 So we were looking at the G equation which is written back here again this is typically belonging to a class of what is called as Hamilton Jacobi equations well not exactly this but we will look at a specific form of this pretty soon which could be looked at as a Hamilton Jacobi equation but the reason why I am mentioning this is yesterday we were talking about the possibility of solving the G equation as if it is a scalar field everywhere in the domain about the G varying from a value that is less than 0 to a value that is greater than 0 through a very thin region where it directly where it drastically increases from minus 0 to plus 0 sorry less than 0 to greater than 0 through 0 and wherever you have this iso surface of G equals 0 will form the flame sheet is what we were saying but it is possible also to solve this equation directly without having to actually assume that is a scalar field so you do not have to actually solve for this everywhere in the flow field it is possible for you to solve for this directly and the only problem there is you will find that you require you have to make sure that you are not creating any spurious numerical oscillations if you are trying to solve this numerically because you are looking at something like a very thin sheet and it is very difficult to resolve this therefore typically numerical solutions will have errors that accrue and try to cause oscillations so there are these what is called as essentially non-oscillating solutions that are two schemes that are employed like typically the more advanced versions are like weighted essentially non-oscillatory schemes and also for the time marching in an unsteady case you should look at something like total variation diminishing schemes and so on so there are some special ways by which you can solve this numerically directly instead of having to think about it as a field so having said that let us now try to see if we can work with this in the classroom on a simple problem right so we do not have to do heavy duty numerics and have to learn a lot of advanced schemes and so on so the simplest example for us is the Bunsen burner so the Bunsen burner Bunsen by the way is a German he was working in Heidelberg so that there is if you go to the Heidelberg town square you will find a statue of Bunsen right across Krzegorzkow's house and apparently Bunsen devised the burner for Krzegorzkow to do spectroscopy with it so the Bunsen burner is a simple idea here you have a you have a field flow through an orifice and you now allow for air to get entrained in this jet of fuel that is coming out so you have like locally sub-atmospheric pressure over here and then the air and fuel mix with each there over here and then issue out as a premixed mixture and you now light up the flame in the in the at the rim and then you now have a almost like a conical flame that is established a conical blue flame that is established for most typical fuels so this is a premixed mixture in fact most of our kitchen stars work on the same principle so you do have a orifice at the bottom and you and it it entrains the air and so on so let us now try to see if we can come up with a simple version of how this will this can be done so let us just enlarge this picture right there so this is the burner rim and of course what we first do is to fix your coordinate system so you can have a R Z coordinate system and you now have a flame and for simplicity well I should I should say this at this stage let us assume that it is a axisymmetric flame that is fine and then what we should actually be able to show that show is this is the conical flame and then of course it is not a perfect cone you have some deviations from a cone at the tip and at the base as well but let us not worry about all those things at this stage what we should be interested in is to get the shape of bulk of the flame right which is what is called as the shoulder of the flame so let us let us aim to do that and so we are now looking at a steady state this is reasonable in fact many times and you now light up a Bunsen burner we the flame hardly shakes so you do not even know if there is any time dependence at all so you have a very good steady state situation in a laminar Bunsen burner so that means the unsteady term in the G equation goes to 0 right and therefore if you now have the steady state then G of let us say X V, T X is the position vector just being becomes G of X vector alone which in our case has to be G of R, Z because it is axi-symmetric that means it is not even depending upon the azimuthal angle ? so it where depends only on R and Z so you want to keep this a bit simple so you do not want to actually have a G equation that is G expression that is so general so rather what we want to do is let G of R, Z be written as Z- Z of R what does that mean we want to have G equal to 0 at the flame right so if you now set this equal to 0 you are going to get a flame shape as Z equal to Z of R that is what we are looking for okay so this is a two-dimensional picture graph in which you want to have the flame shape and a two-dimensional function in a XY plane would be like y equals f of X that is what you are used to but because it is a it is a three what you call cylindrical polar coordinate system we are using a Z or plane in which we are now going to assume the flame shape to be of the form Z equals Z of R right so if Z is equal to Z of R at the flame then Z- Z of R should be equal to 0 at the flame so that should be your G right so what is essentially means is any location on the flame is measured by a vertical distance Z along Z okay so this is Z of R for a given R okay the moment you now have your G written out like this then we can we can write this as grad G right so grad G is dou G by dou R E or cap plus dou G by dou Z E Z cap and this is going to be so dou G by dou R is going to be negative d Zeta by dr er cap plus you just have Z over here to take a partial derivative with respect to so that is going to contribute to 1 times E Z cap and grad G therefore is simply going to be square root of 1 plus d Zeta by dr the whole square right so what we have done is to find out