 afternoon session is devoted to topic 4 problem solving in thermodynamics and then exercises for the work interaction and there were some questions in the morning. So, I will take those en route and I will generally be online. So, as the day progresses and as you solve problems if there are any questions etcetera that I should be able to take from here. So, our topic now is topic 4. So, this session is going to be topic 4 problem in thermodynamics plus exercises on one work interaction. We have WI I think 9 I suppose 7. There are additional exercises at the end of the exercise list they are from the text book specified. We will discuss them only offline through Moodle because there are well to a large extent they are more of the same and apart from this I am going to have discussion as needed. Now, before I go to problem solving there was a query in the morning and what people wanted was illustration of the operational definition which is thermodynamic definition of work where we had those contraptions C 1, C 2 etcetera. I will take a few illustrations perhaps the simplest illustration is and our base illustration is expansion of a gas I will begin with that. So, let us say this is our system A and it goes from state A 1 to state A 2 meanwhile expanding. What is system B? Let us say that the system B could be I am just imagining a system B. System B may be other part of the piston another fluid which goes from B 1 to B 2 and may be the pressure of A is higher or A is being goaded in such a way that this piston moves by this distance into B. And now we want to set up a contraption whose net effect would be the rays of a weight. So, let us try in this case it will be a shorter thing. So, I will again sketch this is our system A and let me go back here I will say because let me extend B by including the piston. So, that there is a common boundary between A and B. So, now this is the common boundary between A and B and it is going to move we will make it move the same way as it moved earlier. And can I imagine some contraption which will finally, raise a weight. Now in this particular case it is rather simple let me see all I do is put a piston here of course, leak proof frictionless all that. Then I will have may be a rod on this rod I can have a linear gear I will put a small pinion here with appropriate my sketching is no good. Then on this I will have a small drum and on that I will have a rope thin massless in extensible unbreakable all our standard physics assumptions to which will be fixed a mass m. So, as this piston moves this will move linearly gravitational field is proportional to it is perpendicular to it. So, its gravitational potential energy will not change it is moving slowly. So, no kinetic energy change of this contraption the pinion and the drum fixed on to it will rotate slowly as this translates. And this mass m will be moved up by some appropriate edge. And by remember that initially we want a state of equilibrium. So, if this is the pressure p if this is the area a we want a force acting on it. So, that it remains initially in mechanical equilibrium and through the process this force is adjusted. So, that this equation is always satisfied. So, that it remains a quasi static process. And so, we will now have to adjust our mass and the diameter of the drum and the pitch of the pinion in such a way that the mg on this is appropriately linked to force equals p a. Too much of m a too large an m then this force will push the gas in which we do not want to small an m then this force not will not be sufficient to overcome this p a. So, we should adjust m in such a way that force remains equal to p a. And if as the gas expands if the pressure changes suppose it reduces as the gas expands then naturally the force has to reduce. So, this mass will have to be you know big by big reduce to maintain it in equilibrium throughout the process. So, this is our contraption which is c 1 contraption it exists or can be demonstrated that we can set it up. So, this means this interaction is work because it is expansion a moves into b a is doing work on b. Let us take another example that of a stirrer. Let us say this is moving by d theta and it is overcoming a torque which is trying to restrain it tau. Our system is this this is system a. What is system b? Let us say what is inside this boundary including this part of the rod and the stirrer is part of system a. Let us not complicate it by the very complicated argument beyond this is our system b. But how is it being stirred? Let us say that we have a drum here with a rope wound on it rope goes over a pulley may be there is a mass here which is going down all this is our system b. So, this mass in the gravitational field g provides the torque to overcome the restraining torque. But now if you look at it from the system a point of view the external effect is that of a lowering of a mass. So, that is not useful for our definition. Remember that the definition should be such that the mass has to be raised then work is done the direction is decided on whether it is the C 1 that we can set up with a or C 2 we can set up with b. Do not make the mistake of setting up something where the mass goes down and then you say negative work is being done that is not part of the definition. The mass should always be raised. So, what do we do here? If you look at it I will not spend time. But if I look at my system a going from some state a 1 to a 2 and if I try to set up C 1 you will notice just from here that C 1 is not possible from the same thing now we will look at the other way. I will sketch the b part I have a drum here shaft going up to the boundary the same rope mass going down d theta tau. This is b going from b 1 to b 2 and as it goes from b 1 to b 2 the mass gets lowered. So, I replace a by C 2 and what is my C 2? My C 2 would simply be a drum similar to