 to the last session, 3 time is over. Now let us look at, we have looked at work interaction so it is time for us to look at some exercises. So the last session will be devoted to some discussion of problem solving in thermodynamics and we will take some exercises from the set of exercises particularly that on work interaction, actually only on work interaction. Now why have I given an hour and a half for this session? This is our first without saying so the first of the exercises plus interaction session. But I think it is important to consider the discipline of problem solving in thermodynamics. I do not know the scheme of evaluation or the type of papers which you have. But I feel that a quiz or an examination in thermodynamics should essentially be based on solutions of problems, solutions of exercises because in thermodynamics there is hardly any formulae to derive. There are no theorems to prove. There are only 3 laws, a few relations. So everything is about understanding and being able to solve problems. And problem solution requires many skills from a real life situation to a reasonable idealization with appropriate assumptions. Then setting up the thermodynamic systems model and then applying our laws of thermodynamics to it and calculating things out. So while solving the problem we should not just be able to just be concentrating on what the answer is. Quite often for many problems in many of these exercises in principle you can write a formula. If you have a good enough calculator you can simply enter the numbers in the calculator itself, get an answer and write it. It would not even take one third or one fourth of a notebook page, a typical A4 size page. So by the time this particular part is over your solution is done. We are more interested in the way a solution is obtained, the way an answer is reached rather than the answer by itself. And I would like all of you to follow this discipline and I would like all of you to try to inculcate this discipline in your students. I know it may be difficult because many of you are affiliated to some universities and you have absolutely no control over the type of papers which are set and the type of questions and the format of questions which are answered. But now that I have a large set of the thermodynamic teachers in front of me, I hope that many of you as you grow and as you absorb this, I hope that many of you think that this is the right way to solve problems in thermodynamics. And naturally a few of you will grow into positions of positions that matter like memberships of board of studies or chairmanship of board of studies. So try to then slowly change the scheme of teaching and scheme of evaluation in an appropriate way. So when it comes to problem solving in thermodynamics, remember the following. Procedure is more important than the answer. We expect that one must be able to demonstrate understanding of the subject matter solving a problem. And hence to begin with I recommend that even thinking as you proceed leave nothing to imagination leave nothing by default. Now let us say something some required structure. For example I would say typical process of solution something like this. First read the problem, appreciate note what is specified also what is missing. Perhaps what is missing is sometimes more important than what is specified. System diagrams diagram or diagrams process diagram or diagrams are always necessary. There are only a few exercises in this where you do not need a system diagram or we do not need a process diagram. Property relations for example that is one such subset of exercises. Otherwise each exercise will require at least one system diagram and will require at least one process diagram. Then remember that the specification will not be complete. If a complete specification is given there is nothing to solve except substitute. So the specification will not be complete and hence assumptions. For this we have to look for hints which may be embedded there. So how to look for things not specified, things missing, look for hints by appropriate words. In fact we are teachers so we not only solve problems but we also have to create new problems or modify existing problems for new exercise sets or new examinations and new tests. So remember that for setting up the question we have a large amount of time provided to us. But for solving that question the student only has a few minutes or maybe one hour or so of that kind. So we have to be faithful to the students. We should not unload it, load the student unnecessarily. So that is why we have to provide hints and as problem solvers we have to look for those hints. If you do not provide any hints the student load goes up but that is not the issue. You do not want to load the student, you do not want to make him rush against time then he will make silly mistakes and we will be evaluating his ability to work under pressure and not his ability to understand basic thermodynamics. Then after making appropriate assumptions actually these assumptions you will have to make as you progress. In fact this is not the order in which things go except you know reading the problem appreciating and note what is specified. That will have to be there at the first but what is missing may be realized as you proceed. A system diagram and a process diagram you may not be able to complete at the beginning. Part of the process is something to be computed out. So as you proceed you complete or maybe even modify that system diagram