 Welcome back. We are going to continue our discussion of sprays. Now, up until now we have looked at droplet formation processes, droplet breakup processes and droplet transport transport models involve different kinds of multi-phase flow models. In all, I mean in most spray applications if not all, droplet evaporation is the next physical process that is of relevance. You could take any process say starting with humidification. So, if I want to let us say humidify a certain space in you know where I am right now in Chennai of course humidification is not a problem. But if you go to a very dry area let us say sort of a very hilly mountainous region in winter you would need to humidify a room and the way that is typically done is you have a small spray in the in one corner of your room the drops that are generated in the spray evaporate and cause water vapor to diffuse into the air causing the relative humidity in the room to increase slightly. You could have other processes like spray drying which is a process that we discussed earlier as one that is used to make granular materials. So, you create a slurry of whatever granular material you want to ultimately manufacture in where the slurry is essentially the granular powder embedded or in liquid in suspension in a liquid. This slurry when it is sprayed the water has to evaporate out or whatever or whatever is your solvent water or whatever is your solvent has to evaporate out and you are left with the granular material that settles down. A third and one of the most important uses of sprays is in spray combustion. Whether it is your typical aircraft engine or a car you know with an IC engine internal combustion engine or power generation from liquid fuel sources typically involve generating a spray evaporating the droplets that make up your spray. So, you produced vaporized fuel whether it is vaporized gasoline diesel ethanol or aviation fuel the reaction the combustion reaction that creates your heat is in the vapor phase that is the typical steps leading to energy release leading to this chemical energy release. So, one of the important processes in this whole chain of physical processes is droplet evaporation and we will look at different models for droplet evaporation in today's class ok. So, like we said applications are in humidification spray drying and of course spray combustion which is a big part of why we discuss evaporation ok. So, we are going to consider an idealized model of a drop to start with we will take a drop that is in suspension some drop of diameter d 0 that is suspended in an infinite medium of some air or gas or some material ok. How does this liquid get converted to vapor and how is this drop what are the processes accompanying this drop evaporating. So, the first before we going to discussion discussing how drops evaporate we need to set the ground straight in our understanding of diffusion as a process ok. So, let us just look at diffusion more specifically Fickian diffusion. So, just to understand Fickian diffusion I am going to take a little one-dimensional simplified model where let us say I have some species A and species B two species separated by a little membrane and at some time t equal to 0 this membrane is removed. So, what you have happening is essentially the species A diffuses that way and the species B diffuses the other way ok. The transport of these species is defined based on a based on what we call mass flux and this has units of kilogram per meter squared second. So, this is mass flow rate which is kilogram per second per unit area. So, the two primes are basically they denote per unit area and we will define YA as the mass fraction of species A at a given point. So, if I define a coordinate system X this YA is the mass fraction at a given point. If I now write down at any given point X what is the mass flux. So, notice how this problem is a one-dimensional problem and if I ask a question what is the mass flux at a given point there are two reasons why you would have mass flux one even if I had one is just pure movement of the species this is what we call advection or convection advection is actually a better word to describe movement of mass. So, pure movement of mass. So, in other words if this whole two species mixture A and B were all moving at some velocity in reference to O, O is my origin. So, think of it this way I mean if this whole pipe containing the two species was moving at some velocity V in as observed by somebody stationary at O that would amount to a mass flux at a given X that would have to be accounted for. So, that is my first mass flux that is my first component of mass flux and that is given by m dot AA times YA YA being the mass fraction of A at that point in space. But there is also an additional component if the whole pipe was itself stationary why would this species move to the right and why would be move to the left it is essentially due to concentration gradients. This is a model due to thick and this comes in in the form of an additional term DYADX is the concentration gradient it is actually the mass fraction gradient of the species A DAB is called the binary diffusivity. So, that is the diffusivity of species A in another species B rho is the vapor density of species A and DYADX is the mass fraction gradient of species A. So, this part is now enough to tell me how a certain species is transported. So, I could either have advection which is a first term on the right hand side or I could have diffusion due to gradients in concentration and the total mass transport is a sum of these two. The negative sign here is only there to tell you that gradient diffusion happens against a gradient. So, diffusion of a species happens from a region of high concentration to a region of low concentration. So, if YA at x equal to 0 is higher than YA at some x plus at 0 plus delta x the flux that means the gradient is negative in the x direction flux is positive in the x direction. So, you essentially require this negative sign to make sure that