 So, we are now familiar with the three simplifications of the boundary layer flow model. The first one was the Stefan flow model, second one was the Kuwait flow model and the third is the Reynolds flow model. In today's lecture, I shall develop the Stefan flow model further for variety of mass transfer problems that we encounter in engineering practice. The types of problems that we encounter are as follows. First is the inert mass transfer without heat transfer or chemical reaction. This means something like simple evaporation of water, where the water and the environment are all at the same temperature. So, there is no temperature gradient and as a result, there is no heat transfer nor is there any chemical reaction because water simply evaporates without any chemical reaction and the mass transfer however takes place because of the concentration gradients. The next would be where let us say the water droplet or water itself in a lake is at a different temperature from that at the infinity state or the environment, in which case there would be heat transfer either to the water or from the water to the environment depending on which temperature is larger. Then we will move to the situation in which there would be chemical reaction or combustion in which let us say liquid droplet or a would be burning along with heat transfer from the environment to the droplet and finally, but there we would use what is called as a simple chemical reaction and I will explain what that means, but many a times particularly when dealing with burnings of solids, the reaction mechanism is so complex that we have to deal with that situation somewhat differently. So, I will deal I will develop the forms of mass transfer relationships that evolve for the four types of mass transfer problems. As you will recall the under steady state, the mass transfer equation would be something like this and psi y a equal to all that and psi can stand for mass fraction or element mass fraction or enthalpy. The radiation term is of course, neglected in this energy equation and m dot double prime y k as you will recall is the diffusion mass flux as per the fixed law of diffusion. So, these are the source terms in each of these equations R k would be finite when there is a chemical reaction if not it will be 0. So, let us consider the first type inert mass transfer without heat transfer. So, let us say I have a tank here with water in it and the water evaporates because the in the infinity state the air is either dry or has a relative humidity less than 100 percent and therefore, there would be mass transfer from water to the infinity state. The column height that I have considered above the water is l and the air in the column is stagnant. Stefan flow model as you know is primarily applied to diffusion mass transfer both water and air are at the same temperature and therefore, there is no heat transfer air also does not dissolve in water as said there will be no transfer of air from the considered phase into the water phase. Steady state prevails that is water is supplied at the bottom at the evaporation rate. What this means is that somehow we have constructed an apparatus in which water is supplied at the same rate at which it is evaporating and therefore, the column height l would remain constant or the water level will remain constant. Now, in this problem we have two spaces one is the air A and the water vapor V and therefore, the governing equation for this because there is no there is no chemical reaction as you will see in the previous equation n v y A would be equal to 0 for both air and water. So, that is what I have written here this means that the mass transfer rate m dot w in kg per second would be area A w into n w is equal to A times n v y plus n A y equal to a constant because this is gradient to 0, but in stagnant air m A w is equal to 0 because there is no mass transfer of air in this it is stagnant and therefore, therefore, omega A plus omega V is equal to 1 of course, and we note this and therefore, in this equation you will see that n A y will be set to 0 and m dot w would be simply A times n v y and that is what I have written here. So, m dot w will be m dot of vapor only equal to A times n v y and n v y as you know is the convective mass flux plus diffusion mass flux and that would be equal to A rho m v or if I rearrange this notice that A rho m v is simply m dot w. So, therefore, m dot w into omega v minus rho m d A into d omega v by d y would be equal to m dot w or if I transfer this on the right hand side then you will see it will be 1 minus omega v equal to m dot w by rho m d d y by A and if A is equal to A w is equal to constant that is if I take the constant area model then simply n w which is equal to m dot w by A w and integration would give me rho m d by l ln 1 minus omega v infinity over omega 1 minus omega v w equal to g star m into ln 1 plus b m this bracket 1 minus omega v infinity into 1 minus omega v w can also be written as omega v infinity minus omega v w over omega v w minus 1 and g m star would be rho m d by l it has the same units as the mass flux n w and this is a constant and a m n written in this form b m is called the driving force for mass transfer to occur and it is given in this fashion. So, you get a very simple logarithmic formula for mass flux or the evaporation rate instantaneous evaporation rate in a of