 So let us now start doing what is called as Rankine-Hugonio relations, so you are not talking about the relation that Rankine has with Hugonio, but you must have come across Hugonio when you are doing gas dynamics for example, so you would have come across Hugonio, so what did we do at the time when you if you did a gas dynamics course, what you typically do is you are looking for what is called as jump condition across shocks right. So you want to now treat a shock as a discontinuity and then want to conserve equations across this saying so much mass came in on this side and it should be equal to so much mass that side and momentum here equals momentum there and so on, we try to do something similar except that we want to now consider a reaction zone instead of a shock necessarily the reason why we want to do that obviously is we find that combustion is too complicated right, so is it possible for us to actually live without combustion right or without having to worry about combustion without having to solve combustion problems is it possible for us to actually deduce some gross features for the flow of reactant gases into a reaction zone by looking at how what comes out of it in relation to what went in right without having to actually look at what happens inside, so we want to now treat the region where the reactions are happening like a black box and then simply want to look at what is going in and what is coming out right and see if we can learn something about that. So the justification for this is that most reactions happen in a very thin zone and so it is possible for us to actually you do not necessarily want to say that we are adopting this is like a surface of discontinuity we do not want to really have like a jump condition across a surface but it is jump condition across a region, so it could be a little thick alright but that is okay but you know if it is going to fill up the room then you are not going to be able to do this, so obviously you want to have it confined to some region which across which you can apply jump conditions, so we notice that most reactions take place in a region, so we can treat these regions as discontinuities, discontinuities between conditions corresponding to what we would call as reactants in one side and products on the other side right, so as a simple case of course let us consider a 1D steady situation right, so consider a steady 1D flow right and take a flame fixed coordinate system that means you are now going to say if you now had a flame that is occurring over here we are simply going to say that whatever is coming in is coming in at a velocity u0 and what is going out is going at a velocity u0 and this has conditions rho0 p0 t0 yi0 and this has conditions rho infinity p infinity t infinity yi infinity right I go I goes from 1 to n for n species right, so what that means is so how did you draw this boundary how do you now decide that what is outside of this region outside of this region outside of this reaction zone outside the reaction zone on either side no gradients are present no gradients in any quantity right is present that means all the variations are happening only here that means you started out with a uniform rho0 p0 t0 and yi0 all the jumps that are happening right are happening only within this and then you are now ending up with a t infinity p infinity rho infinity and yi infinity on the outside which is which is now uniform on the other side, so all the variations are happening there what that means is this is a this is now a essentially a stationary combustion wave stationary combustion wave involving not only reactions but also all the heat transfer viscous effects diffusion everything right because all the all the diffusion heat transfer viscous effects all of them base depend on gradients right and then it is all the gradients are contained in this, so diffusion viscous and heat transfer, so maybe I should just write here that is okay, so that is what is going on within this very small region is it okay can we can we do this we have a feel for what is going on if we were to deal with gas dynamic shocks and treat that as a discontinuity we will find that the shocks are only about a few mean free path stick and that is very very very thin strictly speaking for few is less than 10 mean free path stick over here it is not going to be so bad or so so so thin right, so if you now think about like a flame like in a Bunsen burner you are expecting it to be less than a millimeter thick or around that okay some conditions it could be around about a millimeter or so that is that is still quite small quite then right that is that is for a flame is it possible for us to now think about like a flame that happens along with a shock wave right like reactions happening along with the shock wave or in other words whatever shocks that you have come across were because they were actually propagating in non-reacting unit mixtures what happens if you now were thinking about shocks propagating in reacting mixtures what will happen we have reactions that happen along with it right, so is it possible for us to deduce those when will we have what kind of a flame okay but in in in general what you are basically thinking is whatever we are thinking about it is reasonably thin it is okay you can think about