 In this 31st lecture, we now begin with convective mass transfer. As you will recall, the first 20 lectures we spent on laminar flows in external boundary layers and ducted flows. Then we spent 10 lectures on how to tackle turbulent flows and heat transfer. Now, we turn to convective mass transfer. So, we have 10 lectures to do that. First of all, I will explain what the importance of the subject matter is convective mass transfer and what is our main task in convective mass transfer? What are the governing transport equations? I will then convert them to the boundary layer forms of that. But unlike what we did for momentum and heat transfer, where we actually solved the boundary layer forms of equations in their differential form, what I am going to do is here develop very simplified models of boundary layer forms of equations, which are coupled and partial differential equations. But you will see in the simplified forms that I shall develop, you would have essentially one-dimensional equations are good enough to represent what happens in a real boundary layer flow. Finally, there are three models of this type. The one is called the Stefan flow model. The other is the Couette flow model. Both these rely on one-dimensional ordinary differential equation, whereas the Reynolds flow model goes even beyond and says that even an algebraic model is capable of predicting whatever would be predicted by the full boundary layer flow model. So, to that extent, our approach to the entire convective mass transfer is not based on solving the partial differential equations of a boundary layer, but is actually based on solving the simplified model equations. I will be showing that it turns out that the solutions produced from these models accord very well with the boundary layer solutions that are obtained. Several applications such as evaporative or transpiration cooling in combustion, in ablation, in condensation or in drying, convective mass transfer assumes importance. Mechanical engineers need to deal with all of these in engineering practice. The convective mass transfer always takes place between two dissimilar phases, such as liquid to gas or a solid to gas. So, many a times convective mass transfer is referred as interface mass transfer. For example, evaporation is liquid to gas mass transfer, ablation on the other hand is solid to gas or combustion can be of a wood particle or a coal particle would be solid to gas mass transfer problem. Now, in the absence of fluid motion, we can regard convective mass transfer as diffusion mass transfer. You will see that it is very analogous, but not exactly same as heat conduction. Diffusion mass transfer takes place within a single phase gas solid or liquid as you will see as we go along. The examples of diffusion mass transfer are several. For example, leakage of hydrogen by diffusion through solid container wall. For example, if you store hydrogen in a solid container, then hydrogen is very leaky as they say. So, even through the solid wall, hydrogen would diffuse through the wall and come out and it could be dangerous from fire point of view. Penetration of carbon through iron would be an example. We use this method for carburizing iron to harden the surface of the iron and penetration of carbon through solid iron would then be an example of diffusion mass transfer. Similarly, diffusion of water vapor through solid during drying would also be an example of diffusion mass transfer. Imagine that you have a tea cup and I have tea here. So, I will drink that, but you have a tea cup in which you have put sugar and the liquid is stagnant. After a while, we will see the tea turned sweet. That is because the sugar dissolves and diffuses through the tea and you taste the tea sweet. Of course, in order to hasten this rate of diffusion, what you do is you stir it, which essentially means you are bringing about convective mixing rather than diffused mixing. So, that is the difference between convection and diffusion. The convective rates are much higher than the diffusion rates. So, let us turn to convective mass transfer. As I said, it takes place due to concentration gradients of a transferred species. Let us say this is a flat surface. It can be water, for example. Let us say wind is blowing from left to right and the air is relatively dry compared to water. Then there would be evaporation of water from liquid state into the vapor state and the air would simply take away the vapor. In mass transfer problem, we always draw a figure like this to understand what we are doing. The air into which mass transfer is taking place is called the considered phase and the water from which the mass transfer is taking place is called the neighboring phase. So, the mass transfer always takes place from neighboring phase to considered phase or from considered phase to neighboring phase. In the latter case, the mass transfer flux n suffix w kilogram per meter square second would be negative and that is the interface. That is the interface which is like the wall state. We always do think of three states, infinity state, the wall interface or wall state and the transferred substance phase or the neighboring phase which I shall define very shortly. Our task is to show that n w is actually directly proportional to the driving force of mass transfer b. b is a dimensionless driving force and the constant of proportionality then is called the mass transfer coefficient g again equal to kilograms per meter square seconds. So, unlike in heat transfer where heat flux is related to a temperature difference and heat transfer coefficient has units different