 So we were looking at ignition and then we were looking at ignition as a somewhat like a reverse process of quenching not exactly but you look at the numbers to be slightly different but the processes basically have the same idea therefore we could now think about again like a parallel plates what I was saying yesterday is instead of having like a global energy balance we could actually have a local energy balance so let us look at a different approach to do this to probably get the same results again so or similar results so consider a gas between two infinite parallel plates and now what we are saying is at ignition that is one of the criteria that we listed so at ignition we need to have D by DX of – K DT by DX equal to WQ where this is actually the should produce a different symbol here may be the QR so this is the chemical heat release or the heat of reaction and therefore you could say D square T over DX square which is this is the conduction heat loss that is Q small Q and that is equal to the chemical reaction rate right so that is equal to so if you now assume constant K then you are going to get – QR divided by K we can now write this a W dot as AE into the – E over RT YF power P YO power Q right keep in mind that this Q is not the same as that so we run out of symbols pretty soon because we use kind of pretty much the same kind of symbols most of the time it could say M and N but that is fine now the reason why we have to write the W dot like this is because this is a differential equation in temperature and the temperature shows up here so it means when you now integrate this strictly speaking we should think about like bringing this down here and then taking it up there and integrating and so on now that is of course an involved process so let us actually try to see how we can manipulate this a little bit more so introduce non-dimensionalization eta equals X over capital R that is non-dimensionalizing the length and now let us suppose that nu stands for T – T0 that is a departure from the wall temperature for the for the gas and let us say theta equals E times T – T0 divided by RT0 square this is not the same as our Zelovich factor because we used to have a TF keep in mind and so obviously then this can be written as E nu divided by RT0 square now note that we were saying nu is quite small when compared to T0 that means we are looking at small departures from T0 for the temperature near the wall and if you have this notion then e to the – E over RT can be expanded about T0 as exponential of negative E over R T0 plus nu and that can be written as exponential of – E over RT0 times 1 plus nu over T0 and then you do a binomial approximation so this would be exponential of E over RT0 1 – actually should have a parentheses here 1 – nu over T0 and this should be equal to e to the capital nu over RT0 square times e to the – E over RT0 right and that means this is equal to e to the theta times e to the – E over RT0 okay. Now let us say let delta equal to qr over k e over RT0 square r square well maybe again I am mixing up symbols so we should go back and write this as R u and R u, R u you have plenty of places where we need to make this correction so and qr over k e over R u T0 square capital R square a e to the – e over R u T0 y, yf power p yo power q so just in case maybe we can just go back and put q dots there okay so if you now do this go back to your original equation so this is where we were and we did all these things and so finally what you are going to get is we get d square theta over d eta square equal to – delta E theta that is really what I like about these non-dimensionalizations you know kind of like you have been to a haircutting saloon and got yourself a nice haircut so the governing equation now becomes so looking compact right so for the boundary conditions we have x equals in our new coordinates eta equal to 0 we have to suppose that dT over dx equal to 0 by symmetry and x equals R which implies eta equals 1 we have to have T equals T0 if T is equal to T not then nu equal to 0 if nu is equal to 0 then theta is equal to 0 right theta equal to 0 now let us now do something like a linear linearization so take e to the theta as 1 plus theta that means you do a Taylor expansion and chop off things that are going a square and so on so if you do this then that is pretty straightforward what this means is e to the d square theta divided by e to d eta square plus delta theta equals – delta so you just go back and plug e to the t theta as 1 plus theta in this and expand out and take the term with the theta to the left hand side so that you have a differential equation that looks lot more familiar when compared to what we are used what we have been having and therefore theta equals let us say a1 sin square root of delta eta plus a2 cosine square root of delta eta – 1 right so you have the complementary part and the particular integral there and therefore you get d theta over dx from here as square root of delta times a1 cosine square root of delta eta – a2 sin square root of delta eta so if you now supply the boundary conditions the symmetry boundary condition implies that you now plug your eta equal to 0 and the sin term goes away and then you now say that d theta by dx should be equal to 0 which is the same as dt by dx equal to 0 that means that the cosine is actually going to 1 therefore a1 should be equal to 0 and at the wall theta equal to 0 this means 0 equal to a2 cosine is coming from there cosine delta cosine delta – 1 right so from here we get a2 equal to 1 over cosine square root delta and theta now becomes cosine square root delta eta divided by cosine square root delta – 1 right so with this what this means is we need look at what is theta s theta is t – t0 times e over this right so for the solution what we are looking for is ignition that means we need to have some ignition energy that is there that is going to raise this temperature from t0 to whatever value that will create the reactions to happen significantly therefore obviously t has to be greater than t0 so theta has to have positive solutions right so one of the things that we have to look for is of course eta is a variable so that is that is varying