 We have been looking at the derivation of the Schwab-Zelda which formulation leading up to formation of these coupling terms of a linear operator working on both the energy and the continuity. So the steps that we have followed so far is to have the overall continuity taken in the account in the simplified energy equation where we are having the relevant terms based on the assumptions that we have made and then we add this term which is having a essentially amounting to 0 to the energy equation so that we can now group these two terms together and we now have a v plus vk so this is actually amounting to the species velocity so this is sort of like saying recall this is actually the mixture averaged convection of enthalpy of the mixture averaged enthalpy this is actually the enthalpy convection due to diffusion so if we now try to put these two together this would actually be like the species velocity convecting the species enthalpy right so effectively we are having like a species specific convection of enthalpy grouping together that is actually happening at the from these across the surface of a control volume this is conduction happening across surface of a control volume so obviously we are having like a divergence of both equal to 0 because when you now have surface the facts the Gauss's divergence theorem shows up this divergence here so you can clearly see the physics of the manipulation also that is going on but here what we then do is well if you now have something like a species specific enthalpy convection can I now use the species specific mass consumption a mass convection right that is equal to the so since a speed if you now look at this is like a species velocity small vk vector then the diffusion is embedded in it so we do not have to worry about diffusion explicitly so then it would be like species convection equal to reaction okay the other production or should have actually had a wk here right so of course you do not have the unsteady term because you assume steady state so whatever is convecting in and out should be based on how much is being produced or consumed inside the control volume that is how the species conservation looks like which is now possible for us to plug this into this okay and then we do the expansion of the enthalpy species enthalpy and say this is like standard heat of formation plus the sensible enthalpy you now plug that in there and what you will find this for example for that term with the standard heat of formation alone you now try to use this use this use this equal to wk if you did that then you would get this to be taken to the right hand side with a negative sign and for the remaining you now keep it as the test but then notice that this is going to be mixed or averaged anyway therefore you could go ahead and some sum over all CP case and then you get a CP right so here CP is equal to sigma equals 1 to n yk CP k right so that is what being is being used over here but you cannot do that here because you have a capital vk that is waiting it so you have to have a yk vk CP k within this integral so you cannot get this yk alone here and then have the summation vk is interfering there and then we had the k gradient T coming from here as it is and I told you how we got this right hand side now what I would like to point out which I did also previously is this is the chemical heat release rate term that has now been identified explicitly for you well like to think what we should like to think about is so how does this very work ? h of k not actually means the standard heat of formation of species k right so that is going to be in terms of something like joules per kg or joules per kilogram of species k and wk is actually the net rate of production or consumption of species k in terms of something like kilogram per meter cube per second right and if you now add up this product over all of all of the species you are now going to get the net heat release that is been produced out of this chemical reaction there is an alternative way of doing this so this is actually per species right so you now take heat release per species and then add up across species of course in the in the Schwab-Zeldovich formulation the I think I guess the ninth assumption is assuming a single step chemical reaction but what if you had multi-step chemical reactions right so we know that for a chemical reaction which is not necessarily a formation reaction any chemical reaction you will have a heat of reaction even let us say standard heat of reaction so if you now had a standard heat of reaction for one reaction but you now are looking looking at like a multi-step reaction scheme each of these reactions having a standard heat of reaction associated with it which could be endothermic or exothermic right so if you now have all these reactions happen and each of these reactions has a reaction rate associated with it right which is not specific to any particular species that is participating in the chemical reaction in those reactions so on the one hand you have a heat of reaction on the other hand you have a reaction rate so from there you should be able to find out what is the rate at which heat is released in a particular reaction and then you can now look at all the reactions together and find out what is the heat release the net heat that is released per unit time there is a heat release rate per unit time for all the reactions happening so there you sum over reactions so one of the exercises that you would like to work out is it possible for you to show that the heat released from the chemical reactions on the whole is the same regardless of whether you actually summed over the species and their heats of formation or summed over the reactions and their reaction rates right so that is something that we will not really pursue here because we make an assumption about single-step chemical reactions but that is not going to actually stop us from writing another expression for multi-step chemical reactions in terms of reaction rates and heats of reactions as opposed to heats of formation of species that we are doing here so where where are we actually going to use a single-step reaction assumption is yet to be seen we have not really utilized it yet okay so that is one thing that I wanted to point