this on which there will be a rope. So, as this rope sort of unwinds from this drum this rope on this drum will get wound and there will be a mass here say m 2. Let me say that the diameter of the drum inside b and this diameter although it looks different they are the same. In that case you will notice that to maintain the equilibrium that means this mass at one place and this mass at one place these two masses will be equal. If the diameters are different these two masses will be different and there is a simple ratio which you can derive from mechanics, but now as the process goes from b 1 to b 2 this mass will slowly go up in the gravitational field and this now is our C 2 sorry this now is our C 2 whose only external effect is the rays of a mass in a gravitational field. So, C 2 can be set up and that means the interaction is work and b does work on. I will leave it to you as an exercise to show that here I have assumed that the stirrer is being sort of stirred by a mass pulley drum arrangement, but instead of this just put an electric motor here and the supply from a battery and you should still be able to show that thing like C 2 can be set up, because I am supposed to keep you busy. So, I will give you some homework show that charging a battery electrical battery is work show that say discharging a battery and using it in a mobile phone is also work and say using a an alternator or a dynamo and charging a battery is work interaction. Then consider the following let us say I have a drum and I have break shoes on either side there is a breaking force acting on this, there is some friction on the drum and hence although the drum is rotating at a constant speed omega it is overcoming a torque tau. Now, consider two combinations of systems I will draw one dotted line inside the drum and I will draw another dotted line outside the drum. Now, consider this to be p consider this to be q and you consider two cases case one its system a is inside p system b is whatever outside p you can put the outer boundary anywhere you feel like. So, in the first case system a is inside p that means inside the wheel system b is outside p. So, the wheel break interface is inside system b case two system a is inside the boundary q and system b is whatever lies outside the boundary q. Now, the interaction between the drum and the breaks break linings there is going to be some interaction let us say the interaction is I of course, there are two different systems. So, there will be an I in case one and an I in case two what is I in this case is it work is it purely work or it is not purely work and similarly what is I in this case is it purely work or is it something other than pure work. So, that is going to be the homework you do not have to submit it, but be clear about what you are going to do. Now, that brings us to the main theme you will notice that almost every day afternoon we are going to devote to problem solving sessions and although our students and we ourselves are used to formula oriented problem solving that does not provide us the ability to demonstrate to the one who sees our solution that we have understood the subject matter properly. And the purpose of solving a problem and submitting a solution is not just to provide a correct solution, but also to demonstrate that we have understood the subject matter properly. So, the first principle is how a solution is obtained that is methodology is more important solution itself that is something we should emphasize on our students and for that there are a few things we should emphasize. So, important components in some order system diagram or diagrams process diagram or diagrams you will find that no question or no problem will have the complete detail available. So, we will have to make some assumptions governing laws and equations simplified form and then the algebra numericals to the solution and finally, insist on students and insist on ourselves that we will do this every time. I receive a solution from students and student equivalents I will leave it at that I continue receiving them and in spite of coding them our habit is may be that is from the setup we are in we all typically write a one hour or three hour examination we are pressed for time and may be the evaluators too are pressed for time because they have to evaluate so many does not answer books every day. So, they simply look at the final answer solution and give marks and the students also know that that is what is going to happen. So, the student does not waste time in writing all these things neatly, but I think we should provide enough time and enough answer book space to our students. So, that the students can demonstrate this and provide weight edges for all these components in the solution or the whole process of solution. So, that typically may be for the algebra and numericals leading to the solution only about 30 percent of the weight of a question is provided the remaining 70 percent is to the actual process by which the solution is obtained that is the solution methodology. If you look at you notice that under problem solving in thermodynamics I have 6 sub topics method of solution specification of processes how to read things not specified which is linked to how to make assumptions then how to treat numbers dimensions units etcetera with respect many of the basic errors come up without because we are not able to take care of this we do not treat numbers dimensions and units with respect and how to sketch diagrams for systems and processes. If you goof up 5 you can go typically numerically wrong quite often horribly if you do not take care of 1, 2, 3, 4 and 6 you may not be able to obtain a solution at all in your exercise cheat all of you should have an access to this if not ask the coordinator to print out copies for you this available on Moodle. The first set is exercises on