow a state which is involved in flow and with this now we will see that this can be written down as w dot using this w dot max because of flow from inlet to exit without any heat obtained from a source other than the environment turns out to be minus m into b e minus e i so this indicates that it is something like a potential this has to be m dot the crease in which represents the max power that can be obtained when i is the inlet and e is the exit decrease because of this negative sign now before we go to exercises there are a few things to notice that when we say maximum power this is always the maximum power delivered by a system so w actual will always be less than or in the limit equal to w dot s w dot s max and again before I notice before I say notice here that s dot p represents the amount of power which has been lost because of non-ideal behavior and similar to the closed system situation the factor which converts it into lost power is T naught the environment temperature now this is always less than or equal to whatever is the type of equipment you have for example I will draw a line like this with a 0 here a positive here and negative here this is the simply the real number line if you have something like a turbine which is supposed to deliver power and on the other side you have something like a compressor which is supposed to absorb power in either case this is true this thing is true in case of a turbine w dot s max would be something like this w dot s would be a positive number less than this so this much is the power lost because of irreversibility in a turbine here both w dot s and w dot s max are both positive like 100 megawatt 90 megawatt or 10 kilowatt 6 kilowatt but for a compressor both are negative but the relative position remains the same w dot x max w dot s and the lost power is this so here both are negative so let me put the illustration here for example this is plus 100 kilowatt whereas w dot s could be say plus 86 kilowatt so 14 kilowatt is the lost power 100 minus 86 so w dot lost is 14 kilowatt out here both are negative so let us say the actual power consumed by a compressor would be w dot s would be minus say 50 kilowatt consumed so it is negative with our sign convention w dot s max will be higher than this algebraically so for example this could be minus 40 kilowatt it has to be to the right side of w dot s on the negative to positive number line just the way w dot s is right side of w dot s here w dot s max is on the right side of w dot s here consequently minus 40 minus 50 minus 40 minus minus 50 so here w dot lost will be 10 kilowatt whether this or this the w dot lost will always be this will be always positive or in the limit 0 this is what one should realize do not make a mistake by saying the turbine the ideal power is higher than the actual power delivered for compressor the ideal power is less than the power it is less numerically if you take the amplitude or magnitudes of this 40 kilowatt is numerically lower than 50 kilowatt but because the compressor is a work absorption device actually the power delivered by the actual compressor is minus 50 kilowatt we can say that it consumes 50 kilowatt but the power delivered is always negative for a compressor so it is minus 50 kilowatt the ideal power delivered will be algebraically higher than this say minus 40 kilowatt so w lost in either case is in this cases plus 10 kilowatt the w lost cannot ever be a negative number and of course you can use such a number line even for a non flow process in which case you will have w u max and w u here and here also you will have w u max and w u now let us come to exercises and here in fact the confusion is wider and that is why we will now come to the efficiency by different laws for example the you take a turbine when we came to open systems we define what is known as the isentropic efficiency of the turbine it was defined as the power output of the turbine divided by the ideal power output and the ideal power output of the turbine was defined as the power that would have been delivered by the turbine if the inlet and exit states where to be isentropic that is s e star would have been equal to s i in that case the w star w s star is the maximum or ideal power that goes into the denominator the actual power which is for the inlet state s i and the exit state s e where s e is higher than s i was denoted by w dot s and the ratio of these two w dot s to w dot s star was the isentropic efficiency of the turbine when you apply this exergy principle or use this exergy analysis for a turbine the exergy analysis the standard exergy analysis considered fixed inlet and exit states notice situation considered is fixed inlet and exit states I have written it somewhere here fixed inlet and exit states so the exergy analysis if you were to apply this to a turbine would give you the maximum power that can be obtained by a turbine if the inlet and exit states were fixed and in that case what would be the maximum power how obtained however if you talk to a turbine engineer the turbine would say that look I am not so worried about the exit state I want a better turbine which is as near an isentropic turbine as possible so he is looking at a turbine not from the exergy analysis point of view but from his or her point of view and when we come to an exercise here that difference would be very clear to us for example let us look at the exercise which is cl4 this is an important exercise because it has both the aspects which are asked for we have here let us read it air at 1200 k and 8 bar expands in an adiabatic turbine to 1 bar with an isentropic