the flux is positive in the x direction while gradient is negative in the x direction. Another way of thinking about it is to ensure that your diffusivity is always a positive number in any equation whether it is your heat conduction equation mass transport equation or momentum transport equation which is basically viscosity all these diffusivities can only be positive numbers because you already taken care of the fact that if the gradient is negative flux is positive by putting this negative sign. This comes from your essentially your second law of thermodynamics you take your Carnot statement of the second law of thermodynamics and when you rephrase it in a diffusion process mass momentum or heat the diffusivity has to be a positive number. Now, if I take if I now come back to the situation where I have a drop I will place my coordinate system here RS is the current radius of the drop and in that RS formulation I can look at what happens in this R coordinate system. So, instead of having we are still looking at a one dimensional problem but one dimensional in a spherical polar sense where everything only depends on the radial coordinate not on the theta or phi coordinates theta is like your latitude coordinate phi is like your longitude coordinate in a typical R theta phi spherical polar coordinate system we reduce that down to a one dimensional problem where everything all quantities are only functions of R your radial position from the origin O. So, in this formulation we are going to make certain assumptions we will first understand what the assumptions are the first assumption we are going to make is that the evaporation process itself is quasi steady, quasi steady means that I do have let us say some evaporation of this species A due to I have some flux of species A due to evaporation. This flux divided by the vapor density is essentially like a vapor phase velocity it gives you an order of magnitude of the vapor phase velocity that velocity is much larger than the rate of reduction of this species the rate of reduction. So, there is as this drop evaporates the drop boundary is going to shrink. So, there is a certain velocity of this droplet boundary shrinking that velocity is very small in comparison to other flux based velocities and that is what we mean by quasi steady and what it another way of looking at the same thing is that say for example I take a drop and I put a small source of liquid at the center of the drop. So, I have a small needle that I am injecting liquid into the middle of the drop in order to maintain this radius constant at some value Rs. Now, I am continuously going to facilitate evaporation of this liquid ok. So, this is common observation if I put a drop of liquid on a table like this it is going to be gone in say a few minutes. So, if I have a little needle protruding into that drop from the bottom of the table a continuously replenishing liquid I am going to keep such that I keep the liquid at that same Rs constantly I am going to create a concentration of the liquid vapor away from this liquid drop and that concentration profile away from the liquid drop is going to be steady that is not going to change with time. This is pure steady in the case of droplet evaporation model that we are going to do today we are not going to remove that we are not going to have a source of material in the center of the drop, but we say everything else happens as though there is a source of material and that source of material is replenishing it that is our meaning of quasi steady. So, it is like I have a source. So, I assume everything else is steady, but based on that steady process we are going to remove mass from the liquid ok. Another assumption is that droplet temperature is uniform inside the droplet and a third part is the mass fraction at the surface. So, this is the vapor mass fraction at the surface is at the saturation value. We will come back I want to make sure we understand this number 3 assumption very clearly. Number 4 we are going to assume that properties such as rho diffusivity are constant. So, just that is a simplification that does not cost us much at this moment. Let us understand what 3 is and to understand what 3 is we need to understand how a droplet evaporates in the first place. So, if I take a drop suspended in air this is a liquid drop the immediate and let us say this drop is in a room at some temperature T infinity everywhere. So, the drop plus the air around are all at the same temperature T infinity. How does this droplet evaporate the air in the immediate vicinity of this drop is going to have to be at the saturation value at this temperature. So, air at a given temperature is capable of holding a certain amount of moisture that is given by your Clausius Clapeyron equation in thermodynamics. So, that value of mass fraction that air can sustain at this given temperature and pressure actually also is called the saturation mass fraction. So, the air in the immediate vicinity of this liquid drop will always be at the saturation mass fraction will always hold saturation mass fraction amount of vapor. But the air far away is at a mass fraction has water vapor at a mass fraction slightly less than this or it could be 0. If it is a if it is perfectly dry air the mass fraction of the vapor far away is 0, but it certainly cannot be more than this saturation mass fraction it has to be less than this saturation mass fraction. So, whatever is the profile going from this saturation mass fraction value to the mass fraction value at R e core tending to infinity sets up a gradient that gradient in turn creates a diffusive flux any time we have a mass fraction gradient you have to have a certain diffusive flux. That means water vapor in the immediate vicinity of this drop is going to try to diffuse away from this drop and that diffusion in turn creates I mean I am now talking of this is sort of an equilibrium process, but you can think of this diffusion of vapor away from the liquid interface creating a need for more