 Good morning and welcome to the next lecture in our course on chemical engineering principles of CVD processes. In the last couple of classes we looked at sort of an unconventional non-traditional CVD process which is what happens in tungsten filament incandescent bulbs where the tungsten evaporates from the filament and deposits on the bulb walls and we saw how by taking a simple PVD process and by providing appropriate chemistry you can convert it into a CVD process and actually in this case prevent the deposition of tungsten from taking place. And I also mentioned that there are what are known as filament CVD reactors where you actually try to provide conditions which promotes the evaporation of material from the filament and deposition onto a substrate and we will discuss this design later on in some detail. Today I want to take you through another situation where chemical vapour deposition happens in a somewhat unintended fashion and this has to do with combustion of fossil fuels in gas turbines and power plants and so on. Many of the fuels that we use today are getting increasingly of you know lower quality because we are running out of reserves. So for example with coal you keep digging deeper and deeper and as you go further down the quality of the coal is not as good as layers that are closer to the earth surface. So when you are burning coal you have a problem with sulphur contamination, ash particles contained in the product gases that deposits. So you have problems like fouling and slagging and corrosion and erosion but these are associated with the particular debris that is present in the combustion gases mostly the inorganic alumina and silicate ash particles. So that is a different problem and you know in order to deal with that problem you have to know something about particle technology, how do particles move around, how do they deposit on surfaces, how do they stick to surfaces and so on. But when you talk about systems where a cleaner liquid fuel is employed typically a liquid hydrocarbon you would expect that the situation is better that there will be fewer impurities in the fuel that survive the combustion process and that is true. You know when you burn for example jet fuel or distillate fuel or any kind of liquid petroleum or hydrocarbon fuel it is true that if you look at the products there is hardly any particulate matter. However what does happen is if the fuel and or the oxidant that is used in the combustion process contains impurities, the impurities can react and they can form condensable vapors which can then result in the formation of a condensate on particularly on cooled surfaces that come in contact with the hot combustion gases. Now these deposits unlike the C V D deposits we have been dealing with earlier which have all been solid deposits are typically molten deposits and one of the implications of that is the deposit once it is formed does not stay where it is formed it has a tendency to be mobile. So for example if the condensation is happening on the rotor blade of a turbine then as the blade rotates there is a centrifugal force on the molten layer that deposits on the surface and simultaneously there is also aerodynamic shear because as the combustion gases flow past the molten deposit they are exerting a shear force. So the local thickness at any point at any instant in time is a balance between three processes the deposition process that is causing the film to form the aerodynamic shear which is trying to move it away from where it is formed and the centrifugal shear which is also imparting essentially a stream wise motion. So it is really a balance because as at any location as the film is getting depleted it is also getting replenished from an upstream location right. So it is an interesting problem in terms of analysis and modeling and simulation and so on but certainly the precursor what initiates this whole thing is the deposition of that film on the surface. The impact that this film has is twofold one is these deposits are sometimes corrosive in nature. For example if you have molten sulfates depositing on a surface the sulfur has a tendency to participate actively in initiating corrosion on the surface. So the presence of the film can lead to corrosion effects even when the surface on which it is depositing has a protective coating. For example in the gas turbine industry the turbine blades are typically made of steel for cost reasons and also for structural properties but steel has a tendency to easily oxidize and corrode. So what they do is they put a protective oxide coating on top which is supposed to shield it from particulate matter as well as you know corrosive liquids and so on. But this oxide has a certain finite thickness and if the film the molten film that deposits on that oxide starts to dissolve the oxide eventually you can have excessive thinning of the oxide layer at that location and at that point there is like a threshold thickness. Once the oxide thickness falls below that value the molten film simply penetrates through the oxide and directly attacks the steel alloy material that is beneath. The other approach that the industry takes is to use what are known as super alloys. A super alloy is one that has much better rust and corrosion prevention properties compared to a more conventional steel or even stainless steel alloy and if you use a super alloy and you use an oxide coating on top of it essentially you are providing double protection but there is a huge cost impact. Super alloys are typically 10 times as expensive for the same amount of material as a more conventional