 1, 2, 6, 1, MET is institute, Nashik, over to you. Good afternoon sir. Good afternoon, go ahead. Hello sir, good afternoon. My question is that, in many of the recent petrol vehicles, there is a provision for extra one inlet and exhaust wall. Means sir, right now there are four walls there. So what are the effects of that extra inlet and exhaust wall in auto cycle? Over to you sir. So, the earlier old engines used to have a single inlet and a single exit. And now we have either one inlet, two exhaust or two inlet and two exhaust in most of our modern engines. The reason for that is, remember in the auto and diesel cycle, there is an induction stroke and then exhaust stroke, the first and the last one. In my cycle diagrams yesterday, these were the strokes 0 to 1, processes 0 to 1 and 5 to 6. Now, these were assumed to be constant pressure processes of flowing in and flowing out. In actual practice, they are not so. During suction or induction, there is a pressure reduction. The internal pressure is below ambient because of the resistance to flow. Similarly, during exhaust, the internal pressure is above ambient. So, some work is required to be done for sucking in and for exhausting the gases. This depends on the resistance and by providing two valves, you are providing larger area of flow. So, the mean velocity through these openings reduces and hence your pressure drop during induction and exhaust reduces, increasing the effective power output of the engine and also the efficiency of the engine. That is the reason. Over to you. There is another question that on Mollier chart on ethyl diagram, the constant pressure lines are diverged from each other. So, why such happens? Okay. This is a basic property relation. If you have an H S diagram, if you take a constant pressure line, you should notice that the tangent to this is partial of H with respect to S at constant pressure and you should be able to show that this equals the temperature T and that is the reason why at constant pressure, if you go to higher and higher entropies, the temperature usually rises except you are in except when you are in the two phase zone. So, the first thing is the pressure in the vapor phase or gaseous phase, the H S line goes to higher and higher slopes. Now, when you go to, when you compare two pressure lines say P 1 and P 2 and when you compare their slopes, either you compare them like this or you compare them like this. Either compare this point with this point or this point with this point. All that you have to do is compare the temperature at these two points and you will find that the slope is exactly the temperature to some scale. However the temperature is higher, the slope is higher because if you see the basic property relation that is d H is T d S plus V d P. So, the partial of H with respect to S, partial of H with respect to S at constant P is temperature. This is the basic relation and that is why you will notice that in the two phase zone the wet zone of the enthalpy entropy or Mollier diagram, the isobars are straight lines and you just take an isobar, take certain length, measure the delta H, delta S that means determine the slope and you will find it is exactly equal to temperature and remember in this course when we talk of temperature by itself, we always mean temperature as measured on the thermodynamic scale Kelvin, over to you. Hello sir, there is another question for the Bandarkar sir, Bandarkar sir, now during combustion fuel and air is combined, now how the, during incomplete combustion, how the distribution of oxygen is take place and what is the sequence of formation of the products? So, the question was regarding incomplete combustion and normally what is assumed is that if oxygen and fuel are not mixing properly, then incomplete combustion takes place and since the fuel is mostly made up of hydrocarbon, incomplete combustion of carbon would lead to either CO formation or just pure suit formation where there is only carbon, but definitely hydrogen has more affinity to oxygen than carbon. So, nearly all the hydrogen definitely gets burnt up with oxygen and water is definitely formed. So, the first usually it is assumed that hydrogen is completely burnt, but carbon may or may not get burnt depending on how much it is mixing and hence you will form CO or just plain carbon and not CO2. I assume this is how you were talking, what you were asking regarding distribution of the product. Thank you. Hello sir, there is one more question to Puranik sir. As we had seen the convergent divergent nozzle, in case of diesel injector, if we see the back pressure in the cylinder at the end of compression, when we see that the nozzle injects, why it does not choke? So, the question is related to a diesel injector nozzle and related to its operation. See, I think the one basic difference between what we have talked about and what you are referring to is that the fluid in a diesel injector nozzle is most likely going to be in liquid form when it is moving through the nozzle, whereas what we have talked about is gas flow in the nozzle. So, fundamentally these two are going to behave differently. I think my understanding