 So welcome back today. Now let us come to the main topic today that is topic 12, Combined First and Second Laws. Fortunately I have not yet been asked a question about why is it that I am talking about combined first and second law. That is because the words used for this topic have been varied and there are a large number of terms which have not been formally defined in thermodynamics in the sense that for example idea like temperature, entropy, enthalpy, other properties are formally defined in thermodynamics. I do not know the real reason but may be the international union for pure and applied physics has defined those topics but in any case you can take any book on thermodynamics particularly the one published in English because those published in other languages some confusion gets into the language when you translate from one language to another. Particularly if the translation is not made by a person whose primary language is the language into which the document is translated and because of this words like availability, exergy and there are not so common words irreversibility, energy. In fact I once came across a word energy, these are all confusing out of this perhaps these two words are more common than these two words. However in spite of that there is enough confusion about this to such an extent that I can show you papers, reports and books where in one set what is defined as availability is meant as exergy in the other set and what is meant by exergy in one set is used the word availability is used for that set. Not only that even when you consider two particular words which use availability in more or less the same meaning when you come to the detailed definition of availability there are differences. Of course one can always say that availability being an energy function should also be referred to only in terms of it differences. For example although we define energy we know that difference in energy between two states is important. Similarly when we talk of enthalpy differences it is only enthalpy. Enthalpy differences are of importance the value of enthalpy by itself means nothing. Similarly when it comes to entropy there are questions is there a process in which the entropy reduces. I mean the value of entropy is meaningless. So sorry the question was can we have negative entropy. Now negative entropy does not mean anything because you take even your steam tables and I think I have said this earlier the values of enthalpy, entropy and internal energy the specific value stabulated here are absolutely arbitrary they are with respect to a reference. One reference for internal energy that is the value of internal energy is defined as 0 for the saturated liquid at triple point and at the same point the second reference is the value of specific entropy is defined as 0. We do not have to define a reference for specific enthalpy for the simple reason that enthalpy is always defined as U plus T. So the moment you define a reference for U H is automatically D. So in your steam tables if you so wish replace all values of say internal energy by the tabulated value plus a fixed constant which could be positive or negative but a fixed constant all through the steam tables. Your choice of the constant it can be a large negative constant in which case all values of U in the table will be negative. What changes the so called tabulated value? I would only say tabulated value of U will change I will not use the absolute value because there is nothing absolute about the value of U. But in any exercise we will always be using U at one state minus U at the other state. So we will be using data U or U2 minus U1 or Ux minus Uy and that difference will not undergo any change whatever you do to the so called tabulated value of U. Provided the relative values of U between any two points remain the same. The same thing is true for entropy. So a negative value of entropy does not mean anything you can have as negative a value as tabulated according to your beams. Not only that entropy difference or entropy decrease for a system also does not mean anything that means the system goes from an initial state to a final state where the entropy of the final state as tabulated or as computed is lower than the entropy of the initial state. Only if the system is an adiabatic system then the second law says that the any process executed by an adiabatic process should be one in which the entropy will not decrease. So if you ask me for an adiabatic system can the entropy decrease? I will say no because if it decreases it will violate the second law of terminology. When it comes to availability or exergy analysis there are you know different definitions which leads to confusion. Particularly for example I have seen exam papers where you say what is the availability of this state and I am sure the student has within books in which availability would be defined in different ways. So then here the question becomes too much of a text specific or the teacher specific or the student learning specific which should not be. In fact whatever thermodynamic exam papers which I have seen one of the main flaws has been using terms which are not standard definitions and because of this reason instead of calling this lecture availability analysis or exergy analysis I have called it combined first and second laws of course of thermodynamics. For the simple reason that whatever you say here what you are essentially doing is applying first law and second law in a combined form. All the exercises that we have done so far except one and I will soon come to that. Are situations where we use first law to determine either the end state given interactions and the initial state or given the initial state the end state and some details of the process determine the interaction or interactions. Then we have used the second law to determine whether the process is possible or impossible or it is a limiting case of a reversible process that is about all. The only quantitative things we have ever extracted out of the second law is the entropy produced and that too we have