 Now, let us come to some problem solving, where we will be, at least I think that we are having fun and I hope you will also seem so. We will be soon be looking at the exercises on property relations. So, please open your exercise sheet on page 9, exercises known as property relations P R 1 to P R 14, out of which P R 14 we will not discuss just now, that is and I am giving you a advance warning that will be part of the final exercise at the end of the course, not just that its modifications and now, let us begin with P R 1, it says consider S as a function of T and P show that T D S is something, this is absolutely a basic exercise even basic than what we have, before we do that there are some preliminary stuff for example, we must know how to manipulate and obtain Maxwell's relations, we should be comfortable with partial derivative. And then for fluids, we have shown that C v which is defined to be partial of u with respect to T at constant v is has been shown to be T partial of S with respect to T at constant v. Similarly, you should show and not hesitate to use this relation C p which is defined to be partial of H with respect to T at constant p is temperature into partial of S with respect to T at constant p. So, wherever we see C v or partial of u with respect to T at constant v, we should not hesitate to put it and wherever we see C p, we should not hesitate to put this. Now, let us look at it says consider S as a function of T and P, let us go to the next page. So, that everything is over in one page, this is P r 1, we are told to consider S as a function of P and T or T and P, turn it around does not matter. Let us consider S as a function of T and P yet partial of S with respect to T at constant p plus sorry into dT plus partial of S with respect to P at constant T dP. This is just expansion of S as a function of P and T, this is the differential of S. Now, the moment you say partial of S with respect to T at constant p, it should remind you of C p, but there is a factor T missing. So, let us multiply this whole equation by T and we will get T dS is T partial of S with respect to T at constant P dT plus T partial of S with respect to P at constant T. This we know and I said we should not hesitate to replace that by C p or C p by that, here we see this. So, we will replace it by C p. So, this is C p dT plus what about this, again let us open our Jacobian box, this is partial of S with respect to P at constant T, partial of S with respect to P the constant is T. So, we have a T S in the numerator replace that T S by P v. Now, we have V p P T, P is on a diagonal. So, that means this partial derivative partial of S with respect to P at constant T would be a derivative which is in which P is constant, but because that is on a diagonal there is a negative sign and partial of V with respect to T partial of V. So, this we derive the first T dS relation, T dS is C p dT minus T partial of V with respect to T at constant P and again notice, if you want you can differentiate, you know derive divide this by T. So, you get dS is C p by T dT minus partial of V with respect to T at constant P dP. Again you will notice that there is one term containing C p which will require the knowledge of C p and another term containing just P V T data as we saw in case of the U. So, when you integrate this again you will have to integrate it along an isobar and then along an isotherm. When you integrate along the isobar you will require the knowledge of C p as a function of temperature at one pressure, but you will require the knowledge of the P V T data all over the state space or at least over the required part of the state space. That takes care of T R 1, P R 2 says consider S as a function of T and V and derive another expression for T dS similar to that in exercise 5.1. I will take a 5 minutes break, but meanwhile I would like all of you to show that follow this exercise I will keep this slide here, but instead of S as a function of P and T consider S as a function of T and V, otherwise follow the same procedure and you should be able to show that this is C V dT plus T partial of P with respect to T at constant V dV show that and there are these two are very famous equations usually these are known as the T dS equations. For some reason this is known as the first C dS equation or the equation T dS equation based on pressure and T dS equation based on volume. The name itself is T dS relation, it is over, but anyway this was not a test and I hope you have been able to derive this and you have to follow essentially the same method and you will notice the similarities between the first and second T dS relations. And I also recommend that when we derive this dU also it is recommended that you derive an expression for dH it should be of the order of here we considered everything to be a function of T and V, we considered U as a function of T and V consider H as a function of T and P, you can and obtain dH as a function of C P dT plus something into dP and this should be true for any fluid and if you are interested just substitute P V T data for an ideal gas here and show that whatever is in the square brackets is 0. And you should be able to show that this is very similar to this rectangular bracket here P will be replaced by V, V will be replaced by P and there will be some changes in signs that is about it, otherwise it should be very similar to that because most of the relations are like that. Here also you would notice that between the first T dS and second T dS equation wherever you have P it is replaced by V, wherever it is V replaced by P. So, P and V are interchanged and sometimes there is a change in sign, here there was a negative sign, here there is a positive sign. So, let me do one more problem and then I will take up, I will begin the discussion session. The problem which I am going to take up is PR 3, it says using the results of exercises 5.1 and 5.2 show that C P minus C V is something, something, something. Now, remember that these are the property relations for any simple compressible system or for a fluid. We know that for an ideal