the work interaction there are 9 exercises sorry there are 7 exercises plus additional ones and I will take an illustration to show you how to proceed by spending a reasonable amount of time on exercise 3 and then I will assign some exercises for you to attempt in the classroom before we leave at 5.30 and as you do that some questions will come up and I will be periodically every few minutes ready here to tackle them but let us look at exercise WI 3 very careful reading is in order we have a system containing 5 kg of a substance there is no mention of anything else about the system. So, all that we know the mass of the system is 5 kg let us not look at the numbers to begin with but read the main text it is stirred with a torque some value at a speed of 1000 rpm for 24 hours. So, we have the value of the torque the value of the speed and the time over which the stirring take place the system meanwhile expands oh so not apart from a stirrer we must have a cylinder piston type of arrangement because it expands it expands from 1 meter cube to 2 meter cubes against a constant pressure of some given value determine the network done in the specified units of kilo joules. So, the first thing we understand is there is a system no flow is mentioned. So, it could be a closed system that is one assumption that we are making or indirectly it is involved it is told to us that it is a closed system no mention of any flowing in flowing out. So, let us sketch there is a stirrer there is an expansion. So, one way of sketching it is showing a cylinder piston arrangement instead of that it could be a strong rubber bladder in which there is a stirrer through some nozzle and it is expanding by just the way a football bladder expands when you blow air into it, but that is a an open system because something is flowing into it. So, let us show a cylinder piston arrangement although I have shown this with a gap we need not show the gap unless otherwise specified we will be assuming all our pistons to be mass less leak proof friction less the standard physics stuff. We have a stirrer and we have this expansion possibility that means it is given that two interactions by two different modes of work are there one is W stirrer another is W expansion our system is this it contains 5 kg of something we will note down all the data may be some of this data will be extraneous not really necessary. For the stirrer we have a torque of 0.3 kg F meter we have a rotational speed of 1000 rpm and the time for this is 24 hour. For the experience expansion we are told that it expands against a constant pressure of 4 kg F per centimeter square. So, P and the initial volume is 1 meter cube final volume is 2 meter cube. So, expansion is from 1 meter cube to 2 meter cube determine the net work done in kilo joules. So, where the system diagram the first thing we have already noticed that there is a stirrer work involved and there is an expansion work involved. So, first thing we will write down is the net work W would be the sum of two components stirrer and expansion or expansion and stirrer whichever way you want to write it. Now, how does the process go the expansion work can be shown perhaps on a P V diagram. So, we sketch the diagrams P V 1 2 meter cube constant pressure of 4 kg F per centimeter square. So, from the P V diagram point of view initial state will be 1 final state will be 2 this is state 1 and state 2 these are values of volumes and since the pressure is constant we can make use of this specification to assume that it is a constant pressure or isobaric and hence quasi static process. So, assume quasi static and that is important because I do not know how vigorous this stirring is at 1000 rpm. But, because of that stirring and churning it is possible that this pressure is not really that uniform at 5 kg. But, since it is given sorry 4 kg F per centimeter square but, since it is given we will assume that it is quasi static and the effective pressure acting on the piston at any time during the process is 4 kg F per centimeter square. So, this is one part of the process diagram and notice that because pressure and volume the properties are involved this is also a property or a state space diagram. So, this is an illustration of a process diagram which is on the state space. But, when it come to the stirrer we notice that neither the rotational speed nor the torque are properties of the fluid. So, we cannot represent this on a state space diagram or on a p v or some similar diagram. We will show it on a process diagram against say time and the time goes from 0 hours to 24 hours. So, we can show 0 to 24 hours. And on this axis we will have tau as well as omega. We can say that look tau is to some scale 0.3 kg F meter. We do not show an arrow here because time means that the time progresses like that the process progresses. It is not a state which was initially here going finally there as in this case. And for omega we can again show another uniform line saying omega is 1000 rpm that to a uniform. So, we have taken care of the system diagram the process diagram. We have to the extent possible written down the information. So, that we do not have to refer back to our question again and again. And we have shown that we have understood part of the problem saying the network is made up of two component expansion component and the stirrer component. The next step we have to now evaluate the stirrer work and the expansion work. So, the expansion work by definition is p d v we noticed can be integrated since the process is quasi static. And then this becomes p into delta v since p is constant. This is the algebra now and this will become p into v 2 minus v 1. We stop at the algebra there is no need to immediately put in numbers. Now, we are at a stage where we have to put in numbers. But let us take the other one W stirrer. This is integral