water in the spherical system supposing I have a droplet then the surface area A would go on changing with the radius and A would be equal to 4 pi r square as we go along and A would go on changing. So, A would be function of y if you like. So, in that case you will see d omega v 1 minus omega v would equal m dot w rho m d into d r by 4 pi r square integration from r equal to r w which is the droplet radius to r equal to infinity gives ln 1 minus infinity over 1 minus omega v w equal to this quantity with r w in the denominator. Remember 1 over r infinity would be of course, 0 and therefore, only r w survives. If I rearrange this equation it would be written in this fashion n w will be equal to m dot w divided by area of the spherical droplet 4 pi r w square would be equal to rho m d r w into this quantity which again can be written as 1 ln 1 plus b m where b m is again as before omega v infinity minus omega v w into omega v w minus 1 and g m star which is the coefficient if you like is rho m d divided by r w instead of l in the previous problem when the area is constant. So, we get mass flux in this again logarithmic form when area changes. So, both results show that in diffusion mass transfer you get n w equal to g star m ln 1 plus dub b m, but as b m tends to 0 you can check out on your pocket calculator that ln 1 plus b m tends to b m for both positive or negative b m. Thus the linear variation n w g into b m holds only for very small mass transfer rate b m or n w whichever way you want to look at it. Typically when b m is of the order of 0.02 or less this relationship holds very well and of course, when b m is negative then of course, it will imply condensation when b m is positive it would imply evaporation. Therefore, in general we may write n w equal to g b m where g over g star m is equal to ln 1 plus b m by b m and where g star is the value of g when b m tends to 0. Now, although this result has been found from for diffusion mass transfer we shall later on show that the result has significance even in convective mass transfer. Let us now consider the case of inert mass transfer with heat transfer in which case let us say the water and the environment are at different temperatures. So, let us say the temperature of the environment is greater than that of the surface where T w is the temperature of the water surface and then under steady state besides species conservation equation we must now invoke the energy equation and that for as you will see from our first slide here we are invoking this equation. So, d n v y h m a divided by d y would be d by d y of a k m d t d y which is the conduction heat transfer plus this is the fixed law of diffusion multiplied by h k which is in this case would be omega v by d y h of vapor plus d omega a by d y h of air and the mixture enthalpy in this case would be the mass fraction of water vapor into enthalpy of water vapor plus 1 minus omega v which is the mass fraction of air into enthalpy of air h v of course, would be given by specific heat of vapor into T minus T ref plus lambda ref which is the latent heat at temperature T ref and h a would have only the sensible part C p a into T minus T ref and the mixture specific heat would be simply omega v into C p v plus omega air into C p a. So, that is what this formula and that. Now, let us look at this term k m d t d y now k m can be written as rho m alpha m C p m d t by d y and that is equal to rho m alpha m and if I absorb C p m d t d y and put C p m equal to all this you will see that you get omega v into d h v by d y plus omega a into d h a by d y this result is an important one which we are now going to substitute here. So, you will get d n v y h m a by d y equal to d by d y of rho m a alpha m omega v d h v by d y plus omega d h a by d y plus the second term which is the fix law of diffusion term which carries with it enthalpy of the species. Now, in mass transfer problem it is common to define Schmidt number as nu by diffusivity and we define Lewis number as Prandtl divide by Schmidt number which is equal to diffusivity of mass divided by diffusivity of heat. Now, for gaseous mixtures Lewis number is very close to one for example, as you know Prandtl number is about 0.7 Schmidt number would be about 0.67 to 0.68 in this kind of system. So, in effect Lewis number L e can be taken as very nearly one which implies that d is equal to alpha d is equal to alpha. If I make that assumption then you will see that and replace these rho m into d into rho m into alpha as gamma as I have defined here gamma m h equal to rho m d equal to rho m alpha m. Then you will see this becomes gamma m h into a plus d by d y of simply product of omega v h v plus omega h a which is nothing but the mixture enthalpy d h m by d y. So, we essentially get d by d y n v y h m a equal to d by d y of this. Now, from species conservation we have learned that n v y remains constant which is equal to n w the mass transfer at the surface itself remains constant throughout in a constant area problem. Then hence the last result can be written as d by d y of n n w h m minus h m t minus gamma m h into d by d y h m minus h m t where h m t is simply the enthalpy of the transferred substance and it is a constant. So, all I have done is really added or subtracted h m minus h t and here h m minus h t and therefore I made really no change. So, h m t would be the C p of the