 What you have seen so far is the Rankine-Hugone relations where we are assuming that a premixed set of reactants will approach a flame zone or a reaction sheet at a certain velocity and get converted to products as it leaves the reaction zone or the flame zone, either to follow the reaction zone itself as it propagates into still reactants in a LAPFIX coordinate system if it is a detonation wave or move away from the reaction zone if it is a deflagration wave. Now one of the major problems with that is we never really knew what is the speed at which the reactants would approach the flame fixed approach to flame in a flame fixed coordinate system right it is like if you knew the velocity you could stabilize the flame in a flame fixed coordinate system and this is what will happen but what is the velocity it is only if I knew the velocity I would know my m dot and only if I knew the m dot I could construct the Rayleigh line and only if I constructed the Rayleigh line can I look at its intersection to the Hugone curve and then look for the solution but what gives me the velocity how do I know what is the velocity with which it is supposed to travel all I know from this is it can travel over a range of very low velocities or a range of very high velocities right and these two ranges do not intersect at all there is like about two to three orders magnitude difference in these velocities but what is it we never knew we just write a we just put an arrow and then say you not and then question is what is it how do you how do you find out well it turns out that you cannot you cannot stop with treating the reaction zone like a black box the way we have done right there is obviously more to the story then what happens outside of these is a reaction zone it is not like you have some reactants that is that come in into a black box and magically turn into products and we can say well let me just do the mass and momentum and energy conservations across this and try to understand you could understand to the extent that we have done but not anymore right if you want to actually find out how the flame propagates or how fast the flame propagates we have to get into the flame we have to get into the open we are to open up the black box and see what is happening inside because that is what is actually causing the flame to propagate whatever is happening inside inside this region with with all the temperature and concentration gradients and therefore heat transfer and mass transfer and so on is the one that is actually causing the propagation to happen and that is what is dictating what the u0 should be in a flame pick point system and correspondingly the m dot and then the Rayleigh line and then the whether it is a tangent Rayleigh or not and all those things whether it is you could then you get a weak dead deflagration and so on and of course what happens inside the black box that dictates the propagation velocity depends on whether it is a deflagration that is along the lower branch or in the upper branch because in the lower branch we now have the deflagration that propagates in a subsonic speed that means it is actually seeing what is happening upstream there is upstream influence whereas in the case of detonation wave it is propagating at supersonic speeds so there is no influence on the upstream right because information can propagate only in one direction therefore we have to treat these two cases separately when you now get into the black box and what we want to do is to start with deflagration sure enough we want to be we want to note that the Rankine-Hugonio relations that we have studied so far corresponds to premixed flames okay so we are now talking about a reaction zone that is happening for reactants that are approaching the reaction zone together with the with the with each other that is reactants as in fuel an oxidizer together right so premixed flames all right but we are actually looking at deflagration so let us let us confine ourselves to deflagration now certainly premixed deflagrations and of course we want also talk about laminar in the sense we do not want to get into turbulent effects at this stage right so this is what we want to do so they so we want to actually see what happens inside the flame what we mean by inside the flame whatever we have neglected so far what did we neglect so far we neglected anything that had any gradients right not only did we neglect chemical reactions but we also neglected regions that were corresponding to having a temperature gradient or a concentration gradient or velocity gradient and so on all gradients were all now shout into this this this this region right so question is if you now are thinking about a deflagration how does it really propagate anybody can we can we think about this so let us do a thought experiment that is kind of destructive so let us suppose that this room is filled with methane as well as air of course right however you are sitting here right and then I go and ignite one corner and what is what is going to happen now the flame now begins to propagate to the rest of the room yeah the course with windows open we cannot expect pressure rise to happen but if it is not if the windows are not open you could have a pressure rise and you could have a detonation transition and so on so let us not get into that yet right so let us suppose that we have the windows open and the pressure remains constant more or less or there is like a mild pressure rise across the flame per pressure decrease sorry across the flame to market a different market market as a deflagration then now you have deflagration wave that is propagating towards you how does that look like would you be sitting there wondering why is the deflagration propagating in the first place or you want to get up and run okay well fortunately there is just a thought experiment so can you think about why we are actually having a flame propagate why does it propagate so you have let us say still reactants out there and then the flame is approaching it right and I have ignited it somewhere so let us suppose that that has happened quite a while back it is not having any bearing on the flame propagating at the moment here you now have a flame it is