 So, I was supposed to take 45 minutes of combustion and 45 minutes of kinetic theory and let us see how much I can finish. Combustion is something that you know sometimes you will have to touch upon in thermodynamics at least in basic thermodynamics because it may or it may be a part of what you normally have with a cycles course that is if you are running a cycles course like applied thermodynamics or something you will need to teach some amount of combustion definitely and I am covering only the basics here and I think the real key thing that you need to first explain to students is that this is not you know really out of the way and it is something that they need to know just because when we run any cycle especially power cycle we use some fuel and we need to know at least something very basic regarding the fuels that we are using and this is why we need to know at least something about combustion and these are very reasonably rudimentary stuff that we teach and a lot of it is based upon 12 standard or 11th and 12 standard chemistry. So, I think that is something that you must emphasize. So, before you know get into combustion the a lot of times you need to tell them about fuels and you know this can need not be some detailed thing because we are not doing detailed properties of fuel, but as long as you know and this is something which is very common in knowledge to them you just need to reemphasize that fuels can be solids, liquids or gases and typically what we would normally use in solid fuels is only coal. So, if they are of course you know going to do a further course on combustion they would probably be interested in biomass and wood, but you know you can tell them that this is what exists and then you can tell that coal of course you know they would have studied you know a reasonable amount of materials sometime in may be 10 standard or something about how coal is formed you know that wood gets into the below the ground and there is pressure and so on, but you can just tell them about various types of coal that exists. So, for example, lignite they it is a word which people would have normally heard and then you know anthracite or regular bituminous coal that is found in India and you can emphasize that Indian coal is mostly a big chunk of it is has ash in it and that is one of the reasons it is not so good for good metallurgical processes where you need really good coal which is made purely of carbon. Whereas, if you have something which is which is lot of ash or lot of residue it is a pretty good fuel for what we call as external combustion engine. That is whatever you combust that is not combusted in the working fluid. So, that is I mean you will of course whenever you are teaching cycles and engines you will teach what is external and internal combustion and the most common example that you will give for an external combustion engine is the steam turbine. Because the steam is the working fluid and the combustion is always outside and whenever it is outside you can use any fuel you want because that is not going to contaminate your working fluid. So, I mean you can use coal, you can use oil, you can use gas anything to heat the working fluid. Then, so you should emphasize that the most common solids that we will look at are coal and then you can say that as far as we are concerned the other engines that we will look at are mostly gas turbine cycles or automotive engine cycles and these are going to be internal combustion engines. If there are internal combustion engines you know you cannot use a solid fuel there and then you would rather use liquid fuel and you know the regular internal combustion engines which are automobile engines will use either petrol and diesel they can also use gas. Similarly, the gas turbines can also use petrol and diesel kind or liquid fuel or they can use gas because you can spray it atomize it and there will be very less residue it will go over the moving part may be sometime here and there there will be some black carbon residue but overall you know it will not spoil the working parts of the engine and that is the key. So, at this time you know and definitely sometime in 11th or 12th somewhere they would have heard this word called fractional distillation and they know you can just reemphasize that you know there is something called crude oil which you basically bring from you know drill from below the ground and this is the whole mixture of things and you do what is called as distillation that you warm it up and then start cooling it down in different temperature then depending on the molecular weight of the compound you will you know get distills at different different levels and whatever does not really liquefy will be just purely gas and anything which is gas at room temperature you can say roughly hydrocarb, see most fuels that we will use will be most fuels that we will use will be hydrocarbons and you know one thing you can point out to them mostly is that if I let us say I take alkanes which are the quite common hydrocarbons that they have already learned in 11th and 12th then the lower alkanes that is methane with CH4 or ethane C2H6 all the way till at least around C5 these are low molecular weight compound they will normally be in gaseous state at room temperature. So, they would not liquefy at room temperature whereas the moment you go to C6, C7 they will be in liquid form and roughly you can tell them that petrol is formed of hydrocarbons which are