 I will now visit a few centers for discussions. I am trying to pick up those centers who are who have which have not been visited so far to begin with. So, my first stop is at 1093 LDRP Institute of Technology and Research Gandhinagar, over to you. Sir, is there any example according to non-quasi-static process? Example of a non-quasi-static process. Consider SL 0.5 that is a typical example of an extremely non-quasi-static process. We have a chamber half of which is evacuated and half of which has some gas and the intervening partition is opened or destroyed. So, bursting of a balloon is a typical day to day example of non-quasi-static process. Why? Even you open your tap very wide and let the water fall into a bucket. You know it falls with so much twist and churn that at any time till you stop it and let the water in the bucket settle down, you do not know what the level of water in the bucket is. That is another example of a non-quasi-static process. Over to you. Thank you sir. And there is a second question. According to the critical pressure definition in the case of the classification of boiler, so the case of the critical boiler. So, critical boiler is based on that condition sir? No, see unfortunately the word critical is used in many different things. So, critical temperature and critical pressure are purely thermodynamic entities. Whereas when it comes to boiling, we have something called a critical heat flux in pool boiling and flow boiling. That has nothing to do with critical temperature and critical pressure. But generally in technical terms when I say that a boiler is super critical that means the pressure in the boiler, particularly the pressure where the actual boiling takes place or the boiler drum pressure is at a level below the critical pressure. A super critical boiler usually means a boiler where the steam is generated at a pressure higher than the critical pressure. So, there is no such thing as a two phase phenomena which may occur in that boiler. But again I am warning you that for example a nuclear power plant, if it is a boiling based on a boiling water reactor, the reactor itself is a boiler. But when you say such a boiler is critical or super critical or subcritical that means it entirely means different thing. A boiler is critical means that it is sustaining a chain reaction at a steady level of power. Subcritical means that there is a chain reaction taking place but the power is decreasing with time. And super critical means that a chain reaction is taking place and the rate of reaction and hence power is increasing with time. That is so the word critical is a very commonly used word in different aspects. So when it comes to boilers and reactors associated with them just now we have seen three different meanings of that. From the critical heat flux point of view, from the critical pressure point of view thermodynamic, the third one is the criticality of a nuclear chain reaction point of view. Over to you. If we want to calculate the work done in case of the steam turbine and steam turbine is designed based on the two pressure, high pressure and the low pressure. So in that case we can calculate as a mixture of fluid. We will consider this tomorrow when we consider open systems. We have a reasonable number of exercises on steam turbines. Sir, is there any property which is not considered as a intensive or extensive? Yes, there are properties which cannot be shown to be extensive, which cannot be shown to be intensive. There are illustrations. I took an illustrations of the surface area of a system. If you partition the system into two parts, then essentially you are adding surface area. Surface area of part 1 plus surface area of part 2 is not the surface area of the whole system. And of course the surface area of part 1 does not equal the surface area of part 2. So an area is one simple example of a property which is neither intensive nor extensive. Over, 1-2-2-0 Kurukshetra Institute of Technology, Haryana, over to you. My question is that we measure the performance of engine in terms of efficiency while performance of refrigerator is measured in terms of COP. Why it is so? See the performance parameters, the only thermodynamic performance parameter which we have found useful to us is the efficiency because our Carnot theorem etc. is stated in terms of an efficiency. It turns out that even for real life engines, the efficiency or what we call thermal efficiency or energy conversion efficiency is important not because it is an important thermal parameter but it is important because it is an important economic parameter. See, any economist wants to know whatever for any economic entity economic process, what is the value for money and that is what we do when we select a particular type of item, may be sub G, may be piece of clothing or may be a chocolate, we want to select something which gives us most value for our money because what we are obtaining is value, what we are paying back is a resource which we have called the money. Now look at an engine point of view, the job of an engine is produce power. So the value addition by an engine is the delivery of power. That power can be sold by a power plant and it can make money but to produce that you have to supply heat. Heat comes usually out of some fuel and that fuel has to be bought from somewhere, bought somewhere and then brought over to the place where the engine is located. So the efficiency in that case is important because in the numerator you have the power which is what we want, what