gas C P by C V is a constant R, for a general fluid it will not be so and this is the expression. If you want to check even before deriving this check that if you substitute on the right hand side P V equals R T. You should get that thing to be equal to R, but let us first derive this and for deriving this we must be using these two T dS relations. So, we will rewrite those two T dS relations and we will see what we get. What I am going to do is rather than rewrite the two T dS equations, I will equate the right hand side of the first T dS equation and the right hand side of the second T dS equation. So, the right hand side of the first T dS equation is C P dT minus T dP by dT at constant V dP. Let me check whether I have written that rightly. So, it is dV by dT at constant P dP, dV by dT at constant P. The second T dS equations right hand side C V dT plus T partial of P with respect to T at constant V dV. Now, notice that we have a dT here on the left hand side and right hand side and we have a dP here and we have a dV here. Let us rewrite this equation with dT on the left hand side and dP on dV on the right hand side. Finally, I want to consider T as a function of P and V that is not the only choice. I am taking the choice where I want dT to be left hand side dP and dV on the right hand side. You can select any one of dT dP and dV to be on the left hand side and the other two on the right hand side. That means, you can consider T as a function of P and V or P as a function of T and V or V as a function of P and T. The final result will be the same. I have decided to use this. So, if I transpose terms containing dT on the left hand side and everything else on the right hand side, I will get the following and divide throughout by the coefficient of dT which will be Cp minus Cv. I will get this relation dT equals T divided by Cp minus Cv into partial of V with respect to T at constant P dP plus T divided by T. Cp minus Cv partial of P with respect to T at constant V dV. From here to here, from equation 1 to equation 2 is only algebra. Now, we use calculus. Notice that T, P, V are all properties. So, this is our standard formula d phi equals m dx plus n dy, d phi is equal to m dx plus n dy. Hence, this m must be the partial of T with respect to P at constant V. So, I get partial of T with respect to P. At constant V equals T into Cp minus Cv partial of V with respect to T at constant P. And from the second part, partial of T with respect to V at constant P must equal to T at constant V divided by Cp minus Cv into partial of P with respect to T at constant V. Let us say these are equations 3 and 4. Just the way we had a choice here of selecting any one variables one derivative one differential dP, dT or dV in one on the left hand side. Now, we have a choice of either using equation 3 or equation 4. Let us take equation 3. Let me transpose this Cp minus Cv on the left hand side and the partial derivative on the left hand side to the right hand side. You will get Cp minus Cv equal to T partial of V with respect to T at constant P divided by partial of T with respect to P at constant V. And by the property of partial derivatives whatever is in the denominator effectively it is reciprocal can be put in the numerator. So, the numerator remains as it is partial of V with respect to T at constant P. And the denominator gets inverted partial of P with respect to T at constant V. And this is what we set out to prove. So, we can say Q e T. And of course, we used equation 3 for this. You can check that even if you equation 4 use equation 4 you will reach the same result. So, again you can write Q e D. And here I had made a choice of selecting T as a function of P and V. You can select P as a function of T and V or V as a function of T and P. Again in any one of those two other things you will have a choice at this stage. But any of the choice finally, you will reach the same conclusion. So, just check that appreciate this P r 1 2 in a property table 3 consecutive entries at constant pressure are as follows. Note what we have is remember this is enthalpy entropy. And they are at constant pressure. So, at some constant pressure you have these 3 points may not be on a straight line. I think they are approximate straight line, but not exactly straight line. So, let us say you have points 1 2 and 3. You have to determine T 2 and G 2. So, remember that since this is an isobar constant pressure the variation of D H. Remember our basic property relation for D S is T D S plus V D P. And that means temperature is partial of H with respect to S at constant P. Temperature is partial of H with respect to S at constant P. So, partial of H with respect to S at constant P is the slope of the tangent here. And all that you have to do is say fit the parabola between 1 2 and 3 and determine the temperature at 2 as the slope. Actually these 3 points have been taken from the steam tables. So, but may not be this steam table some steam table. And you can now determine the temperature. And since entropy is also known H minus T S will give you the value of G. If you look at the variation in H is 65 to 86, 21 kilojoule per kilogram between 1 and 2 and 22 kilojoules per kilogram between 2 and 3. So, if you approximately assume it equal then the slope of this chord and the slope of the tangent here will be almost the same. So, you can a crude way. So, recommended is fit a parabola that is H equals A plus B S plus C S squared. Now, this A B C 3 points can be determined by using the data at 1 2 3 and then determine the slope. Whereas, a crude approximation for T 2 will be H 3 minus H 1 divided by S 3 minus S 1. So, this will be 3 5 0 0 0 0 0 0 0 0 0 0 8 minus 3 4 6 5 divided by 8.329 minus sorry 8.384 minus 8.329. Units of H are kilojoule per kg units of S are kilojoule per kg k. So, this will cancel out you will get your answer in Kelvin. I think that explains it and now I am ready to take questions. Let me go in order there are 4 people already. NIT Trichy are you ready with asking for asking some questions on what happened today morning over to you. Sir, in your PV diagram