this is a little black box it is quite thin and on either side and all the gradients are in there and on either side you have uniform flow of reactants in one side and products on the other side okay. Now if I if I have this picture this is obviously a flame fixed coordinate system if you want to have like a lap fixed coordinate system what do you think will happen, so we now have this you not coming in here and trying to make this stop right that is because we are we are we are traveling with the shock, so if you had a still set of reactants and the shock actually tries to go I should say shock strictly speaking could be a flame just a reaction zone right, so this this travels like this that means the reaction zone is eating into the fresh reactants maybe still fresh still reactants right, so this is strictly speaking supposed to go like this but I now find out mark my words now okay I now find out at what speed this goes and then try to oppose that with an equal amount of velocity for the reactants to stabilize this to stay there okay and then it will stay otherwise I let it travel with it that means I should know what at what velocity this goes so that I can travel along with that to make it stationary with respect to me, so there is a little problem and in terms of like having to know what this velocity is or shall we get it with this set of equations that is not true what we are really hoping for is if I were to give these I can get these but in this problem I do not even know this strictly speaking keep that in mind okay that is going to come come back and haunt us little later. So our goal is to obtain product conditions given reactant conditions right, so question is are all of that known okay specifically you know like I have this question with me and I have to seek an answer as we go along but let us suppose at the moment I know what this you notice and see what happens, so how are you going to obtain this these product conditions obviously you have these conservation equations that you want to integrate across the shock or the reaction zone, so mass continuity or mass conservation can be integrated right across this discontinuity to give you can integrate this and then find out that all it matters is we have to substitute the n conditions starting from one side to another side if you now travel X this way for example then you get ? not equal ? not U not equals ? not ? infinity U infinity you are so familiar with this that you do not even question how this in how it how we did the integration and got this right, so I am just going to take liberties and say this is how this is how I got it okay, but then if so this is actually the incoming mass flux and we are now saying that is outgoing mass flux and they are equal, so if they are equal then can I now say both of them are equal to some common m dot that is like a mass flux in a system, so the mass flux kind of tells you the size of the or the strength of the system okay, higher the mass flux stronger is your propagation faster is your propagation, so let us call this equation 1 today, so momentum conservation, so we have ? not U not square plus well actually strictly speaking we should be able to integrate and say ? U square plus P minus 4 by 3 µ du by dx equal to a constant, but then we also notice that we want to integrate this, so across this region you do not have this term because you do not have any derivatives when you are now trying to apply this to a left hand side or right hand side, so this can be written as ? not U not square plus P not equals ? infinity U infinity square plus P infinity right, so let us call this to this is also something that you are familiar with, but I should take a little bit credit to say that we did not get this, this is true not just for inviscid flows this is true even for viscous flows okay, so we could actually keep this and try to evaluate it outside the discontinuity and find the du by dx is equal to 0, so it is like you can have viscosity all right, but so long as you do not have velocity gradients you do not have a viscous effect right, so the viscous effects are negligible on either side of the discontinuity and therefore you get it to behave like an inviscid flow that is how inviscid assumptions typically work right. Now we want to know how do you want to do this, keep in mind what we want to do is our goal is to obtain product conditions given reactant conditions all right and the way we want to do this is we want to deal with that in some sort of a thermodynamic fashion that is if I were to give you two thermodynamic properties that are representing the state of the system when they were reactants is it possible for me to actually find out corresponding thermodynamic properties for the state of the system when they have become products right, so it is sort of like a global approach and we are putting a black box around we are only worried about what happens between when we started out and when we ended we are not really worried about how this change happened right that is very symbolic or symptomatic of thermodynamics therefore I should be interested in describing the state the initial state of the system thermodynamically and correspondingly look for obtaining the corresponding thermodynamic quantities for this that describe the final state