we need and in the denominator is the fuel or something related to the fuel is what we pay for. In case of refrigerator it is the refrigeration effect divided by the power consumed. So that is the coefficient of performance because it is defined in a different way it is not called an efficiency and coefficient of performance is of importance because the refrigeration effect is what we want. We want the things inside our refrigerator or the air conditioned room or the deep freezer to be cool. So the refrigeration effect has to be provided to maintain those low temperatures. So that is what we need and to drive the refrigerator we need power so that is what we have to pay for. So that is why coefficient of performance is defined there. Sir my second question is that you have said that universe is an adiabatic and its entropy always increases. Can you explain sir? See this is only a question of nomenclature. If you look at our entropy relation, the entropy relation says delta S is greater than or equal to or dS is greater than or equal to dQ by T. If our system whichever way it is defined is adiabatic then dQ will be 0 in which case dS will have to be greater than or equal to 0. So the neatest way to say and proper thermodynamic way to say is that the entropy of an adiabatic system during a process which naturally has to be adiabatic will increase or will not decrease because it is greater than or equal to 0. It can increase or it can remain constant. Now such a system or a combination of systems which make up finally an adiabatic system is often called a thermodynamic universe. I am not defining bit but I am telling you so because this is the term used in a large number of thermodynamics textbooks. So just to prevent questions like yesterday, sir you never mentioned a perpetual motion machine. Somebody on Moodle says sir you never mentioned anything like a latent heat because our study of thermodynamics we do not have to define those terms. However because these terms are used reasonably frequently by others we should know what those terms mean. So from that point of view it is necessary for us to talk about a universe but we should qualify it by saying that what we are talking about is a thermodynamic universe, not any universe. It is a thermodynamic universe and our definition of a thermodynamic universe is any system or any collection of systems which together form an adiabatic system that is it, over. Just my question is sir what is the physical significance of entropy? A physical significance of entropy is that it is a property, a very useful thermodynamic property and the characteristics of property are derived like this that if a system executes an adiabatic process there can never be a decrease in this property and if a system executes a non adiabatic process then the change in this property will exceed or at most be equal to the integral of dq by t during that process that is it. Apart from this there is no thermodynamic significance to entropy. Yes sir my second question is when the entropy of the universe tends to be maximum? There is no question of tending to be maximum. Thermodynamics only says that if a system executes a process which is adiabatic its entropy will increase that is all. It does not say that its entropy will reach a maximum. A maximum means it will reach something and maybe after that it will decrease. For example we say the function sin theta has a maximum at theta equals pi by 2 that means as you increase theta from 0 onwards it increases but at pi by 2 it reaches a maximum value and then if you go beyond pi by 2 it decreases. There is nothing like that there is no such thing as entropy reaching a maximum or anything. Entropy simply increases whenever a system executes an adiabatic process and I should be more precise in entropy will not decrease when system executes an adiabatic process because if it is a reversible adiabatic process it will be that entropy will remain unchanged. Over to you. Sir my next question is that you have said that entropy is inversely proportional to temperature. Entropy is not inversely proportional to temperature. The change in entropy is defined as dq by T for a reversible process lemma. So T happens to be in the denominator that does not mean that entropy is inversely proportional to temperature of anything like that. It is not pv equals constant where you can say that for an isotherm pressure is inversely proportional to volume. We are not talking of such a relationship. It is defined by a differential relationship and then in a differential there is no such thing as direct proportionality or inverse proportionality. Over. Sir change in temperature will not affect change in entropy. Change in temperature means possibly I mean definitely a change in state and if that change in state requires that the system absorber rejects some heat only then there will be a change in entropy. Change in temperature by itself does not lead to a change in entropy. For example you take an isentropic process and we are familiar with an isentropic process. In an isentropic process it is routine for the temperature to change but the entropy remains unchanged. Over to you. 1, 2, 5, 7, HCTM Technical Campus Kaital Haryana. A property of the thermodynamics like pressure, volume, temperature, internality and so on. We also discuss about the entropy. So can you just give an idea how to explain the entropy? Suddenly I lost your audio. It stopped after the word can you explain entropy but I will start from there. Meanwhile you check your