now going to go from here to there why anything about this well as the flame comes near you what do you feel because of the heat right so as the flame gets near you you feel the heat right so what happens so what does it mean to say some place upstream of the flame is getting heated up what happens downstream of the flame is already hot right so we now got the products right and the products are at ideally speaking the adiabatic flame temperature and recall how we got the adiabatic flame temperature we got the adiabatic flame temperature therefore the products because the products had fairly low negative or actually fairly high negative heats of formation or low that is why that is why they are products right and they had exchanged their heats of formation for the sensible enthalpy and therefore their sensible enthalpy has become high because their heats of formation or algebraically low right and then they became hot so you now have a hot products over there hardly any temperature gradient there that means you do not have too much heat that is getting conducted there so if you now have a wall that is kind of suddenly getting heated up and the temperature is high what do you have for upstream you now have a fairly large temperature gradient that is actually pointing the in the upstream direction that means all the heat really wants to get conducted upstream of the flame and heat up the reactants and what do the reactants do when you when they get heated up they start reacting like crazy why like crazy because strictly speaking if you now follow the Arrhenius law you have the cold boundary difficulty that they should be reacting even as you are sitting there in the cold reactants get any temperature strictly speaking but the reaction rates is so low that the heat that is released is going to be at such a low rate that it is just going to get completely dissipated away most of the time where you are until the flame hits you right but as the flame is approaching and it is conducting heat upstream right and it is heating up the reactants the reaction rates are beginning to look appreciable particularly when your temperature reaches beyond a certain point dependent upon the activation energy so you know that the Arrhenius law is like e to the have goes as e to the exponential of minus e to the e over Rt right where e capital E is the activation energy right so if you now look at that particular expression what happens is larger the e greater is the temperature to which you want to take your reactants but you will also have the sudden rise in reaction rates much more steeply with any further increase in temperature when compared to if you are ever capital E were low right so with capital E being low you do not have to increase the temperature to a whole lot for your reaction rates to become appreciable but it is going to grow a little bit more gradually as you further increase the temperature right so that is the highly nonlinear sensitivity of the reaction rate to temperature because of a high e that is how it happens and therefore you now have the chemical reaction rates beginning to become appreciable therefore the heat release is not going to is also now going to become appreciable it the chemical reactions are going to consume the reactants and how did you decide where the flame was wherever the reactions are happening so if the reactants that got consumed the reactions them now have to move forward and started reacting started consuming fresh reactants at a higher temperature and then they release heat and conduct so essentially the way it is going on is you go through a high temperature patch that you started out with when you ignited let us say I strike a match stick I am creating a high temperature patch that locally increases the temperature so that the reaction rates become appreciable and once the reaction rates become appreciable the heat release rates become appreciable and this actually conducts heat upstream and heats of fresh reactants consumes the current reactants therefore the reaction zone advances into the region which it has heated up upstream in the fresh reactant mixture and therefore it propagates so we are now beginning to look at some two or three things one heat release by chemical reactions to heat conduction upstream three propagation right why did I list these three because these are the three elements that are again coming back again and again in combustion you see the heat release due to chemical reactions is because of chemical reactions so that is the term in an energy balance that is going to come up because of chemical reactions the heat conduction upstream is essentially the diffusive component right that is a transport phenomena that is going there that is essentially corresponding to diffusion of heat right and then finally you do not have the propagation you can think of that as a convection in a flame fixed coordinate system as if you were to ride over the flame you will see that the reactants are approaching the flame at the flame velocity that is that is basically meaning that the reactants are having a convection of their in a sensible enthalpy of course sensible enthalpy along with their here standard heats of formation right so if you now look at this is like a energy balance that we have to think in our minds there we are where the reactants are actually convecting into the flame at the enthalpy getting the heat conducted into them from the flame because of chemical reactions so we now have a energy balance that we need to think about containing three aspects convection diffusion reaction right so pretty much anything that you can think about in combustion will have to deal with these three balances in these three right and that is exactly what we figured out just by thinking about what happens right so let us can we now draw this picture so let us suppose that I still adopt a flame fixed coordinate system so it is not as dramatic as like a flame propagating towards you or something because we are resting the propagation holding it there and what you are saying is if I now have a cold reactants that are coming in at rho 0 u 0 t 0 pressure is not a major consideration here the pressure is hardly going to change