 Today we will start looking at droplet burning, now the context of droplet burning comes up in many liquid fuel applications, many of the popular liquid or commonly encountered liquid fuel combustion applications is in automobile engines where we inject either petrol or gasoline as it is called in different countries or diesel in piston engines or for example some form of kerosene in aircraft engines and so on. In all these things the fuel is essentially in the liquid form and it is atomized into a spray and the spray is essentially composed of a ensemble of lots of different droplets of different sizes. Another application of droplet combustion is in liquid rockets where you typically are injecting a liquid fuel as well as liquid oxidizer and you try to atomize both of them and try to burn them. So in this situation then in general what basically happens is whichever is in liquid form has to evaporate and burn in the gas phase and the combustion that is occurring in the gas phase has to feed back heat to the liquid fuel or oxidizer in order for it to vaporize. In the case of liquid rockets where you have both liquid fuel and liquid oxidizer having to be atomized into sprays and the sprays have to mix and vaporize and burn together you have a problem of multiple droplets of different species which are trying to vaporize based on flames that are existing while the gas phase products of these or gas phase vapors of these are mixing and burning and that is a little bit more difficult problem think about. An easier problem is where you are looking at a fuel being in liquid phase and being atomized into spray and burning in a gaseous oxidizing ambience right. So first of all we will try to limit ourselves to this situation where only one of the reactants namely the fuel is in liquid phase and it vaporizes and its vapors mix with a gaseous oxidizing ambience right. So that is the first limitation that we will set for ourselves the second thing is we will not deal with sprays now we will deal with a single droplet so out of the spray and whatever is the fate of the single droplet is something that we should expect for all the droplets to face in a ensemble of them that is constituting the spray if the spray is a dilute what is called as a dilute spray. So what is meant by a dilute spray is that the spray droplets are not interacting with each other in other words the combustion that is occurring in this in one droplet does not influence the vaporization of another droplet that is nearby and so on this is typically what is called as a dilute spray and many times you do not necessarily have a dilute spray at all you have typically what is called as dense sprays where most of the combustion happens in droplet clouds as opposed to happening on single droplets but for the sake of convenience at the moment we will try to analyze a single fuel droplet burning in a gaseous oxidizing ambience so this is the framework in which we want to do this having said that then we want to look at this essentially in the context of diffusion flames basically a droplet combustion essentially is fundamentally a diffusion flame because you are talking about a fuel that is vaporizing and the vapors are mixing into the gaseous ambience just at the flame therefore there is no way you could have actually premixed the oxidizing vapor into the liquid droplet okay the liquid droplet is originally separate and then the and as it vaporizes its fuel vapors mix at the flame so it is essentially a diffusion flame. However there are in certain liquid rockets typically smaller ones we use something called monopropellant liquid liquids or liquid monopropellants whichever way you want to call it where the monopropellant for example example is a hydrogen peroxide which can actually burn by itself because it contains both hydrogen and oxygen within itself and as it thermally decomposes because of heating from a flame it gives rise to both fuel and oxidizer gaseous species that can react at the flame and that would be a premixed flame right so except for this particular situation we are essentially going to be looking at a liquid fuel droplet that is vaporizing and mixing with a gaseous oxidizing ambience at the flame forming essentially a diffusion flame and since we are going to be looking at a diffusion flame we will actually look at what is happening at the Berkshumann limit which is where the fuel and oxidizer vapors meet each other in the mixing field at stoichiometric proportions and therefore locate the flame corresponding to where the stoichiometric surface is in the mixing field similar to what we did for the Berkshumann flame. The general problem with most droplet combustion is to examine what is called as the what is popularly called the d squared law so the starting point for us to actually look at is what is called as the d squared law what the what this really means is that there is a certain rate of evaporation or combustion of this droplet dictated by how the droplet diameter changes with time okay so you now take a certain droplet of certain diameter at a particular time and ignite it and get it to combust or let us say it starts evaporating at a time t0 then what this really means is if you now start with a original diameter for the liquid droplet dl not squared and then the diameter decreases to dl squared at any time t this is simply given by k times t-t0 this is what is called as a d squared law. Now this is an empirical observation to begin with and certainly for many different droplets you do not necessarily have to have a d squared law it could be d to the sum power which is approximately around 2 okay there are some situations where it could be as low as 1 