liquid into temperature of the transferred substance which is water in this particular example minus T rap is the specific enthalpy of the make up water deep inside the neighboring phase C p l is the specific heat of the liquid. Now, this is again a conserved property equation in h m minus h m t. So, it is exactly same as the equation we had for omega v and therefore its solution 2 would be identical n w equal to g star m h l n h m infinity minus h m t over h m w minus h m t which I would write as g star m h into l n 1 plus b h in this case where b h is now formed from h m minus infinity minus h m w over h m w minus h m t and g star m h would be gamma m h over r w in case of a spherical system and this would be the case in case of a linear system. So, we find that the logarithmic form is again retrieved from the solution of the energy equation because just as we had recovered the logarithmic form in case of mass transfer without heat transfer. Now, since Lewis number is equal to 1 gamma m h would be equal to gamma m and gamma h because thermal diffusivity is equal to mass diffusivity and hence you will see b h which is h m infinity minus h m w over h m w minus h m t would also be equal to omega v infinity minus omega v w omega v w minus 1. Now, this relation provides the important link between the mass fraction of vapor at the wall and the temperature of the wall. Remember, when the temperature of the surroundings is at t infinity and the liquid which is T t is at some other temperature, we still do not know what the surface temperature T w will be and we need to determine that in such problems. So, how do we determine that? We use this relationship that is b m is equal to b h and h m w would be then h v w omega v w h a w into 1 minus omega v w and if I take T ref equal to 0 then h m w will be simply c p a into T w into c p v minus c p a T w plus lambda at 0 degree lambda at T ref equal to 0 into omega v w. Hence, for a given T infinity and T t the b m equal to b h relationship will iteratively give omega v and omega n T w. So, what one does is one simply assumes a value of T w h m infinity of course, can be obtained because you know already omega v infinity, you know T infinity and therefore, that can be obtained. This too can be obtained as we saw on the last slide this can be obtained knowing T w only thing is you do not know omega v w. So, how do we get that omega v w? We can get that by saying that at the surface of the water saturation conditions would prevail and corresponding to R h equal to 100 percent and therefore, that value of omega v w can be noted for either from the sacrometric chart or you can also use steam tables in which case you will get a partial pressure. So, that from which you can recover the mass fraction of the wall. If after assuming T w you have determined omega v w in this way you substitute in the b m expression. If you find that b h is equal to b m then obviously, your choice of T w was correct and therefore, you already have T w omega v w relationship. If I have the two equations do not balance then you must change the value of T w till you get till you get balance between b m and b h. So, iterations are involved in discovering temperature T w. Now, in order to help if you are doing something on the computer then for air water vapor mixture saturation condition that is the saturation curve on the sacrometric chart is given like this where this is the T dry bulb and this is the specific in humidity and you have the 100 percent R h line. So, the values corresponding to this have been correlated here in omega v w as a function of T w for the range of T w minus 2200 and for computer applications to in order to help iterations you one can use this relationship or simply try by hand this is just for your information. So, now of course, when you consider a fuel we do not know the omega v w T w relationship by way of a psychrometric chart or anything like that and therefore, such a relationship must be determined from what is called the clausius Clapeyron equation and in that case you will see that omega v w is related to x v w into molecular weight of vapor divided by molecular weight of the mixture and x v w of course, is equal to p saturation at T w divided by p tau x v w would be given by exponential of minus h f g divided by R g into 1 over T w minus T b p for all liquid fuels typically you will have a boiling point known and h f g known and therefore, x v w can be recovered for a given T w. So, one tries in such cases assume a T w recover a calculate x v w from that you calculate omega v w and check whether b m is equal to b h as on the previous slide or otherwise go on changing T w till you get convergence. Now, I turn to the problem of mass transfer with heat transfer and simple chemical reaction. Now, what is simple chemical reaction? Let us consider highly volatile liquid fuel that burns in the considered phase according to the simple chemical reaction S c r. Liquid is usually burned by first evaporating in the gas phase without any change in composition that is the chemical formula of the fuel does not change when it comes out in the form of a vapor and but then it burns in the vapor phase as a in the gaseous phase as a homogeneous combustion. And we say that simple chemical reaction simply implies that 1 kilogram of fuel combines with r s t kilograms of oxygen to give you 1 plus r s