 So, today what we will do is try to actually make a mathematically rigorous analysis of the laminar premixed flame problem what we have been doing so far is somewhat like a phenomenological construction of energy balance and mass balance looking at the preheat zone and the reaction zone separately in fact we went all the way up to even considering not only upstream diffusion of products and consequently a downstream production of reactant species concentration relative to the far upstream reactant concentration we also taken to an account the effect of non unity Lewis number of the reactants in trying to find out how the reactant concentration profiles are going to change because of that but however when you are now doing the rigorous analysis we start with the Schwab-Zeldawech formulation which means we have to go along with the 11 assumptions that that are attendant to the Schwab-Zeldawech formulation and then we make further assumptions hopefully only a few that is particular to this particular to this problem so further assumptions the first thing that we make is say somewhat a peculiar simplifying assumption which could be relaxed number of moles does not vary does not vary during the reaction so let us suppose that we have n a equals n b equals n and this this corresponds to saying so for example for a I should say n or for a for a let us say nth order or should say maybe just go to n for nth order rearrangement reaction such as like let us say n a equals n b b when you now say n a equal to n b then this actually implies that w a equal to w b so what basically happens is your mass fractions are equal so you do not have to worry about a further as a further equation that relates to mass fractions and most small fractions second is a we assume a calorically perfect gas now notice that we did not make this assumption as part of the Schwab-Zeldawech formulation we could still deal with a integral t reference to T CPT CP DT in the case of Schwab-Zeldawech in general that means you could assume a temperature varying specific heat but here we make a calorically perfect gas assumption that is CP equal to constant also essentially we are saying all thermal properties are constant k is a constant so let us assume that the thermal conductivity is a constant as well number three flame is one-dimensional and planar and steady of course steady state is assumed in the Schwab-Zeldawech but just we want to point out here that we are looking at a steady flame which is one-dimensional and planar that is assumption which is not very unrealistic if you think about for example if you had a slotted Bunsen burner you could actually have a conical flame and locally along the side of the cone or since a conical like a tent shape flame and locally along the side of the tent like the shoulder of the flame is reasonably planar course it does not look one-dimensional but we will notice later on that we should be able to actually take a component of the flow perpendicular to the flame and then treat the flow field across the flame locally like the way we are doing here so that this is this is a very reasonable assumption the other ones are obviously simplifying assumptions. So let us now proceed with how we would do the Schwab-Zeldawech so the Schwab-Zeldawech energy equation goes like take a divergence of rho V vector integral T superscript not to T Cp DT minus rho D gradient integral T superscript not T to T Cp DT equal to minus sigma equal to 1 to n ? hfi not wi is what we had we denote negative sigma i equals 1 to n ? hfi not wi as let us say WQ all right now that is kind of like a summation is equal to like reaction rate times the heat of reaction of course we are assuming a single step reaction as part of the Schwab-Zeldawech formulation course this was not the final energy equation that we derived the Schwab-Zeldawech this is what we had actually termed as the Schwab-Zeldawech energy equation while we do while we were doing the formulation so it is basically taken right from there we even form the coupling terms or the alphas and all those things at this stage but let us suppose that we can just proceed from here and then apply the one-dimensional apply this to one-dimensional situation then here we can now write this as rho U D by DX of Cp T minus T0 minus rho D D squared over DX squared Cp T minus T0 equal to WQ okay are we essentially the divergence has become like a D by DX and for a one-dimensional flow we know that rho U is a constant therefore we can actually pull this out because rho U is a constant we are together as a product we can pull that out of the derivative and then we evaluate the integral for a constant Cp and we also do the same thing over here and since it is one-dimensional we can write it as a ordinary derivative instead of a partial derivative because things are varying only along x which is in this direction across the flame okay then we write the Schwab-Zeldawech species equation for species A let us say that is the reactant there and so we now write for the species now what you will find when you now try to apply the Schwab-Zeldawech formulation is we do not necessarily solve the entire n plus 1 equations that we talked about where n is the total number of species involved okay although it would look like they were all coupled and we have to solve them simultaneously and so on what we are actually looking for is for equations to look identical so if equations begin to look identical then it is sufficient for us to solve one of the equations that are that is a part of the identical set instead of having to solve for all the equations simultaneously so this is how we simplify our lives instead of actually