sometimes and many times you do not necessarily fit in an integer for the power it could be a non integral number like let us say 1.5 1.8 and so on now on the face of it what it really means is d when you say d squared it essentially refers to surface area okay. So what we can understand is more the surface area of the droplet greater is the evaporation right or and evaporation leads to combustion in a in a in a combusting environment but whatever we are talking about will also be valid for evaporation when you have a hot ambience. So what what we are talking about could be thought of even even in a non-reacting sense where you have a hot ambience okay. So like for example you want to dry dry a spray for example or dry droplets like for example if you get hired by a washing machine company and you want to figure out how to design a dryer like for example droplets that are actually trapped in a cloth how they get dried depending upon of course you will have to bring in convective heat transfer and so on. So you will have to think about all those all the issues that we are talking about in that context. So whatever we are talking about now will also be valid for evaporation in a hotter ambience than the liquid droplet temperature. So and when we say d square it basically refers to surface area and as a matter of fact the goal of atomization is to actually try to enormously increase the surface area of the liquid from originally like let us say a flow of bulk liquid you now want to actually spread it spread the surface area for the given same the same volume and essentially then increase the surface area so that you will now increase evaporation and therefore the combustion rate and so on. So the goal of the theory here is to see if we can look at the governing equations and deduce the d square law by applying the governing equations to this particular geometry and this problem. So obviously here dL0 is essentially dL at time t equals t0 and t0 could be thought of as like a ignition delay beyond which the combustion begins and case what is called as the evaporation constant evaporation constant. So the framework here is we now assume that the droplet is spherical it is not a bad assumption considering surface tension effects that try to keep the droplet spherical beyond below should say below a certain size for a given droplet given liquid so this is not a very bad assumption at all like for example if you are looking at droplets that are of the order of 10 microns or even 100 microns they are reasonably spherical even in a gravitational field like what we face on earth and then so this is the fuel and then what we expect is then there must be a concentric flame around it. Now this is a questionable assumption in a gravitational field because buoyancy effects will induce a vertical convection upward and also distort the flame shape in a way that is pointed upwards alright and of course what we are looking at is a droplet to be stationary in a quiescent oxidizing quiescent gaseous oxidizing ambience therefore we could now assume the spherical a spherical flame concentric to the droplet if the droplet were to be moving right so that there is a motion for the droplet then the motion can correspondingly induce a distortion in the flame shape relative to the spherical shape that we have assumed so that is not to be considered in what we are assuming what we are taking up because we are saying it is quiescent ambience so if you now have a concentric flame and this is the droplet right then we can say that the mass flow rate of the fuel that is coming out of the surface of the droplet is – D over DT 4 over 3 pi RL cubed rho D where RL is the liquid I should probably use different subscripts here but I think it is a mix up of subscripts well just keep it as RL okay RL cubed and RL where RL is the liquid radius and RL is the liquid density and now in this of course the liquid density is not expected to change with time and 4 by 3 pi is constant therefore we can pull them out and then differentiate RL cubed with respect to time and that is going to give you – 4 pi RL squared rho L DRL divided by DT the negative sign is because as time increases or decreases okay so and then we want to have a positive mass flux that is coming out therefore you throw in a negative sign to make the make that out so what we now assume of course here we have already mentioned assume spherical symmetry not a not a great assumption for a moving droplets or in the gravitational field or both the other assumption that is going to be quite important and needs to be elaborated upon is what is called as a quasi steady assumption. What that means is the droplet diameter does not significantly change when the droplet evaporates and the gaseous vapor moves away from the droplet up to the flame and mixes with the ambience to form the reaction in other words the typical timescale of convection and diffusion of the gaseous species is much smaller when compared to the typical timescale of reduction of the droplet size appreciably all right. So how do I understand that the answer is essentially coming from the fact that the rho L could be significantly larger than the gaseous density okay so you can have a factor of about two or three so I shouldn't say factor sorry order of magnitude by about two or three like like the gaseous density is like of the order of one kilogram per meter cube the liquid density could be of the order of thousand kilogram per meter cube right. So what the answer is we need to look at mass conservation across the interface right so for a unit mass per volume of liquid to evaporate will now give rise to a much lower density which means as the surface recedes inward