that is the products well I started with mass conservation I had rho 0 and u 0 rho is certainly a thermodynamic quantity but u is not that is a flow quantity right and similarly I can say rho 0 u 0 square plus p 0 rho and pr but u 0 is not that is a flow quantity, so is it possible for me to somehow camouflage the flow and then deal try to deal with only the thermodynamic quantities the answer is so long as I am going to use m dot okay and treat it as a given and keep in mind this is related to u 0 right and then we cannot quite treat it as a given okay we do not know exactly so there is a little problem but so long as I am going to couch my u 0 or u infinity either of them as a matter of fact into m dot and it is so nicely it is so nice that both of them can actually get hidden into m dot okay this equation will permit us to do that. So I am going to use m dot whenever I try to get a velocity right and then I am going to write like u 0 as m dot divided by rho 0 u infinity as m dot divided by rho infinity so I try to do that then so one implies rho 0 rho infinity u infinity squared minus rho 0 u 0 squared as we can write this as m dot squared times 1 over rho infinity minus 1 over rho 0 and that would be equal to so what am I doing I took this rho 0 u 0 squared to the right hand side so I had this difference so I have to take this p infinity to the left hand side and I have a p 0 minus p infinity right so from this I can write p infinity minus p 0 divided by 1 over rho infinity minus 1 over rho 0 equal to minus m dot squared okay it is very interesting that this analysis is amazing in my opinion okay what you are looking for is to go back and look at the goal again we are looking for product conditions given the reactant conditions and we just decided that by conditions we will actually talk about thermodynamic quantities right and then we just went through these and then found that rho and p or the thermodynamic quantities so if I were to be able to give rho 0 and p 0 I should be able to get p infinity and rho infinity this is what I am trying to look for all right and in trying to get this equation I have already used up the two equations okay the mass conservation and momentum conservation what am I getting out of this so if I were to plot and this is the beginning of some lot of fun that you are going to have in the coming days right if I am if I am not going to plot p infinity on the y axis and 1 over rho infinity on the x axis now keep in mind 1 over rho means specific volume so this is like a PV plane right and if I were to now locate numerically the value of 1 over rho 0 and let us say p 0 all right that is to say I am starting with a modest pressure for the reactants yeah and a modest value of 1 over rho 0 that means like a fairly high density okay that is what I am starting out with and this is my initial condition I am trying to locate for this initial condition where the final condition should be and this equation is basically telling me that I can look for my final condition anywhere along a straight line okay which connects this point to the final condition with a slope equal to what minus m dot square right now m dot is mass flux there cannot be negative strictly speaking okay it can be negative only if I if I had the flow going back backwards like this so I am not really thinking about reactants flowing out of the reaction zone I am looking at the reactions always reactants always feeding into the reaction zone so I am thinking m dot is positive that is okay but m dot squared should obviously be positive so that is not even worry about it right and therefore negative m dot squared is going to be negative so the slope of this line is going to be negative so I am going to be looking for a line that always is having a negative slope you know that never happens you want to draw a line that passes through a point and always goes a little bit of way away from that point okay so we now push the point to wherever the line goes and then say this is the line along which I am expecting the final solution to lie right any point along this line is where the final solution should lie we have made remarkable progress in the last few minutes what did we do we now said given the initial conditions describe thermodynamically by two quantities right we have now zoomed in on the possible solutions to lie along only one line as opposed to a huge plane right until now we could have actually expected the solution to lie anywhere in this plane but now it is all confined to just one line that is amazing right we have reduced a plane to a possible set of points in a plane to a possible set of points in a line that is nice but then there are infinite number of points in a line just as well as there are infinite number of points in a plane so what is it is all infinite number of possibilities right so how do I now pick exactly that solution that that is going to be lying on this line I need at least some more information I should now be looking for this line to intersect with some other line maybe or there must be some way by which I can choose a particular point along this line so we have exhausted mass conservation and momentum conservation together to now get a line we now are looking for a energy conservation to help us with trying to locate a point