 So we have the slope of the temperature profile in the preheat zone leading up to the reaction zone is given by a rho u Cp Ti-T0 divided by K let us keep in mind that we are actually hunting for you this is what actually we do not know and this is what we are actually looking for and we are hoping that people be able to actually get this when we try to use the interface matching condition between the two zones. So what we have to do next is of course integrate the second equation and then also get a slope from there evaluated at x equal to 0 plus and then equate these two and therefore we expect to get the u that is how we are going to go about this that is how that is what is looking down the pipeline we are going to be doing but at this stage we now have a problem the problem is that we assume we suppose that we have T equal to Ti in both the boundary conditions leading up to x equal to 0 minus on the one side and x equal to 0 plus on the other side but we do not know what it is all we are basically saying is the temperature has to match and we have to now plug in the value of the temperature Ti over here we do not know what it is so how the question basically is how do you deal with the situation of not knowing the temperature right so there are a couple of tricks about this we will talk about in both the cases we try to now replace this temperature by unknown temperature without incurring too much error so the bottom line basically is without incurring too much error. So what we are going to do from this side looking from the preheat zone towards the reaction zone at the interface between these two zones is to now say that this is equal to rho u approximately rho u Cp Tf minus T0 divided by K why would you do this well basically the temperatures that we are actually given or Tf and T0 Ti is an unknown and we are not really making any bones about trying to find that out in the first place we are trying to actually see if we can replace the unknown by something that is known without incurring too much error. So the last part without incurring too much error is where the crux is so let us look at how Tf departs from Ti so the answer is if you now look at the preheat zone and the reaction zone basically you are having the temperature Ti very close to Tf the only difference between these two is the temperature climbs up a little bit more within the reaction zone to level off at a constant slope at the end of the flame thickness all right. So you could say that you are almost there except there is a slope is not 0 all right. So you have a finite slope a finite non-zero slope that we are trying to look at but it is it is the values is pretty small well if you want to be a little bit more careful about this picture we got to keep in mind that the preheat zone solution is going to actually give you a exponentially increasing solution is what you are looking at. So effectively yet we are talking about a value Ti that is here and the preheat zone solution would go like this and the reaction zone solution would go like that right. So that is what we are looking at so Ti the point is the point here we have to keep in mind is this is T0 so relative to T0 so look at this difference or this difference relative to T0 Ti is as close to Tf it is not as if like Ti is generally close to Tf the question is how close is it is quite close relative to T0 that is like we are kind of looking from here and then seeing that those two temperatures are very close to each other right. So that is the trick that we are playing in the preheat zone and I will show you how to play the trick in the reaction zone where we do not know Ti and we will be integrating in this region and then if you are not anticipating the kind of trick that we are going to play you must be having your eyebrows raised at this stage how am I going to actually how on earth am I going to have Ti be replaced by T0 okay because that is what I need to do okay because I will be in this region I am going to go from Ti to Tf and T and T0 and Tf for what is given Ti is unknown so in this case I am replacing Ti by Tf in that place I will have to actually replace Ti by T0 poof can I ever do that we will we will see how it works huh so it is so wait you wait with beaded breath on how we are going to play that trick so let us let us just go ahead with so we now have this so we will just go ahead with the second equation to integrate it once so the way we want to do that is taking to rewrite it is it has to dt by dx times to which is like if you now multiply this equation by twice of dt by dx all the way through then it is possible for us to now write this is d by dx of dt by dx the whole square equal to negative 2 dt by dx w q divided by k right if you want to just check we can see that if you now differentiate this you get a 2 dt by dx d square t by dx square so and then the 2 dt by dx gets cancelled on either side so if you now do this the reason is we can now have exact integrals on both sides so integrating integrating once right from 0 plus 0 plus to infinity along x okay along x we get dt by dx the whole square evaluated at x equals plus z plus infinity minus dt by dx the whole square evaluated at x equals 0 plus equal to you are going to have this integrated from 0 plus to infinity minus 2 dt by dx w q divided by k times dx so you could actually say this is not really an integral with respect to x anymore it is an integral with respect to t okay so again you do not have to worry about and then of course what you have to do is you have to now realize that the limits of the integral have to be substituted in terms of temperature instead of the x location value therefore we can write this as minus integral ti to tf 2q w divided by k dt now just in case you are one of those people who is impatient about what is going to be the outcome and what are what is it that we are gaining out of this I do not know if you remember I told you that the biggest difference that you are going to get between what you are doing with all these equations rigorously or seemingly rigorously in a mathematical sense when compared to the phenomenological