minus tau omega d t from 0 hours to 25 hours. So, we have to do this. Notice that here I am not writing from initial state to final state. It is from initial time to final time corresponding to the initial state and final state. And then we notice that since tau is constant omega is constant. This becomes equal to minus tau omega minus t. Now, we have two equations one equation for W expansion and second equation for W stirrer and now all that remains is substitute numerical values and compute. But this is easier said than done and this is where we come across that part of the solution process. Item 5 how to treat numbers, dimensions and units with respect. And here I will spend some time on this and also in particular on what are known as conversion factors. You will notice that in this exercise the units are here. We have kilogram force, we have revolutions per minute, we have kilogram force per centimeter square, we have volume in meter, time in hour. But finally, we want the net work done in kilo joules. That means, it would be better if we obtain the expansion work also in kilo joules and the stirrer work also in kilo joules. For some reason our students do not have a good idea of units and dimensions. And in particular it is not very clear what is meant by a conversion factor. Now, let us try to compute. Let us first compute the expansion work. We have decided that W expansion is P into V2 minus V1. General habit is P is given to be how much? 4 kilogram force per centimeter square. They will write 4. Then V2 will be written as 2 minus 1. What is the unit? Then they start scratching the head and creating some problems. It is strongly recommended that we do not do this. It is recommended that we write as if P is given to be something like 4 x. x is the unit. So, 4 kilogram force per centimeter square multiplied by 2 minus 1 volume in meter cube. So, as not too confused, it may be a good idea to put the volume this at least till you get used to it in square brackets indicating that these are units. Now, notice that we have a force unit here. We have an area unit here and we have a volume unit here. Volume divided by area this will give us length. Force into length will give us work. So, that way the units are consistent. But, we have here the unit if you look at it it will be kg force meter cube per centimeter square. What we want is kilo joules and we know that a joule is a Newton meter. So, a kilo joule will be also 1000 Newton meter or a kilo Newton meter. So, it is necessary for us to convert kilogram force into Newton's and this centimeter square into meter square. So, that finally, only meter remains. What I say is that now we should use conversion factors. Now, what is a conversion factor? A conversion factor is nothing but an entity whose dimensions are nothing. No, no dimensions whose value is 1 and whose units can be any appropriate. We generally remember a conversion factor as say 1 kilogram force is 9.81 Newton's. I think that is what everybody knows. Everyone knows that 1 hour is 60 minutes or 1 hour is 3600 seconds. We will not henceforth remember conversion factors like this. We will remember conversion factors in this way. So, a conversion factor would be not 1 kilogram force equals 9.81 Newton. This is school bookish. So, we will write this either as conversion factor is 1 by 9.81 kilogram force per Newton or 9.81 Newton per kilogram force. Both of these are appropriate conversion factors. Notice, what is the dimension of this force by force? So, it is dimensionless. What is the value of this? Value of it is 1. What is the unit of this? The unit is either kilogram force per Newton or Newton per kilogram force. Out here, we want to convert kilogram force into Newton. How do I convert it? I simply multiply this by 1. So, not changing its inherent value at all. All I do is multiply it by 1, but that 1 I will write as 9.81 Newton per kilogram force. And then I have to take care of centimeter square. So, I will use the conversion factor 10 raise to 4 centimeter square per meter square or its reciprocal meter square 1 over 10 raise to 4 meter square per centimeter square. So, that is also with an inherent value 1. So, if I multiply this by 1, nothing happens. So, I will write this as 10 raise to 4 centimeter square per meter square. Now, notice I decided that look these are like algebraic symbols and we will treat them as algebraic symbols. So, what I will do is in the numerator, there is a k g f. In the denominator, there is a k g f cancelled out. In the denominator, centimeter square, in the numerator, centimeter square cancelled out. So, what remains now? Meter square cancels this gives us meter. I have only Newton meter remaining. So, I confidently go ahead. This will be 4 into 2 minus 1 is 1 into 9.81 into 10 raise to 4 Newton meter. And I know that Newton meter means that meter is defined at joule. So, I might as a way replace it by joule or I may even write it as 4 into 1 into 9.81 into 10 into 10 raise to 3 Newton meter. And I know 10 raise to 3 Newton meter is a kilo joule. So, the solution of this turns out to be 40 into 9.81 kilo joule. Compute this out and write the number. And when you do that or t multiplied by 9.81 392.4. So, that is part of our answer. Let us now proceed with W stirrer. We had W stirrer is minus tau omega t. I think if I go back, you will notice. What is this? W stirrer is minus tau omega t. Go back to that formula. And now we will substitute number. What was our tau? 0.3 kilogram force meter omega. 