alloy. So you want to minimize their use and in fact as we discussed in one of the earlier classes instead of trying to change the entire substrate material to a different alloy material you can use surface deposition and diffusion processes to provide a super alloy structure only for the first few surface layers. So you can actually use CVD as a process to achieve super alloy type of characteristics but only up to a required depth of the surface. So you have a super alloy on which you have this oxide coating but the continuous deposition of these molten salts eventually will lead to corrosion and in fact that particular mechanism is known as hot corrosion because the corrosion mechanism is associated with, first it is a high temperature process. The turbine blade is at a high temperature, the combustion gases are at an even higher temperature but the hot corrosion actually also indicates the fact that it is happening because of the high temperature involved in the combustion process itself which leads to CVD mechanisms being set in motion and resulting in the formation of a condensate on the surface. For example if you take a aircraft application let us say that you are burning jet fuel which is probably as clean as it gets. So in a jet fuel your sulfur for example which is a contaminant in virtually all hydrocarbon fuels can be a very low level you know maybe 1 ppm maybe less than that but when you are flying in a for example a marine environment the air that you inject can have significant quantities of sodium and potassium and alkali salts sodium chloride being the most common but and those can be fairly high levels up to 10 ppm depending on the environment. So when the combustion process happens you have the impurity in the fuel which is essentially sulfur reacting with the impurities in the air that is being ingested and forming for example sodium chloride plus sulfur plus oxygen you are going to get sodium sulphate plus chlorine being released you are going to form potassium sulphate magnesium sulphate calcium sulphate all of these are what are known as molten salts and they all have corrosive characteristics because of the presence of the sulfur and by the way sodium chloride itself can deposit as a particle and cause corrosion as well because the chlorine will react to the surface but in terms of CVD we will not deal with that I mean that is more of a particulate transport and deposition process but just vapor transport and associated corrosion itself is quite significant. Now if you have when you look at naval applications ships which also use turbines they tend to use fuels of somewhat lower quality they are called the navy distillate fuels these fuels have a higher percentage of impurities so sulfur itself can be 5 to 10 weight percent in the fuel and so as you can imagine the potential for corrosion is much greater because of the higher level of impurity in the fuel as well and so while the fuel quality as well as the oxidizer quality can vary there is always the potential that when you burn the fuel and the oxidant together and by the way the temperatures you typically achieve under adiabatic conditions can be of the order of 2000 Kelvin plus so there is an extreme degree of dissociation the more that the amount of dissociation in the gas phase the more important CVD becomes because what it implies is that you have taken a finite number of elements and because of this extreme nature of the dissociation they are now incorporated in many many species containing that element. For example the sodium that was in the fuel sorry in the air as sodium chloride after combustion is going to be present in the gas phase as sodium vapor NaO NaOH Na2O Na2SO4 Na2O Na2SO4 Na2SO3 I mean there is going to be literally dozens of species in the gas phase containing sodium as an element similarly the sulfur which was present in the fuel after combustion is going to be present in the gas phase as SO2 SO3 as H2S and so on and so there are many many opportunities for these sodium containing species and the sulfur containing species to react and form sodium sulfate. So that is why this is a classic CVD problem the molten material is typically the alkali sulfate whether it is sodium sulfate or potassium or magnesium or by the way it could be condensate solutions also so you can have mixtures of various alkali salts forming on the surface but the point is they are not present in the vapor phase as exactly the same species sodium sulfate gas is not getting converted to sodium sulfate liquid. So that is why it fits the definition of a CVD process. So now in the industry what you would like to know is if I have a turbine blade I want to be able to predict the lifetime of the blade or alternatively to achieve a certain lifetime I want to know how much coating I need to put at different locations on the blade right. So what they do is something called burner rig testing a burner rig essentially duplicates in a smaller scale the combustion process that is actually happening in the field. So it is kind of a miniaturized combustion environment where you mix air and fuel together and you have essentially models of the various surfaces that come in contact with the combustion gases such as staters and rotors and boiler tubes and so on and you expose these materials to the flow of the combustion gases under well controlled conditions for well defined periods of time and then you take the specimens out and you analyze them to see what has happened to them. So one of the key things that they look for is what is known as a corrosion map. They want to look at the