is that the liquid fuel will eventually atomize in the combustion chamber and then burn, but as far as its flow is concerned through the nozzle, I think it is going to be in liquid form and in that sense I do not really think the phenomenon of choking really applies here at all. Thank you. Good afternoon sir, one more question to you. Sir, actually what is the phase or state of a fire? Means is it a solid or liquid or gaseous or any else? What do you? Why? Why is it a process? So, then something is burning. Are you asking what is the state when something is burning? What did you ask really? When something is burning that flames are coming. So, what is the state or phase of that flames? I think as far as if nothing is ionized, you can safely consider that all of the flames are in gaseous states because they are mostly going to be oxygen, nitrogen, carbon dioxide and something like this and the flame color is usually going to come because of certain excitation, transition that some of these species will be doing and you can. So, radiation. Yeah and radiation primarily. So, you can safely assume that these are all in gaseous form, no problem. Generally, they are not followed all the gaseous means properties of gaseous. So, how can we say that they are in gaseous form? Yeah, so I am not very sure why you are claiming that they would not exactly follow everything about gaseous. At high temperatures and at if you try to extract anything, you will realize that they will actually behave very close to gaseous. So, I am not sure that why you are claiming that they behave much different from gaseous at all. Thank you sir, over and out. Thank you very much. 1-1-3-7 R K college, Rajkot, over to you. Good afternoon sir. Sir, I have read one sentence in one of the books that give an example of any such kind of process which obeys second law of thermodynamics, but it does not obey first law of thermodynamics. Okay, your thing is you have heard, read a statement in a book which asks the reader to provide an example of the process of which obeys the second law of thermodynamics, but does not obey the first law of thermodynamics. If you like the book otherwise, take a big sketch pen and blacken out those particular lines. That is all I will tell you just now. And if there are more than a few sentences of this kind in that book, I think you should use that book as a paper weight rather than as a book, over to you. Sir, name of the book is Engineering Thermodynamics by Yunus Senjal. I will have to check it out. University, sir, this kind of question was asked in university exam also. I will not blame the university for that, I will blame the paper sector for it. Unfortunately university gets their reputation tarnished because of the actions of examiners. One more question, one more question, sir. Well, we have referred many cycles like Stirling cycles or Ericsson and Atkinson. Do you give some examples of applications of these cycles? See, it is not an application of these cycles, these cycles of models of the real life cycles which occur when you use a certain type of fuel and a certain type of mechanism. For example, the typical idea of an Atkinson cycle is a cycle in which the compression process, compression ratio is smaller than the expansion ratio, requires a certain type of mechanism in the crank and the crankcase. And that is, you can implement it like that. It has certain advantages and it has a certain disadvantages. So, you can come up with an idea of a cycle, coming up with an idea of a cycle for analyzing in a thermodynamics exam is one thing. But only those cycles will work which make economic sense and work with real life fuels and somebody, some company can design engines, sell them and make profit. Thank you, sir. Over and out. Thank you. 1080 Gokaraju Rangaraju Institute, Kukatpalli Hyderabad, over to you. We have a doubt like, what is the, when the adiabatic process is there, then in that the constant pressure is also there. We want an example, sir. I think if you go through my exercises, in the first law exercises F1 as well as F2, there are a number of processes which are both adiabatic and either constant volume or constant pressure and I think in one case even a constant temperature. See, we have had this discussion in the morning also and even earlier. You should get rid of the idea that adiabatic means a certain kind of process. It is not. Adiabatic means any kind of process in which the system is not allowed to exchange heat with its surrounding. DQ is 0. That is the only thing required for an adiabatic system. But you can execute it at constant volume, say using a stirrer in a gas. You can execute at a constant pressure again by using stirrer or electrical work input. But adiabatic means DQ is 0. Any mode of work is acceptable. Even more than one mode of work is acceptable. That is it. So, you yourself can create exercises. For example, because you are all teachers, I suggest that you write the first law or say, okay, you can write it as any form you feel like. You write it as delta u. Now, delta e, you write as delta u. Do not complicate it much. You say the delta e potential plus delta e kinetic. On the