audio connection. See thermodynamics seems to be difficult because right from our childhood we are used to ideas like volume which you can physically see. Pressure in some way we can feel although we can directly not feel pressure but some pressure difference we can feel. If we stand in the wind that stagnation effect of the wind on us that we can feel. We can feel velocity as we move on a bicycle or a car directly or indirectly. And similarly we get the idea of just one thermodynamic variable and that is temperature. All of us have somewhat got a thermal shock because of touching some hot vessel and also a thermal shock by touching an ice or something. We have an idea or we develop an idea of what is hot, what is cold and something about temperature but beyond that ideas like internal energy or even the derived idea called enthalpy and then entropy these are ideas which we cannot have a feel. With energy there is not much of a problem because right from school we are talking of thermal energy comparing it with kinetic energy, potential energy and since kinetic energy is related to velocity we develop a feel for it. A ball moving at a low velocity and a ball moving at a high velocity. The one with a high velocity has a higher kinetic energy. While playing cricket if a very fast ball comes to you, you tend to run away from it. And somebody lost a wicket because he did not touch his bat in the crease seeing a fast ball coming towards him last week. So we have a feel for kinetic energy. Similarly at some time or the other some stone or a bird dropping has fallen on us and we have been hit by something which had initially high potential energy. So these components of energy it is something which we are familiar with and hence taking that small step towards internal energy is not very difficult. And particularly when we relate it generally to temperature whatever it may depend on pressure and volume but the main thermal parameter on which internal energy depends on its temperature. So we get a feel but then the further properties which are defined properties like entropy even enthalpy are difficult because it is not possible for us to get a feel for it. For example if this is water I can get a feel for its pressure maybe I can get a feel for its temperature but I cannot get a feel for its energy and I cannot get a feel for its entropy. And particularly so because these two are something which are hidden inside. They are inherent properties of the system and not only that there is nothing absolute about them. It is only a change in those properties which we find is of significance. Over to you 1128 VNIT in Akpur over to you. Sir I want to ask one question. In the steam table sir just you have to see in the triple point that temperature is there in that case the Hf value it is kept in that is equal to the 0.01. How it is kept in that value? At 0.01 degrees C we look up say table 2 in which it is boldly printed. So here we are looking at the table 2 and the first line here pertains to 0.01 degrees C. And when you come to the specific internal energy for saturated liquid it is 0 and so is the specific entropy for saturated liquid. Now when it comes to Hf, Hf is a derived quantity that will be Uf plus Pf Vf. So you calculate Uf to be 0, Pf to be 0.006112 bar and Vf to be the value here. And when you do that and convert it to the unit kilo joule per kilogram and round it up to two decimal places you will find that it turns out to be 0.01. You can check that calculation yourself. In fact at any place in the steam tables the value of H should equal the value of U plus Pv to the required precision. That is an absolute necessity. If it is not so then either one of the values involved is not printed properly over. 1028, Tathya Sahib, Coray do you have any questions? My question is regarding the first law that is delta E is equal to Q minus W. Basically I am from chemical engineering I am using delta U is equal to Q plus W. Can you throw some light on this? It is only a question of sign convention. Our sign convention which I have been using and which is quite often used by mechanical engineers is that work done by a system is positive sorry work done on a system is negative. Whereas chemists and chemical engineers tend to use the other sign convention that is work done by a system is negative work done on a system is positive. So for example if you consider a gas expanding you would be writing DW expansion as minus Pdv rather than Pdv. So you we have a negative sign in the first law you have a negative sign in the expression for DW itself. The net result is the same if you write in case of expansion work minus Pdv. So if you write then finally in some case you will have du equals dQ minus Pdv. We also have du equals dQ minus Pdv. So the net finally the result is the same. The sign convention we use N root is different makes really no difference finally over. Which one is more universally used sir? I think if you are a chemical engineer or chemist you will say that the your convention is more universally used. I being a mechanical engineer we will say mechanical engineers and physicists use my convention so it is more universally used. There is nothing good or bad or better or worse about any one of them. They are just two conventions that is it. We have to be consistent and use one of them but no for example IUPAC or IUPAP has not forced any particular sign convention on any one. Sir, while expressing the laws of things you stress more importance to Kelvin-Planck statement and you say