surface and map it it is a 3D map essentially to describe the thickness of the oxide at various locations on the substrate of the specimen and they also want to look at any evidences of corrosion which you can do by simple microscopic examination and if you want to confirm corrosion you can always do EDS analysis to verify that the composition of whatever you see matches that of a corrosion product. So this corrosion mapping is a very important exercise in determining the but it helps us do two things one is as I said it helps us predict the lifetime of the actual component in operation be it based on this it enables us to suitably design it so that it can achieve the lifetime that we are looking for and thirdly it also helps us to optimize the operating conditions of the combustor. For example the flow conditions you can certainly influence the rate at which deposition and corrosion are happening by influencing the temperature that is achieved the pressure that you operate at the fuel to add ratio is another parameter I mean as a simple measure if your fuel is much dirtier than your air then you want to run under as low a fuel to add ratio as possible whereas if your air is more contaminated than the fuel then you want to have a fuel rich condition right. So you can do some tweaking of your the inputs also based on this type of simulation and modeling and so on and that is really what people do and so this generation of the corrosion maps experimentally is an important step but you also need to be able to do modeling and simulation of what happens and that is where again we get back to the thermodynamic and transport modeling of the CVD process. Now in terms of the thermodynamic analysis as with many CVD reactors because the temperature is so high it is very reasonable to assume equilibrium prevails. So you use your free energy minimization algorithm for a given fuel composition for a given pressure and if you assume that adiabatic conditions apply you can do a free energy minimization simulation to calculate the composition of the product gases. I can do that at various temperatures you can do that calculation at the mainstream temperature of the combustion gases and you can also do the calculation close to the substrate where the film condensation is happening and the calculated mole fractions of the vapor species will then provide you data on the gradient that is going to be driving the diffusion process and similarly your temperature distribution simulation will tell you what is going to be the effect of thermal diffusion. If you remember I mentioned that a temperature gradient can also drive mass diffusion and that is called thermal or soray diffusion and of course the velocity distribution will tell you what are the convective flows present at different locations particularly along the surface because convection is not going to be significant in the boundary layer adjacent to the substrate. So it is not going to have a direct influence on the rate of deposition by diffusion however it has an indirect effect because it is a convection process that dictates the temperature distribution as well as the species concentration distribution outside the boundary layer. So certainly you have to solve all three conservation equations first you have to solve the momentum conservation equation then the energy conservation equation and finally the mass conservation equation to obtain the temperature distribution the velocity distribution and the species concentration distribution and using that you can estimate the rate at which different species are getting transported to the surface. So the typical equation that you would use is the same one that I had written down for the case of the tungsten filament bulb for any now in this particular case what is going on is you have a turbine blade over which flow is happening and there is a boundary layer that develops on top of the surface and so the convective portion the flow of combustion gases at a particular velocity is happening outside the boundary layer however the deposition process itself is happening due to a diffusive process which is characterized by a fric diffusivity of the species I and a thermal diffusivity of the species alpha Ti using the same formula that we wrote down yesterday that the diffusional flux of any species I is going to be equal to minus rho Di gradient in omega I plus omega I alpha Ti times grad T over T. So this is the fric diffusion portion and that is the thermal diffusion portion and by the way you can also write this in terms of the Nusselt number right. You can write this as minus rho Di over some reference length L times a Nusselt number times gradient in omega I can now be written as omega I if you if you call the substrate W and you call the mainstream of the combustion gases as E you can now write this as omega I E minus omega I W plus omega I W times alpha Ti times grad T dT by dL times Te minus Tw over Tw. So essentially you can rewrite the diffusion equation in terms of the Nusselt number from mass transfer. So you can calculate the diffusive flux for every species. Now supposing the condensing material is sodium sulfate liquid you have to write this equation then for all species containing for all I containing Na sulfur because the elemental sodium may be contained like I said in many many species the elemental sulfur may be contained in many many species. So you have to first estimate the diffusion fluxes for each of these species how about oxygen you also have to write a diffusion equation for oxygen you do not need to because oxygen is not a trace species