or modify that process diagram. Again assumptions as you proceed you make the assumptions. It is not that you know the standard geometry stuff. You make the assumptions, you write all the given things in the beginning and then you simply have a once through process in thermodynamics and in real life engineering things are hardly ever like that. Now after all this we come to computation or calculation. Here we have numbers. We have numbers plus we have science plus we have a precision involved and plus we have units involved and we have to respect and of course we also have context work done by work done on that is a qualitative part. And we have to treat each one of these with appropriate respect. Any missing stuff means the answer is incomplete and quite often incorrect. Because incomplete answer is no answer that does not lead to the final solution. So treat all these things with respect and when it comes to units and remember units come out of the dimension which is more abstract. Dimension is algebraic like length into mass divided by time or mass into length square divided by time square and something like that. Units are arbitrary. I can define my own unit. You can define your own unit. There are certain standardized sets of units but there are a large number of technical units and conventional units which are used. For example, length can be measured in anywhere from nanometers to angstrom units to millimeters to meters, feet, yards, furlongs, miles, kilometers, light years, what, what parsec and all that. What have you right from micro to macro. So you need to use what are known as conversion factors and this happens to be a weak point with many of us. So we will spend some time on conversion factors. Let me take one illustration and for that illustration I would select one of the exercises from our exercise sheet. With whatever we have done so far we will be able to tackle any of the exercises in the work interaction. We have WI 1 to WI 8, 8 work interaction exercises plus there are additional exercises from the textbooks which are recommended. Although our standard set of units is the one which comes out of the SI system we will be using some pseudo SI units. For example, pressure would generally be in bar but just to provide some grinding to the students and expose him or her to the appropriate conversion factors in this particular section on the work interaction we have tried to create some exercises with some odd units. Let me discuss this before coming zeroing in on a particular exercise. WI 1 is essentially a integration exercise. A system diagram is not really necessary. You can show a cylinder piston arrangement with a fluid in it. But a process diagram is absolutely necessary because unless you draw a process diagram you are not sure what you are going to integrate. The second exercise is on electric work. Third exercise is on expansion and stirring. Fourth exercise is typical one where you have to determine the force, determine the displacement. Fifth exercise is for the compression of a solid. Sixth and seven are on elastic substances and eight is on surface tension work, extraction and contraption of a surface area. I would like you to attempt sooner or later all of these but just now let me take one of my favorites and that is the exercise WI 3 just to show the process. Now I will not be writing much so listen very carefully to what I am saying because if I start writing I will be writing maybe till 5 o'clock explaining all the thought processes. So let us take WI 3. Let us read it and think aloud. Do not leave any thought unsaid. You know this type of recitation is something which we were asked to do in primary school. After that for some reason that discipline was suppressed. I think we have to rekindle it in ourselves. So I am reading the problem. A system containing 5 kg of some substance. It is stirred. Oh stirred means it is something like a fluid. Maybe he condensed milk or something like that. With a torque of something the value of the torque is given. The speed at which it is turned is given and the time for which it is turned is given. Oh the system meanwhile expands. So oh it is not as thick as condensed milk. It is some fluid which is expand. So something which can be stirred and something which expands. Some funny thing. From 1 meter cube to 2 meter cube. So initial and final volumes are given. Against a constant pressure which is specified. Oh the pressure is constant. So integration seems to be easy. Determine the network done in the unit specified. So what I have? I have some system closed one of course. Because we have not yet discussed open systems at all. And there is a stirring involved. That is the first specification. And there is a expansion involved. And we have to determine the network which was done. So naturally there are 2 modes which are involved. Stirring plus expansion. So I can write W. Need not write net. W means the total work done. This will be W stirring plus W expansion. Then what is the specification for stirring? That is given. What is the specification for expansion? That also is given. So let me be clear by sketching the system diagram. It is an expansion. So I must show a cylinder piston. And this is my system. 5 kg of something. I am not sure what that is going to be used for. So we have a torque. So we have a stirrer. And we have expansion. So we have a W stirrer. And we have a W expansion. Now the specification. It is stirred with a torque of 0.3 kg F meter speed of 1000 rpm for 24 hours. So one process diagram would be something like this. So I will see after this thing is over whether there is something intermediate available. I have the torque and I have omega. And the torque is steady. And what is the value given? 