from the heat flux. In mass transfer, the mass transfer flux and the mass transfer coefficient both have the same units because the driving force is dimensionless. n w is positive when mass transfer takes place from the neighboring phase into the considered phase across the interface and vice versa. So, what I mean here is that if evaporation takes place then n w would be positive. On the other hand, if condensation would take place then the n w would be negative. It is important therefore, to show the direction of n w which is positive when it is from neighboring phase to the considered phase. In mass transfer, we are always dealing with mixtures. For example, in an evaporation problem in the considered phase you have a mixture of air and water only two spaces are involved. On the other hand, in a combustion problem there can be several spaces like methane oxygen, carbon dioxide, carbon monoxide, nitrous oxide and so on and so forth. I mean they can be depends on the temperature levels attained and so on and so forth. So, the considered phase could have several spaces. On the other hand, the neighboring phase itself could be a fuel if it is a pure liquid let us say fuel like diesel or something like that. Of course, there would be only a single space called the fuel space. But if for example, the diesel itself is mixed with water then the fuel itself is not a single space but a mixture of spaces. So, both the neighboring phase and the considered phase are actually mixtures of several spaces and it depends on the problem at hand how many spaces are involved in these two phases. Anything from one to several is possible. So, several spaces are involved the proportion of k in the mixture must however be defined. This proportion of species k in the mixture must be defined and there are several ways to do it. One is called the mass fraction. So, for example, omega k is used as a symbol for mass of species k divided by mass of the mixture. If I divide this by volume of the mixture and multiplied by volume of the mixture divided by mass of the mixture then you will see that this quantity would really transform to 1 over rho of the mixture. What about this quantity? Well this quantity can be written as rho k. Why? Because Dalton's law says that in a mixture each species behaves as though it occupied the total volume of the mixture and therefore this would be rho k and therefore the mass fraction omega k would be rho k divided by rho mixture. Since the sum of the masses, since sigma m k must equal m m mix therefore sigma omega k must equal 1 and rho mix will be simply sigma rho k in a mixture. Now, this is a very important relationship that we will be using to define the proportion of a species k in the mixture by that is mass weighted proportions is defined by symbol omega suffix k rho k by rho mix and it uses Dalton's law. Now, chemical engineers often use mole fractions and combustion engineers also use many times mole fractions to define species composition and it is defined as x k as number of moles of k divided by number of moles of the mixture and in gases if we assume validity of the perfect gas law then this can also be shown equal to the partial pressure of species k divided by the partial pressure of the mixture. So, x k is simply the ratio of partial pressure of k divided by the total pressure of the mixture. So, n is the number of moles. Chemists on the other hand prefer to use concentration and the symbol is square brackets and inside that k and that is simply rho k divided by m k or essentially in a way it is a molar density. And if we use perfect gas law then you can see that can be written as p k by r u t which is nothing but molar volume v bar k. Now, any quantity defined based on in molar units is given an over bar. So, this is kilo moles per meter cube and m k is the molecular weight of species k. Now, of course, in all our calculations we shall prefer to use omega k which represents the proportion by mass. Let us define driving force. Now, unlike in heat transfer in mass transfer we always refer to three states. The first we will call the reference state far into the considered phase which is like the infinity state that we have already defined. We know where the gradients of any quantity in the direction y are 0. So, likewise velocity gradients are 0. Any quantity temperature or omega k are all 0 in this infinity or reference state. Then there is the interface state which is defined as with the symbol w and then we must always consider the transferred substance state which is deep inside the neighboring phase. One of the principle assumptions of the transferred substance state is very similar to the infinity state that there are no gradients of omega k or velocity or temperature or anything for that matter in this transferred substance state. So, all quantities are uniform in the transferred substance state. The mass transfer can be shown to take place from here. It can be shown across the interface and into the considered phase. So, this is the velocity profile for example, and this is the profile of temperature of any other conserved property called phi which I shall define. Now, the driving force is always defined based on this conserved property. It is a ratio of difference between phi in the reference state or the infinity state minus phi at the interface w divided by phi w minus phi and we shall repeatedly encounter this dimensionless driving force phi is called conserved property. Now, I will explain what conserved property is in a short while, but it can phi can be formed from mass fraction of k. It can be formed from mole fraction