defined as one side of the second law equation minus the other side of the second law equation. So we have said data s has to be greater than or equal to integral of d cube by T. So since the left hand side is always numerically higher than the right hand side numerically and algebraically. The left hand side minus right hand side which would always be a positive or zero number that is what we have called entropy produced. A question was asked and that is a very valid question all of us are curious to know is although we have said that all we are interested in in the entropy production or entropy produced is that it be a non-negative number but does the magnitude of that number when it is non-negative and positive does it mean anything? Zero means well it is an ideal process a reversible process but for example entropy production of 3 kilojoules per Kelvin compared to an entropy production of 10 kilojoules per Kelvin. How do you combine? How do you compare the two processes? Of course you can say that the one which produces a lower amount of entropy 3 kilojoules per Kelvin would be better in a thermodynamic sense than the one which produces say 10 kilojoules per Kelvin of entropy but in what way better and even that 3 kilojoules per Kelvin entropy producing process? In what way is it bad compared to the one which is ideal and which produces no entropy? Such questions we have not answered yet and in today's lecture we will find some meaning in those. Not perfect quantitative meaning but something which provides quantitative you know quantitative significance to the meaning of entropy production. In fact as I mentioned we have solved all problems keeping first and second law separate to such an extent that there are exercises in the first law exercise cheat for example F 1.10 I am not sure but in an examination I have asked the students to analyze this from the second law of thermodynamics and I think this turns out to be an impossible process. It is okay from the first law point of view it is not okay from the second law point of view. Then there have been hints regarding something like exergy or availability analysis. One of these is the exercise SL 10 on page 9. Here we are showing that we have to show that any irreversibility in the process gives rise to a wastage of the thermal energy of steam. The expression is given which you have to derive. Similarly when you come to open systems come to OS 8 the nozzle problem. Here we are asked among other things to determine what is the limiting exit state and exit velocity. Similarly in the previous exercise OS 7 where you have a compressor in a ship propulsion system you are asked is the process possible or impossible why and what is the limiting exit state. So these are you know we are moving towards exergy availability analysis without formally doing so and we can do that because in the availability and exergy analysis there is nothing which is different from either the first law or the second law. We are just combining first and second law and we can analyze these problems and even the problems in the combined first and second law part of the exercise sheet which is on pages 18 and 19 CL 1 to CL 6. Without worrying about this lecture except that there are certain terms here and there are certain terms which are used in the traditional treatment of availability and exergy which the students needs to know. So remember that our availability analysis or exergy analysis is nothing but the combined application of first and second law plus lots of definitions of terms. Now let us come back to our thing. We know that we have a system which undergoes the process from state 1 to state 2 while interacting with the so called surrounding and there will be some heat absorption, some work transfer and all that. By analyzing the situation using the first law we will say delta S sorry delta E of the system would be or let me leave it like that delta E of the system must be Q minus W delta E of the surrounding would be the heat absorbed by the surroundings minus the work done by the surroundings. The second law would say under the assumption that the systems and surroundings together form an adiabatic combination that means the super system shown by this dotted line would be an adiabatic system and because it is an adiabatic system the second law would say that delta S of this adiabatic system would be greater than or equal to 0. So that will give us delta S system plus delta S surroundings should be greater than or equal to 0. Now we know that this equal to 0 is a limiting case so this is the limiting so called thermodynamically ideal. Now if you want to reach this thermodynamically ideal case what shall we do? You have to change either delta S or delta S surroundings to make this equal to 0. Now the initial states of the system and surroundings are fixed one of system and one for the surroundings the corresponding one for the surroundings. So naturally the end state of the system and or the end state of the surroundings and since they are linked by interactions actually in almost all cases the end states of the system as well as the end state of the surroundings will have to undergo a change. And if the end states undergo a change then the from the first law point of view the interactions either Q or W perhaps both will have to undergo a change leading to either an increase or decrease in Q or an increase or decrease in W. If you use the so called standard analysis or a default analysis by which the first and second law are combined using the classical availability analysis or exergy analysis availability and exergy are two confusing terms they are treated and defined different ways. We will define it in a particular way just because we have to select one of the many ways. You may come across a text book in which it is defined in a slightly different way or perhaps in a significantly different way but the basic idea behind those definitions would be the same but the definitions would be different so some of the formulae would look different. We will show now by combining the first and