 So, we now know how to calculate mass transfer flux in stiff and flow models, coir flow model as well as in Reynolds flow model and we are ready to solve practical problems. So, what I am going to do now is to begin with the stiff and flow model in this lecture. The stiff and flow model essentially deals with diffusion problems because in the considered phase the velocity is 0. So, we shall look at simple inert mass diffusion, inert counter diffusion, we will take a problem a very simplified problem of catalysis, we will look at inert mass diffusion with heat transfer and then finally, consider actually chemical reaction in which the burning of liquid fuel droplet in stagnant surroundings. What a variety of problems to help you with the understanding of this of the variety of applications to which stiff and flow model can be put. So, here is a problem, first problem a 10 liter vessel contains carbon dioxide at 25 degree C and 5 bar as I have shown here 10 liter vessel 25 degree C and 5 bar. The vessel is fitted with a 20 millimeter cork, 20 millimeter thick cork whose surface area is 300 millimeter square. So, the surface area here is 300 millimeter square, the height of the cork is 20 millimeters. Determine the initial rate of mass loss of CO2 through the cork, remember cork because this is at high pressure CO2 would diffuse out of the cork. So, what is the initial rate of mass loss of CO2? When the mass loss takes place of course, the pressure in the vessel will reduce. So, we want to find out what the reduction in vessel pressure will be after let us say 6 months if this was left unused for 6 months. You have been given a property data, these diffusivity of CO2 cork is as 1.1 into 10 raise to minus 10 meter square per second very very small value and there is another quantity called solubility. See when CO2 is in contact with cork surface then the concentration of CO2 at this surface, the inner surface assumes a certain value not the same as 1, but a certain value and that value is usually quoted in terms of solubility as 0.04015 kilo moles per meter cube bar. We shall use this value of S to determine what the inside mass fraction of CO2 is. The outside mass fraction of course is 0. So, let us begin our solution. So, here CO2 diffuses through stationary cork and hence there is no mass transfer flux of cork and using Stefan flow model therefore, we will have n CO2 at any distance y as minus rho m diffusivity 1 minus omega CO2 d omega CO2 d y equal to constant. In other words n CO2 the mass flux of CO2 will remain constant and since a is constant even the mass transfer rate would remain constant throughout the surface. So, how do we solve this problem further? If we integrate this equation from y equal to 0 where omega CO2 i to y equal to b omega CO2 o which is 0 then we have n CO2 equal to minus rho m d b divided by b l n 1 minus omega CO2 i and we will shortly discover that omega CO2 i is in fact very very small and therefore, l n 1 minus omega CO2 i can be taken as a minus omega CO2 i and that is what I have done here approximately equal to rho m omega CO2 d by b i and what is omega CO2 i that is rho density of omega CO2 divided by density of the mixture of cork and CO2 and therefore, this will be rho CO2 i d by b. Now this is where we use solubility as density of carbon dioxide at the inside surface is equal to solubility. Remember solubility is given as solubility is given as s equal to 0.04015 kilo moles per meter cube bar and therefore, on the next slide I have calculated rho CO2 which is kilograms per meter cube would be given as s kilo moles per meter cube bar multiplied by p which is in bars multiplied by molecular weight of CO2 which is kilograms per kilo mole. So, this will give me straight away of course, value of p must be used in bars s is solubility given here and this is the molecular weight of CO2 which is 44. So, if we use this then you get density of CO2 at the inside surface will be 8.83 kilograms per meter cube. So, what will be the mass fraction? It will be the rho cork is 1.93 into 10 raise to 5. So, essentially it will be 8.83 divided by a very large quantity plus 8.83. So, you get essentially omega CO2 I equal to omega CO2 I will be 8.83 divided by 8.83 plus 1.93 into 10 raise to 5. So, that gives you equal to 4.575 into 10 raise to minus 5 and therefore, the initial rate of loss of CO2 m dot will be the area of