and you have to make sure that you have to have this one for the range of ethos that we are looking at between 0 and 1 should be greater than 1 right but more than that what we have to worry about is so we need theta greater than 1 everywhere for ignition more than that we need theta to be bounded okay it is one thing to actually have a sorry greater than 0 right so more than theta just being positive we also want to make sure that theta does not really blow up right so also also the solution becomes unbounded unbounded for cosine square root delta going to 0 so if the denominator goes to 0 in even in the first term then you are in trouble which implies that square root of delta is pi by 2 right or you can say delta is equal to pi square by 4 right so if this is the case then you do not have a steady state solution so no steady state solution no steady state solution exists exists for this value and above now unlike problems that have this kind of a governing equation at least for the homogeneous part where we would be looking for oscillatory situation and eigenvalues having multiple values right so obviously you can say that cosine square root of delta is it going to be is going to be 0 when delta is equal to pi by 2 3 pi by 2 5 pi by 2 and so on but that is not what we are interested in right so as you now have the delta actually keep on increasing you now reach this value and that is bad enough we do not have to worry about the next value it is going to reach the next value those things are not basically basically physical anymore what we are interested in is like let us say the first value why are we interested in that because delta for you what we are looking for is the capital R squared as your as your size so long as your size is less than a certain critical value you are now going to have more loss when compared to the heating by the chemical reactions and at ignition what we are saying is this equality happens so this equality begins to happen for a certain minimum critical value of capital R over here and therefore the corresponding delta beyond which ignition will happen and again this instead this equality will not hold good because you will now have more heat released to the chemical reactions than the conduction at the walls and then the reaction grows and then you now have an ignition right before the critical value of capital R or below the critical value of capital R and correspondingly a less value of less than the critical value of delta you are going to have the heat loss to be greater than the chemical reaction therefore ignition won't happen right therefore what we are basically looking for is this implies that delta critical has a minimum value for ignition right now if you go back and look at this delta and let us not worry about how this delta is really formed let us now begin to look at the dependencies so looking at at dependencies of delta delta crit right and this is always something that we have been thinking about we always have to have or crit square and since you have these dependencies here that is going to now have a pressure dependence show up through the concentrations so you are going to have a people in show up here and of course we have a e to the e to the minus e over r u t0 showing up t0 and you have a t0 square at the bottom right t0 square now this should be depends on so let us say equal to equal to a constant right now the constant here that we are talking about is 5 by 2 in this analysis and as I said let us not even worry about the next values and so on but in reality this constant is empirical it is not it is not exactly this that is because we are losing a lot of things first of all you are kind of assuming a momentary steady state here that this this equals that just this particular point but they all say it is a dynamic process you have an unsteady process essentially and then we are not really worried about the variation in the temperature with time when when this ignition is happening so we assumed a steady state process and then looked at when it is violated so this this formulation is somewhat limited in the way it is done but the most important thing that we are looking for is the dependencies on pressure and temperature the pressure of the system and the initial temperature of the reactants that is what we are looking at and therefore we simply say this should be equal to a constant or if you now have this this greater than a constant then you will have ignition or if you want to avoid ignition like for example you want to now say I have a very narrow tube and they have reactant mixture that is being sent through the tube and I want to make sure that I do not get ignited then the the or critical should be less than this constant value right so that is typically obtained empirically the actual value of this so here what we can see is that our crit then squared goes as 1 over Pn Pp so repeat of the N keep that in mind and also you now have a slightly complicated the T0 dependence okay now let us also now look at what is the situation if you now have like a spark ignition okay so what we have done is something like a thermal ignition where we are trying to heat up otherwise so but spark ignition is slightly different you need to have electrical energy put in there and spark ignition is something that is very commonly adopted in many ignition situations of practical in practical systems so our goal here is two fold one is looking at something like a or critical similar to this and I will show you that a similar dependence can be obtained by looking at the global energy balance similar to the way we were doing for the for the quenching so find minimum ignition energy right so what we are looking for is if you now have a spark that is coming up and you now have a flame that is now growing from here spherically locally spherically that is how it is going to happen so strictly speaking we should be looking at a is a critical radius for the spherical flame that well that will sustain growing propagating instead of giving away the heat that is liberated during this ignition to the surroundings so if you really think about this let us suppose that as I said long ago you fill this room with