the Clausius statement is not very important. But as we studied it the Kelvin-Planck statement gives us the proportion of the coroner gene whereas the Clausius statement gives us the direction. So I feel both are useful to understand this convention. See when I was a student for example some of my teachers themselves said these are two equivalent statements but later on when I really started assimilating thermodynamics it dawned on me and also was explained in many books that see the Clausius statements talks about a higher temperature and a lower temperature with scale in ideal gas. Then the question comes up why on ideal gas scale? What is so special about an ideal gas? The laws of thermodynamics and the statements of the laws of thermodynamics should not depend on the property of any material. And from that point of view the Clausius statement is a weaker statement than a Kelvin-Planck statement. I am not saying the Clausius statement is incorrect. I am saying Clausius statement is not as strong and as proper statement as the primary statement of the second law of thermodynamics. Question regarding steam table what is the higher limit of extrapolation for finding out properties of superheated and supercritical steam? Our steam table that is the Mathur Mehta steam table go up to 1000 bar and 800 degree C. There are some other steam tables which go up to let me see 2000 degree C and the 1000 mega Pascal okay that is 10000 bar 1000 mega Pascal is 10000 bar right. Going up to higher pressure is not much of a difficulty going to higher and higher temperatures there is a difficulty because water starts getting dissociated. So somewhere in the fine print of these high temperature steam tables there will be a assumption saying these properties are valid so long as the water does not dissociate. It is a dissociation which creates problems. But today's typical for example power plant technology needs properties only about may be 700, 800 bar and may be 600, 700 degree C. So 2000 degree C and 10000 bar is really really a wide range of. For selecting the material of the superheated coils for higher temperatures more than 550 degree Celsius it is difficult to get the material from economical point of view or commercial point of view and metallurgical properties point of view. One more question is there whether these properties follow linear or logarithmic relation within this region? The properties follow for extrapolation. The first thing is extrapolation is never safe particularly when you are already at the edge of the applicability of these equations. So the extrapolation is never safe and hence is not recommended and about properties following linear or logarithmic they follow neither exactly linear relation nor exactly logarithmic relation. Yesterday I put up on our Moodle website the link for the properties for formulation, link for the formulation of steam properties. It is in the NIST steam table page from there you can get a link. If you go to that link you will find that the equation for properties is very very complex. All that it does is provide you with one equation that for the Helmholtz function as a function of temperature and volume and from that one equation by appropriate manipulation like differentiation and differentiation in different ways and combination you can determine all properties including the critical point the liquid vapor equilibrium point properties of dry saturated vapor properties of liquid everything everything can be accepted over to you. 1200 IES college Gopal Madhya Pradesh. Sir I wanted to know there is a critical point between liquid stage and the vapor stage right. Right. Is there any point between solid and liquid stage? I do not think that has been discovered yet. I made a search for this a few years ago and there was nothing on it and that was the situation for quite some time before that. I have not recently searched for it but I do not think anybody has demonstrated a solid liquid critical point over to you. One more question sir, if I want to design a volume of a vessel say boiler shell what properties should be most importantly considered of a liquid say liquid or steam. See if you want to decide thermodynamics will not tell you what should be the size because thermodynamics talks only of processes for the size you will have to determine the evaporation rate and the details of the evaporation that will give you the required height between the liquid level and the top of the drum from where steam is extracted. There is something called a carryover of liquid droplets through the steam which you have to guard against. The liquid volume and its shape is also dictated by something called carry under of vapor into the down comer that also is something you have to guard against. These are detailed technical considerations not really they do not really have anything to do with thermodynamics. After getting these recommended lengths and volumes you fit it in a shell and after that you determine the diameter of the shell and then the thickness of the shell depending on the pressure and temperature. Pressure dictates what is the stress on it and the temperature dictates what is the safe stress it can handle. Combine these two and you will get the appropriate thickness of the shell which will have to be modified considering the you know the number of holes which you have to drill in it to connect down comer and risers and super heater connections and all that. Over to you. 1055, Mar Baselius College of Engineering and Technology, Thiruvananthapuram. Over