the cork multiplied by n CO2. The area of the cork is 300 millimeter square and therefore, this will be 300 into 10 raise to minus 6 meter square 8.83 is the density 1.1 into 10 raise to minus 10 is the diffusivity divided by the height d B this is diffusivity is 1.1 into 10 raise to minus 6 and B is 20 millimeters. So, that is 0.02 and this gives you the answer the initial rate of mass flow will be 1.457 into 10 raise to minus 11 kilograms per second which is the answer of the part A. Now, the instantaneous mass loss can be given by you can see here that the mass loss m dot will be volume of the vessel volume multiplied by rate of change of density of d t this is the density into volume is mass, but the volume remains constant. So, there you get that and rho CO2 from ideal gas law would be simply minus V m CO2 divided by R U t P CO2 by d t that is replacing rho CO2 in terms of m CO2 P CO2 R U t d t and that would equal as we said in the in the previous slide A into d by B into omega 2 I S into m CO2 P CO2 I which is the density density of crop I mean density of CO2 on the inside surface is what we want rho CO2 I which is going to now vary with time and therefore, that would also equal as into m CO2 into P CO2. So, this is what you get. So, rearranging this dividing by P CO2 here you get 1 over P CO2 d P CO2 by d t equal to minus A d S the solubility R U into t divided by B into volume or ln P CO2 by P CO2 at time t equal to 0 would be minus A d S R U t divided by B P into time in seconds. So, R U as you know the universal gas constant is 8314 the time in seconds for 6 months would be 6 into 30 into 24 hours into 3600 the volume itself being 10 liters is 10 into 10 is to minus 3 and the initial pressure is 5 bar initial pressure is 5 bar and therefore, after 6 months the pressure would be if we calculate substitute all these values here you will see 5 into ln of this is this therefore, P CO2 by P CO2 at t equal to 0 will be exponential of all this quantity is equal to minus 0.2637 equal to 3.84 bar. So, that would be the pressure inside the inside the vessel after 6 months and therefore, there will total of some pressure would be 5 which was initial minus 3.1 point 1 6 bar and which is the answer. So, engineers and maintenance engineers are often interested in knowing these values. So, that they know that they would order only the amount that is required for their use and not leave vessels un guarded unused and lose CO2 thereby. Let us take another problem here we consider two large tanks of CO2 which can each contains CO2 plus N2 this is a tank 1 and this is tank 2 both of them contains CO2 and N2 at 1 bar and 25 degree centigrade. So, the total pressure and total temperature of both the tanks is identical, but they are very large the tanks are very large. The tanks are connected by a 1 meter long tube a 1 meter long tube of 5 centimeter diameter. Although the total pressure in tank 1 is 1 bar the partial pressure of CO2 is simply is a low 50 millimeters of a mercury whereas, the partial pressure of CO2 in tank 2 is 100 millimeters of mercury. Now, as a result what would happen is although the total pressures are equal CO2 would diffuse to tank 1 from tank 2 and since mass must be conserved the nitrogen here whose partial pressure will now be bigger than in tank 2 the nitrogen would diffuse in the opposite direction. So, we have been asked to calculate the mass transfer rate of CO2 under steady state. So, here CO2 will diffuse from tank 2 to tank 1 now p CO2 of 100 millimeters of Hg because 760 millimeters of Hg equals 1 bar p CO2 2 in bars would be 100 by 70 60 is equal to 0.1316 bar and in tank 1 it would be 50 by 760 which is 0.06579 bar and N2 will diffuse in the opposite direction. So, that NCO2 plus N2 will be 0 and as you will recall from our diffusion model NCO2 is N total rho CO2 minus d d rho CO2 by dx and since this quantity is 0 we get NCO2 would be simply minus d into d rho CO2 by dx equal to constant. So, now this equation must be integrated of course to obtain our formula. So, rho CO2 is p CO2 divided by r CO2 divided by T from ideal gas law and r CO2 would be r CO2 will be 8314 divided by 44 and that is equal to 188.95 this is r u divided by molecular weight of CO2. So, you get r CO2 equal to 188.95 joules per kilograms per Kelvin and therefore, NCO2 is equal to minus d r CO2 T into d p CO2 by dx equal to constant. Now, from