 Sir, could you please tell us something about some other uncommon specific fuels which are being used in the IC engines. Say, we use petrol, diesel, LPG, CNG normally, but at IIT Bombay you might be undergoing some research and you might be up to some other useful fuels, could you please tell us something about that. General question, new or alternative fuel development is a continuing exercise. In IIT Bombay, rather than work in mechanical engineering, we have experts in energy science and engineering who are looking at fuels or in liquid form something known as diesel, but we also have a laboratory setup with support means India limited, that is the Cummins engines laboratory and in that laboratory we have experiment in which these fuels are tested out. I think that is about all that I can tell you just now, if you go to IIT Bombay website and go to the department of energy science and engineering, you will find a large amount of detail about this. Over to you. Thank you sir. We have next question. Good afternoon, Professor Gaitonde. My name is Dr. Gupta. I have a general question about the overall workshop, that the workshop is mostly focused on fundamentals and lot of mathematics, but little focused on application side. How would you like to address that question? This is a course on basic thermodynamics because and the reason for this course was that I find that I first discovered that our postgraduate students who come from a large number of non IIT type colleges have a rather poor or I would say rather clear odd idea of what thermodynamics is and then slowly through my interactions and visits to colleges I realized that not just the students, but even the faculty from whom they have learned thermodynamics do not have a proper grasp of the fundamental aspects of thermodynamics. I would just call this course as the secondary thermodynamics course also, although the name is slightly different. These two versions are based essentially on basic thermodynamics pertaining to mechanical engineering. If you want applications then naturally two weeks are not sufficient and we call it applied thermodynamics or energy conversion. If there is enough pressure then maybe in the next few years we can have another course on applied thermodynamics over to you. That applied application where the thermodynamic concepts are used in industries and real life systems that some component of that also if it could have been added then I think the conference could have been much more valuable. I agree with your opinion, but then our application and other applications those can be included in a follow up course on applied thermodynamics or industrial applications of thermodynamics if you want it that way. But over and out 1093 AIDRP Institute of Technology, Gandhinagar over to you. It is required for complete combustion of fuel but at the same time it increases the heat logic as well increases the global warming effect. So what are the recent technologies for to reduce the effects of this, how we can reduce the amount of energy? I think none of us are experts in applied combustion. If you really want to know about it, you will have to pose a question. If you are really interested write to me I will pass it on to our combustion experts in the department of mechanical engineering. They should be able to answer you, but they may be able to answer you. See if in the first law once you say delta E equals Q minus W that indicates that you are using the our convention that is work done by a system is positive. So if you say the value of W turns out to be minus 50 kilo joules best way is to write with our convention W equals minus 50 kilo joules. If you want to say it in English you can say the work done by the system is minus 50 kilo joules or you can say work done on this system by some other system is plus 50 kilo joules or you can say the work transfer is 50 kilo joules from some surrounding system to this system. But the best way is to just write W equals minus 50 kilo joules. If you have to say something the best thing to say in this case is to say work done by the system is minus 50 kilo joules over to you. Sir in exam we had one problem that is for one compressor. It was asked that what is the work done on the system. I got answer as 18600 kilo joules 18600 joules, but answer was given as minus 18600. I will pass this on to the paper setters and I will get back to you. Because I have not really looked at the details of the examination I was busy in the lectures here I had another team which created the question paper and put it upon. I will just check it with them over to you. Second question is the area under PV diagram for a process is expansion type of work. Then what is the area under TS diagram for a process called as it is Q or DSP into T or any other type of thing. If you look at in the morning pertaining to cycle representation, but the slide is already in front of you you can see this. The area under a PV diagram is the expansion work no doubt about it. The area so if you have a closed loop that is a cycle representation on the PV diagram that will be the work done of the expansion type during the cycle. If you have the same process represented on the TS diagram there will again be a curve but then you will have some slightly different representation. All that thermodynamics tells us that the area under the curve which by our calculus and analytical geometries integral TDS by the second law of thermodynamics and definition of entropy that area will be greater than or equal to the heat absorbed by the system during that process from 1 to 2 as represented on this diagram. The equal to will hold only if this process is a reversible process. So there is no name that can be given except integral TDS for the area under a curve on the TS diagram over to you. Temperature in enthalpy temperature H D diagram for the reactants under the