what this is find out what this is and we now have to look at what the V is the V we now assume is now going to be basically a U0 or this is okay if you can say U0 typically the subscript 0 gives you a mental picture of something like a steady state okay I just or you could have used U bar also but I wanted to avoid bar because you have this over bar representing an arrow for the vector so essentially what we are saying is V vector is nothing but U0 E Z cap right okay so what happens then is you now plug in these things so this does this term so in the G equation V dot grad G this is of course measure that G equals 0 minus is equal to SL grad G this goes to 0 and we are assuming a uniform velocity everywhere up to the flame okay we are not saying that the entire flow field is uniform the flow could actually do whatever it wants beyond the flame what we care about is the V that is just upstream just upstream of the flame right so we have to worry about how what the velocity is as you look from the flame upstream right and we now suppose this is something that we suppose now that the flow is uniform up until the flame and then let it do whatever it wants it we do not worry about a fate of the flow beyond that as far as this analysis is concerned right now we will start worrying about it a little later but not yet so we should now say that V is uniform so U0 is a constant of course it is strictly not a constant okay we will have to think also about the velocity profile because we have this pipe and you have to think about like a fully developed flow or whatever it is those are things for a little later not yet okay we will at the moment say it is just a uniform constant flow so then what happens is because you have to take a dot product of V with respect to grad G and you know UV has only a Z component which is U0 all you have to do is worry about only the Z component here right so you get a U0 times 1 equal to SL times square root of 1 plus d zeta over dr the whole square now we assume the SL to be a constant for this particular analysis at this stage there is really no reason to expect it to vary because we are not going to worry about how this tip is going to be formed at this stage we are not going to worry about what happens at the base we are looking at pretty much like a constant slope for a bulk of the flame on the shoulder and we suppose that the mixture is uniformly mixed therefore there are no concentration variations anywhere and the temperature is uniform upstream and so on so lots of reasons to believe that the SL is more or less going to be a constant right for a given mixture of a big or of a given stoichiometry pressure temperature etc so this is this is this is assumed to be a constant U0 is also a constant as I mentioned therefore we can just do the manipulation so you just take a square of the entire equation U0 squared equals SL squared times 1 plus d zeta over dr the whole squared and so this is going to give you so we open up the brackets U0 squared equals SL squared plus SL squared d zeta the over dr the whole squared so from here we can now get your d zeta over dr as plus or minus plus or minus square root of U0 squared minus SL squared divided by SL squared that is not too difficult for you to figure out so we now we have to choose between plus and minus at this stage we will do this in the moment so we can now integrate so what we find is that you not as a constant SL as a constant so square root of U0 squared minus SL squared divided by SL squared is a constant right so all you have to do is integrate this so zeta equals plus or minus square root of U0 squared minus SL squared divided by SL squared times or plus a constant integration constant that means we have to supply a boundary condition right in fact when you now supply g is equal to z time z minus zeta of zeta of r and then plug it in here with the n steady term kept in there right so if you now keep this time dependent then the time dependence is now going to go to zeta of r, t so you will have a partial derivative of zeta with respect to t and then you can you can so you will have a U0 times 1 all right or yeah and then you know plug in this grad so square root of 1 plus d zeta by dr the whole squared that would be actually the Hamilton Jacobi equation that you can solve for zeta numerically if you want to but what you we got past that point and we now have to anyway supply a boundary condition so that is what I was beginning to say so in any case you need to have a boundary condition what I wanted to point out is this is this is roughly first order in space okay so this is like square root of square root of 1 plus square of the first order so it is basically a first order in space equation if you now also have a time dependence in a can steady term you may have a first order in time but this is an equation which requires either an initial condition or a boundary condition it is not this is sort of similar to the hyperbolic class of equations where you do not necessarily have to supply everything unlike in parabolic or elliptic equations therefore we now choose to actually supply a boundary condition here in the steady state problem because you now have the variable R showing up so the boundary condition here is we now say at R equals capital R right zeta equal to 0 okay that means the flame is coming and he getting held firmly at the rim of the burner right now this is a bit questionable if you now look look at the base quite closely you will find that there is hardly any flame at all there is a little gap there okay but we are looking at like bulk of the flame it is not as if like this is very big it the bones and bones are typically very small for laminar flames but when compared to the length of the flame which could be maybe a centimeter or two okay that is what this is this small when compared to that whatever we are talking about is much smaller right so we for this purpose now we know that that particular part but if you now begin to take think about that you are worried about variable SL and so on so you get into a lot of trouble at the moment do not do that so you now say the flame is firmly