burdening ourselves with lots of equations to solve so here in the case of the prem explain we will see that for the moment it is actually sufficient for us to take the species A conservation alone we do not have to worry about lots of other species so here you have a divergence of Rho v y a- Rho d gradient y a equal to w a now since you have a same number of moles everywhere we can we can point out that w a equal to minus w b equal to minus w right so w b is rate of production of b w a is the rate of depletion of a and so rate of depletion of a should have a negative sign to denote the depletion if you are now writing it in terms of the general reaction rate w on a mass basis so here if you now try to unwrap the divergence and vector notation and so on fortunately we do not have an integral to evaluate in this case as in the energy so it should simplify as d by dx of y a- Rho d d squared y a over dx squared equals minus w a matter of fact I should point out that we are also assuming the Rho d to be constant by the way that that is not that sort of implicitly stated here because we are assuming d to be a constant and this is like an incompressible flow for the mixture density doesn't change so therefore and then of course we are not even bothering about the mixture density changing with temperature all right therefore we would actually suggest that Rho d is a constant and we can go through with the divergence all the way to the applying the gradient and therefore you get the second derivative here all right now we want to actually try to see if these two equations are looking alike in some way and this is what we also try to do when we did the Schwab-Zelovich formulation anyway right so here if you denote denote ? equal to Cp T-T0 divided by Q and as per equals 1- y a this is pretty interesting actually because for this is kind of obvious we want to get the Q down here I am sorry we want to get the Q down here okay and then you have essentially T is the one that is varying Cp the constant T0 to the constant but you have this T with all this paraphernalia like Cp times T-T0 together and then okay so we want now see that is like the variable now instead of the actual variable temperature and then we get this Q down here so that it will also have a W just there on the right hand side just as well as the species equation but the species equation has a negative sign with for the W therefore if you want to have only a W there all you need to do is to actually multiply this by a negative sign on the on the left hand side that means we should have actually been saying half is equal to minus y a not 1- y a so why were we looking at 1- y a this is a trick okay that that is not really highlighted in textbooks and so on what we are basically looking for something physical look at how we actually proceeded in the phenomenological development of the structure of the premixed flame that was the key we notice that the temperature profile goes like this and then we started asking questions about how would the product profile look like right and then we came to the conclusion that if the most of the products are actually produced here we should expect anticipate that the product concentrations should actually grow only over here but once the products are actually formed they kind of look around and see hey I am actually a lot over here and I am getting convicted down there so I am also there the only place that I am not here not around is here therefore can I diffuse but I am getting convicted down so there is like a balance of how much upstream diffusion the products can do versus the downstream convection that they experience and that would actually give rise to a concentration profile that becomes pretty much identical to the temperature profile in the event of Lewis number equal to 1 which is assumed in the Schwab-Zeldovich formulation right so then we realize that the product speed that the product species also goes through a convective diffusive balance in what we thought was the preheat zone for only the thermal balance earlier and so we anticipate that the product concentrations should look pretty much like the temperature concentrations if you normalised in certain ways right so how would you normalise like this looks like a normalised temperature in some sense not exactly T to the superscript not which is a standard temperature we should have actually put T to the subscript not for the initial temperature then this would have looked like a normalised temperature and I will come to that pretty soon but this one instead of just saying alpha is equal to minus YA which would have done the job for us here to actually make this look similar to that with T replaced by ? through this transformation we are actually saying 1-YA because alpha now begins to actually act like the product species constant concentration right so it is essentially saying whatever so if I had only to react to two species let us say a and b and then you can clearly see that why YB would be like 1-YA right if you had multiple species it will matter finally you are going to actually have something that the out of which you are subtracting YA so 1-YA will do the trick for you to actually have alpha go more like the product see so that is a that is essentially the idea here so we are actually looking for the equations to look similar by choosing a if you want to call it a transformation by choosing a transformation for the species reactant species concentration that will mimic the product variation because that is how we actually develop the the phenomenological understanding on how upstream product diffusion causes a downstream reactant depletion through