energy balance that we did earlier on is we got all the dependencies right the only thing that we are going to gain is a square root of 2 okay so the factor square root of 2 which is different from the phenomenological expression square of 2 is 1.414 so you are about 41% off in the phenomenological analysis that is still within the order of magnitude you see so 1.4 is about of the order of 1 so that was an order of magnitude estimate so for an order of magnitude estimate that was that was a that was a pretty good job so question is why are we getting the square root of 2 the answer is we are actually having this 2 showing up here right because we now multiplied by 2 dt by dx and then keep in mind that this is going to be a square and I am going to equate this to this right so if I now I am going to have to equate this to this then I have to take a square root so I get the square root of 2 coming from right here at this particular step okay. So now notice that this is going to be 0 because you are having your dt by dx go to 0 at at x as x tends to infinity I also want you to point out that we are actually having the x go from here that is the origin and this is very important for us to think about as engineers x equals minus infinity as far as the preheat zone is here whereas x equals plus infinity as far as the reaction zone is here okay so this does not look like very far away when compared to that it is not like we are talking about minus infinity is being here and plus infinity being there and so on no that is not what we are really bothered about okay so infinity is something that is very kind of flexible it is it is very contextual right so the reason why we are saying this is actually going to be a very small infinity when compared to that is because you see that you are going to have a steep variation in the reaction rate within this region it is going to rise and fall within the small region and then you have to now take very small steps to capture that steep rise and steep fall okay. So that means to say we are ultimately trying to actually integrate this w with respect to t and you have to get an area under the curve of w as it rises as the temperature changes only a little bit between ti and tf okay and that rise and fall needs to be done over very small steps and if you now keep taking very small steps you pretty much reach you get tired and then you say well I think I have reached infinity if you have even if you have not gone too far okay so that is essentially a syntax for you so this is what is called as a inner zone this is what is called as an outer zone this zone has a different length scale when compared to that so we are what we what we mean by length scale is a characteristic dimension over which we would match so that is exactly what we are talking about when you are now saying that this is actually going to 0 there so which means we now say dt by dx the whole squared x equals 0 plus is equal to integral ti to tf to q w divided by k dt so q is the heating value k is the thermal conductivity those are like properties of this system so w is the one that is actually varying with temperature okay and so this integral essentially is of w with respect to temperature keep that in mind so let us call this B versus let us call that capital A and we now want to actually equate A and B except A is actually the derivative of at x equals 0 minus whereas this is B is actually square of the derivative at x equals plus so we have to equate this with square of that okay so a squared equals B implies we now say rho u the whole squared Cp squared see how I am actually squaring things through rho u actually would indicate the mass flux so you want to just keep it together Cp squared tf minus t0 the whole squared divided by k squared equal to 2 q divided by k integral ti to tf as a matter of fact we should have thought about trying to get rid of ti in favor of t0 right here because we did that when we were trying to get a while getting B we should actually try to get rid of ti we will do it in a minute okay so that is what we should be looking at but now the cats out of the bag we will now try to figure out if you that it is not going to be very difficult for us to replace ti by t0 with without much error so that is something that I am going to show you I will talk about pretty soon so here we go how to simplify this a bit further so rho u the whole squared equals 2 k q divided by Cp tf minus t0 integral ti divided time to tf w dt now note that q equals Cp tf minus t0 right or this is actually the adiabatic energy balance so this this is the heat released in the chemical reaction which is essentially the difference the negative of the difference in the summation of standard heats of formation of products to the standard heats of formation of the reactants and this is actually the sensible enthalpy rise of the mixture between initial and final temperatures right and in fact to be able to notice this made a mistake I guess I guess so you have a Cp square tf minus t0 the whole squared so it is mainly to actually cancel this off that we try to replace ti by tf in a hurry back then okay so you might you might you might think both ways you should if you if you did if you got rid of ti there you should have got rid of ti here but if you did not get rid of ti here why did you get rid of ti ti there okay so we wanted to get rid of ti there because we want to cancel this and we could still go on with this okay so let us keep going on and see where we where we want to make a change so therefore rho u squared equals will now for the first time introduce this notation rho 0 sl squared this is basically notation we could basically say rho u is a constant in one-dimensional steady state right so rho u is the same as rho 0 u 0 okay rho u is also equal to rho infinity u infinity alright so the reason why we are saying so many things is the way we want to actually look at this as the laminar flame speed sl stands for the laminar flame speed okay is always relative to I should not say always but you have to specify this the laminar flame speed is relative to the unburned reactants okay