it is now going to put out gas that is that has to actually move radially outward at a significantly larger rate. So to give you an idea correspondingly if you now say that the gaseous velocity that is going to go out relative to our frame of reference centered around says centered at the droplet center if that is of the order of let us say a few meters per second right then correspondingly the droplet rate of motion is going to be only a few millimeters per second right. Because you now have to have this three orders magnitude two to three orders magnitude disparity between the rate of regression of this droplet surface inward to the rate at which the gaseous would evolve out of the droplet what this really means is a couple of things one first of all the droplet regression sets epic convection. So while we would like to think that it is purely a diffusion problem right in fact when you think about it as a diffusion flame in the Berkschermann limit where we now adopt an infinite chemistry mixed as burnt approach flame sheet assumption and so on where we do not have to worry about the chemistry and we will have to worry about the mixing so would like to think that this is purely a mixing problem where in the liquid vaporizes and the fuel vapor steps out of the liquid surface after the vaporization and is just sitting there to diffuse no it is not sitting there it is actually flowing out because you have this regression right but the second point that is actually being made is this which is the regression is so slow when compared to the rate at which the gaseous move away from the center with a velocity such that we can now think of a droplet of a given radius at any particular instant as a snapshot that means you freeze frame your droplet and then say for this particular droplet diameter the gases are rather instantaneously moving out and diffusing with the oxidizing ambience forming the stoichiometric surface and hence the flame and and then the heat is rather instantaneously moving back to this droplet of this particular size. So all this balance happening in the gaseous phase is rather instantaneous when compared to the droplet motion that we can freeze this picture right that is what is meant by quasi steady assumption that means as far as the gaseous equilibrium is concerned the mass balance and the energy balance the species balance and the energy balance is concerned we can essentially say it is steady state we do not have to worry about any unsteadiness. So we now stop with the Schwab-Zelovich formulation the Schwab-Zelovich formulation has its 11 assumptions including a steady state right which we just try to articulate now. So the Schwab-Zelovich formulation is the operator L of beta equal to 0 where beta is a coupling function and of course for n species that we have we are supposed to have n plus 1 betas of course sorry for n species that we have we have n betas and 1 alpha so we have n plus 1 equations but if you notice what we did earlier with the premix flame as well as the Berkschum and diffusion flame where we adopted the Schwab-Zelovich formulation earlier in this course we now try to choose this particular alphas and betas in the Schwab-Zelovich formulation that is the most pertinent to what we are what we are doing so we do not necessarily have to worry about n plus 1 equations right. So here what we want to do and this is actually the art of solving the problem so the take here beta equals the energy beta which is should say identically beta t which is equal to alpha t minus alpha o now I could have used alpha f to subtract from any of the other alphas including alpha t or some other alpha some other species alpha right. So the choice of actually using alpha t and alpha o is because I want to actually keep track of the oxidizing ambient species which is purely gaseous all right and then the temperature because the temperature is very important because the energy balance is the one that is actually giving rise to a heat feedback that vaporizes the droplet. So if you did not factor in the heat feedback you are not going to be able to vaporize the droplet so you need to you need to have the energy balance number one and the oxidizing species is taken because it is purely gaseous I do not have any problem with it vaporizing so alpha t minus alpha o is a good combination that I can think about which will now translate to a t superscript not to t integral cp dt divided by let us say the reference q that is based on the heat of heats of formation plus y o divided by w o nu o where of course o I should I should mention here o refers to oxidizer oxidizer in the ambience and L of beta is basically divergence of rho v velocity beta minus rho d divergence beta equal to 0 right there is something that we have written about two or three times now in solving different problems in fact for the for the premix flame problem we had a one-dimensional formulation so it is in did not quite really matter but it is originally one-dimensional Cartesian in the in the Berkshuman problem and we adopted the Schwab-Zeller which formulation we we we wrote it for a two-dimensional cylindrical polar coordinates system it is now going to be back to a one-dimensional situation because the only variation is going to be along the radius right so but we are going to adopt a spherical polar coordinate system alright in a one-dimensional sense where the radial radiation is the only thing that is considered and in a spherical polar coordinate system of r theta and phi we are we are going to have to get rid of phi and theta variations but still the r variation the one-dimensional variation in r is going to be in a spherical polar coordinate framework