reactor mixture okay and then you pray that you are not going it is not going to get ignited and what happens when it gets ignited right or why is it so is reaction going on when it is not getting ignited that was a question that we are asking before the answer is according to the Arrhenius law because it is actually at some temperature the reactions are going on okay but the reaction rates are so low that the heat release rate is so low that it is just getting conducted away to the surroundings from where it is happening right so therefore what we are basically looking for is what is the energy that is being supplied by the spark that will actually ignite a critical size of this flame ball whose surface is so large now that the chemical energy release that is happening in this is sufficient to overcome the heat loss to the surroundings right so that is what we are essentially looking for so the flame will not propagate if the radius of the spherical spherically propagating flame is less than a critical value and the idea is very similar to what we have seen so far so that is at this stage QR dot equals Q L dot and that is to say W heat release so a reaction rate times heat of reaction that is heat release rate times 4 by 3 pi or crit squared sorry Q should now be equal to conduction conductivity times 4 pi or crit squared that is the surface area times Tf minus T0 divided by or crit that is to say we are now basically assuming that locally the temperature is high alright but then this gives rise to a temperature fall just outside this ball and now therefore you have a heat loss in fact if this is your T0 let us do this on this side and this is your Tf you could actually take a cylindrical column let us suppose that you now think about like a hollow cylinder that means a cylinder of wall thickness right so you have a inner radius inner diameter and outer diameter inner radius and outer radius and the temperature inside is uniformly Tf okay and now you tend the outer radius to infinity okay so it is kind of like a hole in the middle of nowhere right and then the hole is with that at Tf and far from here is actually T0 you should now be able to do a conduction problem for this particular situation and then try to find out what is the heat transfer at the wall and you should be able to show that this is what it is right so if you do that then from here of course you can so you can cancel this or crit here to from here and then this from there and get a squared so or crit squared then is like 3 k Tf minus T0 divided by W qr keep in mind previously we had a W dot which we had to expand in terms of temperature because we are using a differential equation in temperature but here we are not really writing a differential equation in temperature the values of the temperature are actually placed as valid as specific limits therefore we should not have to worry about expanding this in terms of the Arrhenius law on the law of mass action here we can just keep it this way and previously we also said that this qr is nothing but CP times Tf minus T0 so we do not worry about the Tf by T0 but keep the CP at the bottom and if you now have a k over CP that can be directly written in terms of alpha right so if you now find this orchard squared is equal to alpha similar as alpha divided by W dot similarly we also had a SL squared is equal to square root sorry SL is equal to square root of alpha W dot or SL squared is alpha W dot right so if you now think about orchard squared is equal to going as alpha divided by W dot SL squared is going as alpha W dot then you can directly get from here we can see that you can see that orchard goes as alpha over SL as before as before is during quenching right now I want to caution you that you now have a number 3 here that is coming from the volume to surface area kind of consideration but in reality we do not have this kind of number it is more than that right and I will point that out that this was related to the penetration distance yesterday when we are doing ignition so if you now say we are interested in the energy so if you will oppose if you now say that the e ignition the energy that the spark is going to deposit into this gas is supposed to go and increase the a critical mass of gas emcret to the temperature Tf-T0 right that is what this is supposed to do that this energy right and in fact if you think about this this is essentially the statement of the second criterion that we wrote yesterday as a rule of thumb and this is actually relating to the first criterion as rule of thumb so we are kind of applying both in stages and this emcret now is essentially like a like a pocket of gas of a certain critical mass and therefore it is going to be density times a critical volume and the critical volume is associated with the critical radius that we are looking at so therefore we can now plug the critical radius there so this is nothing but 4 by 3 pi R crits square I am sorry R crits cube you also have a density this is this density is for the pocket of gas that is got ignited therefore you are looking at products within this sphere right so you need to have a row infinity there and Cp Tf-T0 and so this implies that ignition let us not worry about the 4 by 3 pi let us now begin to look at the dependence and so what we see because we are now looking back again at the R crits going as alpha over SL as the dependence and therefore this depends on keep the Cp all right and write your row infinity as P over R infinity Tf right and keep the Tf-T0 all right and then look at our crit or crit is nothing but alpha over SL as a dependence therefore this is going to now go as alpha over SL the whole cubed right and in fact putting everything together in terms of how the alpha depends on pressure and the pressure difference here itself and SL pressure dependence everything together we will retrieve from here we could deduce that e ignition e ignition goes as P to the – 2 for N approximately equal to 2 this is for if you now go back here you are talking about P to the N is the order of the reaction that means if you are now looking at a global reaction with a law of mass action like this it is essentially P plus Q right so N equal to 2 