diffusion so yes the effect of that is going to be felt in the flow okay so what will happen is if you really have a temperature sorry density that depends on temperature then your mass balance is going to get affected so you are when you say that rho u is equally a constant of the rho changes with temperature your u is going to accelerate right there is something that we noticed in the Rankine-Hugonea formulation that across the deflagration the flow accelerates right so 1 over rho u increases so rho decreases and therefore T should increase anyway and then you will also increase all that stuff is correct but in the in the in the in the Schwab-Zelovich formulation we are actually adopting a given flow field and the given flow field here is a uniform flow field that means no matter what they constitute constituents of the mixture or what is the composition of the mixture is as we go from the reactants to the product through the flame the mixture as a whole is having only the same velocity we are ignoring the fact that the flow is going to accelerate it is as if like we are not interested in it we are not worried about it okay for correspondingly then we can actually go with a constant density situation or we do not explicitly state that we say rho D is a constant and then you start thinking about why should D vary with rho in a way that rho D is constant and so on that is that is not exactly the way it should it is justified okay so the way it is justified as we are basically thinking about like an incompressible mixture and we are violating the gas law the perfect gas law but that violation is more felt downstream of the flame when our business is over so we do not we do not worry about it okay so this is something that I pointed out that right at the beginning when we are doing the Schwab-Zelovich or when we actually did the one-dimensional momentum to show that it will reduce to something like pressure is approximately equally a constant the point there being if you want to take into account the velocity variation in the flow then only then you have to worry about the density variation with respect to temperature and then the flow problem on the combustion problem will get coupled here we are actually looking at a combustion problem that is decoupled decoupled from the flow problem for which we have to make this assumption okay so if you now go back and plug ? and a in the energy and the species equations respectively you now get equations that look like ? D D2 ? divided by DX2- ? U D ? over DX plus W equal to 0 and ? D D2 ? divided by DX2- ? U D ? over DX plus W equal to 0 hey those two equations look exactly the same except that one of them has a ? rather one has an alpha to solve for right now if you agree Q might have a preference of ? over alpha but I am not so I am not going to really worry about which equation I am going to solve is that okay is it okay to say now it is sufficient for us to solve only one equation you do not have to solve both the equations or what is the catch would ? and ? look exactly the same as you solve for one and then expect the other one to behave exactly the same is that okay fine whenever you are confronted with two differential equations that look identical except for the the symbol that is used for the independent variable also the dependent variable I am sorry okay so all the two equations go and actually give the same result or what are we what are we missing there is one more thing that we have to worry about all differential equations have the burden of having to satisfy boundary conditions okay so at this stage we can say that ? is going to be related to alpha if it is it and it is sufficient for me to solve one equation and get and then satisfy its boundary conditions and obtain one of them let us say ? then I should be able to say that ? is equal to a alpha plus B all right and why would I say a alpha plus B because I am expecting to have two boundary conditions for alpha as well with which I can actually evaluate a and b instead of having to freshly solve this equation for alpha and then apply its boundary condition so that would actually give me room for applying boundary conditions for alpha or before we proceed in that direction we are we are probably better off looking at what are the boundary conditions for ? and see if they are also identical right if they are identical then what do we say right so let us see and we and it is not terribly unrealistic to expect them to have nearly identical boundary conditions because as I said this is mimicking the product and in the in the in the in the case of low unity lowest number we expect that these profiles should be the same right if these things have been normalized right all right so hold your breath okay hold on to your seats we are going to have some fun here right so let us look at the boundary conditions cold boundary that is x equals minus infinity alpha is equal to 0 ? equal to 0 assuming supposing T superscript 0 as T 0 without loss of generality because I could have actually written this as T minus T subscript 0 minus my minus of sorry T minus subscript T subscript 0 plus T subscript 0 minus 2 superscript 0 so I could have I could have added and subtracted my actual temperature initially that is essentially saying I have a sensible enthalpy initially above the reference value for the reactants that is now going to be added up over here which is a constant and therefore it is not going to really affect me at all so I could actually say that effectively T is whatever it is and then I and then say ? is equal to 0 and alpha is also equal to 0 because I am assuming that I am going to have all of them as reactants