 this is where we are in the droplet burning problem where we got the solution for beta which is the Schwab-Zeldovich coupling function in terms of a t and a o the thermal the enthalpy and this is the oxidizer mass fraction in the ambience in terms of psi where psi is actually a non-dimensional radius which is actually the inverse of radius like as radius increases psi is going to decrease okay and then we are looking for boundary conditions and also an interface condition because psi involves is m dot which is unknown and that is what we are trying to find out. So the first thing that we want to do is to actually look at the interface flux boundary condition at the droplet surface and we are trying to actually form beta which means we have to look at what happens to alpha t and alpha o what we are doing here is to actually look for what happens to alpha o. So if you now look at how what happens to y o divided by w o nu o will give you what happens for alpha o and then we need to look at what happens to alpha t which means we have to look at a energy flux balance at the droplet surface interface right. So as far as that is concerned if you now look at so this was basically a mass balance at the droplet surface based on conductive flux and a diffusive flux as far as the energy balance is concerned what we can understand is now if heat conducted into the droplet must be just sufficient and I will explain what has been by just sufficient pretty soon but sufficient is what we should start with sufficient to vaporize the fuel leaving the droplet that is to say if you now have a flame with a flame temperature and then the heat gets conducted inwards radially inwards the heat conduction flux at the surface should be equal at least to the rate at which the heat is required for it to vaporize right and that is what has been by saying just sufficient or at least sufficient that is to say if you if you have excess heat that is being conducted part of it in fact bulk of it is going to be taken by the elitant heat of vaporization to vaporize the droplet into vapor and then you will have a further temperature gradient for the temperature to vary within the droplet okay now if you think about very small droplets what happens is you have a very quick thermal equilibration that happens within the droplet so the droplet is nearly at pretty much the same temperature everywhere within with hardly any heat flux that is going on because with if you do not have any temperature gradients you do not have any heat fluxes so essentially the droplet is looking forward to only that much heat that is by conduction that is required for it to vaporize okay and that equilibrium that is a kind of equilibrium that is attained and that this is reasonable so what this means is 4 pi r squared k dt by dr at r equals rl is equal to m dot l now this is in some sense very similar to the the convective diffusive balance that we have gone through this is the diffusive part this is the convective part except what is important here for you is you have this is simply a m dot times y o okay that is like the mass flux of the oxidizer that is convecting inward in the space that is vacated by the droplet regression okay that is what that is what this dictates but here this convection is actually based on the latent heat of vaporization so in other words you need to have latent heat of vaporization times the rate at which the droplet is regressing of all which is the same as rated which the gases are issuing out of the droplet surface so this is this is pretty much the same of course we are we are neglecting a few things here neglecting radiation that is a significant effect wherever you have phase interfaces somebody we we we notice this when we were adopting the Schwab-Zeldovich formulation with the line with the 11 assumptions and we did neglect the heat flux due to radiation and then at that point we we we noticed or we we pointed out that radiation can be neglected in a homogeneous system but not in the heterogeneous system heterogeneous system meaning when you have multiple phases okay physical phases so here we have a phase interface between liquid and solid gas where radiation could be pretty important but of course we neglected for simplicity so neglecting radiation and kinetic energy we are not worrying about this little kinetic energy that we are going to have for the vapors that are coming out at a certain velocity because of the droplet regression that is significantly small so this is reasonably okay to neglect it now L of course is the latent heat of vaporization latent heat of vaporization per unit mass of the fuel fuel at the at temperature at TL let us say that is the droplet surface temperature TL as I said if the droplet is reasonably small and you have a thermal equilibrium within the droplet the temperature is going to be pretty much the same at TL whatever is the surface temperature is going to be the same uniformly throughout within it at the moment we are not really we interested in what is happening within the droplet and that is there is a reason why we are assuming this just sufficient that means we are not going to have any further heat and heat conduction after the latent heat has been