of k. It can be formed from concentration of j, but as I said we would always prefer to use mass fraction of k in all our development. So, note this definition of B which we shall repeatedly encounter. It is a dimensionless driving force of the quantity of a property in the infinity state, in the w state and in the transferred substance phase. Now, in general there are three types of mass transfer problems. The first type is what I shall call inert mass transfer without heat transfer and without chemical reaction. What I mean by that is that supposing I have water at 30 degree centigrade placed in a beaker which is also at 30 degree centigrade and let us say the air around it is also at 30 degree centigrade. In other words, there are no temperature variations anywhere and the dry air is however relatively dry let us say 10 percent or 20 percent or 30 percent relative humidity whereas, the humidity or the conditions at the surface of the water would be of course saturated corresponding to these temperatures. So, there would be concentration gradient in air, but there would be no temperature gradient right from the transferred substance phase to the infinity phase to the considered phase and of course, there is no chemical reaction. Now, will the mass transfer takes place in the absence of any temperature variation? Of course, experience tells you that water will evaporate although there are no temperature gradients and the principal reason is that there are concentration gradients and just like heat conduction is a spontaneous process which arises as a result of temperature gradients mass transfer arises as a result of concentration gradient in this case the concentration of water vapor in air. So, it is a spontaneous process and that is a situation which we shall call inert mass transfer without heat transfer or without chemical and without chemical reaction, but now imagine that I have this water which is at 25 degree centigrade and in the infinity state I have the gas which is at 50 degree centigrade. So, it is a hot gas and cold water of course, I do not know what the surface temperature of the water will be, but presumably it will be between these two extremes and evaporation would take place as you know it will take place, but in this particular case there will also be heat transfer from the gas considered phase to the transferred substance phase which will bring about evaporation of water. In other words, the spontaneous evaporation in the absence of temperature gradient would now be enhanced its rate would be enhanced because of the heat transfer which is accompanying it. Now, think of the other way for example, I have hot water let say water is at 50 degree centigrade and the air is at 27 degree centigrade again I do not know what the surface temperature will be, but it will probably be in between the two and we will find out the ways of determining that temperature in a minute, but the main thing is even if the water is hotter or the and the air is cooler again evaporation would take place. So, positive mass transfer takes place even when heat transfer is to the transferred substance or from the transferred substance to the considered phase. In other words, heat and mass transfer are simultaneously occurring, but they can be opposed or aiding each other. We consider this to be the situation of inert mass transfer with heat transfer, but no chemical reaction. Third type of course is the mass transfer in which let us say liquid fuel is actually burning away. The manner in which liquid fuel burns is that it first evaporates turns into a gaseous phase where it undergoes chemical reaction and burning of the liquid phase liquid fuel continues. This is a situation of mass transfer with heat transfer and chemical reaction. So, I classify all mass transfer problems in three types. In each case, conserved property phi must be appropriately defined. Now, what is a conserved property? Any transport equation of the form d rho m by phi by dt plus d by dxj, a convective flux and a diffusive flux is equal to S phi. This is the well known form that we have seen so many times. Now, suppose for a chosen phi, if it turns out that S phi is 0, then S phi is called the conserved property. In short, a conserved property is one whose transport equation has no source term. It simply has a rate of change term, it has a convective term and a diffusion term. These are the only three terms that would be present in a conserved property equation. The task is to define appropriate conserved properties for each of the three types. We simply recall this bulk mass equation and I have now put suffix m in all this. The equations are written in the considered phase because in the transfer substance phase, there are no variations. So, all these equations are written only for the considered phase and this is the bulk mass with rho suffix m. This is the momentum equation, this is the species equation, the so called mass transfer equation which I had derived in the first few lectures and this is the energy equation in which this is the convective term, this is the conduction term, this is the convective mass transfer due to convective heat transfer due to diffusion mass transfer. This is the total derivative of pressure, there would be chemical heat generation due to chemical reaction or radiation heat transfer or viscous dissipation. So, all these terms may very problem depend and depends on what type of problem you are dealing with. By and large we will be dealing with situation in which that is 0, that is 0 and that is 0. Sometimes we may consider how accounting for radiation heat