therefore in silver in spherical polar coordinates this can be written as we get d by dr of r square rho v beta equal to d by dr of r square rho d d beta by dr okay the r square as a coefficient is special to the spherical polar coordinates here and notice that we have a double derivative here whereas we have only a single derivative here okay and we are using ordinary derivative because R is the only independent variable now so it is not necessarily a partial differential equation now we want to try to bring in this m dot which will open our RL which is the radius of the droplet at any particular time to any radius so R is going to vary from RL to infinity alright and m dot is going to be m dot can be a better way m dot can be defined for any or for the mass flux of the gas but at the surface it is it has to match the liquid mass flux right so m dot is essentially rho v times 4 pi r square small v of course is the radial component of gaseous velocity right and rho is the gaseous density right or the density of the gaseous mixture in your in your in your domain the domain is now for the gas and so it starts from r equals RL to r equals infinity that is how it is going and m dot is for any radius and that is that has to be a constant in steady state okay what is it that is really what we do not know what we are basically looking for is what is the rate at which the droplet is evaporating if I knew what is the rate at which the droplet is evaporating I would know what is the rate at which the mass is coming out of the droplet surface and I would know what is the rate at which mass is flowing at any point or in the gaseous field and up to the flame and beyond and so on but that is what I do not know in that sense this is actually similar to the premix flame problem where what we are actually trying to find out is the rate at which the flame was propagating right we do not we did not we did not know what the rate at which the flame was propagating at and that is what was setting up the convection right here again the m dot the mass flux which is with a constant right from the droplet radius to any other location or in the gaseous field is the unknown and that is related to the velocity V which is setting up the convection convective field for the gaseous flow right so we want to make use of this and therefore what we want to do is so the point basically is you know whenever you see a R squared rho V right immediately what you want to think is can I throw in a 4 pi and then make it look like m dot and then utilize it to be a constant and then I can pull it out and then take it to the other side slip it into the derivative we can do all these kinds of things so this can now be written as d beta by dr is equal to d by dr of 4 pi R squared rho D divided by m dot times d beta by dr that looks better and we will try to make it look even better by introducing a new dimensionless independent variable so introduce dimensionless independent variable psi equal to m dot times r to infinity 4 pi R squared rho D to the negative dr to the negative 1 dr now what does that mean look at here you have a force 4 pi R squared rho D divided by m dot so strictly speaking course we could take this m dot inside the integral because it is a constant and what you are looking at is m dot divided by 4 pi R squared rho D okay since R squared is actually showing up in the denominator right we now have to end up integrating from R to infinity and that is the reason why we are taking up an alpha O or you can put it the other way I took an alpha O therefore I want to actually look at the region between any R and infinity as a measure of R okay so this is a inverse thing as R keeps growing my psi is going to decrease because at infinity psi is going to go to 0 okay so this is actually trying to invert R because R is showing up in the denominator here R squared 1 over R squared right now this is these are some of the things that are puzzling about combustion theory which is you come up with these what should I say innovative ways of non-dimensionalization which are which is not very apparent to a student who is not really well versed with mathematics but that is the game these mathematics is typically so complicated that we will have to come up with these innovative non-dimensionalization methods that will try to simplify the governing equation significantly so if you know without this if this particular independent variable then you now transform your equations in terms of psi as opposed to R and the governing equation becomes the governing equation becomes d beta divided by d psi equal to minus d squared beta divided by d psi squared so simple because essentially all the well this is the only term that has a coefficient within the derivative and that is simply being robbed off by the definition of psi therefore that vanishes and you simply have d squared beta by d psi squared with a negative sign and so essentially this is d squared beta by d psi squared plus d beta by d psi is equal to 0 right so this is a linear second order ordinary differential equation for which we know how to solve how to write the solution the solution is the solution is beta equal to a plus be to the minus psi now let us call this one in fact this is where we are going to start doing the problem as of now we have not really done anything because the problem is always defined by the boundary conditions alright so strictly speaking I have not really said from here on when the mathematics started that I am working on the droplet problem okay so the droplet problem is going to show up only when we start applying the boundary conditions right and the boundary conditions need to be applied in order to evaluate