between C6 and C10 and the most common compound or which is supposed to form a big chunk of petrol is what normally is called as iso-octane and this is something when I emphasize that roughly when someone says petrol. So, petrol is you know roughly C8 plus minus 2 then kerosene is roughly around C10 and diesel more around C14 this is C10 to C12 roughly. So, it gets heavier as you go from petrol to kerosene to diesel and anything about that you know if people normally use as furnace oil and they would not be flow ability is not that good then that is what happened and then of course you can tell them that what occurs naturally in marshes and swamps is mostly composed of methane and if someone says natural gas that means more or less 90 percent methane that is when people say compressed natural gas that is more or less 90 percent methane or more than that and if you have what is called as LPG petroleum gas that is a combination of propane and butane. So, there is C3 HA8 and C4 H10. So, roughly as I said you know you go up to C5 you have gases beyond that it is liquids and much much beyond you know close to tar which is like C50 or C60 really it is not flowing and more like gooey solid around that. So, once you explain to this and where you use all of these and that most of these are hydrocarbons I think that would at least tell them that why some kind of knowledge about fuels is needed. So, the next thing that we need to do is tell them that whatever chemistry we do it is regular thing that they have seen earlier in 11th and 12th standard. So, if they write an equation for you know solving CH4 plus O2 gives CO2 plus H2O then you should tell them that as with any other equation this is CH4 let us say natural gas if you need to burn it you will need to provide oxygen and you will form CO2 and H2O, but there are a few rules that need to be followed and these are all things that they have studied in 11th and 12th standard that is you need to balance this reaction. So, balancing means two things are happening one each element should be equal on both sides. So, you are not destroying element these are not nuclear fusion or fusion type of reaction where some atom is splitting up. So, obviously if there are 10 oxygens on this side there should be 10 oxygens on the other side nothing is disappearing if some oxygen is not burning it will appear as unburnt oxygen on the right side. So, that is something you can emphasize that this is something that they have already studied. So, each element is balanced. So, that means I will need to balance this reaction and hence you know by default you know mass is also balanced again this is not a nuclear reaction there is no mass defect no mass is getting converted into energy each element to element goes here it carries the same mass on both sides no mass is disappearing mass on both sides. So, you will just have to balance and you will have to ensure that you know I will have to put x y a b and x y a b should ensure that my carbon is equal on both sides my hydrogen is equal on both sides oxygen is equal on both sides or just oh I should say I mean this is standard stuff they should know I mean there is nothing new here is just a reemphasis. The thing maybe you can emphasize is that you know moles will not get balanced you may have 3 moles of this 1 mole of this 1 mole of this 1 mole of this. So, there may be 4 moles on the left side and 2 moles on the right side and that will always happen only you are going to balance elements and mass moles need not get balanced in a because you know you are what is basic is the atoms in the mass not the moles. So, moles need not get balanced. So, once this part is done then you know this is all background emphasis then there is something else also they know normally and which is called as Hess's law and this is again something that you probably can tell them now from a new light from the point of view of thermodynamics that you know for every reaction there is something called you know in olden terms it was called the heat of reaction. So, if I have a reaction which gives a plus b going to c plus b then I add up the enthalpies of these and I add up the enthalpies of this and each compound at a particular pressure and temperature it is state property it is a compound at this pressure and this temperature it has its own enthalpy you change the temperature and pressure it will have a different enthalpy. So, it is just a state property. So, now of course, you know this 2 additions they will let us say I have calculated this at one atmosphere and some temperature which is 25 degrees c. If I really combust the temperature changes, but what I am supposed to do is bring it back to the same temperature. So, bring it back to atmosphere and 25 degrees. In this case what happens is that some you will realize that these 2 additions at this pressure and temperature will not add up to these 2 additions. So, the net enthalpy of the reactants is not the same as the net enthalpy of the products at one atmosphere and 25 degrees c and this is what is called as normal temperature and pressure. So, what is happening is that if the products are at a lower enthalpy it means you have actually released some amount of energy which is what we use. So, here is when you should probably emphasize that if I have something which is called a fuel and if it has to burn with oxygen then definitely the reaction that I am looking for should have the products at a lower enthalpy than