is the order of the reaction if you now plug that in there for the SL dependence on pressure you should now get this taking other other pressure dependence system is expression into account and this is for typical hydrocarbons right and this actually pretty much gives a good dependence compared to experiments right so if you now think about how the experimental dependence is so experimentally it is observed that if you now try to plot your e ignition versus pressure and of course if you now do this on a log log scale right we should now get a straight line that goes downwards but what we find is typically this the data goes more like this which means this is the P to the – 2 dependence and the P to the – 2 dependence is actually fairly well satisfied in the low pressure region when compared to the high pressure you do not have this fall so that means you do you do have a critical ignition still needed means that the ignition energy needed is more than what the pressure dependence would say and of course that that implies a kinetics effect as well that means there is the reactions do not behave exactly the same way in terms of the global kinetics is concerned and the other thing obviously we are interested in is the temperature dependence and there is some data to suggest that if the temperature dependence like for example if you now say for increase in the three let us say 300 to 450 degrees C ray sorry Kelvin Kelvin range right the e ignition drops from around tens of millijoules to a few millijoules that means we are looking at numbers like about 40 millijoules to 70 millijoules at 300 K whereas when you now go to like 450 K you need only about 4 millijoules 5 millijoules kind of thing right so the temperature the initial temperature is increased the ignition energy required minimum ignition energy required is less correspondingly but you can actually work out how this exact dependence is yourself from here okay so with this we should conclude ignition and as well we would like to stop talking about premixed flames is that essentially what we have done is they spent significant amount of time talking about premixed flame deflections we have not done premixed flame detonations starting from where we did the Rankine-Hugone relations and we saw that you have one branches actually for different deflagration to the other branches detonation but essentially the idea is for premixed flames right we have we have we were always assuming that the reactants were approaching the flame in a premixed state and then we started asking the question how do you can how do you know what is the slope of the Rayleigh line and that is actually corresponding to the flame speed and so that actually turns out to be an eigenvalue of the system if you now look at how the equations pan out looking at how the preheat zone and the reaction zone is then we constructed the preheat zone temperature profile and the reaction zone temperature profile to to 0th order we just say this is a convective react diffusive zone that is a reactive diffusive zone and then we neglect the other terms that are in unimportant like the reaction zone reaction rate term in the preheat zone and the convective term in the reaction zone and construct temperature profiles match the slopes of the interface and then try to get the flame speed and then we started looking at the dependence and dependence of spins the flame speed on physical chemical properties initial conditions or the reactant conditions pressure and temperature and so on and then we looked at the flame shapes how the flame actually gets shaped over a burner and how it gets stabilized and what happens to it at the peak at the tip when you have a flame curvature and a flow divergence effect and then we started looking at flammability limits and quenching criteria and ignition criteria and so on. So all these things actually we have been talking about or primarily in the context of premixed diff like laminar deflagration right so it is time we started talking about diffusion flames and so now onwards for the rest of the time or at least for quite some time we should now be talking about diffusion flames we are still primarily in the deflagration regime that is we are looking at low subsonic Mach numbers for the for the reactant flow and therefore we have to go back and look at the momentum equation and find that the pressure is approximately a constant because we are now looking at low Mach numbers and low Knudsen numbers and therefore it just reduces to pressure it is a constant which means we can go back and adopt like the Schwab-Zehler which kind of formulation and where we have to look at the species conservation equation and the energy conservation equation with a flow field that is prescribed unless of course you want to take into account the density variation with temperature which will alter the flow field and you can now look at the flow problem loosely coupled with the combustion problem and not necessarily tightly coupled in unlike unlike in a full compressible manner. So we will now begin to start talking about diffusion flames now diffusion flames is actually kind of like the classical term for flames that are essentially non-premixed so the the modern terminology for this is simply non-premixed flames so we have to understand why we are getting these different semantic notions about this what we meant be at what we mean by a diffusion flame is here we are now saying that the reactants are just mixing at the flame right it is kind of like you are rushing into the exam hall and you are just studying for the exam right so you are really not prepared so similarly the reactants are not really prepared and what has been by prepared pre-pared that means you have to actually pre-mixed right so if the reactants are pre-mixed then they are ready to react and all they are waiting for is for a temperature rise to happen and then they are going to react right and at a steady in a steady state situation the temperature rise happens because of the flame that has already been established and so as the reactants get into the flame they get preheated and then react that's what the flame is all about that's why that's how we were talking