the second law that when you do this when you go from the actual state to the limiting ideal situation by constraining the interactions in a particular way you can either show that in the actual situation you could have extracted more work out of this system surrounding interaction or you could have consumed less heat during the system surroundings interaction. Again go back to your second law exercise sheet and the exercise which I referred to just now SL 10 is a situation where you have to show that there is a wastage of thermal energy of steam by some amount. Now if you look at this formula you have Q minus something that something can be shown as the minimum amount of Q which needs to be obtained or absorbed to provide this change of interaction as shown by delta E, delta V and delta S and in the actual heat absorbed minus the Q min is the wastage of the thermal energy of steam. In a similar way is one way of looking at the combination of first and second law but the standard way of looking at a combination of first and second law is to determine the maximum possible work which could have been extracted, subtract from that the actual work which is extracted and that difference will always be positive and will be known as the lost work. Now again I have maybe I do not know whether I have shown it or not. We will now understand the following and I will show that this is the traditional or classical analysis using a combination first, second law. This if you want you may call availability analysis or exergy analysis. Again and again I will keep on warning you that availability and exergy are ill defined terms. I should say ill defined are terms which are defined in many different ways by many different workers and many different authors. The European way or the Russian way of defining availability and exergy is entirely opposite that of the western American or British way of defining availability and exergy and even where you take Eastern European things the availability is defined in two different ways by considering either a particular type of reference state or neglecting that reference state. Let us consider the following situation. This is what we say over the standard model and we will say standard model for a closed system. Later on we will look at open systems. The standard model is a very special case which is considered to be the standard availability analysis. We have a system which executes a process from 1 to 2 and during this change of process the change in energy of the system is delta E, change in volume of the system is delta V and change in the entropy of the system is delta S. The system absorbs first then now let us look at the interactions. The interaction will be the work done W and the heat absorbed Q. Now the work done W and the heat absorbed Q it split into some components and this is all because this is part of the standard model. If it does not fit one can always go back to the first and second law and derive it the way we want. What we say is let Q be split into two parts. One part is a Q which is let us say Q 1 which is absorbed from some reservoir at T 1 and in particular we think of a standard environment which will always be there. Of course this T 1 also forms part of the environment. This reservoir at T 1 also forms part of the environment of the system but we will consider a standard environment which is the surrounding atmosphere which will always be at some temperature T naught and some pressure P naught and the heat interaction from the atmosphere to the system will be Q naught. Now this does not seem to be so general but we can make it reasonably general by saying that look if there is a heat interaction from two different systems Q 1 from T 1 and Q 2 from T 2 wherever you see Q 1 and T 1 terms just sum them up over the appropriate number of interactions. Similarly if the interaction is not from a reservoir but from some other system whose temperature changes as this interaction takes place. We will replace wherever you see Q 1 by T 1 replace it by integral D Q by T for that system and of course wherever you see Q 1 replace it by integral D Q for that during that process. So we split the heat interaction into at least two sub parts. One standard sub part is Q naught with respect to the environment the atmospheric environment. Now what is that that takes care of T naught because this Q naught by T naught will be forming part of the second law. Then what we say is that look if the system executes a process by which its volume changes by delta V then it is necessary for the system to push back the atmosphere by that volume delta V. So this W gets split into two components W minus P naught delta V and P naught delta V. This P naught delta V is the work needed this place the atmosphere and what remains is see moment you have a volume change delta V you have to account for this work needed to displace the atmosphere. If delta V is a positive number then you have to do expand positive amount of work to displace the atmosphere. If delta V is negative number this will also be a negative number. What remains W minus P naught delta V is given the symbol W U and is called useful work because useful because this is the work you can use for doing something and making perhaps money out of it. Because the moment you say from the system point of view W is the work interaction the atmosphere will impose a tax which is P naught delta V. So only the remaining part is the so called useful work do not really ask me why is it called useful but that seems to be the very common nomenclature used. So coming back to our situation we have our system which executes a process 1 to 2 the heat interaction Q 1 from a reservoir at T 1 Q naught from the atmosphere which again can be modded as a reservoir at T naught and W we can now nicely split into a W useful and a P naught delta V and remember that the energy change for the system is delta E volume change delta V and entropy change delta S. Now before saying anything more let us apply first law to this system and let us apply second law to this system if you