0.3 kilogram force meter. 0.3 kg F meter. And what is the value of omega? Omega is given to be 1000 rpm. Although they have not mentioned it, a minor assumption here is that this is a uniform torque of 0.3 kg F meter for a constant torque and a constant speed. And naturally if this is 0, the total process is over in 24 hours. So this is the specification for determining W stirrer. Now let us look at the other part. The system meanwhile expands from 1 meter cube to 2 meter cube against a constant pressure of 4 kilogram force per centimeter square. So let us say sketch that. This is our standard P V diagram 1 meter cube to 2 meter cube. So the initial volume is 1, 2 meter cube. Pressure is 4 kilogram force per centimeter square. The initial state will be this. Final state will be this 1 and 2. Constant pressure. So the default, sorry the assumptions are tau omega constant as given. Pressure is constant. Hence we assume process is quasi static. Now with this our visual depiction is ready. The system, one part of the process. Now remember this is a process diagram which is some parameters against time. So in this diagram the initial state and the final state, this tau and omega are not properties of the system. Similarly 0 to 24 hours, this time duration has nothing to do with the system as such. So there is no question of showing the initial state and final state on this diagram. But the second part of the diagram where we have something to do with expansion work, we are given the pressure volume relationship. We are given the initial state and the final state. And then this diagram is a process diagram on the state space or at least the P V projection of the state space. So we can show the initial state 1, we can show the final state 2 and we can show the process. All that we are given is that P is constant and at least for integration of expansion work as P dV, the to that extent the process can be assumed quasi static. Since no other interactions are involved, we are not relating to the state to anything else. We do not have to really worry whether the temperature and other properties are uniquely defined. We are simply looking at work interaction and hence for the purpose of evaluating work interaction, this quasi static assumption, a minor assumption is good enough. And now the next thing we do is because we know that there are 2 components of work, we now expand those 2 components. We write W stirrer is going to be minus integral from 0 to 24 hours of tau omega dt which is going to be simply minus tau omega t, where this t is nothing but representation of 24 hours. And similarly, our W expansion, did I write E x or E x p? E x p. So W expansion is going to be integral 1 to 2 P dV which is going to be P integral 1 to 2 dV which is equal to P into V 2 minus V 1 and this is since P is constant. And here also from here to here, this is since tau omega are constant. Now we come to the, with this the formulation is essentially over or actually over because we know here tau is specified, omega is specified, t is specified. So W s t can be computed and in the second component we know V 1, we know V 2, we know P, hence W expansion is computed and hence going back to the original formula from this, we can now determine the net work done. In kilojoule it is something which we have to finally take care of. Now we come to computations and this is where we have to worry about dimensions, units and conversion factors. My recommendation to everyone, students, faculty is write in the formula the units also as algebra. For example, when you say multiply 3 into 4, where I 3 into 4 is 12. But now consider units as part of this. So let us say A is the unit for 3, B is the unit for 4. So we should write now 3 A into 4 B as 12 AB, where AB is the appropriate combination of units. We may have to leave it as AB or if AB has been defined at some another unit say G, replace it by G, we will do that. Let us look at the first part. The first part said W stirrer is equal to minus tau omega t. Let us substitute numbers, tau is given to us as 0.3 kilogram force meter. So write this as 0.3. I tend to use square brackets just to keep this separate Kgf meter multiplied by 1000 rpm. So 1000 revolution per minute, multiplied by 24 hour. Now notice the numerical part can be done 0.3 into 1000 into 24. But what about the units part? Consider the units are as algebraic entities. Finally, we want the work in terms of joule or kilo joule. But here we do not seem to see anything there. We have joule we know is Newton meter. Meter is there, but we do not have Newton. We have kilogram force, then we have revolution, we have minute and we have hour. So we use conversion factor. Now what is a conversion factor? We generally remember conversion factor as x of something is equal to y of something else. For example, we remember the conversion factor that 1 inch is 2.54 centimeters or 1 meter is 100 centimeters or 1 minute is 60 seconds. Now this does not give us the real value of a conversion factor. I propose that the actual value of any conversion factor is 1. So we rewrite our conversion factors. Let us look at it on the next page and then I will come back. Conversion factor the value is 1 and the units are something funny, but