our previous lecture you will look or you can check out that diffusivity of CO2 in nitrogen is 11 into 10 raise to minus 6 298 divided by 300 into 1.5 and since we have our pressure is in bar whereas, 1.0 1 to 1 atmosphere is equal to 1.0 1 to 5 bar and our temperature is 25 degree centigrade. So, it is 298 divided by 300. So, you get 11.0 34 into 10 raise to minus 6 meter square per second. So, integration of this equation from x equal to 0 to x equal to L gives NCO2 equal to minus d diffusivity r CO2 T L which is temperature into p CO2 2 minus p CO2 1 and diffusivity being equal to 11.0 34 10 raise to minus 6 188.95 is r CO2 temperature is 298 K and L the length of the tube is 1 meter p CO2 2 is 0.1325 p CO2 1 is that into 10 raise to 5 newtons per meter square and as a result you will get NCO2 equal to minus all this quantity. The negative sign indicates that the NCO2 is flowing in the negative x direction that is from tank 2 to tank 1 and therefore, the mass flow rate M dot CO2 will be NCO2 multiplied by pi by 4 into diameter of the pi 0.05 which is 5 centimeters 0.05 square is equal to minus 2.55 into 10 raise to minus kg per second. So, this is the answer to our problem about the rate of diffusion of CO2 in this pipe in this problem from tank 2 to tank 1. We now look at a third problem you know that in exhaust gas of let us say of an IC engine may contain nitrous NO nitric oxide and you are told that it is the exhaust gas is at 500 degree centigrade 1 bar pressure and it contains the mole fraction of NO in that stream is 0.002 and the mixture molecular weight is 30. Now, in order to extract NO out of this you pass it over a catalyst surface. So, you have a basically you have a let us say catalyst surface and this is the hot gas which contains NO x NO infinity equal to 0.002. The temperature here is 500 degree centigrade or T is equal to 773 Kelvin and the mixture molecular weight in the infinity state is 30. Now, what I am going to do is although the gases are moving I am going to make a few assumptions in order to render the problem tractable in a very simple way. So, it is assumed that chemical reactions involving NO are very very low 500 degree centigrade. In other words over the entire flow of the gas over this surface NO does not react with any other substance any other species in the gas and therefore, NO essentially remains constant the concentration of NO essentially remains constant. It is neither destroyed nor generated in this over this length of the pipe of the surface. So, we can take NO equal to absolutely constant 0.002 which is the volume mass fraction of the of NO as given. Further at the catalyst surface NO is absorbed with kinetic rate N W. Now, of course, absorb means it is in this direction and you are told that N NO will be k times rho m into omega NO W where k is the kinetic rate constant and its value has been given as 0.075 meters per second. Now, it is further assumed that NO diffuses to the catalyst surface over a stagnant layer of 1 millimeter. What it means is that near the catalyst surface the velocity is so low as you can see it will be like that. So, the velocity is so low that we can in fact take this as a very stagnant layer in the extreme case stagnant layer of 1 millimeter thickness. So, that is what we will do and you have been told that take in this very small layer remember there will be some effects of turbulence and therefore, all the very small we are told that take d effective equal to 3 times d the laminar diffusivity and the laminar diffusivity itself is given as 10 raise to minus 4 meter square per second. So, determine the steady state absorption rate of NO and omega NO W. So, here the total N total will be N NO plus N others, but we have been told that in this layer it is a stagnant layer. So, N others is 0 because N O is the only one that is diffusing whereas, the other species simply do not diffuse and let y equal to 0 define the catalyst surface then diffusion rate will equal the kinetic rate and what will be the diffusion rate N NO equal to N NO omega NO which will be the convective rate and this will be the diffusion rate out. So, N NO is as shown here this is the N NO equal to N NO into omega NO minus rho m diffusivity omega NO by dy and that will be equal to the opposite minus k rho m omega NO because absorption is taking place in this way whereas, our formula is for positive N NO