transfer as well, but that we will see as we go along. The boundary layer flow model of forms of these equations can be generalized in this way that you have d rho m psi by dt plus rho d rho m u by. So, these are the convective terms, this is the rate of change term, this is the diffusion term only in y direction notice because of the boundary layer model and SW which is the source term. So, if I put psi equal to 1 it would simply mean I am talking about bulk mass conservation psi is equal to u would mean momentum equation, omega k would mean species equation with a rate of generation of species r k and then H m is the enthalpy of the mixture which with the diffusion coefficient k m effective divided by C p m and then all these source terms that would arise for a boundary layer flow model. Now, remember I have said here effective and so on so forth, this is simply effective viscosity of the mixture, this is effective diffusivity of the mixture and this is the effective conductivity of the mixture. The diffusion rate m double dot y k which essentially means diffusion in y direction of species k is simply from fixed law is simply rho m d effective d omega k by d y this is just by way of reminder. So, the general comments on the boundary layer model the only thing that requires a little further elaboration is that the enthalpy H of the species k, you can see this is where the species k is involved. So, and this is the mixture enthalpy, this is the enthalpy of the species k, then the enthalpy of the species k in a chemically reacting mixture is written as H naught f k plus del H s k where H naught f k is called the enthalpy of formation plus the sensible enthalpy to go from reference state to t state. H naught f k is defined at the reference temperature t ref to t and sensible enthalpy is defined as t ref integral t ref to t C p k d t species do have gaseous species do often have C p k is a function of temperature often fairly weak function, but nonetheless a function of temperature. Now, this H naught f k assumes different values for different species in a when you have chemically reacting mixture, if it was an inert mixture I could assign any value to H naught f k as a datum value, but in a chemically reacting mixture H naught f k is taken as 0 for naturally occurring substances like carbon, oxygen, nitrogen and so on and so forth, but all other compounds like C H 4 or C O 2 or C O all of them would have a certain value for H naught f k from which these are well tabulated in literature. So, there is no problem in getting the values of H naught f k then from fixed law of mass diffusion as I said m double prime y k is simply defective rho k by d y and rho k can be written as rho m into omega k and that is what I have shown here. Now, boundary layer flow model itself is an idealization of the transport equation, but it still involves solution of simultaneous equations. Why? Remember in heat transfer problems we could often dissociate or decouple the momentum transfer problem from the heat transfer problem. We could do that by assuming constant properties, but even if I assume constant diffusivity it is not possible to decouple u and omega k and energy equation. The main reason is as follows. So, in a mass transfer problem let us say the velocity profile is of course influenced by N w which is rho w into V w and N w itself is governed by the gradients of concentrations N w is governed by the gradients of concentration at the wall that is what brings about mass transfer. Therefore, the velocity profile and the concentration profile are actually coupled to each other. So, you could not solve the momentum equation independently of the concentration problem and if the energy equation was present and if heat transfer was also present then you would have to solve the temperature equation or the enthalpy equation along with it. All the three equations the enthalpy equation or the energy equation, the species equation that is for omega k and the momentum equation u are always coupled through this boundary condition through this boundary condition. In a way a mass transfer problem is very similar to a natural convection problem in which the momentum equation is coupled with the energy equation or the temperature equation and the coupling arises due to the buoyancy term in the momentum equation which you have done in your undergraduate work. The coupling arises because through a source term in the momentum equation not through the boundary conditions. In a mass transfer problem the coupling between momentum and the mass transfer equation and the energy equation arises through the boundary condition which specifies N w the mass flux as rho times rho w times V w and V w is a boundary condition for the momentum equations. Remember we always have a mass transfer problem is always a non-linear coupled problem and it requires simultaneous solution of all these and remember there can be several species in a given problem. Evaporation of water is the simplest because it has only two species in a combustion problems where a chemical reaction mechanism can involve as many as 40, 50 species. Then of course, what it means is that you need to solve so many species equations along with the momentum and the energy equation. So, the mass transfer problem can be very big it depends on what kind of reaction mechanism you have we have specified. In the absence of chemical reaction life is simple and you will simply need to solve momentum and energy equations. The boundary layer flow model itself is an idealization of the full transport equation. It still involves simultaneous solution of