anchored at the rim so this is basically what is called as a flame anchoring condition or flame attachment condition again you can talk of this is in many ways so this is like flame flame anchoring condition if you do this then what happens you see if you now plug in R equals capital R and Z equal to 0 C becomes the negative of this with a capital R on this side okay we need now take this there so that simply means that Zeta should be written as Zeta should simply be written as square root of U squared minus SL square divided by SL square times R minus R so I did choose the positive sign in the end because I am sorry I guess I chose the negative sign so if I now chose the negative sign and plug capital R equals capital R get this C equals positive constant so this is the reason is we want to have Z to be positive right we want to have the flame looking like that about the we are looking at the situation where the flame is actually inside the tube so this is what we get and all minus all is always positive because the maximum value that small all is going to take is capital R right okay what do we get out of this what we what we think is well two things one first of all we find that Zeta is linear in R okay that was that was obtained from right here when we found that d Zeta over dr is a constant right so Zeta then becomes linear in R so we first retrieve the fact that you are going to have like a straight line for the description of the flame and we the second thing we find is Zeta is actually going as something times capital R minus small R so the slope of this straight line is negative okay so you are now looking for a flame that is going like that in the 0 to R 0 to capital R domain for the for the small R so as small R goes from 0 to capital R the flame has to start from a peak value and then go to 0 at the rim right so this of course with the axi-symmetric assumption you now have a volume of revolution or a surface of revolution rather in this case you should now get a conical flame so this basically represents a cone okay so once we understand that then we start thinking about it more like a cone rather than like in a coordinate geometry kind of thing so many times you do coordinate geometry you forget to geometry but strictly speaking you should be able to think about the problem in both ways simultaneously so let us try to do that so we note that at x equal to sorry r equal to 0 at r equal to 0 that is at the center line right the flame has its highest location right so that is actually the height of the flame so Zeta equals L let us say so L is like the length of the flame length or height whichever way you want to call this so length or height of the flame now of course we can try to find that here we can we can find that L is equal to so you can plug in r equal to 0 then we can find out what the Zeta is which is which is now denoted as L you not squared minus SL squared divided by SL well numerator alone can be written as power whole power half times r okay now let us look at some trigonometry so if you now have a triangle with a apex angle theta and what we are saying is this is L and this is r right so L over r should actually be cot theta cotangent theta so this implies that you not squared minus SL squared the whole to the half divided by SL is equal to cot theta and from here we get you not squared minus SL squared divided by SL squared equals cot squared theta a couple of more steps SL squared equals SL squared cot squared theta so you not squared equals 1 plus cot squared theta SL what is cot squared theta sorry 1 plus cot squared theta. So cot is cosine divided by sin so you get the sin up here and keep it in the denominator as well so sin squared theta plus cosine squared theta is 1 right so you get a 1 divided by sin squared theta right so this is going to give you U not so can you can forget about the square root and do worry about the plus or minus just keep it as plus because everything is positive here so SL divided by sin theta or U not sin theta equals SL right that is what you are trying to get now what does this tell us what it tells us is if you now have a U not that is coming up here what we should be looking for is its component perpendicular to the flame right as so if you now look at this as the component perpendicular to the flame that is the one that is along the flame I hope I got my oops sorry so this is my U not sin theta the one that was along the flame was U not cosine theta so the U not sin theta is the one that is actually trying to balance the propagation of the flame perpendicular to itself at any point with a flame speed SL that is what this basically says right now if you knew anything about Bunsen-Berners this is what is going on okay and whatever we have done so far is a vindication that this vindicates that okay what we originally said was we wanted to come up with a general flame surface G of our vector, 0 or x vector, 0 sorry x vector , t which is equal to 0 and we wanted to say that represents the flame surface and then we now want to want to have this flame surface be placed in a flow field to compete with the flow field by propagating with a flame speed normal to its local sorry flame speed in the direction along the local normal to the surface against if the reactants that are flowing in so there is like a flame flow balance that is going on with the help of Lagrangian and Eulerian conversions and all that stuff that we did we came up with an equation okay. If that wasn't convincing you know try to apply this to a simple case of a steady Bunsen flame right and then do all the mathematics and then finally come to the moment of truth that we are familiar with which is the component of the flow locally normal to the flame has to balance the flame speed right so this is basically what the G equation gives you when you are trying to do the G equation Bunsen-Bunsen burner so from here we can actually get your L so so L then is essentially R ? so that that comes from here and you can you can easily see what ? is ? is now defined so ? is nothing but sin inverse SL over U0 so when you now say sin inverse SL over U0 then SL is supposed to