the diffusion flux is only relevant for the trace species in the case of oxygen you basically make the assumption that there is sufficient oxygen everywhere in your system to enable the reaction to proceed. So we do not really have to write a diffusive flux equation for the oxygen okay. So let us say you do this and then therefore from this what can you calculate once you have the diffusive fluxes for each species containing these elements the next step is to write the diffusive flux for the element sodium which will be summation over all I of omega NaI times jiw dot double prime. In other words the diffusive flux of the element sodium will be a weighted sum of the diffusive fluxes of all species containing sodium with the weighting factor simply being the mass fraction of the sodium element in each species. And similarly you can write Js dot double prime equals summation over I omega si. So once you know this then how do you calculate the deposition flux of the sodium sulfate liquid you simply do a scaling. So if you know this the deposition flux let us call that some m dot double prime of Na2SO4 liquid which is the molten deposit on the turbine blade surface will be equal to J Na dot double prime divided by omega of Na in Na2SO4. So if you know the deposition flux of the element sodium and you know the mass fraction of that sodium in your deposit then you simply divided by the mass fraction that will give you the total flux of salt onto the surface because this has to be equal to Js dot double prime divided by omega of s in Na2SO4 I mean they have to be self consistent basically. So that is how you calculate the deposition flux of sodium sulfate on the surface and once you know the flux rate or the rate at which the sodium sulfate is depositing you can from this calculate the local thickness of the molten layer some delta subscript L which is going to be a function of spatial dimensions as well as time. So once you know the molten salt thickness and you know its physical properties such as diffused the liquid diffusivity its density viscosity and so on from this once you know this you can extract J O W dot double prime where this O refers to the protective oxide coating right. So far it is a 3 step process first you estimate the rate at which deposition of the molten salt is happening using that as well as the centrifugal and shear and aerodynamic shear forces that are present in the system you calculate the local thickness of the molten layer from that you calculate the rate at which the protective oxide is getting dissolved at any location. From that you estimate delta O at x, t which is the local thickness of the protective oxide layer as a function of again spatial dimensions and time and from this then you calculate the local rate of corrosion because there is a direct relationship between how thick the protective layer is and the rate at which corrosion is happening at that location. So you can see it is a very sequential process that enables us to start by all we know at the beginning is the fuel to air ratio, the fuel composition, air composition including impurities and the operating pressure that is all you need to know. From that by applying equilibrium thermodynamic considerations by applying appropriate transport laws you can calculate everything up to this point and then by knowing the properties of the protective oxide coating including its molecular weight, its dissolution characteristics, its diffusivity in the liquid and its local thickness distribution you can calculate this part of it and once you know that you can actually start generating what we had called corrosion maps which is exactly this rate of corrosion as a function of spatial dimensions and time okay. So I think it is a you know kind of a very elegant situation where CVD is happening and it is once we are able to understand the CVD process and modulate and simulate it and do necessary experiments both lab scale as well as pilot scale to validate your model I mean this is a very real life application and again corrosion problems in industry account for billions of dollars of loss all over the world. So it is a very practical problem in that you have to design components for long life in extremely aggressive environments and you cannot do that without a proper understanding of the CVD phenomena that are involved. Now there is one additional complication which makes this an iterative loop rather than a sequential loop and that is the fact that in order to form sodium sulfate there is an additional requirement the moles of sodium and sulfur have to arrive at a certain ratio. In fact the molar flux ratio of J of elemental sodium to J of sulfur should be equal to 2 otherwise a stoichiometric film will not result. So what you have to do is once you have calculated these two diffusive fluxes mass diffusive fluxes you convert them to molar fluxes by applying the molecular weight and then you check to see if they are in the ratio of 2. If not what you have to do is essentially iterate until you get convergence where the ratio that is being calculated is exactly equal to 2. So how do you do the iteration? What is the unknown? I mean the input conditions are fixed so you cannot iterate those. The mainstream conditions are fixed I mean whatever the combustion process happens that is going to set the conditions far away from the substrate. So what is it that we are assuming here? What is it that we can iterate on? What we are assuming in this calculation is that the elemental ratios, elemental composition of the gas remains the same wherever you are in the gas stream