and this is the expression. So, that is what I have written here N NO is equal to N NO omega NO minus rho m d omega NO by dy equal to minus k rho m omega NO at W. So, if I rearrange the equation then N NO will be minus rho m d 1 over omega NO d omega NO by dy equal to minus k rho m omega NO W. If I integrate this equation from y equal to 0 to l that the height of the stagnant layer which is 1 millimeter then I will get N NO equal to rho m d divided by l ln of 1 minus omega NO infinity divided by 1 minus omega NO W which can be written as G star ln 1 plus omega NO infinity omega N S omega NO W over omega NO W minus 1 and G star is nothing but rho m d by l as you know for a Stefan flow model and that would equal minus k rho m NO W. Now, you know already omega NO infinity because remember omega NO infinity will be x NO infinity into M NO divided by M mix the molecular weight and that is given as 0.002 M NO the molecular weight of NO will be 14 plus 1630 and you already given molecular weight of mixture is also 30. So, this gets cancelled and in other words the mass fraction is same as the mole fraction. So, 0.002 and omega NO at W will be if we equate this and this then by iteration you can determine 1.44 into 10 raise to minus 3 which is in fact very very small and therefore, 1 plus b will actually be equal to b itself in which case no iterations are required. Hence the value of rho m P R mix by T will be 0.466 kg per meter cube. Remember R mix will be 8314 divided by M mix which is 30 in the infinity state omega infinity is 0.002 and in the wall state it is 1.144 into 10 raise to minus 3. So, first of all 1 in values the mean mass fraction and that is equal to 0.466 and therefore, N NO is equal to minus 0.466 into 0.075 which is the k value already given to you into 1.144 which is omega NO W into 10 raise to minus 3. Remember this is rho m is from the property data you have been given you take the temperature as 500 degree centigrade P is 1 bar. So, you can evaluate the density as P divided by R mix T and 466 into 0.075 into 1.144 into 10 raise to minus 3 would give you 4 into 10 raise to minus 5 kilograms per meter square seconds. So, that is the first part of the problem. So, in other words if you want to remove this will tell you what is the amount of catalyst surface you will require to obtain given amount of removal of N O from that surface. The value of B itself would be because you know now the omega N O W you know omega N O infinity and that is the value of B would evaluate to minus 5.608 10 raise to minus 4 and it is very small as well as notice that it is negative because there is an absorption going on. Also, if the gas velocity now is accounted then G star must be appropriately augmented from H cough from the heat transfer situation corresponding heat transfer situation and in which case you will need to evaluate omega N O W freshly and carry out the calculation. But as of now in assuming stagnancy you get the mass transfer flux is known which is the answer A and answer B omega N O W was is 1.144 into 10 raise to minus 3. Now, we look at evaporation problem a 50 micrometer liquid N hexane droplet whose density is 659 kg per meter cube its latent heat is 335 kilo joules per kilogram its boiling point is 69 degree centigrade or 342 k. It evaporates in a stationary pure nitrogen environment at one atmosphere and 850 kelvin. So, calculate the evaporation time using T w equal to d B p and using T w equal to T l and equilibrium evaporation assumption. In each case take A m equal to 0.07478 C p m equal to 2.434 kilo joules per kg kelvin and m mix equal to 57. So, in the part A of the problem since the droplet is at boiling point B m H would simply evaluate to T infinity minus T B p H f g into C p m and it will be 2.434 into 800 minus 342 divided by 335 kilo joules per kg kelvin and therefore, that would be equal to 3.69. So, notice that when the fuel evaporates in a nitrogen environment it is prevented from burning and therefore, 3.69 very very very high driving force compared to the problems we have seen so far and therefore, evaporation rate would be rho l into d w i square 8 times K m by C p m into l n 1 plus B m H equal to 6.7 milliseconds remember gamma H is K m by C p m and that would be 19.39 10 to the power minus 6. So, it takes 6.7 milliseconds when the droplet is initially at its boiling point. Now, suppose we do not we injected