along this line right so this is like a very step by step thing from a plane to a line from a line to a point is what you are going so this line is what is called as the Rayleigh line and the slope of the Rayleigh line is essentially m dot squared I am not saying negative end of end of m dot squared because I have already taken the negative sign and account and drawing the line that is inclined from left top to right bottom okay so the slope of the line contains so what does m dot signify for us it is actually the flow the flow information is hidden in m dot and the slope of the line indicates how fast it is right if the if the line were very shallow that means it is m dot is quite low if the line were steep m dot is high all right so that is going to tell us how fast the flow is or how slow the flow is in turn it is going to tell us how fast the wave would propagate into still reactants or slow okay but there is something that we do not know okay so keep that in mind so now you look at the energy equation to get initial the energy equation the energy equation reduces to H0 plus half U0 squared equals H infinity plus half U infinity squared let us call this 3 now of course you all is you now have to start thinking about what these H0 and H H infinities are so notice note that note that P0 equals R0 R0 T0 T infinity equals R0 infinity R infinity T infinity these are specific gas constants and these are going to be specific to the reactant mixture for R0 and the product mixture for our infinity right so which means we should know what is the molecular weight of the mean molecular weight of the reactants and mean molecular weights of the products which means we need to know the composition of the reactants and composition of the products which means we know why I why I not but we have to actually get why I infinity right so how do you do that we have to integrate the species conservation equation you try to try to do that so trying to relate so well well well well let us just I do not do that immediately so before before we we do that let us let us just say know again also H0 is equal to sigma I equals 1 to n why I not H I not where H I not equals ? HFI superscript not plus CPI T not assuming calorically perfect gas if not you have to write integral T superscript not to T CPI dt okay T superscript not to T subscript not okay CPI dt is what you should write and then similarly H infinity is sigma I equals 1 to n why I infinity H I infinity H I infinity equals ? HFI superscript not plus CPI T infinity okay so in both the cases we are assuming calorically perfect gas here to simplify things for us ourselves here you can see that the fate of H infinity not only lies with T knowing T infinity given T not but also in knowing why I infinity is given why I not okay and the reason why I was talking about why I infinities in relation to why I not was to get the mean molecular weights for the reactants and the products and that is because we wanted to relate the T nots to P nots and rho nots that is what we are comfortable with okay we are comfortable with P nots and rho nots or P infinity and rho infinity we wanted to write T not and T infinity in terms of the pressures and densities rather than keep them as temperatures for which we wanted to use our infinity and our not and that involves molecular weights and involves why I is and so on that is what we were talking about but we also find that HI's require YI's for forming your H nots and H infinity so we will try to relate trying to relate why I infinity to why I not right how do you do this so let us now try to try to integrate the species conservation equation across the discontinuity right so if you now try to equate the species conservation then the species conservation the species conservation equation leads to notice what we had we had we had for the species conservation the convection term and the diffusion term and the reaction term okay the convection and the diffusion terms involve derivatives so they do not really exist here and what what really it boils down to is to say that simply your w I not equal to 0 and wi infinity equal to 0 now you could recall that you can say w I is sigma k x k equals 1 to m w I k therefore or yeah the w w I k and w I k equals capital W I nu I k double prime minus nu I k single prime omega k so what this what this means is so if you now try to take these two together right then this simply means what you are saying is omega k not equal to 0 and omega k infinity equal to 0 this is what it is going to of course you can say that for both of these you can say k equals 1 to m for each reaction right this is for a multi-step reaction now here we have a little problem which is this is alright this is more or less alright because your reactions at the end of the reaction zone are going to come to completion that is how you actually figure out that the reaction zone is over right and how did they come to completion because you have actually consumed the reactants in those reactions so if you now look at reactant profiles they would actually go to 0 or at least one of the reactants will go to 0 and if the reactant concentrations were present in these terms right through the law of mass action you know that the reactant concept one of the reactant concentrations goes to 0 that means