whether you are far away from the substrate or whether you are close to the substrate but that is really not a good assumption because close to the substrate the elements are actually condensing right. So there is actually a depletion in the elemental concentration near the surface which is not being accounted for in your model. So what you can essentially iterate on is the when you do your thermodynamic calculation with the free energy minimization calculation you have to assume that there is a certain quantity of sodium and sulfur that go into the condensate and from that you calculate the vapor phase composition. From that you calculate your transport fluxes and verify that their ratio is consistent with the formation of sodium sulfate. So you keep doing this process until you get to a condition that satisfies both constraints it should satisfy the equilibrium thermodynamic constraint of a stoichiometric sodium sulfate film being viable at the local conditions and B you have to satisfy the transport constraint that the molar fluxes must be in the ratio of 2 is to 1. So you just keep iterating till you reach this convergent condition. So the iterative loop is really here once you have reached a converging condition here then you go to the next step and in fact you know this equality what happen unless you have done this iteration and reached a self consistent condition where the flux of sodium element and the flux of sulfur element are in the same ratio as their stoichiometric ratio in the film. So that is kind of you know how we do this. Now supposing you have a case where you are forming not just sodium sulfate but it is also got a mixture of potassium sulfate in it okay then what do you do I mean can you still go through the same process of calculation and so on but this when you have a solution forming on the surface it introduces 2 complications one is on the thermodynamic side you have to know the phase equilibria between the 2 condensed phases in order to characterize the behaviour of the film on the surface. The second is in terms of transport constraints now in addition to Na by Js being equal to 2 you also have to have J of k by J of s equal to 2 so that you can have sodium sulfate and potassium sulfate condensing as 2 separate molten layers of course you can start forming ternary solutions where you have more than one alkali in solution let us say with sulfur. So you can have situations where maybe you will form something like an mg Na SO4 kind of a condensate. In that case J mg must be equal to J Na must be equal to Js for this film to happen. So just remember that whatever equilibrium composition you are predicting for the substrate for the film on the substrate must be consistent with the rates at which they are being transported. Now supposing that there is a situation where all of a sudden the temperature fluctuates and instead of forming sodium sulfate liquid let us say that you get a transition from this to Na2CO3 and that is typically a liquid to solid transition as well because sodium carbonate is a solid at much higher temperatures than sodium sulfate. So under conditions where you might expect to get molten sodium sulfate you may sometimes get sodium carbonate which by the way is a very positive thing because sodium carbonate is non-corrosive plus the fact that it is a solid essentially implies that it is not going to be chemically reactive with the surface. So in this particular case I mean so essentially as a designer you would try to achieve combustion conditions that favour the formation of either no deposit or a non-corrosive deposit rather than a corrosive deposit on the surface. So how do you get sodium carbonate to happen? What will be the transport constraint in this case? Is there a transport constraint or there is not? Clearly there are three elements here sodium carbon and oxygen but carbon and oxygen are again present in excess so they are not required to diffuse in a particular ratio compared to the trace element. So in this particular case the only transport constraint will be that the molar flux of sodium must be greater than 0. As long as that condition is met the transport conditions will permit the formation of a sodium carbonate film. In addition if the equilibrium thermodynamics also favours the formation of a sodium carbonate film you are okay you will form a sodium carbonate film and corrosion will essentially stop. And the condition where JNA equal to 0 as I mentioned I think in one of the earlier classes is what we call the dew point right it is the it is a condition which just separates the onset of condensation from a previously bad surface and so this condition must prevail if you want to have virtually no condensate on the film. And by the way if your if this ratio is equal to 2 sodium to sulphur is it still possible that J of NA equal to 0 and J of S equal to 0 is that even possible because it is basically gives you a 0 by 0 condition right it is possible in a limiting case and in fact the definition of a dew point is that in this particular case the deposition rate which we have written as m dot double prime sodium sulphate must be equal to 0 but in the limiting case this is in the limiting case of JNA tending to 0 and J of S tending to 0 okay. So the molar flux of sodium and sulphur must tend to 0 but in such a way that the ratio is always maintained at 2 so this is subject to JNA over being equal to 2. So in other words graphically the way you would estimate this is you will draw a curve of T w versus m dot double prime and you will look at how this is changing so you can draw a curve