the reactants otherwise it is not a fuel. I mean if the products turn out to be at a higher enthalpy that means it is absorbing energy from the surrounding to get to wherever it is then you know it is not a fuel you are actually you know cooling down the environment whereas what we really want is a release of energy. So, that you can you know use it to boil water or use it to heat the gas etcetera. So, fuel is something which necessarily once it burns you know that is the property it better have that you know at the end the product should have a lower enthalpy than the reactants and then you define something called the heat of reaction which is just enthalpy of the products minus the enthalpy of the reactant. And this is also something that people for heat of reaction is a common term that people would have seen in 11, 12, 12. It is not something new maybe they would have forgotten it, but at least you can tell them what it means from the point of view of fuels. So, this is something that you need to put in and continuing with Hess's law what people should know is that I can have A plus B here. Let us say I have A plus B on the enthalpy scale here and C plus B here at a lower enthalpy. So, this pictorially you can show. So, this is like an H scale here this pictorially is delta H R. So, now you know there is these are all state properties this H is just a state property. Now, I will need to know the H of various compounds and this is where this Hess's law comes in that for example, I know that C plus O 2 can give C O 2 and if it is complete burning it will do this. And I know that I can measure a delta H R by just measuring how much heat it is. But C plus O 2 of course I am not balancing any reactions right now. So, giving C O it is very difficult to achieve this very normally because normally things will burn fully. To get the delta H R of this is difficult because I cannot you know ensure that everything is half all the carbon is half burned to give me C O. So, I take 1 kg of carbon and I ensure that exactly that 1 kg of carbon. So, for example, I take 1 kg here I will require some amount of oxygen to burn it exactly to get so much amount of C O I cannot ensure that in any reaction. So, but when such things are done then you normally say that you can go for Hess's law where you say that I can go from C plus O 2 this is some enthalpy state I can go down to C sorry C O plus some O 2 at some other enthalpy state and then finally I will get C O 2. This delta H R is known this delta H R is not known, but if I can get hold of some C O and burn with oxygen then I can get hold of this delta H R. And if you draw this diagram it becomes pretty obvious to them that if I knew this and this then these are all the state properties at this pressure and temperature the enthalpy was known at this pressure and temperature the enthalpy was known hence the delta H R is just the difference. So, you just have to come down so much and then you have to come down so much to get to C O 2. So, if I know this if I know this then definitely I can find out this by just subtracting of course, all of this have to be balanced reaction, but then if you explain to them using this method that this is what is normally done to get hold of certain so called heat of reactions where you cannot easily measure it then you know you would have given them a good idea of what SS law would be. The question now is do we really do all this we use SS law etcetera in our calculations actually no this is just to give them a background. What normally is done is that you know all these values are tabulated. So, for so the enthalpy for all various species that are you know available during combustion are just tabulated. So, whether it is carbon whether it is oxygen whether it is nitrogen of C O 2 H 2 O so many things will occur C H 4 C 2 H 6 everything is tabulated as a function of temperature and the way this is done is that you first have to find out or define a datum and the datum is normally defined is that any element in its naturally occurring form at the standard temperature and pressure you assign 0 H this means oxygen in O 2 form, nitrogen in N 2 form, carbon in graphite form. So, all elements in their naturally occurring form will be assigned an enthalpy of 0 at 25 degree centigrade and 1 at 1. This is a very standard approach that is adopted by all combustion people and this is definitely you know way more standard than many other things are standardized that you just assign this 0. Now you say that if I want to form O 2 O from O 2 then in this case from to go from O 2 to 2 O it actually turns out that you will have to provide energy this is because some energy is required in breaking that O 2 bond. So, if I have assigned an energy to O 2 as 0 then H of products minus H of reactants will be a positive number because I have actually provided energy. So, H products will be more. So, let us say I have to give X energy to go to 2 O. So, now what happens is that to get to O from some 0 state I have provided X energy which means that I will assign if I have formed this O at 25 degree centigrade and 1 atmosphere I will say I will assign X by 2 to each O and I say this is now your enthalpy of formation and this is 0 to tell you that it is done at 1 atmosphere. So, similarly I can get so this H F means enthalpy of formation. So, I will get an enthalpy of formation to get oxygen from some naturally occurring substance which is O 2 molecule. Now, of course I