products of combustion I think we missed we missed the first few words of your question can you repeat the question Professor Bandarkar is here he will try to answer. How to represent adiabatic flame temperature in H D diagram enthalpy temperature diagram for reactants under products of combustion. So the question was regarding how to represent the enthalpy or how to represent the enthalpy on the H D diagram which is the standard diagram that I had drawn to show the adiabatic flame temperature. So let me just draw it again I am just going to draw the diagram here. So let us say the reaction is A plus B giving C plus B or rather you know it could be A plus B A1 plus A2 plus A3 giving B1 plus B2. So N reactants and N products. So if I sum up the enthalpies of all the reactants. So this is where I will put H this is where I will put T. So the standard state is defined at 25 degrees. So let me just put 25 degrees here and let us say I add up and in many cases even the reactant enthalpy may turn out to be negative based on our reference of 0 for elements in their natural state. So once I start heating the each of these compounds and I add up their enthalpies the curve may look somewhere like this. I am drawing a reasonably you know curve I mean there is no particular shape but it will be a curve which will increase with temperature. So if it is an exothermic reaction which is what all fuels would be doing with oxidants then at the standard state the products would have a lower enthalpy than the reactants and this would be what is called as the heat of or the enthalpy of reaction and it should be negative that is the enthalpy of products minus the enthalpy of reactants at the standard state of 25 degrees centigrade would be a negative number. So this is what has come out. Now if I start increasing the temperature of the product they will form some other curve depending on their values of specific heat. Now the adiabatic flame temperature is something which you can imagine in an open system where some flow is coming in and some flow is going out and the enthalpy remains the same at the inlet and exit. So I will just draw a constant enthalpy line and see where it intersects the enthalpy temperature diagram of the product. So this is products, this is reactants and wherever it hits I will see what the temperature is and that is what is my adiabatic flame temperature. So this is how I would normally represent the whole process on the H T diagram and show what is the adiabatic flame temperature. Thank you. Thank you, over and out sir. Thank you, 1040 college of engineering point, over to you. Good afternoon sir. Good afternoon. My question is Puranik sir and the Boundary cursor. Yeah please go ahead. Go ahead. Sir while teaching this subject the student asking so many questions. Yes. But sometimes it is very difficult to give the reply of this question answer to student and every time student think that teacher will give the answer of this question and suppose for each question students say there is a faculty says there is no answer. I have no idea about this topic and this topic is not covered in this session. So what is your opinion? Is it is this question addressed to all three of us in general or to any particular person? This is general question sir, all of you. Okay. See I do not know what is the situation outside but in a may be now in COEP the situation should be like that of IIT. In IIT whatever be the syllabus the teacher is allowed to emphasize, de-emphasize the topics in that provided the teacher does appropriate credit take care of the core or the heart of the course. So out of syllabus etc does not really arise whatever the teacher teaches till the last lecture here is the syllabus. But I understand that if you are in a college which is affiliated to some university then it is possible that professor X teaches the course. Professor Y unknown to professor X may be from some other college sets up the question paper and the answer book of the student may go to another professor Z who does not know anything about what professor X has taught and does not know the thinking process which was involved which professor Y was involved in while setting up the question paper. So I think you will have to handle this in a local fashion considering what the situation is. But I agree that students sometimes come up with really esoteric questions or out of the way questions and I think as good teacher it is our job to satisfy their curiosity to the extent possible. You may not be able to give him or her an answer immediately but I think if we do some homework refer some books go to the library talk to colleagues who are perhaps comfortable with handling such questions. Maybe in a week or so you should be able to satisfy the student. Over to you. Thank you very much. Over and out. Thank you very much. One two eight six UV Patel College of Engineering Kherwa Gujarat. Over to you. Good afternoon sir. Sir in the test number one there was a question that what is the state of water at the normal pressure temperature. So what was its answer because we thought that it was saturated liquid and in the discussion the answer was separated steam. Yes so your question was regarding the state of water vapor in atmosphere at any general condition and the answer is that it is superheated steam that is because if you consider the partial vapor pressure of water vapor at that instant then your temperature is much above the saturation temperature for that pressure and it is in the superheated state. So in fact any of the gases that you can consider oxygen and nitrogen they are all you can consider they are in their superheated states. So all vapor is in their superheated states until you start condensing it out. So if your relative humidity reaches around 100 then you can start condensing it out and then that is when you see start seeing water droplets. Thank