when you now say rho 0 sl that means the laminar flame speed is relative to the unburned reactants that means to say you now have a flame that is propagating into the reactants and the question is you could be actually sitting on the flame and getting the reactants to come into you at the same speed or the other thing that you can think about is you now have the flame that is go propagating and then the products are actually flowing out you see now if you are actually looking at the flame speed relative to the products you will get a different speed when compared to if you are looking at the flame speed with respect to the reactants because the reactants and the products are having different speeds themselves therefore it is always quite important for us to note what the flame speed we are talking about and usually we are talking about a flame speed with respect to unburned still reactants all right but rho u is a constant so a less controversial term is what is called as a mass burning velocity okay so mass burning velocity or mass burning flux if you can you can you can call it different ways laminar mass burning flux is essentially rho u and it did not matter whether you are actually talking about it with respect to unburned mixture reactant mixture or the burnt products because it is going to be rho u is constant you see so we are talking about a flame speed then it is it is it has to be specified that we are talking about it with respect to unburned reactants so there we are so you could you could now write this as 2k divided by Cp Tf-T0 integral Ti to Tfw dt and we are now going to say so SL is equal to 1 over rho 0 k over Cp or 2k over Cp 1 over Tf-T0 integral Ti to T0 w dt now could approximate I am saying could I am not saying should okay we could approximate integral Ti to Tfw dt as integral T0 to Tfw dt without much error because this is an integral and what we are talking about is if you now look at how the w goes right we are looking at essentially the area under this curve well not exactly this curve this is this would be wdx we are looking at wdt this is w with respect to this temperature variation right and you can you can easily see that the w is actually 0 outside this region so you might as well actually start integrating from any temperature upstream you would not really care okay there is no hardly going to be any contribution of w there okay so therefore it is because it is showing up within the integral then it is possible for us to actually replace Ti by T0 otherwise it would have been much harder in this kind of situation like algebraic that would be much harder there is no way you can justify that but this is reasonable this is pretty good why would you want to do this you do not have to do this as I as I will show you soon but why would you want to do this what would we get out of seeing this you see that 2k divided by Cp 1 over Tf-T0 integral T0 to Tfw dt what do we have this is in fact very very similar to the first expression that we obtained phenomenologically except for two things one the square root of two factor that I have been saying we will get and as a and we are we are right there getting it starting from this point onwards we get this two there and then of course it is a square therefore you get a square root of two in the final answer the next is a little bit more important which is originally we simply said we had SL equals 1 over R0 integral 2kw divided by sorry did not have the two kw divided by Cp and the question was what was the W if you simply had a kw by Cp then we assume that the W was actually prevalent over the entire flame thickness delta and as if it was a constant all right but that is not very the case the W varies significantly within this distance in fact it is actually lying very low up until the reaction zone and then strongly increases and then decreases within a very narrow region called the reaction zone. So the major effect of the rigorous mathematical exercise is to recognize that W varies within the reaction zone and take its variation as it is so we are not fully replaced we are not really plugging in a W that is a constant and then the question that we had was what would be the constant and we now came up with this came up with this combination called W not infinity right where W not infinity refers to evaluating the W the reaction rate with reactant concentrations for upstream and the flame temperature downstream right that is what not infinity was referring to and that would actually give you the most the highest value that was there was possible whereas if you now have an integral like this what does this look like the reason why we wanted to substitute for Ti is T not in this integral was to actually see how what this looks like this is now beginning to look like more of an average you see so this is an average reaction rate between the two temperatures the limit temperatures T not for the reactants and TF for the products. So if it is possible for us to actually take an average reaction rate then whatever we got previously with a square root of factor 2 to thrown in is what the rigorous analysis gives us based on the Schwab-Zeldovich formulation right okay. So can we do a little bit better is it possible for us to not necessarily make this approximation okay and then we just plot on saying that let me actually see how this Ti kind of thing is going to work out the under root the limits of integral is Ti to TF up here yes thanks when you have too many T's this is what happens up we want to see if we can evaluate this integral that means we now have to explicitly take into account the dependence of W on temperature and what is that dependence that dependence is essentially given by the Arrhenius law where we are now saying W is now going to be a e to the minus capital E over RUT times concentrations of the reactants now keep in mind that the concentrations of the reactants are actually going to fall like this and the a e to the minus e by RT times the concentrations is now going to give rise to a certain variation with temperature which is a bit peculiar so as a simplification we want to bother