can get similarly for CO 2 I will take C in graphitic state and oxygen occurring naturally burn it to get CO 2 and find out what is the difference in energy is here and here and that I will say is H F of CO 2 and plus at 25 degree centigrade and at 1 atmosphere I will first tabulate H F 0 for everyone. So, this I will do for everything that I think is necessary for me. So, necessary for me would be you know O 2 O N 2 N and this is primarily because I will use air for combustion and if I have to have oxygen then so much percent is already nitrogen. So, nitrogen by default comes in all you know combustion calculation then you know I should be then I should be bothered about the most common products which is CO 2 and H 2 O then I should be bothered about all kinds of fuels common fuels C H 4 C 2 H 6. So, the list is on till then maybe I will do it up to diesel let us say I will tabulate all of these and if you pick up a standard combustion textbook like turns oxygen as O 2 molecule at 25 degrees and 1 atmosphere not oxygen atom. Oxygen atom will not be non it will be non 0 that is what I said here if I have given X amount of energy to form 2 oxygen atom. So, let us say I give 100 joules and now I have got 2 oxygen atom. So, I will say enthalpy of formation of 1 oxygen atom is 50 joules. So, that is a I call that as the enthalpy of the oxygen atom and this is my definition. So, once a set of enthalpy of formations has been tabulated then I go and say now you go to the next step where you say you know H will actually vary with temperature. So, what should I do then you realize that for each substance I will need to plot H with T here you will realize that if you had some knowledge about you know some kind of statistical mechanics then you will realize that for anything which is monotomic a monotomic gas the C P does not change with temperature. So, if C P does not change with temperature then H at any point if I integrate is just integral C P dT and since C P is outside I can bring it outside if it is constant then it is just the C P delta T. So, if I have decided that H is somewhere here. So, now of course mind you you know you will realize that in combustion calculations at least a ton of quantities will have negative H at 25 degree. This will happen by default because I have already chosen that something occurring in its naturally occurring state at 25 degrees and 1 atmosphere as 0. So, some law in enough number of compounds will turn out to have a negative H based on this datum. So, let us say I have some H here at 25 degrees. If I want to measure the H at any point you will realize if C P is constant you will just get a straight line. This is how it should be. If I have diatomic molecules it will turn out that diatomic molecules will have 6 degrees of freedom and ideally for 6 degrees of freedom your final C P should have been 3 R by 2 sorry 3 RT sorry I mean it will be 3 R not just 3 by 2. So, in this case you realize that C P will not be a function of temperature because at room temperature mostly your only 5 of the modes have already gotten totally completely kicked in and which means the molecule can translate in any direction and the molecule rotates mostly in all directions which it is possible and you will realize only 5 of the modes have really kicked in and energy corresponding to only those 5 modes have come in and it will be only 5 by 2 R which has actually kicked in and to get in the extra mode which is the vibrational mode it takes some time and C P is not constant. So, if I have brought C P versus temperature it does not behave constant, but it will behave something like this. So, at a very high temperature it reaches its correct value at room temperature it is lower value and somewhere in between it starts to increase. If I have triatomic molecule CO 2 ammonia then this will no longer look like this it will look like this. So, C P will have start having all kinds of shapes beyond the room temperature and it is now by now it is pretty standard in computational chemistry that if I know the structure of the atom I can know all its vibrational mode and I can calculate C P down to the right value even without having done a single experiment because the quantum mechanical calculations are so good nowadays. So, you take any atom C H 4 you give it to a chemical computational chemist he will tell you what the C P is as a function of temperature without a problem. So, he will tell ok C H 4 there are 5 atoms. So, you know each atom has 3 modes of movement 5 atoms and 15 modes 3 translational modes 3 rotational modes 6 modes are gone you are remaining with 9 vibrational mode. He will tell you the frequency for those 9 modes and give you exactly how C P will vary with temperature. So, this is now that much it has reached the stage and literally every possible C P of combustion products are well tabulated either you go to terms book by terms or you go to this website by N I S T and you will get this tabulation. Usually the tabulation is done in terms of a polynomial N I S T is national some standard institute it is a US based organization. So, you just say probably N I S T dot gov which is a US site and they will give you the C P value for everything. So, that is something that you must go check out tell the students to check out it is available free it is a government site everything is open out there because it is a standard it is not a defense site. So, what happens is C P is normally given as a function