efficiency of 0.85 and the air flow rate is 2 kg per second now in the determine first is power output so up to here up to the end of part a it is simply a first law open systems problem the additional information needed is the isentropic efficiency it is given that we have an adiabatic turbine adiabatic air turbine the inlet is 1200 k 8 bar exit is 1 bar we are given that the isentropic efficiency is 0.85 what is the power output that can be easily determine first let us sketch the H s or T s diagram although I am sketching T s it will be H s to a scale between T and H let us say that this is the 8 bar line and this is the 1 bar line constant pressure line iso bar let us say this is the inlet state I if the turbine where isentropic this would be the exit state E star the actual turbine is not isentropic so the exit state is E actual process would be something like this we will make the following assumption apart from the given data that P 0 is 1 bar P 0 is 300 k so we will assume apart from P 0 is 1 bar which is not needed P 0 is 300 k which may be needed we will assume air to be an ideal gas with constant C p C v some ant moving around this place we will assume that the turbine works under a steady state we do not have to assume that it is adiabatic it is given and we will assume that delta E k between inlet and exit and delta E p between inlet and exit are negligible both of them now this means because it is an air turbine w dot s would be m dot into h i minus h e which would be equal to m dot into C p into T i minus T e here I have used the fact that air is a ideal gas constant specific hits and here I have used the fact that delta E k delta E p are negligible the turbine is adiabatic and it is a steady state now we will have the definition of eta s is given so this is w dot s by w dot s star indicating under isentropic conditions so let me say this is equation 1 this is equation 2 this is equation 3 and w dot s star would be in a similar fashion m dot C p T i minus T e star 4 whereas where T e star is the ideal exit state where T e star is T e but s e star equals s r now because of this and because of the fact that we have used assume air to be an ideal gas with constant specific hits we can write in terms of gamma this expression that T e star by T i would be P e star by P i but P e star is P e so we do not have to worry about it this is going to be raise to gamma minus 1 by gamma this will be equation 5 so here notice that P i is known P e star is known P e star is P e T i is known so from 5 we will obtain T e star this is the first step second step using T e star we obtain w dot s star this is the second step the third step is using w dot s star and eta s we obtain w dot s so this gives you w dot s third step using w dot s and m dot C p if it is given to you is m dot given it is not given but we can assume then m dot to be 1 kg per second or work in terms of w dot s star divided by m dot m dot does not come into operation oh m dot is given 2 kg per second did I write it anywhere forgot to write m dot is 2 kg per second so even the powers can be calculated so from now w dot s is known substitute that in 2 m dot is known C p is known for air T i is known so T e comes out so T e that is step 4 so this is the way it goes although we write the equation as 1 2 3 4 5 first step is this second step is this third step is this fourth step is this and of course in between we have determined w dot s which is the power output A so this one is the answer A now the next part of this and that is maximum possible output for the same inlet and exit condition so this is the standard exergy analysis b w dot s max for the same inlet and exit condition that means the inlet must be P i T i and exit must be P e T e and T e as calculated by the earlier expression this value of T e now this requires because it is same inlet and exit conditions and we are not allowed to have any other heat exchange except with the environment this standard exergy analysis or standard exergy formulation so we will be using this expression w dot s max this term would be 0 so that will be minus m dot into h e minus T not s e minus h i minus T not s i so this will give us w dot s max which will turn out to be minus m dot you can work in terms of b but I will work algebraically so that the significance of various terms comes into play it becomes clear this is m dot into h e minus T not s e minus h i minus T not s i now let us combine these terms so rearrange this m dot into h e minus h i I will write as m dot into h i minus h e so I have taken care of the h terms now I have T not this is there is a negative term here so T not into s i minus T not into s e but there is a minus m dot so you end up with plus m dot into s e minus s i and now notice that this m dot into h i minus h e term m dot into h i minus h e term this one or its equivalent in the temperature mode is w dot s and notice again from our T s diagram that s e and this is going to be s i which is s e star so this number s e minus s i is always going to be a positive or in the limit of isentropic efficiency being 1 which is not in this case a 0 number so this is going to be greater than 0 since s e is greater than s e star which in turn is equal to s i consequently the maximum power which you can obtain is higher than the actual power by this sorry I forgot to write T not here the dimension is higher than the actual power delivered by this term which is m dot into s e minus s i into T not and this term is the lost power so now the maximum power output for the same inlet and exit conditions so that is b so this thing which now can be calculated