just mildly decrease in a deflagration so let us not worry about it u 0 is what is something that we have to find out okay so although I write this u 0 there is something that we need to find out okay but still it is coming with a certain u 0 whatever it is right and so if I now were to plot my t 0 sorry temperature t it starts from t 0 for the reactants and of course we can say why I not for the reactants it starts with this and then gradually increases until it reaches the tf the flame temperature so let us say that that is like t infinity is essentially tf here for stunning standing for the flame temperature or the adiabatic flame temperature specifically for ideal conditions without any heat loss to the surroundings strictly speaking you know you do not really have much heat loss to the surroundings all right because the products are as I said are fairly at a high temperature and relatively uniform temperature therefore there is hardly any conduction going on beyond that point okay most of the conduction is going towards the reactants in fact this is the reason why it is adiabatic so long as you have heat conduction going only to the reactants okay so the reactants take that heat and convert their standard heat of formation into a higher sensible enthalpy of the products okay that is an adiabatic process you do not have to worry about the fact that you had a heat conduction it is not really a heat loss it is after all going to the reactants it is kind of like you amass a lot of property and give it to your kid okay that is not philanthropy right so okay so you know you know have this tf right and then this temperature increases right so what we said was the reaction rate has an Arrhenius dependence on temperature and it becomes fairly sensitive to temperature only when the temperature becomes above a certain point okay so if you were to plot the reaction rate it is not going to really rise a lot until you reached a fairly high point in temperature let us suppose my semi somewhere here okay so this temperature is only badly lower or maybe I am exaggerating maybe let us suppose that I want to mark this point you see this temperature is not very very low when compared to that with the final temperature and that is when the reaction rates are beginning to increase and how do they increase then they just go up right so this is the crazy reactions that we are talking about right and then why did it fall we run out of fuel okay so we just completely consume the fuel so that there is nothing to react right or other reactants any of the reactants you could have one of those deficient reactants typically you are thinking about like a fuel lean situation there where where you have plenty of oxidizer but that is not necessarily to all the time but we essentially looking at the deficient reaction running out and then that abruptly stops the reaction zone okay so this is omega or w whichever way you want to call it mass based on mass balance or on a mass basis or a molar basis okay question what is the flame here you would like to think the flame is essentially a reaction zone and therefore the reactions are primarily happening here because that is where the reaction rates are significant of course I mean the omega could be plotted in any scale it does not necessarily have to go over here but this is sort of like normalizing all of them from something like 0 to 1 or something like that okay so everything goes from bottom to top that is it and so let us suppose we draw this picture and we think we tend to think that this is where the flame is supposed to be but that is not what we decided when we did the Rankine Hugoniot when we did the Rankine Hugoniot our black box was to consider every a gradient that means the flame strictly speaking extends all the way up to here from here right so that is really your flame right and this is this refers to the flame thickness delta so what is happening here upstream of where the reactions are happening essentially you have a preheating right so the reactance is simply getting heated up ahead of when they react here they are just cooking just getting ready to react so we can now differentiate these as the preheat zone and the reaction zone right higher and higher the e the activation energy smaller and smaller is the reaction zone as a fraction of the flame thickness right so rudimentarily many textbooks would simply show like a half the flame thickness is like a preheat zone and half the flame take flame thicknesses as a reaction zone but that is not necessarily true for high e okay so many hydrocarbons have activation energy that is significantly large it is not we cannot say it is infinite but what it is large okay therefore we get into these thin flames in fact what happens is when you now have large e the flame thickness itself decreases because larger the e smaller this region smaller this region higher is the W right because all the reactions have to actually happen within a very thin zone okay the the the the reaction rates became significant all of a sudden and then there is like a huge amount of reactions is going on per unit time right so when you now have a large heat release that means a large temperature gradient has to be set up to dump all that heat upstream so this is something that we have to be very careful in thinking in an idiomatic flame if you have a very large chemical heat release rate in the reaction zone it should correspond to a large temperature gradient so that the heat conduction is happening at exactly that high rate okay and think about the gradient now the gradient is caused by a temperature difference happening over a distance that is what a gradient means right the temperature difference is not too much different if you have a e that is high okay you you control your t0 and that is like let us say 300k 500k something like that and tf is thermodynamically determined okay that has got nothing to do with the kinetics so tf is like typically 2400k or that abouts depending upon the reaction that you are talking about right therefore that difference is not going to change a whole lot at all so what can now change to cause a steeper temperature gradient the only possibility is if this preheat zone