upstream and then alpha y a should be equal to 1 and heart boundary x equals plus infinity right now ? will now give you Cp Tf minus T0 divided by Q and alpha should be equal to y a goes to 0 all the reactants are getting consumed in the flame and therefore you are going to have alpha to be equal to 1 but what is ? you get this funny expression there can we evaluate it that is equal to 1 okay there is because of the adiabatic flame temperature in the case of an adiabatic flame temperature this is the heat that is released in the chemical reaction and all the heat that is released in the chemical reaction goes to increasing the sensible enthalpy of the mixture that is what an adiabatic reaction flame is all about okay so Cp Tf minus T0 is a sensible enthalpy rise of the mixture with which is primarily coming from Q and therefore this is also equal to 1 lo and behold the boundary conditions are also identical wow we meet such a big fuss about you know 4 5 n plus 6 reactions in the combustion problem and then we we we now believe but a lot to get the swap Zelda which formulation going and we still had to start get stuck with n plus 1 reactions n of which were fortunately homogeneous and all those kinds of things and finally we are able to actually get to two reactions that we just pick out of them and then we find that both of them are identical and the boundary conditions are also identical therefore it is sufficient for us to solve only one equation great progress right. So therefore considering one equation is sufficient if you were to still persist and say well theta should be equal to a a a plus b and you can substitute the boundary conditions for a and then plug in values of theta run alpha there you will find that a is equal to 1 and b is equal to 0 alright so and then you will you will be able to show that theta is equal to help so that that is mathematically showing things okay so what we want to do now is get confronted by that one and one equation let us say we take the one that is written first right and then we want to deal with this we want to see how to solve this right for the first time in our lives as combustion students we are going to solve something so far we have been only posing problems okay oh my god how do I solve this what is the problem that is a second-order ordinary differential equation with constant coefficients as a matter of fact okay row D is a constant row U is a constant okay and then you have a it is it is an inhomogeneous equation alright okay what is the inhomogeneity all about you have the W that is the chemical reaction rate and you have the theta there through the temperature sitting on top of an exponent with a negative sign but sitting in the denominator on top of an exponent with a negative sign so that is the ugly looking expression that has been camouflaged by a simple double do there you see and that is been the one that we have been really feeling jittery about all these years and there we are we have to solve so how do you do this chicken out of the problem say well can I look at the flame where you do not have a W yeah sure of course fear you say you get into the flame you know you you now with a lot of trepidation you get into the flame and then still no W oh sure fine this looks great so I go get in there and then no W at all for quite some time so I can solve that can I so what we would do is we now begin to notice the way we were talking about things earlier that you have a flame with a preheat zone with hardly any W where I could actually go ahead and jolly will solve this and then I have to look at this this region where I have to confront the W and then point out that you pretty much have nearly the same enthalpy for the incoming species and the outgoing species so you do not really have a convective enthalpy flux to worry about and therefore I am going to get rid of this right and I am going to look at a reactive diffusive balance and then try to integrate that right why I mean we still have to solve to take this W into account ultimately so what is the big deal the answer is what is it that we are actually looking for out of this equation can we now find out what exactly our problem is what is what is that one single quantity that we are trying to solve for do we know what the problem is you know lives we do not know the flame speed okay so although this is like a very innocent looking constant that has been pulled out we do not know what the U is right so what this actually allows us to do is we can say while I am actually in the convective diffusive balance I keep the U which is what I do not know right and then it turns out to be like an eigenvalue problem so an eigenvalue problem is where you have one of the coefficients that is an unknown and you supply the boundary conditions and then you turn out that you the boundary conditions are not sufficient enough for you to actually evaluate the eigenvalue right. So in this case what you do is you retain this in one of the regions and you get to the fit in the other region where you have to confront the W so you are kind of breaking your problem into two part divide and rule right this is my problem this was my unknown keep the unknown get rid of the problem in one and keep the problematic thing and get rid of the unknown in the other I should be able to handle this but what is the catch I solve for the this is going to have this is going to still need two boundary conditions because of second order right and my zone is only from here to here right and strictly speaking I do not know exactly where to draw this line all right it is sort of like as I keep on entering into this region and then the chemical reactions