removed okay so now let us of course we have Lewis number le equal to 1 which is k over Rho Cpd and of course with the other thing that we also do is take T0 equal to TL for convenience and this is not very difficult to swallow because we have seen this before if you now take a different value and then we still insist on T0 you can have two integrals and we find that these things are actually showing up wherever you have alpha T or beta showing up they solve in derivatives therefore a constant addition is not a problem so that we can we can do this so therefore this condition then is now going to translate to because we can now take for example take the m dot down here and then k can be replaced by Rho Cpd right so it replace k by Rho Cpd and of course Cp can be stuck with the temperature and then you have 4 pi R2 Rho D divided by m dot coming from the other side and then we look at how psi is is defined and keep noting that 4 pi R2 Rho D is in the denominator and the way psi goes is inverse of rho so we should for example we got a negative sign here right and similarly when you now try to write this in terms of the non-dimensional independent variable psi you should you should have d by d psi of integral T0 to T CpdT equals minus L that is the reason why you got the negative sign so if you now try to say I will divide this by Q and this by Wo nu O which means I go back and divide this by Wo nu O and this by Q I will have wherever I have a y O I will have a alpha O and wherever I have integral T0 to T CpdT divided by Q I will have a alpha T and subtract one from the other to get alpha T minus alpha O which means we get the beta right so putting putting the two flux balances at the droplet interface together we get d beta by d d psi of course I should go back and write this at L so d beta by d psi at L equal to minus L divided by Q minus y O L divided by Wo nu O alright alright. Now this is the interface condition in addition to this we need to have the regular Dirichlet boundary conditions right so the Dirichlet boundary conditions at the boundaries to the domain or the Dirichlet boundary conditions domain or we should say at r equals infinity psi equal to 0 okay at r equals infinity psi equal to 0 right and what would that mean we would say what would be a beta a beta would be whatever is alpha T and alpha O at r equals infinity which means whatever is the temperature T infinity right at far away and whatever is Wo infinity so for example if you have air as your ambience your Wo infinity will be like 0.23 because only about 23% by mass is going to be oxygen the other one the rest is going to be nitrogen so you have a diluent present in your ambience in oxidizing ambience which can be taken into account you can also take into account the ambient temperature and keep in mind if we do not have to worry about a combustion heat release then we could still solve the evaporation problem if your T infinity is higher than TL alright and that was that is going to basically mean that you have heat conduction simply from the ambient not necessarily from the flame so the T infinity is going to play a role and as a matter of fact you know this is actually the sensible enthalpy term and this is the heat release term so this is already showing up as a competition between the two okay so we will see these things lot more clearly as we go along but essentially what it means is beta can be equal to beta infinity which is alpha T infinity minus alpha O infinity which is nothing but T0 to T infinity Cp dt divided by Q minus or should I say plus plus y o infinity divided by Wo nu O that is the one boundary condition at r equals rl we are kind of coming inwards because that is how psi increases from 0 so psi is now going to be equal to some psi L which is saying psi L is equal to rl integral rl to infinity right and so psi L and beta will have some value beta L which is nothing but alpha TL minus alpha O L which is integral T0 to TL Cp dt divided by Q plus y O L divided by Wo nu O we have already seen that if you now take T0 equal to TL for convenience this would not have any contribution but that is alright let us let us just at the moment worry only about beta L let us not worry about how it is exactly defined we will ultimately see that what matters is not exactly each of these but the infinite beta infinity minus beta L in which case we can now put these things together and these integrals will now go together and then you will get an integral goes from TL to T infinity okay so the T the T superscript not gets out of the way anyway so good that means we are now ready to attack the solution better we were armed with these so let us do one by one so let us say how do I number this yeah I think I should mention what is psi L here for the record where psi L is m dot integral RL to infinity 4 pi R squared rho d to the negative 1 dr okay and I think we should probably call this as 2 right so we call this is 1 and okay now so the first thing we can do is at r equals RL beta L equal to a plus b e to the minus psi L at r equals infinity beta infinity is simply equal to a plus b because psi equal to 0 so e to the 0 is going to be 1 and subtract to get rid of a and actually you know by now you