the dimension is nothing. For example, when we say for example, we have kilogram force here. We know 1 kilogram force is 9.81 Newton's. So the old style way of writing is 1 kilogram force is 9.81 Newton's. But instead of that, I would like to convert it in a conversion factor which is 9.81 Newton per kilogram force. Its reciprocal is also a conversion factor 1 over 9.81 Newton per kilogram force per Newton. What is the value of this? The value of this is 1 and since it is the value of any of these conversion factors. For example, other conversion factors could be 100 centimeter per meter. Then 60 minutes per hour. Similarly, 60 seconds per minute and any such. So let us use a few of this. Another thing which we will be using is pi radiance per revolution. And actually radiance actually does not have any unit. So you can say pi per revolution is itself a conversion factor. So going back, let us start doing proceeding further. Let us say 0.3 or we need not even do this because all that we are going to do is multiply this by an appropriate number of 1's. So we are not changing any value anywhere. A number or any quantity multiplied by 1 is that same quantity. But for these 1's we will now use the appropriate conversion factor because the value of each conversion factor is 1. So for kilogram force to convert it into Newton, we will replace one of these by 9.81 Newton per kilogram force. Now once you do that, you can algebraically cancel kilogram force. Now you have Newton meter which is joule. So we are on the right track. But out here we have revolution which we have to get rid of and this hour and minute which we have to get rid of. So we multiply further by 1 but replace say one of the 1's here by converting revolution to radiance. So we have, I think I made a mistake, this should be 2 pi. Am I right? 2 pi per revolution. So this is 2 pi, radiance is nothing but name for a unit. So now we have got rid of revolution and then this minute per hour. So another one, another conversion factor which is 60 minute per hour. And when you do that minute cancels, hour cancels. And now we know the only units remaining are Newton meter which is joule. If you want it to be converted to kilo joules, well you can be a purist and say multiplied by 10 raise to minus 3 kilo joule per joule. Instead of joule we can write Newton meter. So now Newton gets cancelled with this Newton meter gets cancelled with this meter and you will end up with minus 0.3 into 1000 multiplied by 24 multiplied by 9.81 multiplied by 2 pi multiplied by 60 multiplied by 10 to the minus 3 kilo joule. Now use your calculators and write this out. But when you do that, do one more thing. Generally our habit is to look at our calculators and see whatever we see in the display window of the calculator and dump it on the paper. Again this is a great tool for us to discipline our students. What the calculator shows in the display has nothing to do with the problem at hand. It is just a result of some multiplication, division, whatever operations. What is the real situation? You will notice here that none of the factors here which are specified 24000 0.3 have more than 2 or 3 significant figures. There may be this is very precise but that does not add any information. So whatever you do is you calculate this out with minus a number and write it in kilojoules. I am not calculating this out but 3 or at most 4 significant figures. I leave it to yourself to put the appropriate number here. Note I am not saying decimal figures but significant figures. Now you may laugh at it. Why do we have to do this long winded thing? We may not have to do but we should force our students at least in the initial part of their study of thermodynamics and for that matter any subject by using this long winded method because that way they will appreciate how to do calculations and they will appreciate how to treat numbers with respect and they will appreciate what exactly is meant by conversion factor, why we use them and how to use conversion factors properly. Now this happens with W stirrer and you can get the actual value. The next thing is W expansion. Do it in a similar fashion. W expansion we have determined is p v2 minus v1 and p is given to us as 4 kilogram force per centimeter square. So 4 kgf per centimeter square multiplied by v2 minus v1, 2 minus 1. Actually if you are a purist you should write here 2 meter cube minus 1 meter cube but since meter cube is common I will take it common otherwise suppose I were to tell you that it goes from 1200 liters initial volume to 2 meter cube as the final volume in which case you will have to write 2 meter cube here minus 1200 liters and now of course you cannot subtract liters from meter cube. So you will have to first convert that liters into meter cube by using a conversion factor which is 1 by 1000 meter cube per liter and then everything comes in meter cube which you can take out as common. Now here you will see that finally we want in joule or kilo joule but we do not seem even to have meter. We have a meter cube and a centimeter square and we have a kilogram force. Although the dimensions match the units do not match. So naturally we will use appropriate number of conversion factors and replace each one of these by the suitable conversion factor. First we will need kilogram force to Newton. So 9.81 Newton per kilogram force something to convert centimeter square to meter square. So you write 10 raise to 4 centimeter square per meter square. Now this way