several coupled differential equations and the coupling as I said arises mainly through the boundary conditions and therefore, you cannot separate the momentum fluid flow problem from the species equation as we could in a pure heat transfer system. Essentially, I said we are going to describe to you through three simple models. The main reason for that is these models have been developed essentially to circumvent the need for solving the full mass transfer problem involving several differential transport equations. So, we will postulate simpler models to serve a limited purpose of essentially getting N w, but the solutions from the model equations can reveal tendencies inherent in the complete boundary layer model. So, we do not claim that everything conceivable can be captured by the simpler model, but the major tendencies would certainly be captured by the simpler models. So, that is our task. Before I make a pouring into this simpler model many a times the reactions take place in such complex manner that it is simply not possible to deal with species k or the mixture, because you simply do not know what the concentrations are of the species in different states of that is the neighboring phase, the wall phase, the interface and the independent infinity state. So, it becomes very difficult to know the exact concentration or the exact mass fractions of all the species involved in these three states. In such a case what one does is to invoke the element conservation that is in a chemical reaction although the number of moles may not be conserved or may be conserved elements which form these compounds must be conserved. You cannot destroy or generate elements consider a mixture comprising let us say methane, oxygen, hydrogen, water vapor, carbon dioxide, carbon monoxide, nitrogen, nitric oxide and atomic O. Now, we define a element mass fraction symbol uses eta unlike for the species we use omega for the element mass fraction eta and what are the elements involve C, H, O, N, N. So, we have four elements these are the typical elements and any few of C, H, O, N systems are well known. Then element fraction is defined as eta sub C would be molecular weight of carbon divided by molecular weight of methane multiplied by mass fraction of methane. Molecular weight of carbon divided by molecular weight of CO2 omega CO2 basically what it says is the element fraction in methane would be this much element fraction of C in CO2 will be this much and element fraction in CO will be this much. Of course, other species do not have C and therefore, we need not worry about them, but hydrogen for example, is involved in H2, H2O. So, therefore, N in methane and therefore, hydrogen element concentration or element fraction would be 4 by 16 into omega CH4 2 by 2 into omega H2 and 2 by 18 into omega H2O. O is to be found in oxygen in water vapor in carbon dioxide, carbon monoxide, nitric oxide and elemental all. Then you will see this would be the case 32 times because O2 with 32 times 44 into omega CO2. So, in other words element fractions can be formed from the species fraction. Finally, the eta N which is the nitrogen element fraction would be 14 by 30 in nitric oxide and 28 by 28 in nitrogen itself. These element concentrations they can be convected and diffused, but they cannot be generated or destroyed which means elements are always a conserved property and that is a great advantage because you do not need to know what is the rate at which they are being generated or destroyed. Therefore, in general the mass fraction I have just simply generalized what I said on the previous slide that eta alpha the mass fraction of element alpha is eta alpha is the mass fraction of element alpha in species k. So, N alpha k omega k and N alpha k is simply the molecular weight of alpha the element divided by molecular weight of the species. Now, just as species are convected, diffused, generated and destroyed elements can also be considered to have been convected and diffused, but they can never be generated or destroyed as I said earlier and because of the principle of element conservation. Therefore, the transport equation for element alpha will have no source term and it would simply have a rate of change term, a convective term and a diffusion term where it is assumed that diffusivity for the elements is same as that for the species N alpha for the species k and N alpha is always a conserved property. Now, let me turn to the three simpler model. The first one among these is the Stefan flow model and it is essentially a boundary layer flow model in which we say u is 0. We also say dp dx is 0 and now included the area change. So, it would essentially be a one dimensional model as you can see from this equation. If I say d by dx is 0, u is 0 therefore, no question of that term being there. I am also saying that the pressure gradient is 0, there is no body force either. So, all these are simply 0 and rho m into v would be the mass transfer flux and that would be essentially now one dimensional model and it would simply say d by dy of n psi y into a which the n psi y is the mass flux at any position y multiplied by area a. I am allowing for area change here. It would be equal to d by dy of rho m v a psi minus the diffusion term would be equal to a times the source term which of course if psi is a conserved property s psi would be 0, but otherwise it would take the meaning that is associated to here with psi. At the moment I am using psi as a variable proper then that would be 0. Now, in the Stefan flow model you do not need to solve the bulk conservation equation or the momentum equation