right hand side, you will have Q and then minus w, you write as w expansion plus w stirrer plus w electric. If you are a bit adventurous, consider a bubbly system and put surface tension. And now decide that, look, I want a problem in which this term, this term, this term is going to be significant. That is how one of the problems was created. We said that, look, I do not want delta u to be significant. I do not want delta e to be significant. I want delta e p to be significant. So, that is why some sack of sand was thrown from a height of some 20 meters or 30 meters. Naturally, in that case, there will be no work interaction. I said initial temperature, final temperature is the same. So, there will be some Q interaction, if at all. You can set up the problem yourself. For example, you can create a problem in which delta e kinetic is significant. All that you have to do is take two bogies, two heavy bogies of a railway truck, good strain and throw them at each other from a distance of 100 meters initially travelling at some 5 meters per second. Finally, if they are equal mass, equal velocity, the momentum conservation will see to it that they latch into each other, let them latch into each other and then they will be at 0 velocity. So, you have a large kinetic energy initially, 0 kinetic energy at the end. So, you have a significant delta e kinetic. And then you decide which one of the other terms are going to be significant and set a problem accordingly. I suggest that you set such problems, not very complex to begin with, but go on increasing the complexity or go on changing the significance of various terms. And you yourself will realize that adiabatic is nothing but this term being 0. You have a large number of options even with the other terms. Over to you. Sir, I want to have some, like know about some text book orders for refrigeration and air conditioning apart from aurora domukundwar and manohar prasad. Any text books that you can recommend for good understanding of refrigeration and air conditioning. Over to you, sir. You have got one. There used to be a book by an old author called Stoker, STOECK S-T-O-E has books on refrigeration and air condition. There was a time when before aurora came onto the scene, this used to be the book. And there was a very old book, not much of thermodynamics, but reasonably good on classical refrigeration and engineering. I think that was Jordan and Priester. But I am not sure, I may be confusing something. Thank you, sir. My colleague is having one more question, sir. Good afternoon, sir. Good afternoon. We are developing a solar energy based equipment and in that we are taking the evaporation and then condensing and then taking the water. And it is a closer jumper with a blackbody radiation. What are the best conditions to get the better evaporation regarding the pressure and how the surface of the water behaves? You should talk to somebody who is involved in solar stills or solar distillation. I would not be able to help you on this. Over to you. Sir, it is regarding the water surface, how it behaves under closer jumper. So that is what exactly I want to know. Even then, this is a process of distillation with solar radiation, etcetera. I have not studied that process in any detail. Over to you. Thank you, sir. Why specific heat? Our value is 1 for liquids and more than 1 for gases like diatomic and triatomic gases. This is an experimental thing. There is no question of it being equal to 1 or anything like that. Depending on what your units are and what the material is, the value of specific heat will be different. For example, at water at room temperature and pressure, it is approximately 4.18 or something like that kilo joule per kilogram kelvin. For air it is at normal temperature pressure, it will be about 1.005 or 0.5 kilo joule per kilogram kelvin. So these are experimental values. The numbers do not mean anything to me. Maybe she is asking for CPC of gases. Sir, one more question. Sir, why water increases in volume when it freezes when compared to the other fluids? I think this has something to do with the structure of ice. So you have to go to either you have to study solid state physics or material science to appreciate why that happens. Over to you. Thank you sir. Over and out. 1 to 1.5 Bajaj group Akbarpur Uttar Pradesh. Over to you. Greetings sir. Greetings. My question is that is it always necessary for a cycle to have four processes? In textbook of engineering thermodynamics by P.K. Nag, I have seen a cycle with three processes. Its name is linear cycle. Do this cycle have any physical significance or any practical application? See it all depends on what you define as a process. If you define a process as something you know standardized like isobaric, constant volume, isentropic, isothermal or something like that, then you can have a cycle with minimal three processes because two processes will not close the cycle. So generally we have four processes because that is how cycles in real life are implemented. But if you go to really implemented cycles for example you just go to a nearby steam plant and get their cycle diagram you will find maybe roughly