of temperature a naught plus a 1 t plus a 2 t squared plus a 3 t cube plus a 4 t 4. This is the most common format that you will find C P given it is called this 5 constant polynomial format and this is normally called as the NASA format and all this C O 2 H 2 O C H 4 everything if the site will give you a 1 a 2 8 sorry a 0 a 1 a 2 up to a 4 everything will be given. So, this is if you go to this N I S T site you should get now if this is given then if I want H is C P d t d H and if I integrate this you will realize that I will have to integrate it from the 0 condition which is 25 degrees to let us say let us say I want some t then here you will realize I will get H t minus H 25 is if I integrate this whole thing it is a naught t plus a 1 t squared by 2 plus up to a 4 t 5 by 5 this I will get by direct integration. You will realize now you need this other constant which is your heat of or enthalpy of formation at 25 degrees and that will give you the sixth constant which is normally in their terminology given as a 5. So, you can again go and check I am assuming that they are calling it a 0 a 1 a 2 a 3 a 4 a 5 and again in the most common format that is out there you will see that the this was because the NASA people were worried about lot of reactions occurring not only in their combustion chamber, but in the reentry vehicles and most temperatures are noted up to most C data are fitted up to around 3000 or 4000 Kelvin and you should realize that you know whenever I use a combustion chamber we would come to adiabatic flame temperatures now. So, temperatures reach around 2000, 2500 maybe you know regular adiabatic flame temperature and maybe something higher depending on what the situation is and people have tabulated all values up to around 3000, 4000. So, this fit if I use one polynomial fit that fit does not go the entire range. So, they split the range they will give you one set of coefficients from let us say 25 degrees to 1500 and then 1500 to 3000. So, two sets of coefficients will be given and you will always when you go to the anaesthetic site find two sets of one for the lower range and one for the higher range. So, that is something that you must go there and see once this is done you can calculate the H of any of the compound that you are interested in at any temperature because this is all now very standard this is all standardized and you can just get. So, if the compound you are interested in is A plus B you give C plus D if I want to find out the heat of reaction find H at 25, B at 25, C at 25, D at 25 get the standard heat of reaction that is what you do. So, now the next thing that probably people should know is that if I use oxygen by default I will get nitrogen and that is because I am using air for combustion and you know roughly that if I go molar wise then there is 21 percent to 79 percent is this correct is this some number that people remember. So, if I go volume or mole fraction wise then if I take air if I get 21 percent oxygen and there are 79 percent of nitrogen. So, if I want to write a reaction for burning something for example, C H 4 plus O 2 gives CO 2 plus H 2 O. Now, I will balance this because first I need to there is one C on both sides C is balanced there is 4 H here and 2 H here I will make it 2 there are 4 H now then there is one O 2 and another O 2 here this. So, there should be at least 4 oxygens on this side. So, probably this is 2. So, is this I hopefully this is balanced. So, now if this is balanced then you will see that if I have used air to burn it then by default I am getting 79 by 21 times N 2. So, how much is that number? It is 3.76 and it is some number which combustion guy is just offensive to remember that this is around 3.76. So, for 3.76 into 2 sorry I should say because there are 2 moles of oxygen. So, for every mole of oxygen of O 2 it corresponds to 3.76 moles of N 2 which means it corresponds to 4.76 moles of air because 3.76 moles of nitrogen and 1 mole of oxygen together will give you 4.76 moles. So, this now has to be drilled into them and you know told what is to be expected. So, this will help them to do combustion calculations and what probably can be told next is you know something about calorific value. So, if I burn 1 unit quantity of fuel and go for complete combustion then what is the amount of energy that I really and this word complete combustion is necessary because if I start with carbon and go to CO it is not called complete combustion. So, complete combustion of any hydrocarbon fuel means that all carbon has been converted to CO 2 and all hydrogen has been converted to H 2 O. So, if I need complete combustion then I will have let us say H X and C Y as my hydrocarbon that must go to H 2 O completely which means that if I put here let us say Z then twice Z should be equal to X and I should have let us say A times CO 2 which means Y should be equal to A if I want to balance. Now, if you calculate how much oxygen is there oxygen is Z plus twice A that means if I add here O 2 that O 2 should be Z plus 2 A by 2. So, this is Z plus 2 A by 2 because here everyone has 2 here. So, it is Z plus 2 A by 2 and then this much multiplied by Z plus 2 A by 2 multiplied by 3.76 times nitrogen will be on this side as well as on this side. So, this is what would occur in complete combustion and you must find out how much energy is released and that of course, you can find out if you know the H of each species. Now, as I said all standard H values delta