are happening somewhere I suddenly feel too too hot and then say I think I have got into the reaction zone it is a feel right I do not know exactly where it is and I need a boundary condition there while I do not even know where the boundary is okay and look at this this part right if I now want to keep this and that I still am not throwing out the leading order term which means I need to have it is still second order equation and I need to boundary conditions and the boundaries are this fuzzy boundary somewhere here and the far downstream boundary okay but all we have stated so far is I knew what the boundary condition is here I knew what the boundary condition is there so what do we have to do here we have to now come up with two boundary conditions because it is second order which are shared by these two so that is what is called as interface conditions so that is an interface between two zones and we have to supply two conditions that are shared by these two okay and whenever we are looking for two conditions it is because it is second order and if it is second order then we are looking for two conditions one in the value in the other way in the first derivative and in fluid mechanics the first derivative mostly has some significance in the sense the first derivative is going to actually deal with the heat flux the value itself is going to is going to be at the temperature okay so we will go through these steps rather quickly in mathematical way so region one is our convective diffusive balance convective diffusive zone which implies that w is approximately equal to 0 and then we can write d square we can go back to T do not worry about ? anymore that was just to show that ? and A are the same we can go back to an equation that is over there so d square T over dx square- ? u Cp over k dt over dx is approximately equal to 0 now when you say this is approximately equal to 0 this equation is what is called as the zero-thorough equation so what we are actually doing is a part of a larger approach called asymptotic matching okay or what is called as matched asymptotic expansions strictly speaking what we should say is the temperature profile can be expressed as an asymptotic expansion all right and we now try to take the what is called as a zero-thorough where whatever is actually a small quantity in the equation is set identically equal to 0 it is not exactly 0 right you do have the temperature rise and you strictly speaking have a reaction rate that is beginning to increase fortunately for us if the e is large it is going to get confined to a smaller region okay so that is what we are really counting on but it is approximate so to leading order to zero-thorough we now set it identically equal to 0 get rid of it from the equation in the first order we have to refine the equation with keeping that small value you see we will do that fortunately okay we won't do that we will just do only the zero-thorough matching so let us let us call this equation 1 and the boundary conditions here as we talked about we want to retain the cold boundary conditions which is t equals t0 and x equals 0 minus so we are now locating our x equal to 0 at the edge of the preheat zone reaction zone match and we want to call this x equal to 0 minus t equals ti many times we attempted to think that I stands for ignition temperature okay and then we are always looking for what is called as an ignition temperature that is like a popular notion and it is a popular engineering notion to think about like an ignition temperature mathematically speaking we have to confront a cold boundary difficulty that you have reactions happening even there okay it is just then it is just simply not significant enough as far as here ti is unknown actually it is not like we are going to plug in a value there that is above that is somewhere in between the room temperature and flame temperature that is not what you are going to do it is an unknown it is a it is a temperature that is going to be shared as the upstream boundary condition for the reaction zone alright and it is it is an unknown so that is that is region 1 for you region 2 is reactive diffusive zone here what we then say for the 0th order reaction is d square t over dx squared plus wq over k is approximately equal to 0 that means this is the diffusive part this is the reactive part the balance is essentially that you hardly have any convective flux that is going on that is changing so you do not have a essentially what that really means is you do not have a dt by dx significantly okay that is a very small quantity so the boundary condition here we can easily write the hot boundary condition which is x equal to infinity we can say t equals tf that is like what we had started with but then the interface condition is x equals 0 plus we have to write the same t equals ti which is an unknown right now if you are really bit a little bit more careful you would probably want to write this equation for the region 1s d square t1 over dx squared minus rho ucp over k dt t1 over dx and then write these temp these is t1 equal to t0 and t1 equal to ti you can say here you can plug in a t2 and then t2 t2 and then finally say at the interface t1 should be equal to t2 and so on okay but we understand that ultimately it is going to be a single temperature profile that we are interested in and then we said that we wanted to have one more boundary condition at the interface that needs to be shared and since it is a second order equation we can go up to the first order the first derivative is signifying heat flux and therefore what this really means is matching heat flux heat flux at the interface between the two regions