and TS diagram you said in the reversible on both curve shown in the diagrams are equal. So, can we equate both area under the curve are equal so that the cyclic process is become 0 that du from the first law am I right over to you. See the question is about the PV diagram and the TS diagram. The question is about the PV diagram TS diagram it is only when you have a reversible cycle. That the two areas are equal when you have a reversible process in a reversible process which is not a cycle d e may not be 0 or will not be 0 in general. So, your heat interaction may not equal work interaction. Remember here any process you have delta e equals q minus w and if you consider a fluid at rest you have delta e equals q minus w expansion minus w other and even if you have all this plus reversible all that happens is delta e is q minus w, but w is then w expansion because w other cannot be there it is only, but this delta e is not equal to 0 it is only when it is a cycle that you have 0 equals q minus w. So, this implies q equals w. So, that is why that reversible cycle for a fluid at rest is necessary over to you. This is regarding the gaseous behavior. So, we have this kind of Maxwell relations the several equations have been derived for gases gas can also be considered as a fluids. For example, in fluid mechanics we have several non-dimensional numbers comparing all the properties. So, why do not we have a simple non-dimensional numbers instead of having this kind of complex equations over to us. You asked about dimensionless numbers. The dimensionless numbers are they are used in fluid mechanics they are used in heat transfer and all that. Remember those dimensionless numbers are essentially for with a transport phenomena that means you have state plus transport properties. So, take Reynolds number you have a geometric factor diameter you have a velocity factor which is the actual process dependent which is the velocity then you have properties like rho and mu out of which rho is a thermodynamic property, but mu the dynamic viscosity is not a thermodynamic property. Whereas, in when you mention Maxwell's relation Maxwell's relations are with variations. The only way you can make them dimensionless is perhaps by dividing them with appropriate numbers. As far as I know the only dimensionless number or the most common dimensionless number used is what is known as for some unfortunate reason the compressibility factor and that is for a state it is defined as p v by r t for any fluid. So, this is the compressibility factor, but if you want to take for example partial of s with respect to p at constant t and you want to make it dimensionless the only way to make it dimensionless is maybe you will have to divide p by something say critical pressure. So, I will write here multiply this by p ref and this will be by some something delta s ref and then you will have this I will call s plus as a dimensionless s and then this will be dimensionless pressure or reduced pressure and then constant this is maintained constant. So, there is no question of something like this artificially we will have to make them dimensionless, but that only will just add some numbers here the basic characteristic of Maxwell's relations will not change and will not be simplified if you use dimensionless parameters over to you. Thank you very much sir over and out. Thank you K. K. Wag Nashik over to you. Sir, how for a reversible process w other is equal to 0? The question is for a reversible process how is w other equal to 0? Now remember that when you want a process to be reversible it is needed that interactions take place say whatever interactions take place from the initial state 1 to final state 2 they can be reversed when you come back from final state 2 to the initial state 1 and of course the process has to be quasi-static there are other requirement, but one requirement is as you go from 1 to 2 whatever are the interactions those interactions must be turned back must occur in the other way by exact method exact quantity. Now we have a fluid and fluid has one mode that is expansion compression mode which is a two way work mode and which gives us w expansion it could be w compression also in one way it will be expansion another way it is compression, but if you have something like a stirrer then you can stir the fluid, but the fluid cannot stir the stirrer. So, this is a one way mode of work. So, if you have a stirrer work in the process then it cannot be a reversible process. Hence for a process to be reversible there should not be any stirrer work or any other stirrer type of work for example, electrical work for a fluid remember which is a simple substance. So, there is a simple system. So, there is only one mode of work two way work that is expansion work. So, for a reversible process since there cannot be any other mode of work for a reversible process for a fluid w will be w expansion I hope that explains it over to you. Thank you sir. Mepko Sivakasi over to you. Sir, I asked one question during the coordinator training program that whether we can write C v from the T d s is equal to C v d t plus P d v from that whether I can write T into dou s by dou t at a constant volume is equal to C v. You told that no to that questions, but now you have derived one equations based on that type that is d t during the derivation of C p minus C v we have the relations d t is equal to T by C p minus C v something like that. From that you have written dou t by dou p at constant volume is equal to T by C p minus C v dou v by dou t at constant pressure similar equation you have written today. I think what had happened during the coordinators workshop and what has happened and what generally happens quite often is something like the following people take short cuts. For example, if you have say let me take d u equals T d s minus P d v. So, from here people simply divide by something say divide by temperature and they the thought process goes like d u by d t d s by d t d v by d t and then you say let us put this as partial derivatives with something as constant. So, then you will say