because we know the exit state we know the inlet state this is already calculated all that we have to do is calculate s e minus s i and we can determine w dot s max so this is answer b and again if you are not sure how s e minus s i is calculated we should note here that s e minus s i is s e minus s e star because s i and s e s i and s e star are equal i and e star being isentropic states by definition and hence now s e and s e star are on the same iso bar s e and s e star are on the same iso bar so under the assumption which we have made that the working fluid is an ideal gas with constant specific heats this turns out to be C p l n T e by T e star and you will work the problem by assuming the certain C p for air and we have already calculated T e and T e star that is done here for example here T e star is calculated here T e is calculated so we can determine s e minus s e star which is s e minus s i which is s e minus s i and from that we can determine the lost power it has not been asked what is the lost power but that can also be determined now this is okay from a thermodynamics point of view but how will this power be obtained what should we do to this turbine to keep the inlet and exit states the same but obtain more power and for that you will have to see that to obtain more power you should apply first law to the turbine under the ideal situation what has happened is your inlet and exit states remain the same h i and h e but your inlet and exit states remain the same mass flow rate remains the same so the first law says that if you just have an adiabatic turbine from inlet to exit you cannot change this W dot s but then where will you obtain the additional power from the additional power would be obtained by making it a non adiabatic turbine and having a 2T reversible heat engine working between the environment and the turbine may be complicated adjustable reversible heat engine which would provide you the additional power which is shown in this expression as W dot lost now the turbine engineer says look I do not want anything complicated for me an adiabatic turbine would be an ideal turbine when it is reversible and adiabatic that means the inlet and exit states would be isentropic a turbine engineer is not interested in keeping the of course he does not have a choice of the inlet state that comes from the for a gas turbine from the combustion chamber exit but he says there is nothing special about the exit state I would like to make it a adiabatic reversible that means isentropic turbine by bringing e down to e star and that is the question which is asked maximum power output if the turbine where to remain adiabatic what is the exit state in this case in this case we have already solved the problem the part C of this if the turbine where to be adiabatic that means q dot 0 also equal to 0 then ideal situation implies a reversible turbine under the condition adiabatic so adiabatic turbine reversible turbine this implies that the ideal situation in this case is exit s e star and power output equal to w dot s star and both of these have been as computed now you can do this computations I am not going to get into numericals remember the two definitions we have isentropic efficiency eta s is defined as w dot s is w divided by w dot s star where w dot s star is when s e star is s i this is what we have seen and this is the maximum power output under isentropic conditions adiabatic reversible means isentropic condition whereas using the traditional exergy analysis we have determined a w dot s max and which has in this formulation the isentropic efficiency does not enter it has maybe it is entered because s e is different from s e star so the s e depends on what the isentropic efficiency is but now you will notice that the actual power output divided by this w dot s max gives you another definition of efficiency this is known as the second law efficiency or sometimes it is also known as the exergetic efficient and this is defined as w dot s divided by w dot s max whereas where this comes out of this equation I do not have a number for it but I have equation 1, 2, 3, 4, 5 let me say this is equation 6 so w dot s star comes out of equation 4 whereas w dot s max comes out of equation 6 so this is equation 4 this comes out of equation 6 the two numbers are different which can be higher which can be lower that I will leave it you to find out maybe they are higher or lower depending on the particular case at hand but the two things are sure that when the isentropic efficiency is 1 the exergetic efficiency will also be 1 and as the isentropic efficiency reduces below 1 the exergy efficiency is below 1 now before I take up question I would like to tell you the following things sometimes the situation is such that it becomes too cumbersome to apply over standard availability analysis or the standard exergetic analysis even then somebody ask you to determine what is the lost work in case of a closed system or lost power in case of an open system in which case remember that our first law and our second law forget about definitions of availability exergy anything if you apply first law apply second law then you can always determine the entropy produced sp for the rate of entropy production s dot p forget this all that we need then to do is for a closed system determine lost work as sp into the environment temperature and determine the lost power as s dot p into the environment temperature so of all this the safest thing to do is determine s dot p and then determine the lost power or the lost work as either t0 s dot p or t0 sp as the case may be and if you I will leave