to you. I have some basic doubts one is regarding already you have explained still I want a little more explanation in first law of thermodynamics for a process DE is equal to minus DW adiabatic. Is it applicable for compressible as well as reverse expansion process as well as the compression process? Yes this DE equals minus DW adiabatic is the first consequence of the basic statement of the first law of thermodynamics and it is applicable to any process in a closed system. It is also applicable to any process in an open system but the model here followed is a closed system. Tomorrow when we consider open systems we will see how to apply this to open system. When it is applied to open system no new principle of thermodynamics is involved only some bookkeeping and change of coordinates is involved. So your question was that whether this is applicable to any process for example expansion compression he said yes it is applicable to any process but of course the condition here is adiabatic process. If the process is not adiabatic then the applicable equation is DE equals DQ minus DW. Sir if it is so there are some engines which are working on cycles like diesel petrol engines. So in that case when we evaluate the work done so we can see we are taking some ratios like cutoff ratio expansion ratio so is there any proper rule for setting that expansion ratio as well as compression ratio like final volume to the initial volume or initial volume to final any. Some of these aspects will be considered when we come to cycle analysis but our discussion of cycle analysis would essentially be from the point of view of basic thermodynamics. If you really want to discuss this apart from thermodynamics many other things like heat transfer fluid mechanics mechanical dynamics of that engine properties of the fuel properties of the oxidant the pollution levels which are allowed all those things come into operation. So it is not such a simple issue to realize an engine a real engine is a very very complex open thermodynamic system and if it is so complex that you cannot analyze it using purely the principles of thermodynamics principles of thermodynamics will not be violated they will have to be applied but along with that there will be principles of heat transfer fluid dynamics combustion physical chemistry mechanical aspects all those things will also have to be included over to you. So there are times like available energy unavailable energy irreversibility so is there is it that irreversibility is what is called the entropy. Not really in the next week when we look at next week we will be studying combined first and second loss and in that we will be coming across terms like entropy production we can define irreversibility we can define exergy and things like that. In the next week notice that on a Wednesday 19th we have a session or two sessions on combined first and second loss. So out there we will be discussing such aspects but unlike the first law and second law for which we have more or less standardized treatment in thermodynamics we do not have such a standardized treatment when it comes to availability analysis and exergy analysis that is the reason it tends to be very confused and that is the reason why I have called it combined first and second loss rather than exergy analysis or availability analysis over to you. 1058 Goa College of Engineering Pharmagudi over to you. Sir I want to clarify my doubts sir regarding the two process one is a polytopic process and another is adiabatic process sir. So in case of polytopic we are writing PV raise to N is called constant but in case of adiabatic we are writing PV raise to gamma is called constant. So in case of adiabatic process gamma is a constant which is a CP divided by CV. So in case of perfect gas the CP and CV are constant. So adiabatic process is only suitable for the perfect gas it is not available for the ideal gas sir. There are quite a few things in your thing in your question see first thing you talked about a polytropic process a polytropic process is defined as a process which can be modeled as PV raise to some constant let me call it N is N is some constant. Now this process is a process specification or path specification and this is a model used for modeling some processes just an equation which is fitted. It is used for processes in the IC engines it is used for processes in the steam turbines it is used for processes in refrigeration it is used for processes in the if you look up old books like that by professor D. A. Lo and Rangham where you have detailed analysis of steam engines reciprocating steam engines. Nowadays we do not teach those things at all out there polytropic processes were used as models for real life processes okay and N the value of N came out of experience or came out of measurement there was nothing thermodynamic about N. The only thing that helped us do after this was with the help of this we could integrate and say that in one stroke this much is the expansion work done or this much would be the compression work required. Adiabatic process means a process in which the interaction is work transfer only and since we have defined heat interaction as something which is not a work interaction this means essentially the heat interaction of any kind is 0 that the only meaning of an adiabatic process. I have been saying so many times that adiabatic process does not mean PV raise to gamma is constant adiabatic process for an ideal gas with constant specific heat in which only PDV work in done