thickness came down you see so because you had a high e and therefore got a fairly low thickness for the reaction zone correspondingly the preheat zone will become thinner alright but as the e increases the fraction of the flame thickness that the reaction zone occupies become smaller that is what I was beginning to say okay so as the e increases the reaction zone becomes smaller the preheat zone becomes smaller okay the flame becomes smaller on the whole but the fraction of the flame that the reaction zone occupies become smaller alright so this is what is going on as you now think about how the reaction rate increases with temperature through the activation energy okay fine so can we do the energy balance that we were talking about okay can we now start doing like a let us suppose like a order of magnitude order of magnitude analysis or whatever we just know danced about for this flame propagation in the room and thought about how this exactly propagates can we now put it in some sort of mathematics right so can we can we write like a energy balance so we talked about an energy balance of convection diffusion and chemical reaction okay so the way we want to actually do this is one of the major problems as I said with combustion is we have to actually take into account these three things convection diffusion and reaction and typically the reactions are the ones that are that pose the major hurdle in dealing with this so you will find this time and time again that we try to avoid dealing with all three of them at the same time that is what is called as a mixed problem okay because you are now mixing up all the all the three issues of convection diffusion and reaction is it possible for me to isolate only two of those at a time that means can I deal with like convection and diffusion alone without reactions in a region where no reactions are hardly happening right is it possible for me to actually deal with only diffusion and reaction where you do not have to worry about convection it is just a passing gases they do not really contribute to convection of enthalpy alright so that is what I am going to actually do now think about think about so when I now look at this region I find that the reactions are happening only here out here there is some energy balance going on you have species that are coming in with a certain certain enthalpy and they are all actually feeding into this feeding into this right but the rate at which this is coming it should actually balance the rate at which the heat is being supplied so that the temperature can rise from T0 to Tf right so there are two or three different balances that we can write for example when we now when we now say if you now look at the rho 0 u0 that is your m dot Cp Tf-T0 that is the rate of rate of change of enthalpy right that came primarily from the heat that was given in the chemical reactions or let me take a step back that is primarily coming from the heat conduction that is happening from here to there right the first place so as the reactants are approaching the flame the first thing that they notice is the heat okay so and then that heat is the basically getting conducted so if I were to write this as k approximately all these things are approximate we will will do only like an order of magnitude idea so the temperature gradient here can now be written as Tf-T0 divided by ? right and this heat conduction has to match the rate of heat conduction has to match the rate at which the heat is released right so that is equal to I could write a wiggly equal to but I have already done that I mean this is this is the wiggly equal to is to basically mean we were supposed to write k DT by DX but DT by DX is approximately approximated by a temperature difference divided by the distance as if you had like a straight line all the way okay we ignore this little little gap here where this curving is happening and we suppose that this is about the same as this we said that this is about the same as that we did a lot of approximations there okay and so how does you how does this heat get conducted because you have a heat that is being released due to chemical reactions and that is happening when you now have so much mass of react reactants that are actually getting consumed per unit volume per unit time per unit time all right this is per unit time okay this is per unit time because this is actually m dot all right so ? u0 this is essentially corresponding to m dot so this is per unit time this is per unit time also so w corresponding to a volumetric mass change mass consumption per unit per unit time okay we have to multiply this by the heating value of the reaction so this is like joules per kg this is per kg per meter second per meter cube okay so this becomes like kgs per second per meter cube we now have to multiply by a volume right so the flame thickness is ? over which the volume over which let us suppose as a first approximation the reactions are happening I do not I am not even going to distinguish between the chemical so the reaction zone and the preheat zone as a first step but we know better we know that this is where the reactions are happening I still do not know exactly how the portion this this this thing okay I said that this is going to be thinner if you had a high e so obviously depends on e in some sense okay I do not know how okay let me not worry about it at the moment and ? times 1 1 is like for a unit cross sectional area all right perpendicular to the boat if I did this what am I looking for I am looking for you not okay but you now have additional things that are coming up there is like a ? that is coming up tf is something that I can find out I can give you t0 and cp bro not all these things I can take not a problem but there is a ? that is coming up fortunately we have now two equations two unknowns you see this is one equation and this is another equation all right so the two unknowns are you not in ? okay and of course the nice thing is the ? does not show up in the sorry you not does not show up in this you can actually try to find the ? over here and use that in this expression for finding the you not so that is how we are going to go before we just proceed the way that I just