it to you to do exercise 6 a few years ago this was an exam question which I set up I got the idea from some statement in a book where this something like this was an illustration and it is a good exercise and being true to the laws of thermodynamics the questions I have started at 0, 1, 2 and 3 I will leave it to you to discuss this among yourself and determine the lost power and all the other answers now I am ready for let me wait a minute then I am ready for 1, 1, 2, 0 sent a live pierce institute of technology Jabalpur over to you sir you had given very good example for heat transfer like money in the bank and money in the purse are valid similarly I have got a question that what is the difference between and between and entropy and availability can we explain the physical significance of these terms okay the first thing is the physical significance of many terms in thermodynamics is very difficult to say because unlike velocity which we can physically see or matter at a height which we can physically see you know temperature is something which cannot physically see but perhaps at least within a safe range of temperature say between 0 degrees C and 50 degrees C we can touch something and get a feel for that but beyond that when we come to you know internal energy or entropy we cannot have a physical field and you will notice that of all the thermodynamic properties only the temperature which is measured by some means more or less directly using a thermometer but internal energy or entropy is never measured it is always deduced from some equations or some tables hence providing a physical meaning to quantities which involve internal energy and entropy is why difficult it is almost impossible. Number 2 your question was the difference between entropy and availability well as I have said earlier during these lectures today that it is really the first law and second law which matter so it is only the internal energy change and the entropy change which matters so if you go back to our derivation it is only the change in energy and entropy which matter from all that we can determine what is the entropy produced and we can determine what is in T0sp which is the lost work everything else it is simply algebra which has been done in a particular way and my purpose in doing all this thing for you is I am always pestered by saying you did not teach availability you did not teach exergy I do not tend to teach availability and exergy except as combined first and second law because there is no unique way that you will find it defined all this thing is you can say algebraic manipulation to get finally in this form and then what you see here they say is the change in availability some people will call it change in exergy of course change in exergy is related to for a closed system change in energy change in volume change in entropy so it is a combination of all this and if the major component of energy which is thermal internal energy as well as this another component of availability which is data s or something which we cannot talk about with any physical significance in mind availability is also something about the physical significance of which we cannot talk in words it is something which is defined in terms of this equation and I am showing it to you just because there are other people who have presented it in this way and called it availability analysis or exergy analysis. In fact the only thing which we should remember out of this is SP here represents lost work and if you were to do everything in a perfect reversible thermodynamically ideal fashion you could have extracted some additional work which is T0 into the entropy produced where T0 is the ambient temperature over to you. Siddiqui from SAIT Javalpur, I want to know sir the thermodynamic meaning of exergy destruction is again a funny word there are words like destruction of an exergy or loss of exergy or irreversibility the lost work sometimes is known as irreversibility the exergy destruction is only thing the following way come here. This is the no exergy destruction that means if everything were reversible so they say that this is the exergy reduction which should give you power now if that much work is not obtained that is because of exergy destroyed because of some entropy production. So entropy production is an abstract idea which is the imbalance in the second law of thermodynamics that sometimes is known as exergy destruction these are only terms which are defined there are no new concepts involved the only concept which is really involved is and the only neat definition is that of entropy produced or entropy production which we can properly define as the differences in the two sides of the second law of thermodynamics over to you. Good afternoon sir, this is Abhishek Vishal and I would like to know that the example of reversible processes in thermodynamics. Okay, well reversible process is an ideal process and let me take this opportunity to answer a few questions which I have turned up on the discussion forum there still seems to be some confusion about the distinction between a quasi-static process and a reversible process they are distinct and one can argue that for a process to be reversible it needs to be quasi-static but let me make the thing clear that for a system to execute a quasi-static process it is required that from the initial state to the final state the system goes through a set of states all of them can be asserted to be states of equilibrium or can be demonstrated to be or in a approximate situation can be approximated to be or assumed to be states of equilibrium. So if we have