you. So this was about water vapor but in question it was written state of water the word vapor not index. I think I will intervene here I have been saying right from the time of the steam table that our nomenclature for all three phases of the water substance is simply water. So what is the state of water does not mean that it has to be liquid. If the water happens to be at a pressure low enough and temperature high enough to be in the vapor phase we will call it superheated steam. But it finally it remains water at may be 300 degrees C and 2 bar. So what it is superheated steam. Before knowing the answer I cannot say it is vapor or liquid or ice. That is why the proper name technical name given is ordinary water substance. Okay, second question is regarding what is the question please. Can you throw some light on sonic boom what is the condition of Mach number. Yes, so the question is on what is meant by sonic boom and what is Mach number. So sonic boom is simply a terminology used when something like an aircraft starts flying at just over a Mach number of one situation. So you can assume that let us say the aircraft is flying subsonically initially but slowly it accelerates and it reaches a Mach number of one. What happens at that time is that we have seen a sound wave which will be generated at Mach number of one can be considered as an infinitesimally weak shock and that situation is simply called as sonic boom. So sonic boom is nothing but a really really weak shock wave essentially a sound wave which is generated when something like an aircraft just about crosses the sonic speed limit. Thank you. Thank you sir over and out. 1055 Mar-Beselios College of Engineering, Krivan Ram. I am using the old name I hope you do not mind. Over to you. Hello. Good afternoon sir. Good afternoon. I have two questions. The first question is that is there any correlation between the cycle used in an IC engine to the properties of fuels used like if we can whether it is possible to use petrol in a diesel engine without any considerable changes in operation and if not why that is the first question. The second question is something regarding the dual cycle what are the conditions when we have to go for a dual cycle or change maybe a constant pressure heat addition process into a combination of constant pressure and constant volume process heat addition in an engine. And what are the types of fuels used in especially when we go for dual cycle? That is the second question. Over to you. I think here the question is that how do you select fuels for auto cycle and fuel for diesel cycle. I think the answer is it is not the cycle first and fuels later. We had the fuels first engines were somehow developed to run using those fuels and then when people or theoreticians let me say started modeling and looking at the details of those engines then they realize that if you have petrol type of fuel which is easily mixed before going into the engine with the oxidant which is air in this case then you need a spark to ignite it because if you compress it too much it will ignite on its own before reaching the top dead center. So in that case the engine will not work properly or engine may not work at all. So for such volatile fuels which can be premixed and which will ignite the moment a trigger is provided through a spark we use what are known as spark ignition engines of the petrol type and the idealization of that cycle is the auto cycle. Similarly diesel and similar fuels like diesel, LDO, may be kerosene these are fuels which are not that volatile the vapor does not mix with air so easily at room temperature or low temperatures. So what is done is but it is realized that these will burn with appropriate air quantity with them provided the temperature is raised to a high value. So these are compression ignition engines and hence what you do is you compress air and only air till the air temperature is high enough so that as I said yesterday when you inject say diesel or a diesel like fuel the high temperature will first make the liquid heat up then evaporate and then mix and then burn. And since there are this is a multi stage process you can see that there will be time required for heating, time for evaporation, time for appropriate mixing of diesel vapor with air because combustion cannot proceed unless a diesel molecule finds the required number of oxygen molecules surrounding it and finds itself at the appropriate temperature and pressure. Now when diesel engines were run and people started experimenting with them and plotted the indicated diagram they realized that it does take time and during that time the piston is not going to sit tight at the top dead center the piston will start moving. So there is a complicated combustion process which is neither at constant volume nor at constant pressure. In petrol engines also the combustion process is neither at constant volume nor at constant pressure but if you want the first level approximation we model it as a constant volume pressure giving us the Otto cycle. Similarly in the diesel engine the cycle actual cycle is not exactly a constant pressure process but the first approximation is modeling it as a constant pressure process. So it is a fuel first then the engine which works on that fuel then a study of that engine and then idealization of that cycle as either Otto or diesel cycle and or a dual cycle because dual cycle is a model for better modeling of the combustion processes in IC engines. Now the next question was that suppose I have a diesel type engine and if I put petrol in it the if you just put petrol in the diesel tank what is likely to happen is air will get compressed but when the petrol is injected it will suddenly find itself above its ignition temperature it will immediately evaporate and burn perhaps uncontrollably fast already the pressure is high the pressure will still