described I want to point out that this is what is now called the convective diffusive balance right this is primarily happening in the preheat zone we could say that the enthalpy flux that is coming in to the preheat zone primarily goes primarily this is coming from the heat conduction in the preheat zone because there are hardly any chemical reactions happening there so if you now think about only an energy balance that is sufficient for us to consider only the convection in the diffusion right whereas here this is what is called as the diffusive or to have it rhyming with this we can say this is reactive diffusive zone or reactive diffusive balance so this is the reaction zone right so many times the technical way of actually talking about a preheat zone essentially is to say that this is a convective diffusive zone and this is the reactive diffusive zone why did not we have to worry about convection here because we are now not talking about any appreciable change in temperature in this there is hardly any change in the enthalpy flux there is a so much amount of enthalpy that is coming in at this temperature it is pretty much leaving at pretty much the same enthalpy there is hardly any change there was cost in the enthalpy and enthalpy flux convection right so we did not have to worry about that all we had to worry about was as the chemical reaction heat release is happening it is getting dumped by conduction upstream all right so that is the reason why we had we could live without a convection showing up in this balance and so effectively what it means is we are dealing with two of these three processes at a time here we are not really dealing with reactions here we are not dealing with convection all right. So that is a split that we do in dealing with chemical so we dealing with combustion we look at the look at how this zone clarifies isolate where the reactions are happening try to look at what happens elsewhere and look at balances without the chemical reaction there and then try to match it up with the balances that are happening in the chemical reaction zone taking the chemical reactions in account so this is how we do this so let us let us just try to find out how to how to go about this so there is also a global energy balance that we can this is a per unit time okay so this is energy balance per unit time right there is a global energy balance that we have already done to begin with global energy balance which is basically to say that the reactants had a standard heat of formation and sensible enthalpy the products have standard heat of formation and sensible enthalpy these two sums should be equal in an adiabatic situation right which means the differences in the sensible enthalpy of the reactants in the products should be equal to the negative of the differences between the form standard heats of formation of the products in the reactants right and that is what we are calling as Q the last thing that I said the negative difference of the heat standard heats of formation of the products in reactants is essentially the Q right so we should now be able to say that CP Tf-T0 equal to Q this is on a per unit mass basis okay per strictly speaking I should basically take a certain amount of reactants a mass of reactants and say m Cp ? t is equal to m times Q because Q is essentially heat release per unit mass k joules per kg therefore I can I can I can say this per unit mass and so how do I how do I now factor factor these things into count so if I now say that rho 0 Cp rho 0 u 0 Cp Tf-T0 is approximately k Tf-T0 divided by ? the Tf-T0 gets cancelled it does if it did not matter what the temperature difference was as far as the convective diffusive balance was concerned and what you then get is you are looking for u 0 so therefore you get u 0 equal to k divided by rho 0 Cp ? and then we say that if I now say rho 0 u 0 Q is equal to W Q ? right then what do I get I can cancel the Q and I get ? equal to rho 0 u 0 divided by W so plug this in here to get u 0 equal to k divided by rho 0 Cp all right and rho 0 u 0 divided by W this will be you can take the u 0 to the left hand side you get a u 0 squared equal to W k divided by rho 0 squared Cp all right which implies which implies u 0 equals 1 over rho 0 square root of W k divided by Cp this is the first cut estimate that we have had for the flame speed so this is essentially the laminar flame speed okay first approximation what is mainly the first approximation here is mainly the first approximation is we have assumed that the W is happening over the entire delta we did not know how to split the flame thickness into a reaction zone and a preheat zone we just said that the reactions are happening everywhere okay so obviously we now get a we should actually get more like an over prediction of what is going on interestingly this this gets all the dependencies right okay so what we can understand is if the reaction rates were very high right if the reaction rates were very high then your flame speed will be high okay you can also get an idea of what the thickness looks like we get we get like a ? equals rho 0 u 0 divided by W right and then what happens is when the reaction rates are very high the flame thickness decreases so this is also coming from saying that if I had a large e I would have this zone decrease and then the reaction rates would shoot up because all of them all of it is supposed to happen within a short distance therefore you have a high reaction rate so that means a thinner flame essentially is going to propagate faster that is what this means this really means right if you have a high thermal conductivity it is going to propagate faster because the upstream heat conduction can happen faster right if you have a lot of CP that is bad news for the flame propagation because as the flame is trying to conduct heat upstream into the cold reactants the reactants are simply soaking up all that heat without a rise in temperature that is what CP means okay CP basically means the amount of heat that is required for a unit rise in temperature if you have a lot of CP that means you take up a lot of heat without showing up in terms of temperature right later you are a very calm person you are pushed