a system let us say some gas so in PV diagram if you have a process which can be shown to go from initial state one so anytime you see the process you will either find it here or find it here or find it here not only that when it starts at one and suppose the whole process takes some few minutes to be executed after a small amount of time you will find it somewhere near one after some more amount of time you will find it slightly further down it is not that you will find it somewhere once here once here so it is an order against time also there is a neat variation. But notice that this is quasi-static this is a representation of a quasi-static process and here we are looking only at the states of the system from the initial state to a final state we are not looking at the interactions we are not looking at what the interacting systems are doing whereas a reversible process cannot be shown just by the state space of one system in fact a reversible process means all involved systems execute properly defined and reversible processes and reversible interactions now the moment you say interactions more than one systems are involved suppose this system while executing this process this system executes an interaction with respect to some other system of the thermal kind then that interaction also has to be reversible that means that thermal exchange heat exchange has to be between two surfaces which are negligible which have a negligible temperature difference in principle 0 suppose it is doing some expansion type of work then the pressure difference between the system our system which is doing work on the neighboring system should be negligible always there should be a force balance and this makes a reversible process difficult to show in practice quasi-static process can be shown like this but if I say that this is a quasi-static process it is also a reversible process just by showing it you cannot say that it is reversible I must then know what is happening to the other systems and what are happening to the interactions a quasi-static process is easier to understand because I can demonstrate it by showing a process line like this on state space a reversible process cannot be shown like this because more than one system is involved quite often and also interactions are involved in interactions are something which I cannot show on a diagram I can show them only by arrows and say that this heat transfer must be across negligible temperature difference and this work transfer should be across negligible pressure difference for example earlier in the day I saw that problem of vacuum notice that that ideal work was obtained by making the process of the vacuum vanishing a reversible process by not only making it quasi-static but arranging it in such a way that just this small impulse would execute it in one way and another small impulse for stimulus would execute it in the other way. So reversible that is what I said a reversible process is something to be thought about say thought process not a real it is just a limiting situation that we want to derive and for that we use that thought process called reversible process over to you 1140 Sastra University Tanjavur I am I am I am sure from Sastra University I would doubt in the irreversible adiabatic process while representing the irreversible adiabatic process in the TS diagram what is the area under the curve represents okay the question is the TS diagram and area under the curve remember that Tds is greater than or at most equal to dq ds is greater than or equal to dq by t that is the second law of thermodynamics okay. So area under the curve that means you are looking at a quasi-static process so quasi-static process means you will have a representation of that now this is the area is integral Tds and as the thing represents it will be greater than or equal to dq so this is the area under area of this is greater than or equal to dq that is all you can say you cannot say that it is equal to dq you will be able to say it is equal to dq only if it is reversible and I will say here equality holds only for a reversible over to you where the area is equal to 0 heat transfer whether it should be equal to 0 or not no if the area turns out to be 0 that only means that your q has to be less than or equal to 0 that is all it means you can have any sign positive or negative so for example if this area represents 5 kilo joules q has to be less than or equal to 5 kilo joules if the area turns out to be 0 for example in an isentropic process area turns out to be 0 that means q has to be 0 or negative my question is because of the irreversible adiabatic process because of the adiabatic process whether the q should be always become 0 because in that representing your T s in the T s diagram there is some area under the curve so that represents the q is not equal to 0 for irreversible adiabatic I want to clarify that see you can be anything let me come to this remember I wrote this that ds equals dq by T plus dsp multiplied throughout by T and you will get T ds equals dq plus T dsp now in all this this dsp has to be greater than or equal to 0 so if the area is 0 that means this is 0 this will always be positive if this is 0 and if this is 0 that means this has to be positive generally that means this has to be generally negative in the special case where it is reversible this will also be 0 this will also be 0 so the equation will be satisfied at 0 equal to 0 plus 0 in case of a irreversible process it will still be satisfied at 0 equals some negative number plus corresponding positive number over and out sir okay thank you very much.