only about variation with respect to temperature which means we will as a first step we will now say for a 0th order reaction W equals A e to the minus E over RUT and that is it of course all along we have been using W instead of Omega Omega is the number of moles per unit time per unit volume okay that is how the reaction rate is going to be whereas W is actually the number of kilograms per unit time per unit volume and therefore the difference is you know you know you know have to have a molecular weight thrown in here but we suppose that A contains that as well so let us not worry about that explicitly but whenever you are using W or Omega or any of these things you have to be very careful about the units okay and this is this is many times where you get into troubles with numerical calculations so keep this in mind we are still working on a mass basis and this is this is all we are going to have for a 0th order reaction that means we are not going to worry about how the concentration varies along all right which means we are what are we actually looking for since we have done the phenomenological expressions right we have we have some physical intuition about why what is what to expect if you are going to do only the 0th order reaction not worry about the reactant concentration did decrease into the reaction zone then all we are basically looking for is this is going to look like the average is the peak value W0 infinity divided by beta that is the correction that we are basically aiming for okay so is it possible for us to actually show that this average reaction rate is going to be like W0 infinity divided by beta that is what we did before when we said if we wanted to keep in mind the variation of W with respect to the flame thickness and say that it varying mainly within this small region and therefore we scale this distance by as delta divided by beta that is what we did okay then we got S L equals 1 over rho 0 integral k W0 infinity divided by beta Cp so we so that essentially we are looking for this average integral if you evaluate it as it is without replacing Ti by T0 is it possible for us to show that this will be somewhat like W0 infinity divided by beta is what we are basically looking for we are not looking for further corrections which are taking into account the reactant concentration decrease with the into the reaction zone and still for the corrections with non-inteluous numbers which we did before okay because those things are still not possible with whatever we are doing now okay. So now in this in this case see that E over Ru T Tf particularly Tf is typically a large value which is around 10 which means it is one order magnitude more than unity keep this in mind because of most hydrocarbon fuels you have ease of the order of 30 to 40 kilo calories per mole that is again a mixture of units you are not talking about Joules per kg it is calories per mole there so you got you have to make that conversion this is typically how the values are given and Tf is anywhere between 1500 to 2000 Kelvin or even upwards and Ru is equal to 1.98 calorie per mole Kelvin right. So if you now take these values you can you will find that this is of the order of 10 which is quite larger than 1 and we are going to use that fact so what we want to do is if I equal to integral Ti to Tf W DT right then this is equal to Ti to Tf A e to the minus E over Ru T DT what we want to do is we want to basically be looking at a variation in temperature only within the small region right that small variation is what we what we want to do so might as well actually look at the temperature as a departure from Tf rather than the actual value okay so when you now talk about departures then you start magnifying that value rather than look at high values and then varying well over a small range there. So set a new variable sigma as Tf minus T and then sigma i becomes Tf minus Ti and of course sigma is now going to go to 0 when T becomes Tf so sigma equal to Tf right and we also have to find out what T is like so T is Tf minus sigma equal to Tf times 1 minus sigma divided by Tf and you want to note that sigma divided by Tf is much less than 1 right. So sigma is a very small value it is a departure from Tf it is not going to be large because T itself is now going to it is going to vary only from Ti onwards to Tf and Ti is quite high very close to Tf in the first place so this difference is going to mean that T sigma is going to be very small when compared to Tf. So then how is your e to the minus e over ut go and get changed or transformed in terms of sigma is exponential minus e divided by Ru Tf times 1 minus sigma divided by Tf so that is approximately exponential minus e over Ru Tf times 1 plus sigma divided by Tf now these kinds of things are all sort of tailor made for approximations we should we should sort of get an intuition about how these analysis are done so if you now start thinking about a sigma is a small quantity then that is like a sitting duck for a binomial approximation okay so whenever you have a chance we approximate gay we do that in our lives most of the time by the way so we do a binomial approximation and then it also gives you this factor which we have a feel for from here right. So we get all these things isolated then then I is equal to you can pull out these things Ti to Tf e to the minus capital E sigma divided by Ru Tf squared dt whereas our integral is with respect to sigma so therefore we should write this as e to the minus e over Ru Tf integral sigma i to 0 e to the minus e over e sigma Tf squared d sigma now that is a little bit of a cumbersome thing over there so what you can do is let not very cumbersome it is possible to deal with this but still if you want to now call this as e over e sigma over Ru Tf squared and beta i is e sigma i divided by Ru Tf squared then i is a e to the minus e over Ru Tf Ru Tf squared divided by e its cup pulled out and 0 to beta i e to the minus beta d beta now this is so what are we expecting what are we expecting to happen what you basically saying is that we are looking for this integral which is what we started out with to