around and you still smile okay you do not really heat up and show your anger okay that that means you are a high CP guy okay and then that is bad news for the chemical reactions because the temperature needs to rise for the Arrhenius rates to increase right therefore if you have a high CP you are going to get a fairly low you not and therefore sorry and then similarly for the drone on drone artist essentially like a mass inertia CP is like thermal inertia there is like mass inertia you have a lot of mass per unit volume to heat up alright so obviously it is going to take a while for the flame to go when you now say you not is getting low so you get all the dependencies alright let us now start correcting this a little bit from what we what we can think about so corrections I will finally lead up to a point where we can actually construct the exact expression for you not taking a lot of a lot of corrections and therefore additional effects into account which is only about the square root of two factor away from the analytical result okay so and I would like to go through this because this is lot more physically intuitive so we think about what happens and then construct these equations in some sort of a phenomenon what is called as a phenomenological manner okay that means we think about the phenomena and then set up these equations as opposed to well the equations are already derived you know what the equations are you have 5 and plus 6 equations to deal with okay and then we can we can appropriately simplify and pick on choose some of these equations to deal with and all those things to do the Schwab-Zeldovich and then have n plus 1 equations to confront and all those things we know all that stuff okay so we will do that and we will find out that finally we have only about a square root of 2 difference between what we end up with all these corrections that are physically employed okay so how would you want to do this it turns out that what we want to do is we want to make sure that the W is factored in only in a small region that means you cannot put a delta over here but it has to be like a fraction of the delta question is how do you develop that fraction we know that the fraction is going to depend on the activation energy right so how the activation energy we have thinner this region right so there must be like a measure of activation energy that is factored into this right so so define beta as e times tf minus t0 divided by Ru tf squared the way this is defined is pretty important we could have simply said ef divided by Ru tf and that that ratio is important because it shows up that way in the Arrhenius expression right but we also want to factor in like a temperature difference right so the temperature difference is tf minus t0 for a given tf okay so if you now say e divided by Ru tf times tf minus t0 divided by tf you get this right so this is actually factoring in what happens in the Arrhenius expression and also the initial temperature as a fraction of the final temperature final temperature so this is typically referred to as the Zeldovich number or many times in this particular context it is also called the Zeldovich scaling factor because you are going to try to use this as a factor to scale our length scales okay so when you now say you want to have the reaction zone so we suppose that the reaction zone thickness gets scaled relative to flame thickness as so the means if they react if the flame thickness were delta then the reaction zone thickness is like delta over beta so typically you know for hydrocarbon fields beta is about 10 typically I will try to give you exact number some of the time okay on what is the e is like and so what will be the tf like and so on so we can actually factor in and find out that beta is out of the order of 10 exactly means that it is about one order of magnitude more than one that is when you are now doing an order of magnitude analysis we are looking at things like 1 2 3 4 5 all these things are like of the order of 1 okay even like 6 7 is of the order of 1 but anytime now you exceed 10 10 to 100 counts is like order of magnitude of 10 so this is one order of magnitude more than one so that is how that is where you want to look at it okay so this means that we are talking about the flame thickness typically being about a more of magnitude more than the reaction zone thickness or vice versa okay so for high activation energies essentially you get your reaction zone thickness to be quite small but as a fraction of the flame thickness itself and now let us go back and do what we wanted to do so we now say the now this is fine okay so if you want to now do this so redo energy rate balance energy rate balance the rho 0 u 0 Cp Tf – T0 that stays as it is and k Tf – T0 divided by delta you might quibble here you should say very wait wait wait it cannot be delta it does be a little smaller it is like going to be like 1-beta okay and since 1-1 over beta that is how it is going to be okay and since beta is 10 1 over beta is like 0.1 0.1 is much smaller than one on an order of magnitude basis so let us not worry about it essentially what you are saying is if this is going to be so small the temperature is risen almost close to Tf almost at the end of the flame and therefore this is like roughly how the heat conduction should still behave and that is now going to be equal to Q sorry W Q delta divided by beta this is exactly where I want to really bother about it because it is not showing up in comparison with one gates showing up as it is and I have to I have to take that there if I now do all these things so I go through the same rig morale right and what should I what should I get I should now get my u0 to be equal to 1 over rho 0 square root of W K divided by beta Cp so that means that the the second estimate that we have for the flame speed is little less how much is how much is it less it is like square root of 10 less that is about so this was actually predicting this about three times more still within an order of magnitude three is like all the order of one okay so this was this was okay but if you want it further refinement here we have okay let us let us go on to the further refinement tomorrow at 10 o'clock