kilogram force cancels out, centimeter square cancels out, meter square cancels with meter square here. Only meter remains. So now we have Newton meter. Since we want kilo joules you can even write 10 raise to minus 3 kilo joule per Newton meter. And now Newton cancels with this Newton, this meter cancels with this meter and you have the answer as 4 into 2 minus 1 is 1 into 9.81 into 10 raise to 4 into 10 raise to minus 3 kilo joule. Again write this and be conscious of this as plus something kilo joule. Again remember 3 to 4 significant figures because that is what our specification is. And finally again W is going to be W stirrer plus W expansion. So you calculate out the value and write the units kilo joules. Again while doing this you can write the it is possible that you end up with a large number and a small number in which case take care of the significant figures and you have to write the appropriate number here. Now with this I hope I have impressed on you the detailed procedure of solving the problem. And now I will do the following. We have about 45 minutes. So I recommend that you attempt problem the following problems. I expect different centers to take up different problems first and I am available here for any interaction. I will visit a few centers as you solve the problems. Although all 8 exercises and the additional exercises are recommended what you attempt just now are W i 4, W i 5 and then W i 8. You will do something like this divide your center number by 3. If the reminder is 0 attempt to begin attempt with 4 first. If the reminder is 1 attempt with 5 first the reminder is 2 attempt W i 8 first. And of course later on you go to the other exercises. 1, 2, 3, 2 Jayalakshmi Institute at Thoppur Tamil Nadu over to you. Sir in conversion factor for converting joules into kilo joules you write their kilo joules per Newton meter. The 1 Newton meter if you go to joules is it right? Then you are always writing kilo joules per Newton meter instead of kilo joules per joules. Is there any difference between kilo joules per joules and kilo joules per Newton meter? Part user. Good question I think I am happy that we have started looking at nitty-gritty. In fact when I write something kilo joule per Newton meter actually those are 2 conversion factors combined into 1. If you are a real purist I should write 10 raise to minus 3 kilo joule per joule multiplied by 1 joule per Newton meter. Because 1 joule per Newton meter is not a conversion factor that is the definition of joule whereas 10 raise to minus 3 kilo joule per joule would be the definition of the prefix kilo. So what you are saying is right in principle we could do that but we need not insist that we go to that extent. If you get the idea of how to use conversion factors then you have won the case. Thank you. Sir our son family the son is going to be died because of no fuel. Could our engineering thermodynamics help for making our son alive? You are asking a question which is I think beyond the scope of this course. Yeah I think you are saying something about the death of the son right? Sir I have a problem one of the major issues of the aircraft because son is going to be died because without fuel. Yeah what I suggest is you go to perhaps an old library and look up a book by what is his name George Gantt. It is called the birth and death of the son. You will get enough information there and then you should actually discuss this with somebody like professor Nardikar who is an astronomer cosmologist rather than in a course on thermodynamics. Thank you. 1, 2, 1, 9 prestige institute Indore over to you sir. I am Srikant Tare from Indore prestige institute of engineering and thank you. Engineering science. I am trying to solve this W at 4. Yes. And the pressure is rather the area normal to this pressure it is x into b. The width of the channel is b normal to the plane of the paper on which this figure is printed. Yeah so p0 is the pressure exerting on the area that is x and b. Pressure p0 is pressure exerted on the area which is x by b you are right. Yeah so when I am seeing that f equal to it has to be proved that p0 h into b. Force will be p0 into x into b x into b the force because of air on the water. Yeah the force yeah so the force exerted on the piston what it is written is h into b. Yes that formula is right this is the force exerted by water on the piston. This is right you have to derive it properly using some principles of hydrostatics. Thank you. Thank you very much. Over and out 1, 1, 9, 4 Gandhi engineering college. Orissa over to you. Sir we have used expression to calculate the force pressure into area that is the pressure exerted atmospheric pressure p0. Here will be base into height w i 0.4. Yes the part a I suppose you are working on right. Yes what is the question? What is the problem there? Actually the question is that the force exerted by the water on the piston is f is equal to p0 plus rho g h divided by h into height. Yes that expression is right. There we are using equations that is the force exerted on the piston will be exerted by the water will be equal to the force exerted first because the water has to do work against the atmospheric pressure plus the pressure that is exerted by the piston. See for determining the force the idea of work does not come into operation. This is simple hydrostatic. First thing you do is you take the point in the water which is at the height h and touching the piston. The atmospheric pressure which acts downwards