apply the first law that will simply say Q equals delta E plus W and Q is going to be U naught plus Q 1 equals delta E plus W which is W U plus P naught delta V. Now apply the second law the second law says delta S must be greater than or equal to sigma or integral as appropriate of that d cube by T if it is integral d cube by T if it is sigma it is sigma of Q j over T j. So we will write it delta S so delta S will be on the right hand side that is going to be I am putting properties like delta E on the right hand side nothing wrong if you write it on the left just flip it. Delta S is going to be greater than or equal to Q naught by T naught plus Q 1 by T 1 and now what we are going to do is we are going to introduce SP and the second law becomes let me say this is equation 1 this is equation 2 in terms of SP we will now have Q naught by T naught plus Q 1 by T 1 plus SP this is incorrect delta S should be greater than or equal to now there is no greater than or equal to this will be delta S this is equation 3 and this is second law whenever you write 3 you should realize that the entropy production must be greater than or equal to 0 equation 3 now becomes the definition of SP. Now let us look at the two equations again we have Q naught by T naught plus first let me write the first law which from here Q naught plus Q 1 equals delta E plus W useful delta S is plus P naught delta this was equation 1 by S and equation 3 is Q naught by T naught plus Q 1 by T 1 plus SP equal to delta S and we should never forget that SP is greater than or equal to 0. Now what I am going to do is I am going to multiply 3 by T naught because why am I going to do that the question which is going to be asked is see what happens the actual process may be reversible making SP greater than 0 the ideal process would need SP to be equal to 0. So I have to reduce SP from its positive value to a 0 value for that in the standard analysis it is assumed that you are not going to change the change of state. So aim reduce SP from its positive value to 0 a real process will always have SP to be positive value. So to 0 without changing states 1 and 2 and without changing Q 1, T 1 and T naught which anyway is fixed T naught, P naught is the atmosphere absolute surrounding so you won't change. So that means only change allowed SP to 0 is in Q naught. So that means to bring SP to 0 change Q naught and since delta V is not going to change, P naught is not going to change, delta E naught is not going to change if you change Q naught your W U is going to change. Now let us see how this changes by multiplying equation 3 by T naught. So we have let me again write it down so that the process is very clear. First law is Q naught plus Q 1 is delta E plus W U plus T naught delta V this was our equation 1. The second law equation was Q naught by T naught plus Q 1 by T 1 plus SP was equal to delta S. This was 3 and we always had this that SP was greater than or equal to 0. Now I am going to multiply this by T naught. So what happens? T naught goes away from here you have this term becomes Q 1 into T naught by T 1 this becomes T naught SP. So this becomes T naught and right hand side becomes T naught delta S. Now all that I do is subtract 3 from 1 thus eliminating Q naught because our aim is to change delta S which would change Q naught and through Q naught W U would change. Since we want a direct relation between SP and W Q we will get that only if we eliminate the intervening variable Q naught. So when you do that you will get Q naught plus Q 1 1 minus T naught by T 1 in a term which is familiar to us minus T naught SP. On the right hand side you will have T naught by T naught delta E plus W U plus T naught delta V minus T naught delta S. Now we have a direct relation between SP and W U. What we will now do is keep W U on one side say the left hand side say the left hand side and we will put everything on to the other side. I made a small mistake here because I eliminated Q naught in this expression Q naught should not be there because this expression is 1 minus 3 multiplied by T naught 3 multiplied by T naught is this. So Q naught minus Q naught is 0. So Q 1 into 1 minus T naught minus T 1 0 minus T naught SP delta E plus W U plus T naught delta V minus T naught delta S. Now what I am going to do is keep W U on one side and keep this on the other side and even push delta E plus P naught delta V minus T naught delta S on to the other side and you will get this expression W U equal to Q 1 1 minus T naught by T 1 minus. Now this term delta E plus P naught delta V minus T naught delta S I will combine and push it to the other side minus delta E plus P naught delta V minus T naught delta S. What remains? W U has been put on the other side. On one side this has been taken care of the others have been taken care of only this remains that is minus T naught SP the negative sign will remain because it is on the opposite side of W U minus T naught and of course we should never forget that SP is greater than or equal to 0. Now look at these terms what happens if SP tends to 0. We have not changed anything else T naught T 1 P naught delta S delta V delta E Q 1 these are all fixed terms. So, as you reduce SP to 0 because T naught will be a positive number SP will go from a positive value to a 0 value slowly you will increase the value of U and finally you will take W U to W U max. So, I will rewrite this expression which I may call 4 in the following way W U equals W U max minus T naught SP and I will write W U max equal to Q 1 1 minus T naught by T 1 minus delta E plus P naught delta V minus T naught delta S. Now I will take a break for a few minutes and I will be back with you in a few minutes, but till then just absorb all this algebra that we have done so far. In particular realize that we wanted a direct relationship between W U and SP which we have obtained and then these two terms the first two term we have called W U max and rewritten this as W U is W U max minus T naught delta S and the definition of W U max is also seen here just wait a few minutes I will be back with you.