and which is a quasi static process only such a process will be PV raise to gamma is constant that is a very very special case of an adiabatic process. So adiabatic process only means Q is 0 so a process in which there is no heat transfer and which is executed by any system whether ideal gas non-ideal gas steam mercury vapor liquid gold anything which comes about so long as Q equals 0 it is an adiabatic process okay and get rid of this idea that adiabatic process is PV raise to gamma is constant that is a very very very special case of an adiabatic process okay. Now you use an idea what is a perfect some perfect gas what is the idea of a perfect gas I do not know I have always been using ideal gas there are some books of physical chemistry which talk of perfect gas but according to their definition perfect gas is nothing but an ideal gas. So what is your definition of a perfect gas I want to know because you might have read it in some book. Sir hello sir. Yes. Like ideal gas is a gas which obeys the equation of state but here the specific heat is a function of temperature here in case of ideal gas but in case of perfect gas perfect gas is a gas which obeys the equation of state but the specific heat are constant it is a not a function of temperature for example if we take a superheated steam which is a perfect gas it is not a ideal gas. Wait wait superheated steam is not a perfect gas just now you defined a perfect gas as one which is an ideal gas with constant specific heat right and now you are saying superheated. Hello sir. Yes. So we are calculating for the superheated steam like superheated steam specific volume we are applying a perfect gas equation that is a V superheated by T superheated is equal to VT divided by. So we are. No we do not apply that equation who asked you to apply that equation what uses this steam table for if you are using such simple equations as V1 by V2 is T1 by T2 for superheated steam that idea itself is incorrect steam does not obey any such equation we have to use steam tables either this type or this type or write a big program and crank it out the behavior of steam is not as simple as V1 by V2 is T1 by T2 in the superheated zone. Okay one more question sir regarding with the reversible adiabatic process and isentropic process sir. So both cases will apply that the PV restrain is called constant. No all I said is that if a process is reversible as well as adiabatic it will have to be isentropic that is all and there is no link between a reversible process or an adiabatic process or an isentropic process and PV rest to gamma is constant. PV rest to gamma is constant is a very special case of some of these processes very very special case. Subquestion F 2.2 Subquestion E. Wait wait wait. Subquestion E. Right. So we are in the enthalpy and the entropy is given. Right. So without using I mean MOLIA chart so how we are supposed to find this sir. Okay. So using the CHIM table without using the MOLIA charts are we are supposed to classify it. Right. Now first let me come to 2.2 E. 2.2 E and F are two situations where enthalpy and entropy are specified. In case of 2.2 E the specific enthalpy is 2900 kilojoule per kilogram. Specific entropy is 6.2 kilojoule per kilogram kelvin. Now if you look up your steam tables let us look up table 2 in terms of pressure because it is a more detailed one. It is good you should all students of thermodynamics in mechanical engineering particularly all teachers should browse through this table. I would like you to start from page 5 and look at the column pertaining to HG the specific enthalpy of dry saturated vapour. Okay. At the triple point it is 2501.4 kilojoule per kilogram. Now go to higher and higher pressures. You will find that the value of HG keeps on increasing. Go on page 6. Even page 6 it is increasing. Now come on page 7. On page 7 also it is continuously increasing. But then when you come to page 8 and page 9. Okay. You will notice that on page 8 in 0.5 kilojoule per kilogram at 20 bar. Then at 21 bar it is higher at 2800.5 increases up to roughly 2804.2. After that it starts decreasing. So you will notice that the specific enthalpy of dry saturated vapour is never higher than 2804.2 kilojoule per kilogram. Okay. So any state of steam in which the specific enthalpy is higher than 2803, 2804.2 kilojoule per kilogram will be superheated vapour. So this particular case where specific enthalpy is 2900 kilojoule per kilogram will definitely be superheated vapour. You can confirm this by opening up the steam chart, the enthalpy entropy diagram. And you will notice that the dry saturated vapour line crosses just about 2800 kilojoules per kilogram at about 20 or 30 bar. And that is it. After that you go to higher pressure the enthalpy reduces. So any state which is at 2900 kilojoules per kilogram is always in superheated vapour zone. But when it comes to the case F where enthalpy is 2500 kilojoules per kilogram and entropy is 8.8 kilojoules per kilogram, you will notice that it tends to be in the wet zone from the chart. And yes in this case you will have to use the chart, beyond that if you want to determine the detailed properties you will have to interpolate between the steam tables in two dimensions pressure as well as drainage fraction and then you will get the value. But that is not what is asked. All that is asked is classify the following states of 1 kg of water substance as wet, etcetera, etcetera. 