about it but in this case you have a flame all right but the reactants are just mixing at the time they are reacting right so when that is happening then you have to start thinking about which is faster which is a faster process is it the mixing is it going to be quite fast or is it the reactions that are going to be quite fast now typically in a combustion situation what we are interested in is fast exothermic reactions right this is what we started talking about right at the beginning so we are always interested in situations where we have a very high sensitivity of the reaction rate of temperature in turn meaning a high activation energy right and that simply means that the reaction rates are going to be very very fast once you got the mixing to happen so this is a situation where the mixing is the one that is limiting the the entire set of processes in other words if the fuel and air were to mix with the fuel and oxidizer were to mix with each other then the reactions will happen rather instantaneously or immediately right so mixing is the rate limiting process in case of diffusion flames and of course the technical sounding term for mixing is diffusion so we can say diffusion is the rate limiting process and therefore the heat that is released from a diffusion flame is not necessarily limited by the chemical kinetics as in the case of premix flames so you would say there premix flames are kinetically limited in terms of heat release whereas in the case of diffusion flames you should say it is diffusion limited right so you are not going to really experience the as much heating there then premix flames is that is essentially the idea so here the mixing between the reactants or mixing of the reactants is the rate limiting process relative to relative to reaction therefore the heat release in a non premix flame is diffusion limited as opposed to as opposed to being kinetically limited in a premix flame hence the name diffusion flame for these but we will see that there is some exception here I mean we will just keep building this idea and then see where it falters okay so at the moment the idea basically is that diffusion is the rate limiting process and therefore the heat releases diffusion limited rather than or the heat can be released only to the extent but to which the reactants can mix then whether they can actually release by chemical kinetics like what is the chemical reaction rate based on law of mass action and so on so the thing that we have to think about is are these very common in fact these are actually more common when compared to the premix flames premix flames pretty much have to be contrived many times you have to mix somebody has to mix the reactants right whereas diffusion flames are the ones that have happened naturally in nature right so if you want to for example light fire to a block of wood or something like that that is essentially a diffusion flame that is going on right and candle flames are best example right so the candle flame is essentially a diffusion flame so that the laboratory if you want to now establish a diffusion flame take go back to the Bunsen burner that we have been talking about but you now shut down the air entrainment along with the fuel at the bottom right and just allow the fuel to come along and have the air mixed with the fuel outside and this is that is a pretty good interesting experiment that you can think about so if you now had a a Bunsen burner and you now had the bottom ring open pretty much fully and you had this orifice through which the fuel is coming out like a jet and creates a locally low pressure so that the air gets sucked in and in and entrains into the fuel and then mixes into this tube and then you now have a nice conical blue premixed flame and then you now start slowly trying to move this ring so that the hole is actually progressively closed and you are letting in less and less air and as you get in less and less air you are making this flame more and more fuel rich because you just have more and more fuel and less and less air so it becomes fuel rich and we talked about fuel rich premixed flames earlier in the context of stabilization and what we understand is two things one as the air entrains at the base of the flame you are going to have a enrichment of this mixture and it is going to be close to the stoichiometric and actually get anchored much more firmly the other thing is for the remaining shoulder of the flame you are going to have a fuel rich products that are coming out and that is going to mix with the reactants and have a diffusion flame envelope right so ultimately when you now close this ring completely and you know do not let in any more fuel then what is happening is you have only fuel coming out and then you have mixing of air that is happening and typically what you should find is you now have a diffusion flame that is fairly elongated when compared to a more compact premixed conical flame and this is going to now look not exactly shaped like a cone anymore right it is going to have like a tear drop kind of shape and it is not blue and blue any longer it is more orange and yellowish in some pockets and so on this is essentially typically your orange flame. So what this means is you have inside of the flame you have all fuel outside of the flame you have all oxidizer at the flame you are basically having the two mix and produce products of course the products are free to diffuse on either side and they are in general getting convicted upwards and of course the whole set of species are actually moving upwards because of buoyancy that is another that is another matter in a gravitational field around and this and that is partly the reason why you are seeing this this kind of a shape for the flame and also the reason why the flame looks elongated besides the diffusion process itself causing it elongate when compared to a compact conical premixed flame. So these are lots of these kinds of different things are happening as you now change this from a premixed flame to a diffusion flame it is a nice experiment that too for you to think about and do so diffusion flame can be commonly come across and we will now talk about how to analyze this next class.