two regions we get dt over dx x equal to 0 minus should be equal to dt over dx x equals 0 plus well strictly speaking we should say k dt by dx but since it is the same species and as the as you now go to 0 from both sides it is essentially the same location the case at the same so we do not have to worry about it but these gradients are coming out of two different solutions and we are actually trying to force this this is going to say that we are going to force these two solutions to have the same gradient at this point what that really means for you is keep in mind what we were talking about for the length scales characteristic length scales for thermal balance in the preheat zone versus the species mass in the in the in the in this zone when we were looking at non-unit loss number the other day we pointed out that if you were to just bother about the Schwab-Zelovich only here and then take a equation that looks like this with a 0 on the right hand side you can actually solve this but what you will find this you will get a solution that kind of goes go goes like this and then goes exponentially there for x greater than 0 which is not the domain of validity it is outside the domain of validity of this equation if you were to extend the solution that you are going to get from there for your T1 of x you are going to get a solution that goes exponentially right that is what a second order request second order homogeneous equation is going to look like you try to solve this you will get exponents right so you will have this on the other hand what we are talking about in this equation this is also something that you can solve what you will find is this is going to actually give rise to something like that and what we are saying is at this place we want to have the temperatures as well as the slopes match and that is where we want to position our x equal to 0 somewhere there right so this is typically how we want to construct the composite temperature profile in a composite manner by having this asymptotic matching what you are doing here is a 0th order matching strictly speaking because we are identically setting whatever is a small quantity in the equation to 0 and solving for it so this is the problem and out of this we are trying to get the velocity u we will do the first region and then hold on till next week for the second region with baited breath I hope right we cannot wait to get there so of course we say that this equation is 2 and we what we want to do is integrating 1 once right we say dT over dx equals rho u divided by k Cp T plus a constant and we have to supply a constant of the boundary condition to satisfy to evaluate this constant of integration one way of doing this is to actually integrate it twice and then you now deal with two constants and then apply the boundary conditions right but I am not going to do that what I am going to look for is I am actually interested in dT over dx at 0- okay so I am not really interested in going ahead with integrating this twice to look at actually the temperature profile this is a point where the analysis sort of cheats you okay so until now we have been talking about these profiles matching and all those things we are actually going we are going to think that we will we will get this matching and get this composite profile but typically at this stage the analysis does not abandon sets care about the temperature profile it now starts bothering about the velocity which is the eigenvalue right that is what we are looking for so we do not really want to care about getting this and I am stating this explicitly because textbooks do not states is explicitly and then when you go back to laminar flame analysis to actually look for the temperature profile you will not find it there okay and then they but they are like let us just go to go through this and then get the velocity put a box around then we got laminar flame speed let us talk about the laminar flame speed and this keep going on okay so you do not really get unless you go dig up literature further you cannot get this temperature profile so that is what that is a kind of attitude that you are going to get so since that means we now have to actually evaluate what the dt by dx is far upstream so we since we are saying that at far upstream in this region 1 t equals t0 at x equals minus infinity then t equals t0 for quite some distance therefore the dt by dx is going to be equal to 0 x equal to minus infinity t equals t0 dt by dx equal to 0 this is asymptotically approaching that is the reason why we are able to actually reduce this otherwise we cannot okay so the means dt by dx equal to rho u divided by k Cp t minus t0 right now if you think about it this is the first thing that we wrote in the phenomenological balance we said that rho u Cp is essentially your rho u Cp t minus t0 is actually the enthalpy rise that is happening and that is because of the heat conduction k dt by dx and we approximated it as kt minus tf minus t0 divided by delta that is what we did earlier okay so whatever we could actually write by just looking at things if you think about it takes about an hour for you to get if you go through the mathematics all right and there we are so that is what we got now therefore if you want to now apply this to the other boundary we now say at x equals 0 minus this is equal to rho u Cp divided by kti minus t0 here we are imposing the next boundary condition and then what you are going to say here and or maybe this is a good point to stop okay so we will just say this we keep in mind we do not know what ti is that is an additional problem that we have to worry about and we will see how to handle this next class.