t kilograms of products. Now, where r s t is the stoichiometric ratio of the fuel and you must have run from your stoichiometry how to evaluate say for example, a hydrocarbon fuel how to evaluate the value of r s t. Now, in this problem therefore, we have three spaces the fuel spacey, the oxygen spacey and the product which is itself a mixture but it is a product spacey which we take it as a single spacey. Then we will have three equations and because the chemical reaction is present you will have convective flux and a diffusion flux d by d y of would equal rate of depletion of fuel. Therefore, I have said magnitude of r f u with a negative sign likewise oxygen 2 would deplete minus r o whereas, the product would increase and therefore, it has a plus sign in front of here. So, we have three equations because we have three spaces with three mass transfer equations. If we add the three we must retrieve the bulk mass conservation equation because the addition of the fuel oxygen and product sum of these will be equal to 1 and when all these are sum that will be equal to 1. So, d 1 by d y would be 0. So, all sum of the diffusion space fluxes would add to 0 and therefore, you will be having a n w d by d y of n w into sum of all these quantities and therefore, the sum of the r k must also be equal to 0 because the bulk mass conservation. So, therefore, SCR simply implies that r o 2 the oxygen depletion rate is equal to r s t times fuel depletion rate where on the other hand the product generation rate would be minus 1 plus r s t times fuel depletion rate and likewise the diffusion of oxygen would be r s t times m dot f double free view and product will be 1 minus minus 1 plus r s t m double dot f u. So, if I now divide this equation by r s t and then subtract the resulting equation I divide by r s t throughout and which is a constant and subtract that equation from this equation then you will see that r f u minus r o 2 divided by r s t would be of course, 0 and as a result you will get an equation which is like this d by d y of a n w phi minus gamma m d phi by d y equal to 0 and phi will be omega f u minus omega o 2 by r s t likewise if I divide this third equation by 1 plus r s t throughout n add then again the right hand side would 0 and I will have a phi which you will stand for omega f u plus omega product over 1 plus r s t and we said that any equation of this form with the 0 source term implies that phi is a conserved property. So, we have again got an equation of conserved property like the inert mass transfer problem and therefore, we would have n solution of that would be n w equal to g star m l n phi infinity minus phi t over phi w minus phi t equal to g star m l n 1 plus b m where b m would be given now by this quantity phi infinity minus phi w over phi w minus phi t and phi can stand for this group or it can stand for this group then one uses the appropriate group depending on the convenience of the problem at hand g star m would be a course gamma m divided by r w in a spherical system or gamma m divided by l in the plane system. The important thing to note is that even in a problem involving combustion we are able to reduce the problem mathematically to a form just like that of a evaporation of water in the absence of heat transfer. So, let me now turn to energy part of the of the because whenever combustion takes place there would be heat transfer. So, now we would have d by d y of a n w h m minus k m d t d y would equal d of the diffusion mass transfer of d y you remember from slide 1 this is the right hand side and this is the equation that I am writing now this is the energy equation then equal to the right hand side this is the right hand side where h sub k is the species enthalpy and as you will know that whenever you have reacting fuel we write it as h naught f k is the enthalpy of formation plus c p k into t minus t ref or for the short I will write as h naught f k c p k delta t. So, hence making use of definitions of phi let us say h m would be simply omega k h k and therefore, it will be omega k h naught f k plus delta t into sigma c p k omega k then the first part of the term would simply result in omega f u h naught f u plus omega o 2 h naught f o 2 plus omega product into h naught f product and c p k omega k would be simply c p m the mixture specific heat into delta t. Now, from the previous slide we have defined phi. So, I can say that omega o 2 by r s t will be minus omega product over r s t. So, that is what I am going to use here. So, I am replacing omega product by 1 minus r s t over r s t omega o 2 equal plus c p m delta t and omega o 2 itself would be equal to omega f u r s t into all this and as a result you will see I get omega f u into h naught f u plus r h t h naught o 2 h naught f o 2 minus 1 plus r s t h naught f product plus c p m delta t. This essentially means that this is the enthalpy total enthalpy of the reactants one mole of fuel combines with r s t moles of oxygen. Therefore, this is the enthalpy of the reactants and this is the enthalpy of the products. So, the reactant enthalpy minus product enthalpy as you know is the heat of combustion and therefore, you will get omega f u del h c plus c p m into t minus t ref where delta t is written as t minus t ref again as I said there. So, we get enthalpy equal to heat of combustion into omega fuel. Now, of course, I have here replace