is p0 because of Pascal's law. Pressure is the same in any direction. So the pressure acting on the piston because of the water at the upper level of water is also p0 horizontally. Now go down the piston and start diving in the water because of hydrostatic as you go down the pressure increases. At the top the pressure is p0. At the bottom the pressure is p0 plus rho g h. Since there is a linear increase the mean pressure will be the average of the pressure at the top which is p0 and the pressure at the bottom which will be p0 plus rho g h. So when you take p0 plus p0 plus rho g h whole thing divided by 2 you get the expression in the big brackets the first factor. This mean pressure when multiplied by the area on which it acts which is h into b gives you the force acting on the piston because of water. Over is that clear? pb calculate the work done by the water on the piston and by the atmosphere on the water. Now remember that what you have obtained is the force acting on the piston or the force exerted by water on the piston. Now piston is moving out from x1 to x2. Now what is it say channel of constant width? So the process equation is going to be the volume of water is going to remain unchanged. So the height when the piston is at a location h is going to be given by the relation h into x into b which is the volume of water at any x equal to the initial volume which will be h1 into x1 into b. So this gives you a relation between h and x just the way we have for a gas relation between p and v. So you can say this is sort of an equation of state for water in that channel. Use this equation and the force equation and integrate it out and you get the result. The answer is can you verify the answers of this? 1 by 4 rho g h. I do not have the answer here just now. Today late evening I will upload the solution of this. So you can check it out. I will upload the solution of all w i's sometime late evening. So over and out 1182 M K Orchid College, Solaapur over to you. So I have doubt regarding work interaction. So where does the flow work stands? Is it non-cosistatic process? If yes then how to find out the flow work? First thing is we have not yet come to flow work. Flow work is something which is associated with open thermodynamic systems. Just now we are solving exercises from the work interaction part and no flow work is involved because no flow is involved anywhere here. Over to you. W i 2 sir. So in that exercise number W i 2. So in that the answer which I got power is 1150 Watt and number of units of electricity 2.30 units. In that my doubt is that if suppose the environmental whatever the power is generated in the form of heat due to the passage of the current that power if it is given to the atmosphere if it is a convected to the atmosphere then only the power consumption will be uniform. If the power is not conveyed convected to the atmosphere in that case the resistance will go on changing and in that case the power consumption will be different. So regarding that you please clarify. I think your real life situation is all that you have to make is an assumption that the resistance of 50 ohms remains unchanged during the period of 2 hours. Unless you make that assumption you cannot really proceed because what happens to the resistance will depend on how the resistor or the heater behaves. If the heater the so long energy provided to the heater is removed by some means say by cooling etcetera then it may remain unchanged. If not it will change. If it changes then we will have to specify the way it changes ok over and out 1, 2, 2, 2 Chamele Devi group of institutions indoor over to you. I am M. K. Jan and asking one question that in most of the mechanical engineering books T is used for torque and N for RPM but you are introducing other symbols. We will be we may be slightly confused. See there is nothing special about symbols. If you are comfortable with some other symbols just edit the nodes and edit the files with any other symbols you are comfortable with. There are hardly any set of symbols which is standardized. So, I would not insist that you use my symbols. If you are more comfortable with some other symbols go ahead and use them. There is nothing wrong in that 1, 0, 4, 0, COEP. Now, I can hear you I suppose. As per here problem solving procedure we are solving the problems but do you have the answers for this? Yeah. Today evening I will upload the solution procedures on the model. So, you will be able to see them. They will be on the coordinators model. So, you ask the coordinators to download them and give them to you. Ok sir. And one more theoretical simple problem sir. The work instruction and the force measurement in that direction d w is equal to minus t dx. So, why this minus sign? Because if you see you are talking of that t dx for the extensible string. Am I right? Minus tau dx. Minus t dx or minus tau dx. Let me go back and see what symbols I used. Yeah minus t dx. Yes that minus t dx is because if you see when I hold the tension strut the tension is in the minus x direction whereas I am trying to extend it in the plus x direction. Unlike in case of a pressure the pressure is in the same direction in which the piston moves for expansion. So, it is p d v whereas in case of this thing it is minus t dx because the tension is pulling our hand in. A tau string if you leave will run away from you whereas a piston which held against pressure if you leave the piston will move towards you. Yes sir. Yes sir. Sir, while