rise higher and I do not think the diesel engine will work properly. Similarly if you try to put diesel in a petrol engine I think you will have problems even before starting the engine because diesel will not evaporate it will either flood the carbulator or flood your inlet manifold if it is a injection type of engine and the engine will not work over to you. We use a mixture of fuels and I have seen that some of the drivers use them in very primitive engines like two stroke engines they use a mixture of diesel and petrol and will it help in any way. This is trying to save money on fuel while damaging your engine significantly. So this is hoodwinking customers hoodwinking themselves and everybody including customers the owner the operator will lose out in a not in the long run even in the medium term. After having said that there are specialized engines for example during the second world war and around that time a German company called Mann MAN developed engines with specialized combustion chambers those were engines which had absolutely wide multifuel very wide multifuel ability and they were used in their war effort because you never know what type of fuel you will get. So with some minor adjustments from the drivers cabin you could switch over from one fuel to another fuel mixture of petrol, diesel, kerosene anything which comes but those were very specialized engines and of course very costly but then when Germany took over the rest of the world in a war costs were not their consideration at all. Over to you. Yes, I have first question to Gaitanade sir. It is regarding a process expansion or compression process. The compression ratio or expansion ratio is it absolutely relating between the final final volume to initial volume or initial volume to final volume which way it will be? Ok see our in when it comes to engines the compression ratio or the cutoff ratio these ratios are always defined to be greater than 1. So when it is compression the compression ratio if it is the petrol type IC engine. In IC engines we always talk about volumetric ratios. So compression ratio is an IC engine is volume before compression to volume after compression. So that is the volume at BDC bottom dead center to the volume at TDC top dead center. Similarly the cutoff ratio is the volume at the end of fuel injection that is when the fuel flow is cutoff to the volume at the beginning of injection which in the standard cycle happens to be the volume at the top dead center or the so called clearance volume. When it comes to turbine type of engines volumetric compression ratio has no meaning we work in terms of pressures and when you talk of a pressure ratio for a compressor or a turbine it is always the higher pressure divided by the lower pressure. This is unlike what professor Puranic talks about the compressible fluid flow. I think for a nozzle the pressure ratio is always exit pressure to inlet pressure. So since the exit is in the direction of flow invariably that ratio will be less than 1. If it is 1 there will be no flow at the static condition. Over to you. When a nozzle is operating at the sea level and if it is in deep space how single nozzle can operate between these two ranges as you are seeing in rocket as well as in state structure. So the question is on whether a nozzle will operate in a similar fashion if it is operated at the sea level and later at a much higher elevation. So if it is operating at the same pressure ratio which are employing and if the nature of the fluid does not really change then in principle the performance would be similar. Of course what is going to happen is that when you go to sufficiently high elevations the density of the fluid if it is air for example is going to be much less. So eventually the thrust if it is what you are interested in calculating the thrust provided by the nozzle will reduce because the mass flow rate is going to reduce at higher elevations if all other conditions remain the same. But as far as the exit mark numbers etcetera are concerned as long as the pressure ratios are kept the same you would get the same mark numbers. But again remember that the thrust is something that is going to degrade. Thank you. When you come to the case of the hypersonic flight how is the combustion is sustaining? So the question is on hypersonic flights and how combustion is sustained at hypersonic flight. So to be honest with you hypersonic flights is not really fully developed completely people are working on developing a successful hypersonic type flight and the most promising design is what is called as a scramjet or a supersonic ramjet. So if you look at the design of a scramjet what is done is that the inlet that the engine has a certain profile which generates a series of oblique shocks from the inlet up to the combustion chamber. So what is done is that the air which is getting sucked in from the inlet is successively slowed through these series of oblique shocks before it enters the combustion chamber. And if I remember it right the mark number of the air that enters the combustion chamber is managed to be just about two or less and with the air getting slowed down to something like mark two or less then you can actually have a reasonable meaningful combustion process in the combustion chamber. That is at least the idea behind a scramjet engine. But please note that successful scramjet flights have been very very minimal and as I said people are still developing these engines properly. So in that sense I do not think there is a clear answer to what your question is but at least people are thinking that if you slow down the air sufficiently through a series of oblique shocks before it enters the combustion chamber then you have a good chance of a complete combustion type process happening. Thank you. Physically how the flow down is possible in that engine because the length of the