omega o 2 and omega product in terms of omega f u, but I could do it the other way. I can replace omega f u and omega product o 2 in terms of omega product or I can replace omega f u and omega product in terms of omega 2 and I will get different expressions involving del h c which I will show you shortly, but now let us consider the right hand side which is minus d by d y of a sigma k m y k h k. So, here summation of k m double prime y k h k. It would be simply h naught f u c p f u delta t into this is the expression for m y f u plus the same quantity for oxygen and the same quantity for product. Now, we make an important assumption which is always made. Now, for gaseous mixtures specifics are functions of temperature and for different species they are somewhat different, but they are not so different that we cannot make an assumption of equal specific heats. If we assume that this each species has the same specific heat, then it would simply equal the mixture specific heat. So, that is the assumption I am going to make and use the circumvent regulation omega o 2 equal to r s t f u omega f u and omega product equal to minus 1 minus r s t omega f u and if I do that, then this entire relationship can be written as minus del h c rho m d omega f u by d y because c p m delta t into sigma k d d omega k by d y because sigma omega k would simply add up to 1 and that is equal to 0. Also, from the previous slide we have shown that c p m d t by d y is equal to d h m by d y minus remember. So, if I take differential of this equation with respect to y, then the c p m into d t by d y would be d h m by d y into minus del h c into d omega f u by d y and that is what I have written here. So, now I have this expression and I have this expression which I am going to make use of in our in deriving the right hand side. So, the right hand side can now be formed which is as you know is d by d y. So, I can now take everything on this side and you can see that I can transform this equation into a times n w h m into k m by c p m d h m by d y minus del h c d omega f u by d y. This is essentially the k d t d y term. This is the del h c rho m d d omega f u by d y which is the right hand side term. And then if I again make the Lewis number equal to 1 assumption that is alpha m equal to d, then you will see that gamma m this k m by c p m can also be written as rho m into d and equal to gamma h let us say. So, then you will see that this term would get cancelled with that term and I would get a times n w h m gamma h into d h m by d y equal to 0 and again a conserved property relationship has been obtained from the energy equation. So, the solution would be again same as that for inert mass transfer with b h defined as h m infinity minus h m w h m w minus h m t and this is the G star m h. So, again you will see that in a simple chemical reaction we have got the same formula both from mass transfer equation as well as from the energy equation. So, as I said earlier that we can define our mixture in variety of ways for a simple chemical reaction. One is to say h f u is equal to c p m delta t omega f u delta h c which is what I did earlier h o 2 will then be equal to c p m delta t and h product will be c p m delta t because of the equal specification. But I can also associate del h c with oxygen in which as h f u will be c p m delta t h o 2 will be c p m delta t into omega 2 o 2 r s t del h c and h h product and then again you have h q h o 2 and I can associate now with omega del h c with the product mass fraction. So, there are three ways in which you can do it for a liquid fuel burning in air we often choose second type because we often know omega 2 omega o 2 concentrations much better. So, then h m will be omega k h k would be simply c p m into t minus t ref omega o 2 by r s t del h c which is from this three relationships and if I now take for a moment that t ref is equal to t w which I do not know usually. Then I will get for convenience rather than not knowing where t ref if I take t ref equal to t w then b h which was h m infinity minus h m w over h m w minus h m t would transform to c p m t infinity minus t w del h c omega o 2 infinity minus omega o 2 w divided by r s t del h c into omega o 2 w by r s t minus c p l t t minus t w where of course, omega o 2 t is 0 because oxygen does not exist in the fuel and t t is known or knowable somehow. If t w was already the fuel surface was already at the boiling point t w by t b p then of course, no oxygen would survive at the surface and therefore, that would be 0 and that would be 0 because the fuel concentration there would be 1. If t w is not equal to t b p then of course, omega o 2 and t w relations must be established iteratively by balancing b m and b h where b h is given by this. So, you assume a t w evaluate the b h value then evaluate the b m then evaluate the omega v w or rather omega o 2 w and then again get the balance done in favor of fuel convergence is obtained where b m is given by this relationship and you will see phi value now will be known in all states for example, omega f u in the infinity state is 0 whereas, omega o 2 is known. So, phi infinity is known in the t state omega f u will be 1 and omega o 2 will be 0. So, phi t is known omega w will depend on the temperature which will determine omega f u and the remaining will be omega o 2. So, we can determine the relationships in that