if it is if it is come if either of the reactants that are happening in reaction goes to 0 concentration then the reaction is complete is what you are basically thinking about but here as far as this is concerned omega k not the question is is it 0 one of the words are you having reaction rates equal to 0 here I thought so because I thought all these this is the region that contains all the reactions right this is this is region that is not supposed to contain any reactions is that right we have all the reactant concentrations yeah and then we also have a temperature T not at which they can react so strictly speaking there is nothing that is really stopping reactants at like let us say a reactants temperature let us say room temperature there is nothing that stopping them from reacting it is just that they have to circumvent the activation energy and it is take a long time so the reaction rates are very low but it is not necessarily 0 so this is going to be some very low number maybe 10 to the minus 14 but it is not 0 right so while we are now thinking that you do not have any gradients over here that is because we are not really taking into account the heat conduction that is happening because of chemical reaction so there is like a mild gradients possibly but we are we are disregarding that saying that there is like a line that we are drawing and saying no gradients on the other side simply leave you similarly we are also drawing a line saying no reactions there but that is not going to numerically work out so this is this is what is termed as in most combustion problems this is what is called as the cold boundary difficulty. In other words if you want to supply an upstream boundary condition on the cold reactants in most combustion wave problems you will find that it is not really strictly true to say that it is not reacting at all all right but then we suppose and this is this is primarily because of the way the Arrhenius law works the Arrhenius law is essentially an exponential dependence on temperature for the reaction rate constant and that that is not really going to 0 identically as you go to room temperatures of the reactant temperatures it still exists at a finite non-zero infinitesimal may be value so we circumvent this in some sort of an ad hoc manner okay we do not we do not do this perfect way of integrating things across the jump and all that stuff for the species conservation we could do this for the mass conservation we could do this for the momentum conservation we could do this for the energy conservation but the species conservation is not going to actually lead to this the reason is this all these are actually rate equations okay so this is actually a mass flow rate this is the momentum flux rate and this is a energy rate and similarly this is a species flow rate is what the species conservation equations is all about but what we are primarily looking for is only the composition we do not we are not really worried about the rate of change of composition or the rate of flux of composition and so on so we could actually fall back to thermodynamics instead of looking at rate equations like species conservation equation in this case so typically we sort of binker rise in this do not worry about this and then resort to thermodynamics and in thermodynamics we have seen how the product concentrations actually can be worked out given the reactant concentrations but the caveat there is we are assuming equilibrium right whereas in all these rate processes that we are doing we are not necessarily assuming equilibrium that means we are allowing for changes to happen and it manifests in saying reactants themselves are undergoing change even far upstream of where you thought with the reaction zone right so strictly speaking it says no no equilibrium so we we we we binker rise on that we just say do not worry about it we will assume equilibrium we can try to get our via infinities from an equilibrium condition and recall what we did when we when we did that we started out with atom conservation on either side that is like balancing the chemical reactions for additional equations that we needed for extra unknowns on product products we assumed hypothetical partial formation equilibria and so on so we had ways by which you could you could enumerate the via infinities assuming equilibria so that that is what we do so we we overcome this by supposing equilibrium condition for the products and obtain why I infinity given why not under that framework okay already already done before okay so let us suppose that we let us all make a big fuss about this we know how to do this we can proceed on dealing with the energy equation so we have the energy equation integrated across the jump and we call this three and of course there are there are as well we have you not squared and we do not like to see you not we want to replace it with m dot right so we use 1 in 3 which means H infinity – H not equals – ½ m dot squared 1 over rho infinity squared – 1 over rho not squared because we can we can write you not squared as m dot squared divided by rho not squared and then you can also use you now notice that you have 1 over rho infinity squared – 1 over rho not squared can be written as 1 over rho infinity plus 1 over rho not times 1 over rho infinity – 1 over