out so while we have now dealt with this term there are three terms that we want to now deal with in the on the left hand side and let me just number them as 1 2 3 so that we will actually try to work on some of these terms identified by the numbers so so now let us let us first for example take take term 3 right so we now take term 3 that is like K grad T that is the simplest looking term in all this so let us now take this and then here what I am going to do is I am going to write K and then multiply and divide by row CP D with a row D put over there and a CP put over here and then write the grad T hey that looks like a lot of cosmetics that has been done around without any effect but what is what is in here now what do we have so this is our lowest number right this is our lowest number that is equal to K over row CP D and we have assumed it to be 1 right and therefore so this is assumed therefore this is now nothing but row D CP grad T all right so that is what you are going to get for the third term so effectively what you are basically saying is the conduction is going to happen pretty much based on the diffusivity because the lowest number is equal to 1 yeah okay now the second term is a little bit more complicated to deal with so term 3 all right so the term 3 is is is is is row sigma K equals 1 to n YK capital VK vector integral T superscript not to T CPK DT all right now we will have to actually show why what this amounts to and so I am going to I am going to try to show that for you so this is I am going to show that this is like minus row D sigma K equals 1 to n grad YK integral T not to T CP K DT this is actually coming from so I will open parenthesis here multi-component diffusion equation right multi-component diffusion equation simply boils down to D gradient XK equals XK sigma J equals 1 to n XJ VJ vector minus XK VK vector K equals 1 to n this is to say in the multi-component diffusion equation only the first term on the right hand side survives and all the other terms are the all the other terms vanish because of the assumptions that we have made exclusively to get rid of them and we also have made the assumptions that all the binary diffusion coefficients are equal to D so you just have to unwrap the left hand side on the right hand side together in the multi-component diffusion equation to get this now you can further so what you can actually do here is now we are not really interested in having XK we are interested in having YK that is what we are dealing with right so one way of actually trying to get rid of this is divide by XK and multiply by YK right so if you do that then what you will find this you have a grad XK by XK right that gives you something like a grad natural logarithm XK right so you get into those kinds of things so what you whatever I would say is multiply I am going to skip steps here multiply by YK by XK and sigma K equals 1 to n right that means you add over you should okay so you now get a D gradient natural logarithm XK minus sigma J equals 1 to n YJ gradient natural logarithm YJ should be XJ I guess XJ equal to minus VK right K equals 1 to n now so you can show this okay so show as in like you now try to do this and then you can further simplify this if you now try to write it out for all the species you can say further show show VK vector is equal to simply minus the gradient natural logarithm YK K equals 1 to n right this is actually retrieving fixed law so the way we have actually made the assumption that DIJs will be equal to D and also getting rid of all the other terms except the first term on the left hand side of the multi-component diffusion equation should essentially amount to being able to retrieve fixed law for a multi-component system right so if you now say VK is equal to minus D natural gradient of natural logarithm of YK this amounts to saying gradient YK divided by YK so this YK can actually go over here to the left hand side and then you can now see how this YK VK can be written as minus D gradient YK so you also had a row so the negative sign and the row will go there then sigma K equals 1 to n gradient YK integral T not to T CPD so you can we can so use this above okay so that closes this parenthesis then let us now consider but gradient of T not to T right the reason why we want to do this is we now find that gradient YK shows up here right and then you now have a CP integral CP DT let us look at what we have over there we have a divergence of Ruby CP DT right and and I am sorry no no no no not that you see here you now see row DCP grad T okay so there the gradient is actually on the temperature and here the gradient is the gradient is on the YK so is it possible for us to see if we can combine these so if you now consider this particular term out of the blue to seemingly out of the blue okay we just we just consider this term then gradient let me we can we can now write this as gradient of sigma K equals 1 to n YK integral T not to T CP K DT not a problem simply writing what CP is starting from here it is only CP that needs to be integrated with respect to time temperature not YK so you can pull the YK out of the integral but it has to be within the summation so let us now try to get the gradient inside the summation and see what happens so this is K equals 1 to n gradient YK alright integral T not to T CP K DT plus sigma K equals 1 to n YK gradient integral T not to T CP K DT right so this is equal to sigma K equals 1 to n gradient YK integral T not to T CP K DT and then we now have to see what what is going on here T is a variable right so we are looking at actually taking a gradient of temperature this is always been our in our minds alright we always found that temperature is actually the variables sitting on top of this integral so if you want to look at any gradient that needs to be taken derivative that needs to be taken with respect to temperature what we what do we need to do we will have to apply Liebnitz rule because Liebnitz rule is where you now are looking at derivative of an integral whose limits or functions of what you are going to take the differentiation with respect to K so that is where the problem is and so that is going to yield for us plus sigma K equals 1 to n YK CP K of T grad T the way it works is Liebnitz