two dozen processes involved in it and it is not just one loop there are loops within loops. So we should not worry about how many processes in a cycle. For example just the boiler process in a steam generator which takes the water from subcooled liquid state to superheated steam state. I can consider it as one process but somebody can say that look these are three processes. The first process is you know subcooled heating of water up to saturated liquid state. Second process is the process of evaporation from saturated liquid state to tri-saturated vapor state and the third process is super heating process from the tri-saturated vapor state to the exit state of the boiler. So get rid of the idea of how many processes create a cycle. It is for you to define, over to you. Next question is is it possible to apply the study flow energy equation in the analysis of fin? Study flow energy equation for? In enthalpy. The application of study flow energy equation in the analysis of driving the enthalpy of the fin. Enthalpy of the fin. Which fin are you talking about? Yes sir. The projected outwards area which are used in engine cooling. Anywhere where you can consider your system to be an open system and steady situation, steady flow energy equation is applicable. In fact, steady flow energy equation is applicable when you have a steady state and you have no flow at all. All that will happen is the flow terms will become 0 and you will get automatically you will fall back onto the cloth system form. And if the fin is not moving, the cloth system form is applicable. So SAPE is also applicable. But you do not have to start with flow equation. You can start with the first law for cloth systems and be done with it. Over to you. My next question is about the saturation table of a refrigerant which is ammonia. I have seen most of the saturation tables at minus 40 degrees C. The value of entropy assigned 0 that is SF and HF is also 0. Is there any convention or any? That used to be the earlier convention and may be even now it is a convention. But all I would say is the reference state does not matter. So long as you use the same table for your inlet state and exit state or initial state and final state. It does not matter whether saturated liquid enthalpy at minus 40 is 0 or saturated liquid internal energy at triple point or some other temperature is 0. It is a matter of convention and also a matter of personal choice and convenience. Over to you. Hello sir. My next question is can two reversible adiabatic processes can cross each other or intersect each other? I have seen somewhere that two reversible adiabatic processes cannot intersect with each other. Why is it so? If you appreciate the fact that a reversible adiabatic process means an isentropic process then rephrase your question as to can two different isentropic lines, constant entropy lines intersect each other. The obviously thing is they cannot because one will have one fixed value of entropy say 1 kilo joule per kilogram Kelvin. Another will have another value of entropy say 2 kilo joule per kilogram Kelvin. If they intersect that means one particular state will have two distinct values of entropy which is not possible entropy being a property of the state. So two isentropic lines or two constant volume lines or two constant pressure lines or two isotherms they will not intersect each other. That is a basic characteristic of any iso line. It is like asking on a map on which you know constant elevation lines are marked. All of us have seen such maps in our school. Will two constant elevation lines intersect each other? The answer is obviously no because one line will represent say an elevation of say 1000 meters above mean sea level. Another will represent say 1200 meters above mean sea level. If they intersect they would mean that at the point of intersection the elevation is both 1000 meters above MSL as well as 1200 meters above MSL. So that is just not possible over to you. No that is not true. Depending on the type of system it will depend on either temperature and pressure. If it is a fluid system it will depend on two variables. Choice is yours TP or PV or VT. It is only for an ideal gas that enthalpy is a function only of temperature over to you. That is the basic corollary of Carnot theorem. Carnot theorem says that the efficiency if a cycle works between two distinct temperature levels then the highest efficiency is that of a reversible cycle working between these two temperature levels and our Carnot cycle is defined to be a reversible cycle working between two temperature levels. So the Carnot cycle will have the highest efficiency but that is not the only cycle which will have the highest efficiency. Any other cycle which is reversible and 2T will have the same efficiency over to you. Hello sir. If there are two thermal energy reservoirs and one Carnot cycle is operating between them and one Rankine cycle is operating between them you have said that all the reversible cycle will give the same efficiency. It means that efficiency of Carnot must be equal to efficiency of Rankine. Is it so? Yes. What is saying is our Carnot theorem says is that if you