put something like this something like this and convert this into partial derivatives that is wrong that is not what we want to do. If you want to get d u by d t at constant p then you should start with d u equals T d s minus P d v. Consider u as a function of T and P because you want partial of u with respect to T at constant p. Consider s as a function of T and P consider v as a function of T and P expand d u d s and d v using the normal calculus and then combine terms containing d p and combine terms containing d t together. So, quite often in case of simple situations what you get by this and what you get by the proper method may be the same, but that is only accident. So, do not take those short cuts do not it is very dangerous to take mathematical short cuts. We are not mathematicians we are engineers. So, whatever mathematics we use we will be use it on the straight and narrow track we will not go off track that is the domain of mathematics. Let them develop short cuts for us explain them properly and then we will use them. One such short cut we are using the Jacobian formulation which we are using is one such short cut, but mathematicians have developed it properly made it available to us through standard text books. So, we are using it. So, my objection was not to your writing say C p equals T d s by d t at constant p or its sister formula C v equals T d s by d t at constant v. My objection was the you know the short cut and quicky way by which this was obtained over to you the equation d s is equal to C v by t into d t plus p by t into d v is similar to this equation d t is equal to something into d p plus something into d v for that you have used that dot t by dot p that is which is similar to d z is equal to m d x plus n d y. This is also similar to that d z is equal to m d x plus n d y why do not we use for this. Again your question is the same thing it looks similar it may even be correct, but that is not the mathematically proper way to do it. See we are remember one thing we are for simplicity and because this is for undergraduate students we are looking at systems which are inherently simple. That means one two way work mode we have only two properties, but the moment we go to complex systems say for example a fluid which is electrically charged or even magnetically active like a plasma. Out there you will have not just two variables, but you have three or four variables which will be needed to define a state and then our partial differential partial derivatives will not be something like partial of t with respect to p at constant volume, but may be partial of t with respect to p at constant volume, constant magnetic induction, constant electric field and when you have such complications taking such shortcuts is absolutely no no over to you. How do you write dot t by dot v at the constant volume is equal to t by c p minus c v. Into dot v by dot t at constant pressure from that from the equation d t is equal to something into d p plus something into d v. That is t by c p minus c v dot p by dot v at the constant volume one side. The other side t by c p minus c v dot v by dot t at constant pressure d p from that equations you have written dot t by dot p at constant volume is equal to t by c p minus c v into dot v by dot t at constant pressure. Only I applied that concept only not you sir. No, I think you have missed some step our first t d s relation was t d s equal to c p d t minus t d v by d t at constant p d p. This is the first t d s relation. The second t d s relation is c v d t plus c v d t plus t d p by d t at constant v d v. This is the second t d s relation. Now, I think what I said something and which I think a few of you have missed out. All I do is since the left hand sides are equal and if the r h s s are equal, then I equate them. Now, all I do is transpose this to the left hand side and this to the right hand side and I am combining factors of d t together. On the right hand side, I will have t d v by d t at constant p d p plus t d p by d t at constant v d v. Now, I am just multiplied by d p by d t at constant v d v. I am multiplying everything left hand side and right hand side by 1 by c p minus c v. So, in that case, this will get cancelled out. I will get c p minus c v here. I will get c p minus c v here. I am not dividing by any differential. I am not multiplying by any differential. This is very proper mathematics and beyond this, it is just calculus of partial differential equations over to you. So, after this equation, how do you write the dou t by dou p at constant volume is equal to that part t by c p minus c v to dou v by dou t at constant pressure. After this step, consider this part to be m. Consider this part to be n and then. So, now, I have d t equals m d p plus n d v. Compare this. This is equal to by definition or by partial differentiation v d p plus partial of t with respect to v at constant p d v. Just by comparison now, my d t by say here or if you want d t by d v, I will take d t by d v must equal n. So, d t by d v at constant p must equal n, which by our previous slide is t by c p minus c v partial of p with respect to t at constant v. So, that is t divided by c p minus c v partial of p with respect to t at constant v over to you. So, now, you have the d t is equal to m d p plus n d v. So, for my equations d s is equal to c v by t. So, it means c v by t will be m plus p by t into d v p by t will be n here. So, instead of d v here also I am having d v. So, instead of d t I have d s instead of m I have c v by t instead of d p I have d t instead of n I have p by t then d v I am also having d v. This is similar to the equations over to you sir. Professor, I suggest you write your argument down and mail it to me because you are saying something and I do not know where the numerator and denominator is and there is likely to be some confusion. So, you write it down and mail it to me and I will take care of that. I think there is some issue some misunderstanding between you and me over to you. Thank you sir. So, let us break for lunch and we will see you soon after lunch at 2 p m. Thank you.