explaining extension intensive property that time understood. But after few minutes there is very difficult to understand what is extension and intensive property. Very simple but very difficult to remember. See it is the extensive the main confusion that happens about extensive or intensive is the following. One should understand that a property there is a students thing that a property is either extensive or intensive that is not true. There are properties which are neither in extensive or intensive. Although most of the properties which we come across will either be extensive or intensive. For example, there are generally we come across in thermodynamics only two properties which are inherently intensive. Those are pressure and temperature. You know all other properties like energy, enthalpy, entropy, volume, mass these are by their very nature extensive properties. But when we define them as or redefine the specific properties. For example, volume per unit mass entropy per unit mass that is specific volume specific entropy or the reciprocal of specific volume which is density mass per unit volume. These turn out to be ratios of two extensive properties and hence they turn out to be intensive properties. But you can for example, mass is a property but square of mass can also be a property nothing wrong in it. It is not much useful but the square of mass will neither be an intensive property nor an extensive property that you can check out. But this is a pathological illustration but we have to define the properties at extensive and intensive because everybody has been talking about. Otherwise there is nothing very special about that. But tell the students that there are properties which you cannot classify as extensive or intensive. An area is a typical example. You take a loaf of bread. The loaf of bread has an area which is L by B all those things 2 into LB plus BH plus LH whatever. Now partitioning means convert it into slices. Now what is the total area of the partition system? It is much larger than that of the loaf of the bread. This is the simplest example I give quite often to those who get confused. I hope that satisfies you. Over and out 1, 2, 2, 3 RBS engineering Bichpuri Uttar Pradesh over to you. Good evening sir. Sir I want to ask is internal energy a property of system or not? If it is property of system is it an intensive property or extensive property? We have not yet defined internal energy that we will do tomorrow when we come to first law. But to satisfy your question an internal energy is the property of a system. The internal energy is an extensive property of the system. But the specific internal energy that is internal energy per unit mass happens to be an intensive property of the system. Over and out or is there another question? Over and out. Thank you. 1047 SDM college of engineering and technology Darwad over to you. Sir my doubt is with WI-5. Sir this is a problem related with solid mechanics saying that a block which is subjected to a I think hydrostatic pressure. And this that is bulk modulus is given and it has been bulk modulus is defined as hydrostatic pressure divided by change in volume. And how to relate this change in volume with this work done on the that is blocks. You will get the details when I upload the solution today evening. I just tell you that this is an extreme case of a fluid. Now a fluid we consider very compressible so that whenever there is a change in pressure there is a reasonably significant change in volume. Whereas for a incompressible liquid or a solid like this although we consider solids to be more or less incompressible they are not absolutely incompressible. And now you will see that the bulk modulus is has a very large value here. Isothermal bulk modulus is 2 into 10 raise to 12 dine per centimeter square. Now this means that a large amount of pressure is required to create a small change in volume. But the formula here it is still integral p dv. The only difference is that dv will get related to dp using the bulk modulus relation. You write the bulk modulus relation b which is defined as minus partial of p with respect to v at constant t. Expand it as minus v dp divided by dv where dp is a small change in pressure dv is a small change in volume both at constant temperature. And then whenever you come to p dv replace dv you will get to replace dv by something like minus into v into dp divided by b. And you will finally get an integral in terms of dp rather than in terms of dv. Although it looks like a solid mechanics problem but it is actually an absolutely basic thermodynamics problem. Don't just think that because you get terms like bulk modulus and things like that it is immediately moved to solid mechanics. It is as much thermodynamics as anything else. And of course thermodynamics does not say that this problem is unique to thermodynamics. Thermodynamics works very well with other branches of physics. So, if solid mechanics people want to look at it they are welcome. We have no fight with them. Over to you. Any other question? Hello sir, Myself Manjuna sir. Sir, my problem is with WI4 sir. Right. Go ahead. Here you have taken hydrostatic pressure rho g h WI4 sir. Yes. So, in this you have taken hydrostatic pressure rho g h divided by 2 as the mean pressure sir. On both side of the piston and the wall sir. Why can't you not taken this atmosphere pressure P naught sir? No, I have taken both. As I think I have explained this to some other center earlier I do not know whether you heard it at that time or not. See if you look at the water level the top of the water the pressure by atmosphere on water is P naught. So, at that point because of Pascal's law the pressure imposed by water on the piston at the level of water is also P naught by Pascal's law. Pressure is the same in all directions. Then you go to hydrostatics as you go down the pressure increases linearly and at the bottom the pressure will be P naught plus rho g h. So, at the top the pressure is P naught at the bottom the pressure is P naught plus rho g h. So, the average pressure because of the linear pressure profile is P naught plus P naught plus rho g h whole divided by 2 which turns out to be P naught plus rho g h by 2. 