1000 revolutions per minute multiplied by 24 hour. Now, we know that we have to take care of this minute and hour business. We have to take care of kilogram force. We must go into joule and we have to take care of revolution. Revolution is a dimensionless thing, but we must have it in terms of radian measure. So, we will use a conversion factor there. So, we will end up with there is no need to write this. Let us have conversion factor straight away. Kilogram force to be converted to Newton. So, I will simply multiply that by something which is Newton by kilogram force. So, 9.81 Newton by kilogram force. Revolutions to be converted into radians. That is 1 revolution is 2 pi or 2 pi radians per revolution. Radians does not have anything. So, I can simply write 1 over revolution. Now, there is an hour by minute here. I must multiply it by a conversion factor value 1, but written down as something minutes per hour. So, we have 60 minutes. Now, let us see kilogram force has been taken care of. Revolution has been taken care of minute and hour have been taken care of. So, the units remaining are now Newton meter. So, we write this as minus 0.3 multiplied by 1000 multiplied by 24 multiplied by 2 pi multiplied by 60 Newton meter or I will take care of this 1000 Newton meter as a kilo joules. So, this becomes 0.3 into 24 into 2 pi into 60 1000 Newton meter which I know is a kilo joule. I missed out on 9.81. So, multiplied by 9.81 multiplied by 9.81. Am I right now? Now, we have decided notice that we have treated so far units dimensions and conversion factors with respect. Now, it is time for us to treat even numbers and the results of numerical computations with respect. Let me try this and typically I will see what a student does. I take my calculator 0.3 multiplied by 24 multiplied by 2 multiplied by pi multiplied by 60 multiplied by 9.81. And what is the number I get? Oh, I will write down what I get. I will not miss the negative sign. This is what a student will write down. What is the problem here? The problem here is all that I know is that the student's calculator has a display showing 8 digits and the student and the solver has not thought about whether any all these digits are significant or not. Notice that in the space of the calculator, the student has a display showing 8 digits and the student and the solver has not thought about whether any all these digits are significant or not. Notice that in the specification, we have just 1 or 2 significant digits may be 3, but they also come out of the conversion factor. So, having 8 significant digits here is meaningless by default unless otherwise required 4 or 5 significant figures is all that is needed for most of our thermodynamic calculations. And impress this on our student and impress it on ourselves that that is going to be like that. So, out here it is more than sufficient to write this as minus 26630 kilojoules or you could better write it as minus 2.663 into 10 raise to 4. And now you will notice that I will not going to do this for you, but you better complete it yourself. This is 26630 or 26600 of that order of kilojoules. Other component which we are going to add to it is only 392.4 kilojoules. So, out here we have decided that even the single digits of kilojoules are not going to be of any use. So, for example, here 392.4. So, we are going to round this off to 390 kilojoules and then add to our answer to get the final answer. I will leave that to you as an exercise. I have spent about an hour so far in taking care of some initial explanation talking about solution methodology. And I have provided the solution methodologies one illustrative example. Now, before I assign you for the rest of the sessions till 530, you will notice that each problem here has certain peculiar characteristics. The first problem WI 1 straight away specifies the process. So, we will see something about the process specification. Take for example, expansion work. We know that for a quasi-static process it will be evaluated as integral of p dv. So, we must know p as a function of v. And one possibility is that there is a direct specification. By direct specification we mean p is straight away provided as a function of v. For example, in WI 1 it is given that p is some constant into v some constant divided by v. The constant is specified. Sometimes it is specified that p itself is a constant that is a direct specification. But quite often you end up with an indirect specification. We have not studied equations of state in any detail. We will study that after a discussion of the zeroth law tomorrow and then properties where all the equation of state will come up. But usually for example, a fluid will have a specification called the equation of state where you will have a relationship some say p to the power of v. And p will be a function of v and t for a fixed mass. And then t will be specified or another relations will be specified. In our exercises you will notice take for example, WI 7. Here there is a specification linking force, length and temperature. And then you are asked to do the work calculation when temperature is constant. Later on we will have situations where temperature also varies in a funny way. But that is after that is during the exercises on first law. Then I would also like you to notice that almost any problem cannot be proceeded with unless appropriate assumptions are made. Sometimes there will be hints in the problem specification which will help us make those assumptions. Sometimes there will be no hints, but we will find that we are stuck unless we make those assumptions. When it comes to work evaluation, you will notice that things like integral p d v can be evaluated only for a quasi-aesthetic process. So, if a process is not given to be quasi-aesthetic we will have to assume that to be that is quasi-aesthetic. There will be sometimes some hints. If you look at the exercises here in WI 3 there was no hint, but you come to WI 5. It is very clearly given that it is increased quasi-aesthetically and isothermally. So, it is directly given that it is quasi-aesthetic. But you come to WI 4. Here you have a water basin with a movable wall called the piston on one side. Bring it to the elbow