be less than U0 okay if SL is less is less than U0 then the flame so essentially what it means is the flame is trying to eat into the reactants and the flow is trying to blow it away okay so it is sort of like the flame cannot propagate just as fast as the flow so the flame sort of like tries to give way to the flow as much as it as the flow wants but it starts eating into the reactants that are coming as fast from the side so it kind of like instead of actually trying to propagate exactly this way you now say okay you want to go fast let me give you way but then as it goes it sits on the side and then starts eating into this and survives right why does it turn like that because it is anchored at the base right so the flame is not going any way going any far it is just trying to give way to the flow or so it seems but it is eating from the side eating the reactants from the side because it knows that it is actually anchored at the base right so this is a condition where clearly the flame speed is less than the flow speed that is one you are going to get a flame that is inclined otherwise it going to progressively try to become flatter so that it is normal is in the direction of the velocity so that you do not have to worry about a component of the velocity the actual velocity itself will suffice then the catch is what is happening at the base because the base is where it is actually being held and from there you can now get this flame to tilt that means there is this dynamics so long as the flame is anchored the dynamics is always between SL and U0 that dictates the orientation of the flame or the shape of the flame so you can call this in the several ways shape of the flame you can say is conical but what we refer to is in the angle what is the length of the flame how tall is the cone and so on so lower the SL when compared to U0 taller the cone all right so this takes us to what we were doing the other day with trying to tinker with SL know if you now say well I am going to send an air which is a mixture of nitrogen and oxygen and you have a certain flame angle now what if I actually progressively remove the nitrogen then send an organ right so if the SL now begins to increase then the flame is going to become shorter and then if I now progressively withdraw my organ and send in nitrogen sorry helium right the SL is going to still get it get higher and therefore the flame is going to get still shorter and so on so there is a point when the flame speed could exceed the flow speed so one of the easier easiest ways of doing that is to try to shut down the flow without putting off the flame right so typically new graduate students who do not really care about safety they would just try to shut down the air first before shutting down the fuel all right then what happens you might have been working in fuel lean you shut down the air first you are actually progressing towards torque a metric right so if you now progress towards torque a metric the flame speed is increasing and the flow speed is decreasing because you are shutting down the air so this is a very very good combination for what is called as a flashback to happen that means SL increases the U0 decreases the flame begins progressively shorter and becomes normal to the flow and then it can begin to propagate inside and that is a bit dangerous right so if you now have allow for the flame to propagate inside it can go all the way up to the point where the premixing happens in the case of a Boonson burner it is not very very serious because the premixing is happening just upstream a little bit and it cannot go beyond that but the flame can actually get attached here as a diffusion flame and so on okay if it can but this is actually a fairly high velocity but in typical experimental apparatus you are trying to mix with in a controlled manner because you want to meter your air as well as oxygen as well as fuel so you cannot just let air and train from atmosphere in an uncontrolled fashion so you do all these things and then you have like a mixing chamber in which this mixes and so on so the flame goes back to the mixing chamber in a confined region it can lead to a detonation right and typically grad students blow up a couple of mixing chambers before they start doing their experiments properly but it is very dangerous so it is where premix flames are somewhat unsafe to deal with you have to be very careful because of this reason on the other hand if you now had a flame which was already fairly elongated and you now do things that will actually decrease the air so further and further right or increase the you not further and further right it progressively becomes steeper and steeper and at some point it blows off so on the one hand we talk about a flashback and the other one other the other hand you talk about a blow off in both these cases the anchoring is lost okay so long as the flame can get anchored it will now try to adjust its angle in such a way that the flow of balance the flame balances the normal component of the flow all right so this is what is going on now there is a very simple idea so for example if you now think about something like a ramjet combustor or a afterburner all right so what do you what do you expect you know typically the combustor is like you now have a fuel injection manifold ring which is essentially a circular tube of circular cross section right so you have like a circular tube which is now coil like a circle and then it has holes to inject fuel and this is typically liquid fuel but it vaporizes within some distance and that you have what is called as a V gutter so the V gutter is again like a ring but if its cross section is a V and the flame is actually stabilized or anchored at the edges of this V and it is now going to actually begin to look like that I am of course drawing it in a wiggle for two reasons one is to tell tell the flame from distinguish the flame from the rest of the drawing and the other thing is it is actually a turbulent flame okay so here what is going on is the flame is trying to propagate normal to itself and the flow is coming very fast so these velocities that you are talking