liquid to evaporate to saturate the immediate interface interfacial air. So, it is a continuous process as although I made it look like a stepwise process it is a continuous process wherein this vapor is diffusing away and liquid is evaporating right at the interface to saturate the air in the vicinity. This continuous process is pumping liquid into vapor, vapor diffusing away and in some due course of time you do not have any more of a liquid drop remaining. So, you essentially evaporated all the liquid out and that liquid mass has gone into the air as vapor. So, how do we so essentially what we are saying in number 3 here is that during this entire evaporation process the air in the immediate vicinity of the liquid interface liquid interface is always at the saturation mass fraction value which is given by the thermodynamic with which is given by the Clausius Clapeyron equation from standard thermodynamics. Now I use water as an example to illustrate this, but the same argument works with every liquid in every other gaseous species ok. So now if I have a certain mass flux of vapor so m dot m double prime dot a is a mass flux of vapor species a given that the m dot is nothing but 4 pi r squared if I take any radius r the actual m dot crossing that sphere of radius r is given by 4 pi r squared the surface area of that sphere times my m dot double prime a. So, this m dot double prime and this m dot is constant for all r ok. This is our meaning of steady process that I have some liquid being replenished to the middle such that I have a constant mass flow rate of species a at all radial locations are. So, first writing Fick's law in spherical polar coordinates in one dimensions in one dimension will find I can rearrange a few things here. So, for a given steady state concentration profile that I want to sustain I can replenish. So, if I use R s which is my radius of the drop if I put in this much mass flow rate in the center of my liquid drop I will I will evaporate the liquid subject to. So, if I know my mass flow rate that I put in if I know the mass flow rate that I put in if I know m dot this equation essentially gives me a way to calculate y a the mass fraction as a function of r. So, I am going to assume I know the mass flow rate I am putting into the middle of this drop. So, I am going to assume for a moment that I know m dot and calculate y a as a function of r. So, if I take this that gives me let us make sure we understand the boundary conditions at r equals r s is a saturation mass fraction. So, notice how this is a one dimensional if I rearrange this in the form of a differential equation it will say d y a d r equals minus m dot into 1 minus y a by 4 pi r s squared rho d a b or 4 pi r squared. If I say this is valid at all r at any r this is the mass flow rate flowing across the sphere m dot is a mass flow rate flowing across the sphere at any radial location r. So, if I this is a 1 this is a first order ordinary differential equation I only need 1 boundary condition, but I know I have to use I have 2 boundary conditions I will choose the first one and we will see why the reason we have why we have this sort of additional boundary condition we will see the reason for that in just a moment. But if I use just the first boundary condition and solve this ordinary differential equation I can find y a as a function of r and that comes out to be notice. So, essentially the concentration profile or the mass fraction profile is a function of the mass flow rate itself, but I am thinking the mass flow rate is determined by what is the saturation mass fraction at the drop and what is the what is the mass fraction far away from the drop. These are the only 2 parameters that have to determine the mass flow rate I cannot determine the mass flow rate independent of those 2. So, if I now use the second condition to find m dot given y a as r tending to infinity equals y a infinity. If I use that notice how I have m dot divided by 4 pi rho d a b r. So, I will make sure I put this in parentheses to be exact. So, this r goes to infinity which means m dot divided by that whole thing goes to 0 I have an exponential of 0 1. So, y a at r tending to infinity is 1 minus d a b r s and I know this is equal to y a infinity. If I rearrange a few things all I have done is move the y a infinity to the right hand side and move the rest to the left hand side from here I can write out m dot sorry this would be s on top. So, this is the mass flow rate across any sphere of radius r for a given y a infinity and y a s y a s is the mass fraction in the immediate vicinity of the drop y a infinity is the mass fraction far away from the drop. We will just look at a couple of limiting cases if y a infinity is equal to y a s that is if the air outside the drop is saturated with vapor then you can see that that ratio comes to 1 log of 1 is 0. So, you have no mass flow rate you can take another sort of an example if y a infinity is actually greater than y a s that is I have a drop that is slightly cooler because of which the mass flow rate because of which the saturation mass fraction is less than what I have available outside. I am going to have condensation otherwise this is in general a negative number which means I have evaporation. If y a infinity is less than y a s sorry m dot being positive amounts to evaporation m dot being negative amounts to condensation. So, as of now we really have no distinction we have not made any distinction between evaporation and condensation as far as this process is concerned. So, this is the rate of mass loss from the drop for a given y a infinity and y a s. Now if I take a drop the drop itself is losing mass at this rate m d is given by rho l pi d cubed over 6. Now at some point in time we are going to make the change to diameter this r s is nothing but d over 2 in the next equation r s is the is the radial location of the surface of the drop that is equal to the drop diameter divided by 2. So, if I now substitute what I have for m dot into this equation