1, 2, 2, 1 NPR college Natham Tamil Nadu. Over to you. Sir I have one question regarding ideal gas equation. Sir when the hydrogen goes to plasma state whether it will obey ideal gas equations or not. When it goes for a Quark-Bulkon plasma state that means QGP whether it will obey ideal gas equation or not. Over to you sir. See it will obey any plasma will not obey ideal gas equation because the ideal gas equation is for a fluid without any other influence. So it is the particles in that fluid are not supposed to have any electrical charge or any magnetic properties or even any dielectric properties. So the moment a fluid goes into the plasma state that means it has got dissociated. You have positive ions and negative ions all mixing together. So it is a complicated situation. It will not obey the simple ideal gas equation of state. Over to you. One more question sir. In the aspect of thermodynamic properties in what ways the supercritical steam is far better than the superheated steam. The question is in what way is supercritical steam better than superheated steam. There is no such thing as supercritical steam is better than superheated steam. You will notice that our aim is to set up engines with a high value of efficiency that makes sound economic sense. Now assume that although we do not set up 2T reversible heat engines it is impossible but even the formula for efficiency of a 2T heat engine tells us that efficiency is 1 minus T2 by T1. T1 is the heat absorption temperature, T2 is the heat rejection temperature. Although this formula is not applicable in general it is true that if you reduce the heat rejection temperature or if you increase the heat absorption temperature the efficiency of any engine any cycle which produces power will tend to increase. And because of this and the fact that it is virtually impossible for us to reduce the lower temperature the heat rejection temperature because the environment dictates that temperature all our efforts are towards increasing the upper temperature the heat absorption temperature. And as you go higher and higher temperatures from the point of view of sizing and operation it also makes sense to higher and higher pressures. So that is from that point of view supercritical fluid boilers or supercritical fluid steam power plants become economically more attractive as the size increases and as the cost of the fuel increases because then there is a significant pressure on us to improve the efficiency. Even a small fraction of a percent increase in efficiency makes a significant economic difference in the balance sheet for that plant and that is why you will find slowly people are using supercritical steam for large plants over to you. Baramati 1015 VPCOE over to you. Sir what is the heat death of a universe? What is the heat death of the universe? Okay this is an idea proposed by people who want to tell you did you hear that somebody has said that on December 21 or December 22 the day this course ends or the day after the universe is going to end that is the last day of the universe have you heard that? That is because there was an old civilization called the Mayan civilization I think in South America they had a calendar which in those days was considered very perfect calendar and their calendar was set up in such a way that that calendar ended or could not count or note any date beyond 21 December 2012 which is today is 14 December so it is only 7 days. Many people say that the Mayans were so intelligent that they set their calendar up to end with the end of the universe so there are some cranky people esoteric people who have started doing all sorts of cranky things to same themselves from the end of the universe. So the heat death of the universe is one such idea it is not really heat death nothing is getting heated up or nothing is getting cooled down there are big arguments going on as to whether the mean temperature of the universe as a whole is going up or going down because many theory says that the universe is expanding and hence it has to be it has to be getting cooled and things like that so the idea out there not perfectly valid idea is that their idea is the astronomical universe is an adiabatic thermodynamic system. I am not sure that is true because we do not even know whether the astronomical universe can be modeled as a thermodynamic system in the first place if it cannot be modeled we are not sure that if it is not modeled then there is no point in calling it adiabatic if we can demonstrate that it is a thermodynamic system and then we can demonstrate that there are no heat interactions with the entity outside that universe what that I do not know then we can say that the entropy of that universe will be increasing at time progresses because some process of the other will be irreversible process in the universe almost all processes will be irreversible and since the increase in entropy can take place in only in one direction they say that at some stage it will reach entropy beyond which it cannot increase and then entropy increase they claim automatically means an increase in the heat content whatever they mean by that or increase in temperature. So these are all funny arguments heat death there is nothing like that if you are interested in the all sorts of possibilities of end of the universe please read collection of essays by the late Isaac Asimov that collection is known as a choice of catastrophes highly recommended reading over to you. Sir entropy of universe at absolute zero temperature we assume that it is zero but as you said because sometimes that it is not possible to gain that absolute zero