rho not and then you can find out if now you say this taken over there – m dot squared times 1 over – rho infinity – 1 over rho not is p infinity – p not so you could try to use that so H infinity – H not is equal to ½ 1 over rho infinity plus 1 over rho not times p infinity – p not this is good because this completely gets it gets rid of m dot right previously we had this Rayleigh line that was that could actually change its slope depending upon the m dot but now we have a truly-truly thermodynamic relationship no flow information at all that is great because this is actually related to thermodynamics in some sense okay if you start out with set an energy content in your flow in terms of a mixture of heat of formation and sensible enthalpy right you will get your reactants your products to now have so much enthalpy which will have another combination of heat of formation and sensible enthalpy right that is essentially what this means so we can now further say well let us now write this as in fact this is this before we do that let us now in fact this is what is called as the Rankine-Hugonew relation let us just try to do this a little bit more next couple of steps is going to be a bit clumsy but let us just bear with it and see what happens so if you now write your H infinity as this and H not as why I why I not H I not and then you have to write this as sigma i equals 1 to n why I infinity delta Hfi not plus sigma i equals 1 to n why I infinity Cp infinity T infinity this should be Cpi minus sigma i equals 1 to n why not delta Hfi superscript not plus should say negative again sigma i equals 1 to n why not Cpi T not now let us call what happens now so let us now call this as our H infinity not and let us call this is H not not why would you call this H infinity not and H not not because they are actually standard heats of formation of species I weighted by the product composition here this is weighted by the reactant composition so this superscript not in both the cases actually indicates standard heats standard conditions and the the infinities and infinity and not basically means you are waiting with respect to product composition and reactant composition respectively so this is equal to H infinity T not minus H not not plus Cp infinity T infinity minus Cp not T not why would you say Cp infinity because T infinity can be pulled out of the summation so sigma yi infinity Cpi is Cp infinity right here you are going to get so this is Cp infinity and this is Cp not so it is essentially the compositions that are desiring these terms rather than the the heats of form a standard heats of formation of the temperature there so this is what this is what you started out with the left hand side the right hand side can be written now so this is equal to one half P infinity minus P not times 1 over rho infinity plus 1 over rho not right okay now I told you we do not we do not like T's temperatures we want to write this thing write things in terms of pressures and densities so we use our equations of state applied to the N conditions and if you do that and in one more step we will hit a point that is ready for us to draw a curve that will intersect with the Rayleigh line and try to get us a solution so ? infinity R infinity T infinity divided by ? infinity minus 1 so Cp is being written as ? R by ? minus 1 basically minus ? not R not T not divided by ? not minus 1 minus one half 1 over rho infinity plus 1 over rho not times P infinity minus P not so you try to take this to the right hand side to the left hand side and this term you now take to the other side with a negative sign so we now call this equal to minus H infinity not minus H not not and we want to call this Q right so the negative of the differences in the standard heats of formation weighted by the product composition and the standard heats of formation weighted by the reactant composition is essentially what we are saying as the heat release due to chemical reactions in this right so this is exactly what we talked about previously as well so that is the heats of formation the difference between those is the one that is actually giving rise to the chemical heat release so we will we will just write the next step and stop for the day so if you now write your R R infinity T infinity as P infinity divided by rho infinity because that is what we are looking for so P infinity divided by rho infinity minus ? not divided by ? not minus 1 P not divided by ? not minus course keep this equal to Q this is now clean because it is now coming back to within pressure and density we are not really having any temperatures anymore now for just a simple simplify if ? infinity equal to ? not this is not very bad because ? is a typically about 1.3 1.4 1.2 whatever so we get ? by ? minus 1 and P infinity minus sorry divided by rho infinity minus P not divided by ? not minus half 1 over rho infinity plus 1 over rho not times P infinity minus P not equal to Q we will just start from exactly this point when we meet again this is the equation 5 and what I want you to think about is how is this curve going to look like in this plane that means given your P not and 1 over rho not look at this curve to find out how in this plane of P infinity to 1 over rho infinity this curve going to is going to look like have a good weekend.