have two reservoirs say one at temperature T1 and another at temperature T2 then any reversible engine working between the two will have the same efficiency. This is a general reversible engine. This is a Carnot implemented as a reversible engine. You could have an Ericsson but must be reversible. You can do some cranky things in any cycle. This could be Sterling implemented as a reversible engine. So long as they work between these two fixed temperatures all of them will have the same efficiency. You can create your own very complicated cycle. So long as it is a reversible cycle and works between a given set of temperatures all of them will have the same efficiency. If you give up the idea of reversibility then you are losing on the efficiency. And if you compare two engines both reversible but the T1 and T2 are not the same pair of temperatures again then the efficiency is likely to be different. Over to you. Thank you sir. Over to you. Thank you. Vignan Institute, Nalgonda 1126. Over to you. Question is for Bandarkar sir. Good. Sir regarding entropy, when we define entropy, entropy is a property which increases with respect to temperature. When temperature increases, entropy increases and similarly temperature decreases, entropy reduces with respect to the definition. But if you go to the formula of dS equal to dQ by T, when temperature increases dS should be reduced that is entropy reduces. Similarly means vice versa no sir. How can you explain about this thing sir? See although you said this is of Bandarkar, let me talk to tell you that your basic idea that entropy increases with temperature increases that itself is not true. Entropy and temperature are two distinct properties for a system. In Carnot cycle you have a situation or on a TS diagram you have a situation where you can lay out a constant temperature process where entropy can have any value. Similarly you can have a constant entropy process where temperature can have any value. On a TS diagram you can select a process for example this is TS. So if you consider a process like this, this is temperature increasing entropy increasing. A process like this will be temperature decreasing entropy decreasing. A process like this will be temperature decreasing entropy decreasing. Process like this will be temperature increasing but entropy decreasing. You can have any combination. It all depends on your initial state and final state. Get rid of the idea that as temperature rises entropy has to rise. That is not true. Over to you. Another question regarding Carnot cycle sir. Yes. In Carnot cycle that is a constant that is reversible compression process and adiabatic reversible compression process then comes constant pressure heat addition process. So here adiabatic process is a very fast process. No, no, no. Get rid of this idea that adiabatic process is a very fast process. Adiabatic process only means dq is 0. It has nothing to do with the fast or slow or anything. If you have dq equals 0 that is an adiabatic process. It may be fast, it may be slow. It does not matter. Thank you sir. Is exergy of a fluid at a high temperature more than at a lower temperature? I did not get your question. What was your statement? Is exergy of a fluid at a higher temperature more than at a lower temperature? See exergy, if you say the flow exergy it is defined as H minus T naught s. So depending on the at a high temperature calculate the values of H s, low temperature calculate the values of H s and compare the exergy. There is no direct relation between exergy and high temperature or a low temperature. Exergy is completely destroyed at that state. Is it correct? I do not know where exergy is just a related term. When the exergy difference or the decrease in exergy during a process we have shown tells you what is the maximum possible work you can extract when heat exchange is allowed only with the ambient. There is no such thing as exergy destruction or anything. Exergy is just a property or pseudo property because T naught is involved. So that property can change but that property it is not proper to say that it is destroyed. Exergy in minus exergy out is exergy destroyed. We call it exergy different. Why should it be exergy destroyed? When temperature at the initial state is 100 degree C, final temperature if it is 50 degree C, that temperature we say it decreases by 50 degree C. We do not say 50 degree C of temperature is destroyed. I do not like the word destroyed. There is no creation, there is no destruction. It is just a change in some quantity from state 1 to state 2. Over to you. I think I will stop here now because we are into the T time. General question which I think I should answer. Some centre I do not know the identity says I think our number will not come. I am sorry for that because there are large number of hand raises. So we are asking here please convey to professor the following question. Why Pt are taken on y axis and other quantities on x axis or so for any plot? Well it is just a question of convention. Usually P goes with V and the area under a curve is what we usually talk about. Not the area to the left hand side of a curve. If you are comfortable