the constants of integration a and b which are coming up because we are we have solved we have just solved a second order equation so we have two we have two boundary conditions that need to be provided in order to solve for this and keep in mind we do not quite know psi as well it involves m dot that is the unknown okay so in that sense this is kind of like an eigenvalue problem and as I said it is a it is a it is very similar to the premix flame problem where we do not know the propagation speed and that was an eigenvalue in that problem and in order to evaluate that besides applying boundary conditions there we also applied an interface condition between two regions the predes zone and the and the reaction zone and the interface condition was the one that actually fetched us the eigenvalue namely the flame speed here we will also have to have an interface condition at this at the liquid surface which is now trying to say so much of evaporation is happening for so much heat that is being sent right so in addition to applying boundary conditions in the geisha's phase that r equals RL which is the liquid surface and r equals infinity we will also have to supply an interface condition in order to evaluate the m dot this is what we will try to do so need the boundary conditions for beta and psi right so the first boundary condition is so you have to go back and see what what is what is beta beta is alpha t minus alpha o alpha t is this which pertains to with energy balance right and alpha o is this which pertains to y o which is the mass fraction of oxidizer right so if you are now able to actually generate boundary conditions at r equals RL and r equals infinity for y o and temperature or energy balance then we subtract the one from the other in order to find the boundary condition for beta so this is the process that we are going to do so as far as the species is concerned what we know is that the oxidizer cannot penetrate the liquid surface all right now this is something that need not necessarily be true but the moment you now are able to suppose that you have a liquid droplet which is intact and it is going to produce geisha's fuel means that the oxidizer does not really the geisha's oxidizer does not penetrate the liquid okay now there are more complicated situations where typically one of the products of combustion so what is going to happen to the products of combustion so you are going to have the products being formed at the flame right and just like how we thought about the flame as a flame sheet in the Berkschumann problem and we tried to actually locate how the temperature profile and the product concentration profile would vary away from the flame on either side and of course the fuel concentration oxidizer concentration plunging in at the flame and so on we could think about it like that so what we would expect is as the products are being formed at this flame they are at a high concentration relative to the neighborhood and therefore they would have a tendency to diffuse right so they would diffuse inward against the current in this case as well as outward right when they diffuse in and of course the temperature is also going to be at its peak at the flame and it is going to come down so it comes down on both sides and as well it goes out there it is going to match the oxidizer temperature in the far field but it is going to match the surface temperature of the liquid which could be like the boiling point of the liquid right and then of course inside the droplet you could have a constant temperature if the droplet is too small so thermal equilibrium must happen significantly considering the liquid thermal conductivities are an order more than the gaseous thermal conductivity and so on or let us say if you have a still a profile that is going in but at the temperatures of boiling point of the liquid the product could also condense into a liquid okay so examples of this or for example if you now think about aluminum which is used widely in solid rockets where you have a molten liquid aluminum droplet that is burning vaporizing into vapor phase aluminum or aluminum vapor and mixing with oxidizing species course in the solid rocket you do not really have atmospheric oxygen kind of thing you have oxidizing species in the form of water and carbon dioxide and so on which carry oxygen atoms in them and they mix and then form a flame and the product of combustion is Al2O3 right and in gaseous state because the temperatures are so high that it would be even higher than the vaporization temperature of Al2O3. So you form aluminum oxide vapor at the flame which now tries to diffuse out from the flame radially inward and radially outward and a little bit further out the temperatures are now low enough for the Al2O3 vapor to condense into Al2O3 liquids the vaporization temperature of Al2O3 is significantly higher than the melting point of aluminum therefore as the Al2O3 is diffusing backwards or inwards it is now going to begin to condense and then as it comes down closer it now starts reaching the aluminum droplet surface keep it aside look at another example let us think about things like methanol or ethanol droplets right and in this case you could easily think about an oxidizing Indians made of just atmospheric oxygen right or pure oxygen or air whatever and then you have a flame and the products of combustion or carbon dioxide and water which try to diffuse but as the water one of the products comes back diffusing in it now condenses because when methanol exists methanol it is quite volatile right so it can vaporize but the temperatures for the combustion are so high somewhere in here as the as the water vapor diffuses back in it