should get the idea of what what what the drift is we are not really interested in solving for alpha I mean betas at all okay that is that is not really our problem our problem is to find psi L so that from psi L we can get an idea of m dot and m dot is what we are basically looking for we are not looking for the actual profiles so the real problem is what is the rate at which the droplet is shrinking everything else is a detail and so we will solve the we will keep our focus on the problem all these details like how exactly beta is varying that means how is the temperature going to vary how is oxidizer concentration going to vary and all these things are all exam questions PhD qualifying exam questions and so on okay so that is for you to figure out but we will we will focus on our problem that means at the moment we can just subtract these two get rid of the a we are not going to bother about evaluating it for ourselves and we are simply going to get b equals beta infinity minus beta L plus v e to the minus psi L right that is true so b equals beta infinity minus beta L plus b right then our question is what is beta beta beta infinity minus beta L but beta infinity minus beta L is nothing but you just have to this is this is beta infinity this is beta L so this is beta infinity this is beta L subtract one from the other so you get sigma k equals 1 to n y yk infinity integral T L to T infinity Cpk dT let us say let us say a fancy way of writing Cp simply you could have taken the yk inside and but but you want to show the infinity very clearly divided by Q and the reason why we are not really worrying about ykl is because as I said here T of T L is equal to T not this is not going to contribute right and plus yk yk sorry y y o infinity minus yol divided by w o nu o right so that is what we get then we have to use the interface condition so what we have now is we have b is equal to this plus b e to the minus psi L with which you can evaluate b all right but you we are not really bothered about as I said evaluating b we are we are not interested in evaluating a as well we are not interested in finding betas what we are really interested in is to apply the interface boundary condition so that we can even eliminate b okay so fortunately the interface boundary interface flux condition put together in terms of beta has it in terms of derivative so if you have it in terms of derivative a automatically drops out you try to take a derivative here a does not even show up b shows up okay and between that and this we will eliminate b as well okay that is what we are going to do so using the interface flux balance right we get b e to the minus psi L equal to L over Q L over Q d beta by d psi is going to have a minus b e to the minus e to the minus psi right and evaluated at L right so we should have a negative sign but you also had a negative sign here which I am going to turn it as positive right so we we now have this as L over Q divided by plus yol divided by wo nu o now if you call this as your three I guess this is your three and this as your four then what you do is we now use four and we now use four and three together in order to get e to the minus x psi L and use what is the definition of psi L okay that is what that is what we do so putting four and three together and using two right we get psi L equal to natural logarithm you got the natural logarithm because between these two if you now try to get rid of b you will be working with e to the minus psi L okay and if you now try to extract psi L out of it you will get a natural logarithm okay so natural logarithm of 1 plus sigma k equals 1 to n yk infinity integral tl to t infinity cpi dt divided by Q plus yo infinity minus yol divided by wo nu o so that is coming from here divided by L over Q plus yol divided by wo nu o that is coming from here okay and this is 1 plus was coming because we had a b and a b there you could have a 1 plus so that is that is the mathematics and then in this you now use two because psi L as this definition so you now get a in dimensional form in dimensional form m dot is going to be 1 over integral rl to infinity 4 pi r squared rho d to the negative 1 dr natural logarithm just for the record got to write the whole thing again 1 plus sigma k equals 1 to n yk infinity tl over t infinity cpi dt divided by Q plus yo infinity minus yol divided by wo nu o divided by L star we could do a little bit of manipulation if you now try to multiply Q throughout both numerator and denominator you could write this as plus Q and this would be plus Q yol divided by wo nu o of course in the final videos we can edit out this time it took for it to write okay and then we can just show the final expression written again and here we will basically point out that we have multiplied the numerator and denominator throughout by Q so that the Q shows up here and here instead of at the bottom of this and this so that that is a final expression for m dot and as usual we do not understand this okay so we get some complicated looking expression which just does not make sense we started out thinking we wanted to get the d squared law okay the d squared law is essentially saying that if you now are looking at the rated which