with that there is nothing wrong in plotting V on the y axis and T on the x axis or say S on the y axis and T on the x axis. So the choice is yours it is only a matter of convention. Maybe somebody will write a book in which P will never be on the y axis. If people like it so be it. This one I am going to go to Mahatma Gandhi missing college. Noida Uttar Pradesh 1205. I hope they are there. Good afternoon sir. Sir have you written any book on the subject Tharmodani style? No I have not written any book. I have not written any book to want the book which is nearest but not exactly according to my style. It is the book by professor Achyutan who was my teacher. I suggest I should not use the word recommend I suggest you look that up. But if you have understood the basic thoughts and the ideas the way we have tried to discuss thermodynamics I think you should not worry about a book. You start your own understanding refer to books and our Moodle site on which some discussion is taking place that site will be on for an indefinitely long time. So, so long as you follow the protocol of asking a question properly at the end of the question writing your name and the affiliation where you belong to and then I will I generally do not respond to questions which just say I do not understand anything about flame velocity and there are large number of such you know queries. And you will notice that if there is a need query with proper format and with the final signature line because I must know who I am talking to at least in picture I am seeing you now I know you are someone from Mahatma Gandhi mission. So, I do not care about the name but when everything is only in terms of words I want to know what the name and place is over to you. Yes sir. It is a first law that there is a form of first law. You know I am really unable to understand how this you know adiabatic work is 0 because under adiabatic condition one can expect work done. Yeah the first law does not say adiabatic work is 0. The first law says that the adiabatic work in a process with independent of the path of the process and depends only on the end states and not how those how the process has gone from initial state 1 to final state 2. A corollary of this would be that if a process starts is an adiabatic process which starts from a given state and after executing a cyclic process reaches the same state again then during that process the work will have to be 0 because it depends on the end state. If the end state is not changing there will be no work done but otherwise in a general adiabatic process the work can be 0 positive negative but it depends only on the end state and not on the details of the path. On the Moodle site I have provided link to the works of Joule and Ruhmfold do look them up they are historical the terminology is odd not our terminology but you will get a hint at the thought processes which were involved in formulating these laws of thermodynamics. Another recommended book is by Bridgeman it is called Nature of Thermodynamics. It is a very old book but in a old well established library you are likely to find a copy over to you. Thank you very much. My next question is to Professor Proudhik. You have very nicely described the choking phenomena in the convergent or convergent nozzle. If you apply this phenomena in the centrifugal compressors and the axial flow compressors along with this phenomena there are two other phenomena like stalling and surging. So I want to you know a little explanation or a little light on this subject how they are. Professor I am really sorry but I am not aware I mean I am not really an expert in these aspects at all. If you do not mind you put this up in on moodle this question and somebody who is more familiar with fluid machinery is like what you are mentioning. I will pass on this information or the question and get the answer from them but me per say I am not really aware of these technologies in the machines that you are referring to. I will be able to answer these questions. Okay so thank you very much. Thank you very much. 1139 Rajarambapu Institute Islampur Maharashtra over to you. I just want the hint for the problem CL6 that is combined first and second law. Question number 6. Question 6 in combined law. Oh that is my question. Okay put this up on moodle I will give a very detailed story type of explanation to this do not worry. I will do that. Question is how in Carno cycle how much we can get maximum possible efficiency through Carno cycle? Again let me go back Carno cycle is a reversible cycle. If it is implemented in a reversible way between two temperatures T1 and T2 you will get the maximum possible efficiency. You cannot do anything additional or special to get extra efficiency from a Carno cycle. If you want to improve the efficiency of a Carno cycle all that you have to do is increase the T1 or decrease T2 or perhaps both but then you are you know providing different driving potential that is it. If T1 and T2 are fixed why Carno cycle any reversible cycle working between those two temperatures will give you the same efficiency over to you. Okay more and out. Thank you. I am going to stop here now.