you know again the classic dew point curve at each point on this curve JNA over J S must be equal to 2 if this is giving you sodium sulphate as your deposit. So this curve is essentially a graphical formulation of this you are taking the limit of the of limit of well essentially you draw this curve and extrapolate it till it hits 0 right. Now this dew point again which has been achieved by slowly increasing the temperature and by the way this is the reverse of the conventional CVD situation where as temperature is increased the rate of deposition increases in this case of molten salt deposition it is actually the other way as temperature increases the rate of deposition decreases. So in this particular case the when you have a film that is present and you slowly raise the temperature the first temperature at which you are still satisfying the flux ratio condition but the rate of deposition itself is tends to 0 is called TDP the dew point temperature. However for this system the dew point is not unique because you can also define two other dew points one of them is a purely the thermodynamic dew point you can just run your chemical equilibrium simulations and find the first temperature at which the condensate will stop happening and you can call that a thermodynamic dew point but it will be different from this dew point and it will not be the true dew point because that calculation does not take into account the transport constraints on the system. So this constrained approach to 0 will yield a dew point which we call conventionally TDP minus TDP 0 which may be over here somewhere is what we call a purely thermodynamic dew point and interestingly there is one more supposing you were to initiate deposition by approaching it from the other side suppose you start with a very high temperature and keep lowering the temperature until the first condensate starts appearing now from a purely thermodynamic dew point there is no difference right but when you take transport constraints into account the constraint now is very different you do not have to satisfy this constrained approach to 0. So when you approach the dew point from the other side essentially you will be initiating deposition at a temperature TDP plus where deposition will begin but in this particular case I mean the first point at which deposition happens is the first temperature at which J of NA is greater than 0 and J of S is greater than 0. So it is a very different constraint right you are approaching the 0 point with 2 separate constraints one on the sodium one on the sulphur and the ratio constraint does not have to be met until this particular point. So it is a different approach or limiting case under different constraints so naturally this will yield a very different value. So there are 3 deep dew points that characterize the CVD system depending on if you calculate strictly based on thermodynamics or you calculate for the case of a pre-existing condensate that begins to vanish and the third is if you calculate on the basis of no film to begin with and then slowly approaching a condition where condensation begins to happen each of these will yield a completely different dew point and the spread between these dew points will depend on how disparate the molecular weights of the species are in the CVD system. So the more dissociated the system the more the number of species present in the system the larger will be the range between these dew points and actually the range between TDP minus and TDP plus is a very interesting area because some very non-intuitive things happen. You can have for example the formation of a patchy condensate where instead of a continuous condensate film you have a film that is present in patches. So essentially what is happening is you are so close to that you know dew point condition that the system almost cannot decide if the film will form or not form. So what it does is kind of statistically averages it. So over a certain area about half the area will have a film and half the area will not have a film right. So in this particular region if you blow this up and look at what is happening on the surface you will essentially have a situation that kind of looks like this there will be stripes in which the film will appear with spacing in between. So you can almost get patterns of film by operating in this range. This temperature range can be small I mean for example if it is a purely physical vapour deposition problem then this range is 0 right there is no difference between the thermodynamic dew point and the transport constrained dew points. However in a situation where you have let us say very high temperatures highly reactive highly dissociative environment this spread can be of the order of I would the largest values that have been observed are 20 to 30% of the nominal value. So if your thermodynamic dew point is 2000 Kelvin the range between TDP minus to TDP plus can be all the way from 1600 to 2400 right. So it can have a tremendous influence and if you are not doing your CBD calculations you can completely misestimate the range in which a molten deposit will form and cause corrosion. So you can provide much less protection than you should have or in another case you might over design and provide protection where it is not necessary. So this is again a classic illustration of where the CBD principles both the thermodynamic aspects as well as the transport aspects have to be fully understood for the design engineer to be able to design the system appropriately okay. So let us stop at this point any questions okay so I will see you at the next class.