the evaporation happens the time of evaporation is proportional to square of the diameter and we notice that the square of the diameter is going to be sort of like the surface area and we also then notice that if you now look at the m dot the mass flux the square of the radius shows up which means the square of the diameter shows up and this in fact is the surface area 4 pi r squared of the droplet so all that seems okay and now we got an m dot how in the world are we supposed to get the d squared law from this okay so or how are we supposed to now find out how the d squared varies with time for the droplet regression okay we have adopted a quasi steady assumption by which in the instantaneous snapshot at a particular time the m dot is like this without any time dependence any further okay the time dependence that we have gotten rid of is the small time scale that that that it takes for the gases to just diffuse and convect at a particular snapshot freeze frame time and from this time for you to get to the next time when the droplet regresses significantly takes a lot of time and then we will now look at the picture now and then do this balance and then figure out what the current m dot is and so on that is what we are trying to do so if you want to try to understand this further it is it will be helpful if you do some more simplifications okay so let us try to do that so let us try to do some simplifications here the first thing that we will say is yol is approximately equal to 0 that means we are saying that the droplet the oxidizer as it diffuses all the way to the droplet surface is pretty much of zero concentration or put in another way all the species that you have at the droplet surface is going to be mainly fuel vapor so if all of it is going to be fuel vapor it is completely displaced all oxidizer so you do not really have any oxidizing species this is one way of looking at it in a mostly evaporation situation in a combustion situation it is lot more convenient because we expect that the fuel vapor and the oxidizer vapor are going to diffuse against each other radially and meet its stoichiometric proportions where the flame is and then get consumed there right so if you now took a magnifying glass and looked at this particular sheet we should now be able to find out how these things gradually decline the oxidizer con the fuel concentration gradually declines up to this flame and then rapidly declines because of consumption and the oxidizer concentration gradually declines as you go radially inward because of diffusion and then rapidly decline across the flame right and therefore at at the droplet surface you are not going to get any oxidizer at all so yol can be assumed to be more or less 0 without much of a problem the exception see when it anytime you make a assumption which is simplifying you have to look at how good it is okay or when would it not be held and the answer is this is not really quite true near extinction conditions that is when the droplet is going to actually extinguish right you have a rather weak flame and you have ample opportunity for the oxidizer to diffuse past the flame in spite of going through reactions and consuming getting consumed quite a bit because it is not getting completely consumed right so this is mostly true true particularly in combustion situation except except near extinction conditions so that is one simplification what would that give me that will get rid of the yol here right that would get rid of the yol here so it is kind of like we could send this to a hydroster and get it to be nicely dressed up and look like a very simple expression right so that we are trying to get rid of some of these terms that way let us also assume that TL is equal to TB which is the boiling point of the liquid that is a simplifying assumption in reality we have to actually look at something called a saturation temperature and so on which depends on the ambient ambient what do you call the ambient composition of its vapor and so on so you have to look at vapor pressure balancing the the the actual pressure and all those things now we let us not worry about all that simply say TL the droplet the temperature is going to be the same as the boiling point and let us assume that Rudy is a constant which is equal to case K developed by CP okay because L equal to 1 and suppose CP I equals equals constant equals CP all CP's are the same and it is it is it is constant right so make these assumptions and what do you have we get so with which we can now try to evaluate this integral also right so Rudy gets pulled out of the integral and you can you can you can integrate dr divided by 4 pi r squared right that is not a problem and then substitute the limits you get m dot equals 4 pi k RL over CP that is what Rudy has now become times natural logarithm 1 plus CP can now evaluate the integral inside as well simply T infinity minus Tb plus Q y o infinity divided by I should say we have a W o nu o divided by L right okay you still have some expression but this is little bit more manageable we can we can try to understand something more then let us just think about it for a minute okay what this means is this is the sensible enthalpy okay for the gas going