H R are at 25 degrees centigrade and 1 atmosphere that means I must have both my products and reactants at 25 degrees and 1 atmosphere that means oxygen and nitrogen on both sides will have 0 by definition. So, what will happen is that there will be some H F for the fuel and some H F for CO 2 and H 2 O separately. This minus this totally will give me this delta H R at 25 degrees here. Obviously, this quantity has to be negative if you are using the fuel because you are releasing the energy and the product should have a lesser H. Now, calorific value is just defined as a negative of the sheet of reaction in per unit basis of the fuel. So, if I have let us say I used this using 1 mole of C H 4 then I will get some delta H R. So, let us say it came out to be 1000 joules. So, this 1000 joules is obtained using 1 mole. So, now normally I will convert into a per kg basis in the most common. So, I will say if 1 mole 1 mole of C H 4 has so much kg. So, 1000 joules was for let us say this is x kg. Of course, you know x kg is let us say this is 1 kilo mole. So, this x kg would have been 12 plus 4. So, 16 kgs let us say gave 1000 joules then 1 kg would have been just 1000 by 60 and you know that the calorific value of the fuel is just the delta H R by per unit kg. And the kg is the most common unit sometimes for gases they will quote it in terms of per meter cube. This is assuming that you know the gases can be written in that term that you know it is per meter cube, but per kg is by far the most common way to give your calorific value. So, all you are doing is write a balance reaction find out the delta H R calculate the delta H R divided by the you know ensure that you calculate it on a per kilo basis and take out the negative value because if it is a fuel you better have got a negative value remove the negative value whatever number you get is your calorific value. So, that is something that you will need to explain maybe you can tell them you know quantities various quantities which are tabulated for calorific value. So, this is something that you need to do and finally, I think maybe two other things there is something called as an exhaust gas analysis again I will just take two minutes of this. And what happens is that if I have complete combustion in the products if I have a pure hydrocarbon fuel then in the products I should have only CO 2 H 2 O and N 2 because N 2 is not supposed to burn. So, exhaust gas analysis is important if you have unburned fuel which means you will have some CO or if you have some O 2 that means you have provided more air than necessary. So, also again that will come the concept of stoichiometry this is something that they know for sure. 11th and 12th standard they have been told what is the balance reaction and hence you must be able to tell them what is stoichiometric air required. So, if I just ensure complete combustion of the fuel then the amount of air required per kg of fuel on a kilo to kilo basis will be normally what is called as a stoichiometric air fuel ratio. If I provide more air than required I will have something called as an excess air and then you know I will call it a lean mixture and if I have more fuel than required that is less air than required I will call it a rich mixture. So, this is again something that you can put and you can maybe this will come again when you are teaching petrol engines or diesel engines that when should I put in a rich mixture in a petrol engine when should I put a lean mixture the diesel engines normally run lean. But this is part of cycle analysis at the combustion level you can only need to tell them about you know rich and lean mixtures what is excess air factor what is stoichiometry and finally there is only one concept that probably you need to tell them which is adiabatic flame temperature. So, this is the last thing that hopefully I will and then continue on to kinetic theory. So, again I mean I am not going in detail if the topic is needed each of these things will be done in sufficient detail, calorific value, balancing of reaction, you know problems using exhaust gas analysis I mean if there is so much of exhaust gas how much excess air did you put or how much less air did you put those are some things that most engineers would need to know just because you know it is really necessary in analysis of what kind of combustion you have done especially if you are doing going to power plants and doing combustion. If you can analyze the exhaust gas and tell what is happening and you know if something is wrong with the combustion then you can take care of it. So, the last thing that I said you need to do is what is called as adiabatic flame temperature. So, all this is pretty straight forward I have H products sorry H reactants and H products. So, I told you that whenever you are calculating heat of reactions you bring down the products to the standard pressure and temperature and then calculate what is the heat of reaction. Let us say you did not do that you did not do that and let no energy escape from the system you took a closed system you did not let energy any energy escape. So, there is no queue no work was done all that is going entirely is getting into the products. The products temperature is increasing and it reaches a temperature which corresponds to what it should be that is what is called as adiabatic flame temperature. So, the best way to do this is represented pictorially let us say this is H let me go to another this is