manner and carry out iterations. Now, I come to mass transfer with heat transfer and arbitrary chemical reaction what I mean by arbitrary chemical reaction by arbitrary chemical reaction this typically occur in solids combustion. So, for example, let us say consider burning of a graphite C star given by C star at very very high temperatures t w of the order of 1950 or much higher then there are several reactions taking place. The first four reactions take place at the surface of the graphite C star half o 2 equal to C o whose equilibrium constant is infinity C star C o 2 would be 2 C o whose in again K p is 4000 C star H 2 o equal to C o plus H 2 1230 and C star equal to 2 H 2 equal to C H 4 1 by 790. C o thus generated an hydrogen and C H 4 would then burn in the gas phase. Of course, here the K p is so low at 1950 that hardly any C H 4 would be formed and therefore, we can say that very small amounts of C H 4 will be present. C o would then react with oxygen to produce C o 2 which would then dissociate in this fashion C o 2 will be C o plus half o 2 giving a C o 2 and H 2 o would dissociate to give H 2 plus o 2 equal to again the K p is very very low and therefore, the reverse reactions would be dominant the producing C o 2 and H 2 o. So, for such a complex mechanism it is best to conserve the elements. We have the elements C H and o in this case you can see elements are C H and o of course, nitrogen is not present in the fuel and therefore, we write this and as you recall the element mass fraction equation is always a conserved property equation. So, let the infinity state comprise of C o 2 H 2 o and N 2 only that is the full products that. So, the noting the equilibrium constant K p for each reaction it can be shown that in the considered phase C H 4 cannot survive in appreciable magnitude. Hence, it will comprise the considered phase will essentially comprise of C o 2 H 2 C o and H 2 o only. Similarly, in the W state only C o and H 2 will survive and therefore, since species change in different states it is best to define a eta c equal to omega c star over 12 by 44 omega c o 2 equal to then eta H would be given by that and eta o it will be given by that composition. Thus we have three equations for eta c eta H and eta o. So, instead of solving three equations we need anyone can be solved, but you will find that not all these quantities on the right hand side are very well known in the three states. So, the best thing is to derive a composite quantity which we shall take as eta c minus 3 by 4 n o and from this relationship you will see that I can form a new variable omega c minus 3 by 11 omega c o 2 minus 2 by 3 omega H 2 o. Now, I can definitely form phi w equal to n c minus 3 by n o w equal to 0 in the W state. In the T state only eta c will be 1 or omega c will be 1 and eta c will be all these are 0s. So, they are put to 0 and in the infinity state I do not have any carbon, but there is omega c o 2 and omega H 2 o. So, they are retained and as a result I will get n w equal to l n 1 plus b m, where b m would be 3 by 11 c o 2 infinity plus 2 by 3 omega H 2 o infinity. Both these are known in the infinity state because that is what we said that we know only the products in the infinity state. Therefore, I can calculate the mass transfer rate of graphite burning simply by knowing c o 2 at in the infinity state and omega H 2 o in infinity state. So, in this case because I know the relationship connecting element mass fractions with the species mass fractions, I am able to create a composite phi as eta c minus 3 by 4 simply by observation. This manipulation is simply by observation such that I do not want any of the things any species on the right hand side whose concentration I would not know in the infinity w and t states. And this is what I have been able to achieve so that the calculation of mass transfer becomes easy. So, in summary I would say that we have analyzed all types of mass transfer problems from the by converting every problem to a conserved property equation and psi has to be defined appropriately. So, that n w equal to g b with g by g star equal to l n 1 plus b by b and b is equal to psi infinity minus psi w over psi w minus psi t and a is 4 pi r square or a constant. Now, for inert mass transfer without heat transfer psi was equal to omega phi as you know and gamma was simply equal to rho m d. For inert mass transfer with heat transfer we had psi equal to omega v and H m and we made the assumption of Lewis number equal to 1. For inert mass transfer sorry for mass transfer with heat transfer in chemical reaction we choose psi equal to appropriate phi and mixture enthalpy. We make the Lewis number equal to 1 assumption and also we say that the specific heat of participating species would be equal equal to see specific heat of mixture and in the mass transfer with arbitrary chemical reaction we showed that psi can be simply appropriate phi and gamma m as rho m diffusivity. So, you can see that a variety of problems have been reduced to conserved property relationship through appropriate and justifiable assumptions which makes calculation of the mass transfer rate simple and we are able to derive analytically derived relationship connecting N w the mass transfer to the driving force b and the relationship we found is a logarithmic one. In the next lecture we will see what the Couette flow model has to say.