from the boiling point of the droplet to the ambient temperature so this is the sensible enthalpy rise of the gas that is coming out of the droplet this would actually be corresponding to the heat of combustion so for the moment let us not worry about the heat of combustion okay let us suppose that we are thinking about evaporation that means you have an ambient hot gas that is trying to evaporate the droplet and that means it is simply giving rise to a sensible enthalpy rise by L so the L is what is actually is a latent heat that is required first of all for you to vaporize and then the next thing is you have to rise it have sensible enthalpy okay so the m dot is dictated by how much you are able to evaporate versus how well you are able to rise a sensible enthalpy of whatever we have evaporated if you now throw in combustion you now have this term in a in addition okay so the heat that is available to you goes first to evaporate the droplet and second to raise a sensible enthalpy okay so you can see how these three processes essentially heat release that is available to you during combustion can go go on to raise a sensible enthalpy in competition with evaporating the droplet okay so this is what is actually dictating this m dot now what we are interested is in trying to find out what is the relationship between the or how is this RL going to change in time okay that is what we are really interested in so for for this purpose we can we can see that so note notice that m dot is now linearly proportional to RL right so say m dot equals k prime RL the reason why we are using k prime is k is reserved for the evaporation constant in the d squared law and that is what we are trying to deduce okay now we are trying to deduce the d squared law and therefore we want to see how what is the dependence of the evaporation constant on the system parameters namely like the ambient ambient gas composition ambient gas temperature the boiling point of the liquid the latent heat of the the the liquid and then so on right so we we want to know what the k is and so we now keep this as a k prime so where where k prime is equal to except RL we have to write everything else which is 4 pi k divided by Cp natural logarithm 1 plus Cp T infinity minus Tb plus q y o infinity divided by W o new o divided by L right so keeping there is a k prime the next few minutes we are simply going to ignore all this all this stuff and going to basically say m dot is equal to k prime RL and let us see what that means right so m dot m dot on the other hand is negative 4 pi RL squared rho L d RL over dt right this T is now at the time scale of the droplet regression it is not a shorter time scale of the gases equilibrating in diffusion and convection so this is the kind of time that we are looking for in the d squared law so this is now equal to k prime RL which means we can now cancel the RL squared one of the RL's in the RL squared with this and from this we can now write or RL d RL equal to negative k prime divided by 4 pi rho L dt alright then just go back and write this with let us say RL not to RL any RL starting from T not to any T and that is now beginning to show up as the d squared law so this is one half RL squared minus RL not squared or okay let us just keep it that way and we have a negative sign to fix things so we have a 4 pi rho L t minus t not and of course now this negative sign will allow us to flip things because RL not is actually greater than RL therefore we want to keep things positive and keep in mind RL is nothing but dL over 2 so RL squared will be dL squared over 4 and so the 4 and 2 are going to get together to make an 8 and then of course you have a 4 pi that is going to go away and k prime also has a 4 pi so don't get me in a hurry to cancel that with a 8 okay so dL not squared minus dL squared is equal to 8k prime divided by 4 pi rho L t minus t not so now it is beginning to look like the d squared law boy couple of minutes in couple of ten minutes ago this did not look like the d squared law at all okay but that is the d squared law there and so we have the d squared law so with the evaporation constant k right so what is the d squared law just for the sake of completeness d squared dL not squared equals k times t minus t not k equal to so evaporation constant k right where k equal to these are put things together 8k prime divided by 4 pi rho L and then of course you have a so you just divide this by 4 pi rho L and then put through in an 8 and you are simply going to get a 8k divided by rho L CP natural algorithm 1 plus 1 over L CP t infinity minus tb plus q y y o infinity divided by w o nu o right and as I said all the dependencies that we are looking for over there in the evaporation constant you can also use this for evaporation as well as combustion all you have to do is neglect the heat of combustion for getting the evaporation if you want to keep the combustion then keep the heat of your heat of combustion and you will get an additional additional evaporation rate okay so that means because the combustion it burns faster it regresses faster that is what it simply means so this we should stop doing droplet burning and we will see what what else we need to do next week.