H this is T. Now, H I will say H reactants is let us say H A plus H B where this is some fuel and let us say this is oxygen plus nitrogen. So, I will just add them up. So, now you should realize that the fuel may be at room temperature may be at some other temperature. So, if I add this up obviously each of those three components that is the fuel, oxygen and nitrogen they will have certain enthalpy at certain temperature let us say at 25 degrees they are here. So, I am just putting negative because it could happen that the fuel has a negative enthalpy at 25 degrees oxygen and nitrogen will have 0. Let us say I start here then at some other temperature I will just calculate the enthalpy of each of them obviously at a higher temperature the enthalpy will increase as I go ahead in temperature I will get another point here another point here another point here. So, this line or this curve it is not a straight line it is some curve is the edge of reactants as a function of temperature. So, that means I have added the individual enthalpy of each and you know gotten this curve then I know the products the products would be CO 2 let us say H 2 O and N 2 let us say in this case it is just about complete combustion if it is not I will take care of extra oxygen again I will plot. Obviously, you realize that if I draw it like this at T is equal to 25 degrees H products here would start at a lower value and this would have been your heat of reaction or enthalpy of reaction you go down lower in the enthalpy at 25 degrees you will start here and if I plot or calculate the H of each of these three species I will get a curve like this. So, if I get a curve like this I need the same since I take an open system analysis there is no Q no W H should be same on for reactants and products if H is same then at the constant H line I draw I cut this curve somewhere this temperature corresponding here is what is called as the T adiabatic. So, of course, this is starting at 25 maybe I have put in air preheated air and something else. So, maybe the H is here then I will just draw H constant here come here this will be the T adiabatic. So, depending on the initial state of the reactant you will get a different T adiabatic. If I put no more nitrogen in you will automatically realize that your T adiabatic will go down because the nitrogen is not contributing to any release of energy it is just ensuring that its own temperature increases and you know your adiabatic. So, if I have so if you take most common fuels you will realize that all I need to do is just a balance I will write H reactants is H products H reactants I know the state of each of them. So, this let us say you are given a temperature that you know initially it is preheated air at 200 degree centigrade and fuel is also 200 degree centigrade I will use the curve fit get H of fuel plus H of you know O 2 and N 2 and calculate a number X. Obviously, this number Y has to be equal to that this number Y let us say products are made of only N 2 CO 2 and H 2 O then H products is just you know 25 degrees or whenever you have known it to up to some T C p d T for CO 2 same thing for H 2 O and same thing for N 2. So, what this T is you do not know. So, the only thing you know is that if I put in a T I can integrate up to that T using the polynomial expression and get some value. So, the most common way you have to do this is just go for trial and error use a T calculate the values on the product side and get it. A lot of times people do not do that they will do a trial and error, but they will approximately know. So, for most fuels around the adiabatic flame temperature is somewhere between 2000 and 2200 they will take a guess. If they take a guess they will calculate an average C p rather than you know integrating the C p using the average C p then I will just say this because I have calculated the average C p for each of these I will just get it as H f plus C p T adiabatic minus T wherever it began T 25 degree. This is the product side. So, H f for all these species are known at 25 C p I have already calculated some average value T 25 I know. So, this will be the only unknown and I will calculate it. Now, I see if this matches my guess value if it does not I will do another. But, roughly you should expect the temperature between 2000 and 2200. If you put excess air there is more nitrogen automatically you should know that your adiabatic flame temperature should drop down it may drop down to 800, 1600 anything. If you do not use nitrogen that means you use highly you know enriched oxygen air. So, for example, you use oxidously there is no nitrogen. So, if there is no nitrogen then it is not eating up your energy to lower the temperature in that case your adiabatic flame temperature will go high. So, for example, in oxidously inflamed you can easily achieve more than 3000. But, that is just because nitrogen is playing a role of absorbing the energy and bringing things down. But, as long as you explain these concepts and roughly this diagram here of what an adiabatic flame temperature looks like I think it will give them a good idea of combustion you know calorific value balancing stoichiometry air field ratio excess air factor exhaust gas analysis and adiabatic flame temperature. Of course, I have rushed through this, but this is roughly you know something that needs to be covered and probably in far better detail I would normally take around 3 to 4 lectures when I am teaching this part.