can it can condense right and then further in it now finds the surface of methanol now the difference between these two situations of aluminum combustion and methanol combustion is in one case the product liquid that is condensed is invisible with the parent molten aluminum fuel droplet whereas in the case of methanol the product water that is got condensed is missable with the reactant fuel in liquid form so as the water actually comes back in it starts mixing with water in liquid phase and effectively dilutes the fuel so as the combustion continues to progress you get to your point where you are less and less rich in methanol and therefore as the and so what what effectively the flame is trying to do is not only vaporize the methanol but also tries to vaporize the water that is mixed with it beyond a point it is not getting enough methanol to burn and it is so the flame is going to extinguish at a certain size of this droplet including having a certain amount of methanol as a residue along with water right as opposed to that if you now look at aluminum since the aluminum oxide is invisible in aluminum because of interface tension forces similar to surface tension effects all the aluminum oxide that is coming in cannot really stick at different points so they tend to actually accumulate to one side and they find what is called as a oxide loop on one side it could be any side because we are not looking we are not really necessarily looking at effect of gravity here right so for very tiny droplets so it could be on one side and this is going to first of all spoil your spherical symmetry right but it could still assume cylindrical symmetry about the lobe center right so there is an axis and then the flame is now going to assume the shape that is cylindrically symmetric or axi-symmetric and then as the more and more combustion happens the aluminum part shrinks and the oxide lobe grows and so you now have a to the later part of combustion of the droplet you have a smaller aluminum droplet and a larger lobe and ultimately when all of the aluminum is burnt the lobe closes and becomes a oxide droplet right so that is the final product so in both the cases you get a liquid draw right but in one case it is contaminated by some of the field that has got quenched without the flame being able to sustain in the other case you get this and of course when you have an oxide lobe and if you are now beginning to think about a flow field it becomes unstable and then it can stop wobbling and so on so this is actually in reality a lot more complicated problem the point I am trying to make is there are these problems where we would expect either the oxide to come and collect to one side or penetrate in a liquid form okay penetrate the surface in liquid form and mix whereas there is not necessarily the case with ox oxygen okay if that were the case then even as the liquid exist liquid fuel existed the oxygen would begin to actually get dissolved in it right so the first thing that we have to adopt is that oxidizer cannot penetrate the liquid the droplet surface oxidizer cannot penetrate the droplet surface there is rho v y not minus rho d d y not by dr at r equals equal to 0 this is to say that the oxygen for the oxidizer the the gaseous oxidizer can diffuse only up to the point that the droplet surface vacates by convection. So rho v times 4 pi r squared is m dot and that would be the same as this m dot right so as far as the gaseous convective effect is concerned this is because of the space vacated by the droplet trying to regress in its surface. So the diffusion is trying to match the droplet regression right so the droplet regression in terms of convection of the oxidizing gaseous species is this and this is the diffusion so this basically states that the diffusion can happen just as much as the convection can allow it due to the droplet surface regression so this is essentially the boundary condition here and this is in terms of r but we have adopted a non-dimensionless independent variable psi so this basically means that this is similar to the flux boundary condition that we had maybe we are seeing when we are considering axial diffusion of the Berkshuman problem by the way right quite similar there and d dy o divided by d psi at r equals psi equals psi l is equal to minus y o at psi equals psi l which can simply be called as minus y o l where y o l is the oxidizer concentration at the drop at the liquid droplet surface what we expect is the oxidizer concentration to be in stoichiometric proportions with the fuel concentration at the flame all right and further decrease to a value y o l in combustion you will have the situation where the oxidizer is getting consumed in the flame in evaporation you do not worry about a flame it is just the heating from the outside that there is that is causing the evaporation and you will still have a ambient gas trying to diffuse up to the droplet surface and hold the certain concentration there right so this essentially is the oxidizer concentration at the liquid droplet surface we do not know it quite but we may have to make an estimate we will worry about these things as we try to plug in some numbers or our typical values for these things as we go along but right now the boundary condition needs to be formulated to describe the droplet combustion problem so that is as far as one boundary of one component of alpha o right then we have to worry about what happens to alpha t at r equals rl corresponding once at infinity and so on we construct these boundary conditions and the interface condition plug them in into the solution in order to find out m dot that is essentially what we want to do we will do that tomorrow.