combustion chamber, a total engine length out is very very small in case of scramjet. Yeah the slowing down of the air is actually carried out through the series of oblique shocks as I mentioned which are generated by providing a specific kind of inlet. So if you look at the inlet shapes for these the scramjet engines they are supposed to provide a sequence or series of oblique shocks and each of these oblique shocks will keep on reducing the air velocity until it reaches something like 2 or below. And having said this I am not too sure about the dimensions whether they are small or large but definitely the process that has been thought about to slow down the air flow is really the oblique shocks from the inlet towards the combustion chamber. Thank you. So now my next question is to Bandarkar sir. The question is nowadays you can see some of the vehicles exhaust is water is shedding from the exhaust. Is it because of the complete combustion of fuel? I think the question is about water at the exit of the exhaust. So you realize that if any fuel is combusting most of all our fuels are hydrocarbons and when they burn with oxygen they will form water. Now depending on how much it has expanded some of the water will condense out and some of it will remain in vapor. So whatever condenses out you will see that as water droplets. Thank you. Next question is to Bandarkar sir. Is there any equipment which can directly measure this rich mixture or like any other? Okay you are asking the question about whether if there is a mixture which can if there is something which can measure whether something is rich or not. Directly at the entry I do not think there is any equipment which will do it unless you are talking of some sophisticated mass spectrometer kind of equipment. What is usually done is at the exhaust people try to figure out how much oxygen is there in the exhaust and this gives an idea of whether the mixture was overly lean or if it was rich. So otherwise just to figure out if something is rich or not I do not think it is so easy to figure out unless you are going to measure a volume and a mass related to it. Thank you. Next question is sir what about the normal reactions are takes place in atmospheric condition. Now what about if the same reactions takes place in zero gravity condition? Zero gravity. I think in zero gravity things should not be that different because I mean I am not very sure if there is any role that gravity is playing in any of these. So it just in gaseous phase you know molecules hit each other and in fact most of the modeling that we do is assuming that there are ideal gases in some way the gravity plays absolutely no role in any of these models that we do. So often I do not think gravity would play any role in modeling any of these combustion processes. Thank you. Sir combustion instability is it a part of the combustion technology actually means combustion instability. I am not sure what the question is supposed to imply I mean if you what would you mean by combustion technology. So if you can clarify that then I mean anything which is which people would be interested in to get good combustion I would say that that is combustion technology then if they do not want instabilities that is going to be part of it. I mean if that is how it is I would say you know I do not know what other answer to give in this case. Thank you. Stability, combustion instability some solid fuels and liquid fuels or some rigorous combustion you know something called combustion instability. So what is the question really I did not get it what is the question really. What is combustion instability. So this would be delving into combustion heavily and I am not really that much of a combustion expert. So I would not know often how to describe this. So again you know if you put this question upon Moodle then I would go and talk with the faculty in our department who are more of combustion experts and you know come back to you on this. Thank you. Can you recommend some simple text books. Yes I think you can read the text book by terms that is a pretty good text book T-U-R-N-S Stephen terms I think that is the name of the author. It is. Plains and combustion is that the name? Introduction. Introduction to combustion that is the name of the text book. So it is we have just written it down Stephen R terms introduction to combustion. Thank you. Is there anything else? Can you explain a SL flow? Yes I mean you can go to terms and all the details will be there. It is just re-emphasizing that you know enthalpy is a state function. So instead of going directly using one path if you go via 2-3 path you should still have the difference in enthalpy between the products and reactants as the same. If you use that that is what is really Hess's law in trying to get enthalpy of reaction for which you cannot easily measure the enthalpy of reaction. That is about it. Thank you sir. We really enjoyed the 10 days program and thanks a lot for cooperating for our convenience and all. Thank you very much. 1166 July Institute Doge. Over to you. Sir I want to ask you one question. Can you suggest us the topic so that the research work can be conducted on that particular topic related to thermodynamics? Often I cannot give you it is for you to do some review, search, find out what your interests are, find out the learn about the work being done, find out the gap in that work which you feel should be filled up and that automatically leads you to the research topic. It is difficult for anyone of us to hand out just the way we have a set of exercises. We cannot create a set of research topics by any routine means. Over to you. Currently in your mind? No, I am not doing basic research in thermodynamics. If at all I am worried about how to teach thermodynamics and I do not think that is the type of research you are interested in. Over to you. Thank you very much.