rule so this is actually coming out of Liebnitz rule the way it works is you evaluate the integrand at the limits and then multiply that by the gradient of the limits or the differentiation of the limits so in this case T not is a constant so the second term which is having a gradient of T not is going to vanish so you do not have to worry about evaluating the integrand at T not so the only term that is going to actually and then of course what you then with then can do is you will also have a another term which is like an integral of gradient of CP K okay so that is not going to contribute so what is going to contribute in these three terms from Liebnitz rule is only that term where the gradient of T is there with the CP evaluated to the T that is how it goes in fact in my notes I am a little bit more particular about pointing out that you can say this is actually this is T prime DT prime okay so where we are now saying T prime is like a dummy variable of integration whereas here we are evaluating the temperature that is going to be there as said unknown okay. So if you now are okay with this then what happens is we now get sigma K equals 1 to N gradient Y YK integral T not to T CP K DT plus grant T does not depend on species so that can be pulled out of the summation and all you have a sigma Y K CP K which is simply CP right so this is like CP K CP grant T right so what are we trying to do we are now trying to look for this term in here okay so from the third term we got this by applying the multi-component diffusion equation reducing to fix law to get this term right and this term can be identified here as what we started out with minus this okay therefore therefore sigma K equals 1 to N gradient YK integral T not to T CP K DT let us also do this a little bit better if you want to now look at this as a negative sign with a negative sign therefore minus rho D right then what happens you have you now throw in a minus rho D over here so you now have a minus rho D gradient integral T not to T CP DT that is here and then this goes to the left hand side with a negative sign but you have multiplied by a minus rho D so you get a plus rho D CP grad T good then what happens try to put all these things together right so we have the first term as it is just keep the first term as it is rho V integral T not to T CP DT okay the second term has been the one is the one that is been giving us a little bit of a problem and that is now looking like effectively two terms okay so there is minus rho D grads integral T not to T CP DT plus rho D CP grad T okay the third term is coming with a negative sign K grad T which is also written as rho D CP grad T so we now have this subtracted from the second term you can clearly see that this gets cancelled with this right away okay and this appears with a negative sign along with the first term and of course the last term on the right hand side remains as it is so we now can assemble all these different pieces of the energy equation so therefore the energy equation becomes gradient sorry divergence of rho V vector integral T not to T CP DT that is exactly the first term we need to borrow only this term for the second term the other one is going to get cancelled with the third term so we have minus rho D gradient integral T not to T CP DT equal to minus sigma K equals 1 to N whatever we had on the left and right hand side there we will just copy here so delta H of K not WK all right what is this like so the last 5 minutes 5 10 minutes they are gone through like huh but what have you got finally is it something that we can think about yeah this is the mixture average velocity right this is the sensible enthalpy of the mixture so this basically means this is the convection of the mixture enthalpy all right and this you can clearly see is divergence of gradient of something right with a rho D sticking in there so that is actually the diffusion term it is effectively coming from a combination of species convection of enthalpy and the conduction okay so this is acting like the diffusion term and of course that is a chemical reaction rate term so we now see that there is there are three effects that are now coming together convection of the mixture diffusion and chemical reaction so whatever we talked about right at the beginning of the course okay where we are now talking about a interplay of three processes convection diffusion reaction so after all the shake up that has happened we finally come down to only these three right so that is that is the physics that is involved in in this in this equation we call this the Schwab-Zeldovich energy equation right so we also want to work with the species equation right so let us get back to how the species equation looks like equation B so equation B is now divergence of rho yk v plus vk equals wk we do not like capital vk we do not want to look at capital vk in fact what we decided here was we wanted to throw out capital vk that is how we got this rho D grad work that grad yk sigma right so that is coming from how the fixed law looks like so we want to be able to now substitute vk is equal to minus D grad yk divided by yk or in other words yk vk is equal to minus D grad grad yk if I if you are able to now substitute it over here right then what happens divergence rho v yk plus also the negative sign minus D grad yk right that is what you are going to get so let us try to write that out species equation B becomes applying the simplified form of multi component diffusion equation right we simply get divergence of rho v yk minus rho D gradient yk equals wi this is what we want to call the Schwab-Zeldovich species equation right so here again you can see more clearly this is convection of species this is diffusion of species this is reaction of species right so in species conservation also we are having convection diffusion reaction three terms the other interesting thing is this is rho v times something yk in this case and minus rho D grad that that thing which is yk in this case here we have rho v times something which is a little bit more complicated than that integral t0 to Tcp dt and minus rho D grad is the same thing which is integral t0 to Tcp dt and what did we want we wanted our left hand side to look like some sort of an operator that is