about are of the order of 100 meters per second or so right so you are having a fairly high speed flow and the flame is quite turbulent as well so in these cases you cannot use a SL for the flame speed you have to use a ST that sounds very simple although all you have to do is change the letter of the alphabet to get from laminar to turbulent huh well it is not so easy but it is not very difficult either okay so as a quick rule of thumb for you to estimate what the combustor length should be if you are now trying to design a combustor based on this you see all you have to do is look at a turbulent flame speed as a factor of sorry should say factor as a multiple of the laminar flame speed so what happens when you have a turbulent flame speed the turbulent flame speed is a lot faster when compared to a laminar flame speed I am not going to go in into details about how this happens but effectively you can it is essentially an order of magnitude more for faster okay it is not as fast as detonations detonations are about three orders for three to four orders magnitude faster right we are still talking about the fluctuations right and this is about an order of magnitude faster than a laminar flame so if you are now going to use like a rule of thumb on how to get your ST or like use in order of magnitude appropriately for the for the turbulent flame speed then you can simply use this kind of approach and try to get a length estimate on where the flame is going to hit the wall right or where the flame is going to merge from these two points near the centerline either of these could be longer right depending upon whichever is longer you need to have a combustion chamber that is that much longer or more than that okay then throw in a factor of safety to make it longer you will typically find if you go go around after burners or ramjets the combustor is just a long pipe beyond beyond the fuel injection manifold and the v-gator just to give room for this right and how did you arrive at the length exactly the same as what you would do with the Bunsen burner isn't that kind of funny right so the same idea essentially works in these cases so bottom line the flame tries to propagate against C against the flow so we can do a little bit more on on the flame speed now because what we have to further think about is question this this is approximation of the assumption is SL really a constant okay so why would you want to worry about that because we find that many times the flame is not exactly like like this you don't get a sharp tip over there and yesterday we pointed out that we don't really have so we assume that the flame shape is smooth right and it doesn't really have any kinks and so on so indeed you will find it at the tip you have like a curvature okay and I understand many conditions I will discuss conditions when there are exceptions okay but many conditions we have to actually think about why you have a little curvature at the tip very much near the tip okay later down we will talk about diffusion flames which are much smoother if you want to think about that in your mind right now you have to think about a candle flame okay the candle flame is not really as sharp as a premixed flame cone of course it may be it is made sharper in a gravitational field because of buoyancy effects that that cause the flame to shape itself that way but it is not because of because of the convection diffusion reaction processes right the convection diffusion reaction processes that we have encountered so far gives rise to a fairly sharp cone except near the tip for the for the flame for the for the premixed flame on the most conditions and then we asked the question if it is going to be locally like that if I now zoom in and see the flame is actually coming like that and then becoming like this it is flat at the top right if it is flat at the top that means it is propagating against the flow right that means the flame speed should have somehow increased over there to match the flow speed is that right is it what is going on or is the flow speed really that uniform is it possible that the flow would have actually slowed down because of the flame right so how does the flow know that it is actually approaching a flame so that it can change its velocity and how if that is the case so on the one hand we have to worry about the flame on the other hand we have to worry about the flow is it possible that the flame would have changed its flame speed is it possible that the flow would have changed its flow speed locally there right so in a in effect what we should worry about is two contributions two contributions to what is called as a flame stretch what is going on there is that the flame is actually getting stretched effectively when you are thinking about a increased flame speed over there your thickness of the flame is actually decreasing so if you had a let us say think about a thickness so if you had a thick flame as you now go towards the end you have to actually thin it so for example in a in a PhD qualifying exams question or something if you asked about thickness of the flame make sure that you are not drawing it uniformly thick near the tip as in along the shoulders that is that is pretty important that is what they are looking for for you to appreciate right so the flame stretch effect has two contributions one is what is called as a flame curvature and the other one is called flow divergence the flame is curved all right and therefore is in it there is an effect of the flame curvature on the flame speed and then there is a flow divergence that that that is caused by the flame flame curvature that also slows down the flow against the flame speed right now there is a very very interesting point that I would like to make what is meant by flame speed okay flame speed is that flow speed for which the flame is stationary all right so how would I measure my flame speed like the way we were at least in beginning to think about flame speed right in the beginning was you now try to actually have a flame fixed coordinate system and then allow for the flow to come come in come in here right so we now said whatever is the