approximate W0 infinity divided by a different beta than what we have used here fortunately okay and that beta was e times Tf minus T0 divided by Ru Tf squared Tf minus T0 is already here so we should be looking for something like Ru e divided by Ru Tf squared in the denominator to couple with this in order to get you this Zelda which scaling factor and then we should get a W0 infinity W0 infinity is essentially the W evaluated at the reactant concentration levels and product temperature levels we are now doing a 0th order reaction which does not depend whose reaction rate does not depend on the reactant concentrations anyway so we do not have to worry about the not part of the W0 infinity we do not have to worry about evaluating it at the reactant temperatures what we have to reactant concentrations what we have to be looking for is to evaluated at the product temperatures which is what this is so this is the reaction rate expression evaluated at the product temperature right and then this is there coupled with the Tf not Tf minus T0 in the denominator you are going to get the Zelda with scaling factor in the denominator so this all of all by itself is giving you for what we thought was the average reaction rate giving you W0 infinity divided by beta so we are fine so all we need is a unity out of this integral sure enough that is what we are expecting to get so this is going to be like 1 minus e to the minus beta i where we are now saying that beta i corresponds to sigma i and sigma i corresponds to Tf minus Ti which is a very small quantity right when compared to Tf therefore we are now going to get rid of it and say this is approximately equal to 1 and then from there on we should now be saying therefore SL is equal to 1 over rho 0 integral 2k divided by Cp a e to the minus e over Ru Tf that is W0 infinity for a 0th order reaction divided by Tf minus T0 times Ru Tf squared divided by e so e times Tf minus T0 divided by Ru Tf squared is the Zelda with scaling factor showing up the denominator of W0 infinity so we essentially are getting back the same thing again except for the square root of 2 that has come out again right so this is typically how you are able to retrieve the answers that we were expecting for this case what I want to point out just point out not really go through this because we are just spending too much time on this you should look up so this is looking like what we got before right you should look up how to actually do the rigorous analysis on taking into account the reactant concentrations then what that means is we have to now relax this assumption on a 0th order reaction and notice that W will now depend on the reactant concentrations so for a first order concentrate first order reaction you will have a concentration of one of the reactants coming into picture for second order reaction you are going to have concentrations of two reactants coming to picture so all those things have to be brought into account and then the next thing that we talked about was the lowest number effect so what you will find this you can actually when we did the lowest number for example we found that and also the reactant concentration you got this W0 infinity divided by beta power M for the where M is the exponent in corresponding to the order of the deficient reactant right and then you also had a lowest number in the numerator so we should be able to retrieve though those things if the 0th order reaction assumption is relaxed and non unity lowest number of reactants considered right we can retrieve the expressions obtained earlier on a order of magnitude basis here with a square root of 2 factor right so we can get this and then what would we suppose then so we notice that for let us say for for a nth order reaction right W goes is P to the n right because you will have the radius expression times the concentrations as many as the number of reactants are and the order the reaction depends on and for that much order you are going to have each concentration contributing a pressure and therefore you are going to have a P to the n for the nth order reaction for the pressure dependence right now what you have is R0 in the beginning right so R0 or R0 goes is P I do not want to say P0 because we are assuming the pressure to be constant it does not matter whether it is P0 or P infinity okay we are just talking about a constant pressure lower less a constant pressure deflagration here therefore your SL is going to go as P power n by 2 divided by P and by 2 coming from having W depend as P to the n and then under square root therefore you are going to have this as P to the n minus 2 divided by 2 right or you can say n by 2 minus 1 so n by n minus 2 divided by 2 so for a second order reaction second order reaction SL is nearly independent of pressure keep in mind order is a global quantity it is valid for a global reaction right or if you are able to actually represent a scheme of chemical reaction steps fundamental elementary reaction steps in terms of a global reaction so it does not have to be an integer so typically for most what we call a second order reactions let us say hydrocarbon gaseous hydrocarbon oxidation your order of the reaction is going to be somewhat like let us say 1.8 so it is not going to be exactly 2 right so the reason why I am saying nearly so we find that there is a weak dependence on pressure so if you now think about an n that is like about 1.8 you see you are going to have an inverse dependence on pressure right that means as the pressure increases the SL decreases okay that is in fact true that is what we actually observe but it is a weak dependence so we are now beginning to talk about dependencies of S on various factors so what we just mentioned was pressure which was obvious what we have not really worried about is how it depends on let us say initial temperatures okay and what we will see is this is primarily the dependence there and SL is going to depend on lot of other factors mainly through W okay so if W is going to depend on initial temperature it will depend through TF okay so we will now talk about these things in the next class.