please. Now this is WI 4, the figure in WI 4. It is given that the basin as a fixed width be perpendicular to the plane of the paper, contains water exposed to the ambient pressure p naught and the basin is held in place by the movable piston p. And we have to determine the force exerted by water on the piston. That is actually a problem in fluid statics. You should be able to do it straight away. But then we have to calculate the work done by water on the piston and by the atmosphere on water. When the chamber length is increased mark the word slowly from x 1 to x 2. Why do we say slowly? We say slowly because that is a hint that the process is quasi-aesthetic. And quasi-aesthetic here means that as we move the piston the water is not disturbed significantly. The level of the water remains horizontal as it goes down. And because the level of the water is uniform, the water is not disturbed. This relation between f p naught rho g h and b that remains valid. And WI 4 is another illustration of a problem where you cannot directly integrate. First you will have to determine the force and then multiply it by the differential displacement and integrate it. I recommend that between now and 530 you first appreciate and complete the solution of WI 3. Then do quickly WI 2, 4 and then 5, 6 and 7. 5, 6 and 7 are essentially similar to each other. 6 and 7 are very similar to each other. And every time you attempt the solution of a problem follow the process which I illustrated while showing you the solution procedure for WI 3. These important components you should not neglect and the item 5 that treat numerical values, units, dimensions, conversion factors, etcetera with respect. Sir, this is the coordinator from NIT Kirchay. Regarding the fourth problem, the fourth P naught that is the static pressure acting on both sides of the piston. Is that a case whether the P naught will come to the force equation? About WI 4, take the water as the system. So, the rectangle showing water draw dotted line just around it that is your system. So, work done by the I will sketch it I will go to the my sketch board. Let me say that the extent of water is A B C D, A B C D A. Point B will remain as it is at the piston moves, C will move to the right, A will go down and D will move to the right and become lower in height. When we are asked to determine work done by water on piston, it is the interaction across the C D boundary. Do not get confused by what is here. The boundary is C D. So, work done by water on piston is across C D and work done by atmosphere on water is across A D. The other side of the piston does not come into the consideration over. This just illustrates the importance of sketching the system boundary. If you are not sure where the system boundary is, if you are not sure where the system boundary is, you are likely to get into trouble and you will get confused as is likely to happen in this particular case. If you over to you. Sir, my question is how to do the physical interpretation for any new material. So, if you are not sure where the system boundary is, you are likely to get into trouble and you will get confused as is likely to happen in this particular case. If you over to you. Sir, my question is how to do the physical interpretation for any numerical solution and I want suggestions regarding how to write the comment. Is there any methodology? That is a good question. Actually, in work interaction and in the earlier cases, this question will come up because there is hardly anything to comment about. Take for example, over current problem W I 3, which I took as an illustration. But, here also something you can comment. For example, we notice that here the expansion work is approximately 400 kilo joules. Whereas, the sterrer work turns out to be with a negative sign something like 26.6 mega joules or 26,000 kilo joules. So, the comment you can make here is that the sterrer work is very significant compared to that the expansion work is insignificant. The comments will come on their own when we come to the second law of thermodynamics. Because in second law of thermodynamics, we study the possibility and impossibility of the processes. So, one very important comment, which we make out there based on our calculation is whether the process is thermodynamically impossible, thermodynamically possible or a limiting reversible process, which although thermodynamics says is possible, we human being and engineers are realistic enough to realize that it will be very difficult to implement it in a reasonable time and cost. And if the comment is that it is not possible, we should say why is it that it is not possible, where is it that we are making a mistake and making it impossible. And if it is possible and irreversible process, we would generally like at that time to have a comment on what are the likely or possible causes of irreversibility. And if specifically something is asked for, the question may have a hint that you need a comment of how to reduce the irreversibility and things like that. Over and out, unveil any questions from you. Over. Sir, in the morning session, you talked about the delta E, that is summation of all forms of energy, in that you have not touched the internal energy part. And in the fifth chapter, we are last point, we are having is the internal energy point, point number 11. I did not mention internal energy, it will be included at the end of our zeroth law and when we talk of equations of state and properties of fluids. It is not a deliberate mistake, I could have written it, I just skipped it. Just I wanted to emphasize that energy has various components. So delta E will be delta E component 1, delta E component 2. Depending on the system and the situation, there will be 1 or more than 1 component. Delta U will be one of them, we have not yet defined it, we will define it as we go. Over.