1, 2, 6, 3 Techno India, Salt Lake Kolkata over to you. Actually my question is regarding property again that is if it is true that property can be only defined if the system is in thermal or the thermodynamic equilibrium. Yes property value can be uniquely defined only if it is in thermodynamic equilibrium. In fact the definition is the other way if we have the values of all properties uniquely defined then the state is said to be in thermodynamic equilibrium. Then only property can be defined otherwise you cannot specify a state with the help of property if it is not in thermodynamic equilibrium. No you cannot because unique values of properties are not available. Right ok sir I have one more question when you have defined the property you have underlined the word relevant characteristics to specify a particular state. Actually relevant term is related to a process, but state is an instantaneous thing at a particular position then why relevant term you are adding in the definition of property. No the relevant is not related to property relevant is related to characteristic. See when we see a for example we solve problems regarding cylinder piston right. Now a gas in a cylinder will have a certain volume, but the volume is defined by a certain length and certain diameter. We solve lots of problems looking at only the volume without really looking at the length and the diameter. With the length and diameter we take care of themselves. So far as the problem is concerned at that instant or the solution procedure we decide that although length is a property of the system we can define it at any instant. Diameter also is a property of the system we can define it at any instant. Length and diameter under that circumstance for that problem solution are not relevant characteristics whereas volume is. If that particular characteristic is changing then also the state will change. So in that case that will be relevant. So when you are defining a property then why relevant term should be used. Because if I do not use the term relevant for every system you will end up with a list of thousand properties or a few dozen properties at least. This is true. Yes, but will you do you really consider all property? Yes, but it all depends on to what detail and to what extent you want to analyze the situation. For example that is what I was going to tell you that although we solve cylinder piston arrangement without really worrying about length and diameter but suppose you want to design or analyze the combustion process in an IC engine then only volume will not do. Then you will have to worry about the length and the diameter also. So that is a situation where you will have to consider length and diameter as properties along with the volume. I understood your point but my objection is that in the definition of property this term should be avoided actually because it is a general definition not for a particular process. I am not, I think you are getting confused. I am not using the word relevant for property. I am using the word relevant for a general characteristic. There are a large number of characteristics only the relevant ones you consider as properties. Properties are always relevant whereas characteristic may be relevant or irrelevant and that too under a particular situation. What is an irrelevant characteristic today may become a relevant characteristic tomorrow. Thank you very much sir. The last interaction today is with 1207 Javaharlal Institute Boravan. Sir, I would like to ask regarding question W 1.5 or W I 5. Go ahead. Yes, actually as you have explained the pressure and volume actually we have analyzed that this could be 1 to 2 P dv. So we feel that the pressure is also changing as well as the volume. So are we going to integrate both this? See, there are some crucial words here. Read the statement, assume that the density and isothermal bulk modulus of the metal remain in bracket almost constant. Now that means that as the process goes if you draw the P v diagram the process goes increase in pressure but the change in volume is very small. So as the pressure changes you can assume the volume to remain essentially constant. You can take it out of the integration and write it in terms of density and mass. Mass anyway is fixed. The density remains almost constant that means the volume remains almost constant. So finally the integral will turn out to be only in terms of P dp. Integral of P dp will be the integrant everything else will come out of the integral sign. Over to you and over and out and to all of you we have had a nice but short set of interactions today. From tomorrow afternoon we will have almost three hours for interaction and problem solving. So we will have more fun. Thank you and that is all for today.