temperature then how it is gave that entropy at that absolute zero temperature my mouse theorem it is zero. I never claim that entropy of anything at absolute zero is zero if you cannot have a system with state or whose temperature is at zero Kelvin then you cannot do an experiment between that and any other state and hence you cannot determine what will be the entropy of that state with respect to some other state. Remember that the way we have defined entropy it is always ds or data s. So rather than talk of entropy of state we should always talk of entropy difference between a state and some other state. There are there is some law some people say it is a third law of thermodynamics by which they say that the they provide the something called an absolute value of entropy with that absolute value also is a defined value and that too is they said for a perfect crystal. Now again you are getting into a situation where you are talking about a particular type of material and I do not think that is a good thing to do in thermodynamics. The laws of thermodynamics are independent or it should be independent of the properties of any material over. I lost your video I hope you are still connected. 1108 RC Patel Shirpur over to you. Yes sir I want to ask a basic question that how can we distinctly differentiate between heat transfer and thermodynamics. Okay the differentiation is simple. Thermodynamics talks about interactions the work interaction and the heat interaction. Thermodynamics only says that heat may flow only from a system at a higher temperature to a system at a lower temperature on the thermodynamic temperature scale. It does not say how or at what rate that heat should be transferred or to transfer that amount of heat in a given time which you want to do. What should be provided? What sort of interface? How much area? What sort of material? What its thickness is etc. This rate of transfer of heat is something which is not in the purview of thermodynamics. Thermodynamics is not a science of rate processes. It is a science of processes in which the initial state and the final state are linked to the interactions which occur in between not the rate at which interactions occur. That is the job of for heat transfer that is Q that of heat transfer where we will consider material properties like thermal conductivity, material and fluid properties like fluid characteristics and flow characteristics like heat transfer coefficient and so on. In thermodynamics, you will never ask at what rate a body cools. We will only say that if it is the body is kept like this and allowed to cool what is the final temperature that it will reach. It will not tell you in what time it will reach or what time it will take to reach that temperature. Over to you. It is said that heat transfer principles are based on second law of thermodynamics. Is it true sir? Any comments sir? No, they are not based on the second law of thermodynamics. They have to be consistent with the second law of thermodynamics. And one typical illustration of this is in the theory and derivations pertaining to thermal radiation. We have a set of laws or a law called Kirchhoff's law or what is sometimes called principle of detail balancing. That essentially is an application of the fact that you cannot violate the second law of thermodynamics. This question is regarding the Van der Waals equation. About the critical temperature, the Van der Waals equation is an improvement of the ideal gas law. And for lower temperature, the equation is also qualitatively reasonable for the liquid state and low pressure gas state. However, the Van der Waals model is not appropriate for rigorous quantitative calculations, which makes it useful only for teaching. So, is it really necessary to use or to learn Van der Waals equation? Low gas is approximatable by ideal gas as a perfection. But we use the model of an ideal gas because it is a very useful model. We teach, see, the Van der Waals, even in this course, you will notice that the Van der Waals equation of state is not a part of the main course. It is just in the set of exercises. And I have already said that the Van der Waals equation of state is a useful equation of state because it has a form which is very simple. You can argue that the form takes care of to some extent intermolecular attraction and the volume or space occupied by molecules. But the most important thing about Van der Waals equation is that it perhaps is the simplest equation which shows that a fluid which obeys Van der Waals equation of state will have a critical point. And there are exercises in the properties of fluid section. We will not discuss them now. If you are really into thermodynamics, you will do it. These are exercises P R 20 and P R 21. P R 20 is general one. P R 21 pertains to a Van der Waals gas. So, for such a gas, you can derive what should be from the Van der Waals equation of state, what should be the critical point. We have done that yesterday, but you can also derive what should be the P T saturation relationship. It is very easy to extract that. Of course, you have to numerically extract it because the equations are significantly non-linear. So, you need to use a computer program to crank out the data. Determine, you can show that at the reduced pressure of 1, the saturation temperature is 1, reduced saturation temperature is 1. And as you go on reducing the pressure below P R equals 1, P R also, T set also reduces. And you can even plot that curve till about reduced pressure of approximately 0.8.