operating upon a quantity like alpha i or alpha t we are now beginning to think we are now beginning to see the contours of that both the left hand sides of the Schwab-Zeldovich energy and species equations are actually the on the same operators operating upon two different quantities in this case yk the other one is integral t0 to Tcp dt right so we are we are beginning to achieve achieve this goal but what did what was our original goal our original goal was we wanted to have a operator script L of alpha i is equal to omega and operator script L of alpha t is equal to omega that means we wanted the right hand sides to be the same okay so the case you now have a wk over here right and you have a you have the wk within a summation over all k how is that how do we deal with this so this is actually this is our first of all this is n equations k equals 1 to n right that is one equation summing over all wk weighted by the ? hf if not k so how do you deal with this situation so you now try to form you say you and you are not quite at the alphas yet we have identified yk and integral t0 to Tcp dt as potential candidates for our alpha k and alpha t but the right hand sides are looking good right so let us do something here so for a chemical reaction of the form sigma i equals I can say you can keep it k k equals 1 to n new k single prime script mk gives sigma k equals 1 to n new k double prime script mk that is a single step reaction okay so you are now beginning to stop start thinking about why single step reaction as opposed to a multi-step reaction scheme we may write we may write omega equal to wi divided by cap or should say wk wk equal to divided by capital Wk new k double prime minus new k single prime k equals 1 to n this is how we defined our omega which is the molar rate of reaction okay the fact that we are now bringing in moles is because we are now dividing by the molecular weight capital Wk now this can be done obviously only for a single step reaction if you had multi-step reaction then you had many omegas themselves it is not possible for you to write a single omega right so a single omega can be written only for a single step chemical reaction that is where we have to bring in the single step chemical reaction assumption and what you are essentially saying is let me now try to look for some something that is common for all wk so that I can just have that alone and pull everything else to the right and left hand side and keep only the omega on the right hand side I can easily achieve that with the species conservation equation because I have this definition that relates the wk's to the omega the single chemical reaction rate. So if you know so what do you do if you now were to divide your equation your k species conservation equation by capital Wk times nu k double prime minus nu k single prime right the right hand side becomes omega the left hand side you do not really have a yk in itself you have a yk divided by capital Wk times nu k double prime minus nu k single prime right that is your alpha k so then let alpha k equal to yk divided by wk nu k double prime minus nu k single prime good can you also substitute wk equal to omega times capital Wk nu k double prime minus nu k single prime okay can we now substitute that over here if you did that then you are going to get this term is sigma k equals 1 to n delta h of k not capital Wk times nu k double prime minus nu k single prime times omega omega does not have any subscript k therefore I can pull that out of the summation and keep it as a factor and whatever is there in the summation the summed up quantity I would now try to divide that throughout if I divided this equation throughout by the summed up quantity leaving only the omega on the right hand side I have only omega on the right hand side and I am going to get omega on the right hand side for here as well so now I am beginning to look at a alpha t so alpha t is originally we thought our candidate was integral t0 to t Cp dt but now it is actually divided by sigma k equals 1 to n delta h of not k capital Wk nu k double prime minus nu k single prime so what if I use these alphas now in the Schwab-Zelovich energy and species equations then we can write L of alpha equal to omega right where alpha equal to alpha k k equals 1 to n gives species equation I should say Sz species equation and alpha equals alpha t gives Sz energy equation so what is L for us and here L effectively is L of alpha is divergence of rho v vector alpha minus rho d gradient alpha right so there is divergence rho v alpha minus rho d gradient alpha equals omega and this is convection this is diffusion this is reaction we have pretty much achieved our goal right all we have to do now is to notice that this is a linear operator so it is possible for us to now subtract one alpha from the other and the right hand side is the same for all these equations therefore you should get a 0 there right so notice L of alpha is a linear operator okay so the non-linear chemical source term can be eliminated in all but one equation by subtracting that equation from all other equations right then we have we have L of alpha 1 equal to omega and L of beta equal to 0 okay where beta equal to alpha t minus alpha 1 so let us call this beta t and beta can also be equal to alpha k minus alpha 1 let us call this beta k k equals 2 to n right so the beta is then or what is called as a coupling terms right so you now have linear homogeneous n minus wait a minute this is this is one equation this is actually again n minus 1 plus 1 originally we had n plus 1 equations n species equations and 1 energy equation right and if you now subtract all equations so one equation from all other equations and keep this one equation you should still have n plus 1 equations but n of those equations is now have now become linear have become homogeneous right it is only one equation that continues to have the inhomogeneity therefore you now have a you have now changed your system of equations into n linear homogeneous equations and one linear inhomogeneous equation alright so that is what you achieve by doing the Schwab-Zelovich or adopting the Schwab-Zelovich formulation this is the good point to stop we will start thinking about what to do with the formulation as we proceed.