flow velocity that is approaching the flame should be the flame speed because if this were still the flame would propagate at that rate now we have to seriously think about this what happens is if you now have a flame that is not exactly flat we have been thinking about a flat flame all the time right so if you now think about a flame that is somewhat curved okay just to just do not worry about why we are thinking like that at the moment the suppose that it is somewhat curved and maybe because of this and we will explain this soon maybe because of this as the flow comes in it tries to now get diverged or converge or something of that sort but all that stuff is happening locally near the flame right and because of this what the actual flame sees is a different flow than far upstream but just like how we said the entire flame thickness consists of a preheat zone and a reaction zone for the flame thickness and anything for the flow is upstream of that again anything anything for the reactants that we considered like the flow velocity and so on is all upstream of that this is the flame on the whole with the preheat zone and so on here what we have to think about is okay you had curvature and all those things and the flow changed its local velocity here and so on because of which it will look like the flame can propagate against the slower velocity but still for an experimental point of view I would measure the flow speed here you see so if it is possible for me to actually handle a higher flow with a curved flame so that locally the flow slowed down to accommodate the flame it will look for me apparently that the flame speed is higher because it could handle a higher flow right in fact in some sense this is exactly what is happening in wrinkled if the flame can actually get so wrinkled at a very high flow rate right and then stay there very very wrinkled I now begin to say that is a turbulent flame and its flame speed is much faster but locally it is probably trying to have a flow that is much slower when it is trying to eat into it there because of it changed the flow therefore the effect of the flame on the flow basically comes back at it as a change in the flame speed because of the way we think about and measure of flame speed because this is a relative to the far upstream velocity not what happens in the vicinity of the flame right so effectively these two effects are actually going to change the flame speed that is what we are looking for right so if it is possible for us to think about a modified flame speed that that that takes into account these two effects then we now can go back and say well let me put that SL here and still work with the constant velocity because far upstream the velocity is constant is he certain how to worry about the details what is the point in working with SL we did not have to worry about the details of preheat zone and diffuse reaction zone and all those things in these lines it is all there inside it is all like a package right we are now trying to further package everything that the flow goes through right into the flame speed so that we can now use a modified flame speed against a uniform flow right so what you essentially talking about here is if you now think about a curved flame for some reason let us say it is a perturbation right and I am a little bit hesitant to use the word perturbation because that gets us into stability on whether this perturbation is going to diverge or converge and so on will hold on for some time not thinking about that but essentially what I am saying is let us think about like a slightly curved flame for some reason I would like to talk about two things when you now have this situation one is a thermo diffusive effect that is if you now for example take this part of the flame you can argue the same thing on the other part of the flame so typically what I do is in the classroom I just discussed this and I and the exam I ask you to work on that okay so you now look at a flame that is actually having a concave curvature with respect to the upstream flow right what is happening is this is now actually heating the reactants this way the conduction is happening radially inward so the reactants that are coming in or not just getting heated up right here right but it is also getting heated up from there you got to be a bit careful when I draw a film like this it is assuming that the preheat zone and all those things are within this and we should be looking at only the reactants that are going locally normal to that point and then getting heated up there until then it is not supposed to get heated up so we are actually blowing this image up to a length scale that is comparable to but maybe a bit like higher than the flame thickness right and when would I really have to think about that if I have to go to a region that is so small there we are talking about a curvature which is of the order of the flame thickness all right so under those circumstances we will now find that the heating is not one dimensional the heating is now multi-dimensional so you have now an increased heating that tends to increase the flame speed as we see from far upstream okay but there is another effect the reactants will have to get spread out then this is particularly critical for it the deficient reactant let us say if you are now looking at a off stoichiometric condition right say a fuel lean mixture that means the fuel is deficient right and that particularly is now going to actually spread out linearly and it is going to feed the flame less with the lesser concentration than if we if it were to feed a straight flame or a flat flame right that weakens the flame so what you would expect is we are now beginning to talk about like the fuel diffusion versus thermal conduction upstream so fuel diffusion downstream versus thermal conduction upstream right what comes in what comes into our minds conduction versus diffusion the Lewis number right so Lewis like a god of combustion you can't forget him so the Lewis number has to be has to come in a picture there is no yet to begin to think about what happens as a function of Lewis number okay we will stop here