here for m d what we find is that d dt of rho l times pi d cubed over 6 pi d cubed over I am sorry pi d cubed over 6 is the volume of a sphere of diameter d. So, if I take this rate of change of rho l times pi d cubed over 6 to be equal to essentially I have this 2 pi d rho times the diffusivity times log of 1 minus a infinity I want to integrate this in time. This d is diameter and the script d is our diffusivity let us be clear about that. So, in the term in the square parenthesis we want to take the rate of change of the term in the square parenthesis with respect to time the only part which is changing with respect to time is my d rho l is of course assumed to be constant. So, if I invoke that I will find rho l times pi over 6 times 3 d squared times d d dt equals minus 2 pi d rho d ab. I can scratch out one of the diameters and when I do this simplification what I will find is this equation that d d squared dt. This d d d can be written as half of d d squared and I have one half coming from here and when I rewrite this whole thing I have d d squared dt it is customary to define a term called b y or the transfer number given by b y equals by 1 minus y as and now I can show that d if I now notice how all these all this part in the inside my yellow oval is constant in time ok. So, what this is telling me first of all is that d squared varies linearly with time ok. So, if I now complete this process I can write this as some d 0 squared minus k t I will call this k q I will tell you will for quasi study k q equals you can easily show that with this definition of b y what is inside the argument for the natural logarithm is nothing but 1 plus b y. So, when the saturation humidity and the humidity at infinity are equal saturation mass fractions and the vapor mass fraction far away are equal your transfer number is 0 ok that is like saying I do not want any mass transfer to happen ok. And this k q is given by this equation and d 0 is the initial drop diameter if I subject an initial drop if I subject a drop of initial diameter d 0 to an evaporation process that evaporation process is going to proceed such that if I plot d squared versus time the initial value is some d 0 squared then it is going to decay linearly such that the slope is given by k with the value of this k q and you have a finite lifetime of the drop there is a finite point at which the drop reaches a diameter 0 and that t l is given by d 0 squared over k q ok as simplified an analytical procedure as what we followed to get to this kind of an equation. It has been verified empirically in many different instances that this t l the lifetime of a drop is very reasonably predicted very accurately predicted as a matter of fact in many different liquids and gaseous species. So the assumption that we made that you have a quasi-steady evaporation process is actually quite reasonable in even a realistic process where a drop may be moving slowly ok. So let us go back to our initial assumptions and see which ones are like the worst culprits for a real spray we are assuming quasi-steady that means inherently we are assuming that the droplet is stationary with the air around being stationary there is no movement of the air itself there is no convection of the air that is going to enhance this mass transfer process. So if I take a liquid drop and have some flow of air around pass this liquid you are going to tend to evaporate the liquid faster at the moment this analysis does not account for that but as far as quiescent single droplet evaporation is concerned this is reasonable but what has been found empirically also is that even if you have a superimposed air velocity the phenomenology that the diameter squared decreases linearly with time seems to hold very well ok. So diameter squared still decreases linearly with time so if the air is not quiescent but u infinity is the velocity of air we can define a Reynolds number based on rho air u infinity d0 divided by mu air and there is a very famous very famous work by a German author by name Frussling who developed this Frussling correlation that says that k in the presence of a u so k in the presence of some advection pass the drop divided by k quiescent is of the form 1 plus a times re power n ok a is I think typically about 0.3 and n is also about 0.25 to 0.3 so it is a relatively weak dependence but you can find a correction for the k q that you measure k q that you estimate from pure analytically from pure thermodynamics and mass transfer k q is only a function of the diffusivity vapor density rho liquid density rho l and your transfer number by so you are able to correct for this k q in the presence of a finite Reynolds number by using this Frussling correlation but your phenomenology that your d squared the surface area of the drop still decreases linearly with time is a very good is has been empirically shown to be reasonably accurate assumption. So all you have to do now is use this k u for your given Reynolds number calculate from the Frussling correlation what your k u will be and d squared would be d 0 squared minus k u times t. So if your Reynolds number is is of some finite value k u would be higher than k q that in turn means the lifetime of your drop is lower. So if I am looking at whether this whether I should use the Frussling correlation or the other way is is sufficient all I have to do is look at the time scale associated with the evaporation in comparison to the time scale of transport of the drop itself. If the drop is moving slowly in a otherwise quiescent air the time scale of movement of the drop may be small in comparison to d 0 squared by k q that means the drop is evaporating much faster than it is being transported. So I can just ignore the Frussling part and use k q alone but if I have some finite time scale of motion that is comparable to k q comparable to t l that we showed for the lifetime of the drop this is this has to be accounted for. We will stop here we will continue our discussion of spray combustion in the next class.