because u itself is 0. Therefore, we simply have to solve psi can only be omega k or element alpha or it can be the enthalpy H m and it would have an appropriate source term. So, of course, because u is 0 which means it is a stagnant medium there is there can be no turbulent contribution to effective exchange coefficients. So, essentially it would always be a laminar diffusion through stagnant surroundings. Now, notice the nature of the model. I have said this is the interface. Let us say this is water and this is the stagnant air over it. In the infinity state I may have air of any relative humidity less than 100 percent. So, that the mass transfer will take place, but as the mass transfer take place the level of the water would go on decreasing or the length L would go on changing. But what is implicit in this model is that somehow water is being injected in the transferred substance state T state at the rate equal to the evaporation rate. So, that L remains constant and therefore, you see no transient term written in this equation. So, Stefan flow model is for a steady state mass transfer in which implicit is the assumption in the transferred substance state. The mass transferred amount is somehow made up by supplying the transferred substance. Coed flow model goes closer to the boundary layer flow model, but instead of elaborate velocity profile which is like this and which is actually influenced by the mass transfer rate. This shape is taken to be simply as linear. So, in the Coed flow model as a rule we say u is going to be a constant multiplied by y at all positions x. The pressure gradient will be 0. In fact, all other gradients in x direction will be 0. Therefore, the equation would look remarkably similar to the previous equation, but psi can be equal to u because the flow has been allowed now and u is equal to c times y. Unlike u equal to 0 in the Stefan flow model, you have u equal to cy, but we say dp dx shall be 0. So, there is no body force term or anything like that in the considered phase. Now, this model, the merit of this model is, because it deals with the flow I could allow for variation of omega psi with respect to y. Now, that can occur because sorry variation of gamma psi which is a diffusion coefficient or exchange coefficient. It can be a function of temperature, it can be function of mass fraction in a turbulent flow, it can be function of the position as indicated by the Prandtl's mixing length. For example, the exchange coefficient, the turbulent viscosity for example, goes on varying with y. So, the great merit of the Coed flow model is that with an assumed linear profile, you can actually study effects of property variations. In fact, that is what the purpose of this limited purpose of this Coed flow model is. It is not to predict the mass transfer coefficient g or to predict mass transfer rate so much as to really examine the question of what would happen if the properties were to vary from the wall state to infinity state. It turns out that the model shows that the nature of variations that are captured by this model are actually found even in the full boundary layer flow model or very close to it. With very simple algebra, you can show what the likely effect of property variation shall be and that is why the merit of the Coed flow model lies. Finally, I come to the algebraic Reynolds flow model and it was postulated by Spaulding D B in a book called Convective Mass Transfer published by Edward Arnold in London in 1963. This is an algebraic model, very simple and therefore, there are no differential equations to be solved. What one says is that there is a considered, there is the interface, there is the considered phase, there is the infinity state and then there is the t state. So, t state has the same meaning as before. This is the interface and what the model postulates is that there is a fictitious inward g flux in the infinity state which will bring with it properties of the infinity state. So, the model simply postulates a fictitious flux with a symbol g which flows towards the interface from the infinity state and carries with it properties of the infinity state. On the other hand, in the infinity state there is n w plus g flux which flows away from the w state, but it carries with it from the values of the w state, the properties of the w state. So, n w g flows away from the interface, g flows towards the interface and therefore, there is no net generation or depletion of mass as a result of postulation of these two fluxes. Now, at first sight this might appear to be highly arbitrary. The model claims to account for all effects produced at the w surface in a real boundary layer flow. That is the claim. In the lectures to follow, we shall show how that claim is justified. So, in summary let me say like in heat transfer, the greater part of the resistance to mass transfer is confined to near wall region and therefore, boundary layer flow model should suffice, but this requires computer solutions to govern equations. In fact, a very large number of governing equations which are coupled and to avoid this and to establish the validity of n w equal to g times b simplified models are invoked. The Stefan and fluid flow models are one dimensional and hence analytically derived closed form solutions become possible. Reynolds flow model on the hand is the simplest being algebraic. I will develop these mathematics of this model, very simple mathematics which will convince you why all these models capture the real flow effect which is n w is equal to g the mass transfer coefficient multiplied by the dimensionless driving force b.