at some other temperature T l in which case we will have to find out what is the value of T w by from equilibrium evaporation assumption that is by iteration. So, T w equal to T l is is not known and hence we must adopt iterative solution. Now, since inside the droplet there is no temperature variation Q l will be equal to 0 and therefore, our formula would be m dot w equal to 4 pi r w gamma H l n 1 plus C p m T infinity minus T w divided by H f g and that would also equal B m which is the mass transfer formula from B m 4 pi r w gamma m as you know is rho m multiplied by diffusivity gamma H is K m divided by specific heat 1 plus omega B infinity minus omega B w omega B w minus 1 and where from lecture 32 T m if we must assume a T w and evaluate T m. So, presently I am taking it as let us say 596 K and diffusivity of therefore, I will calculate as from our previous lecture where diffusivity of ethane in nitrogen is also given as 8 into 10 raise to minus 6 T divided by T naught phi 96 by 300 into 1.5 the pressures are equal to 1 bar. So, that is one atmosphere and therefore, nothing to correct for pressure and therefore, diffusivity would be taken as 22.4 into 10 raise to minus 6. So, if we now equate these two equations then I get gamma H l n 1 plus B H equal to gamma m l n 1 plus B m and that would essentially mean that this quantity is equal to 1 plus C p m T infinity by T w divided by H f g raise to Lewis number which is gamma m by gamma H as you can see that would be the Lewis number. And omega V w will be from Clausius Clapeyron equation would be m v by m mix which is m v being the vapor minus H f g m v R u into 1 over T w minus T b p. So, what one does is a Schumer value of T w and hence since you know T b p you know H f g you know the molecular weight of vapor you know what is m mix. So, therefore, from T b p you evaluate omega V w substitute that value here and calculate the left hand side as well as the right hand side and see if the two are in agreement. Of course, in each case you must calculate gamma m and gamma H for each temperature because the mean temperature would change for each choice of T w and T infinity is 850 as you know already. So, one needs to iterate on these two formula on these two equations in our case m v is 86 or n x n rho m is 1.165 kilograms meter cube Lewis number is gamma H by gamma m is 0.743. So, to carry out iterations we assume T w and evaluate omega V w and properties until the first relation on the left hand side on the previous slide is satisfied within tolerance. So, after iterations the answer is convergence is obtained when T w is 322.1 Kelvin or 49.1 degree centigrade. This is 49.1 degree centigrade for which omega V w is 0.7794 and the properties are mean properties turn out to be 200. So, 2009.5 joules per kg per Kelvin that is a specific heat conductivity would be 0.045 Lewis number turns out to be 0.888 and the mixture molecular weight would turn out to be 43.5 and B H would be 3.167 whereas, B m would be 3.53 because omega V infinity is 0. So, you do see that the effect of Lewis number would be to have B m and B H are different. So, I can use any of the formulae of the previous slide. I can use this formula or I can use this formula only thing is I have to use either omega H and B H or omega m and B m. I am evaluating here from omega H therefore, K m by C p m and for these conditions of T w which is 49.1 degree centigrade I evaluate K m by C p m equal to 22.4 10 raise to minus 6 and 1 plus 3.167 which is burnt. Now, you get the answer as 6.44 milliseconds whereas, by injecting at T b p you had 6.7 milliseconds. Now, you have 6.4 milliseconds so that our assumption of T w equal to T b in part a is reasonable although the droplet temperature is now nearly 20 degree centigrade less than the boiling point. Now, this is as far as evaporation goes, but in actual engines this difference of 0.3 milliseconds can be significant because the total burning time itself is in milliseconds. Therefore, 0.3 milliseconds for evaporation is a significant quantity for engine applications. But here in order to study evaporation of a volatile fuel what we have done is we have taken the environment at 850 K to comprise nitrogen. Now, let us take the same liquid droplet of n hexane of 50 micron diameter and allow it to burn. Again the properties are given and it is burning in one atmospheric pressure and 300 K. We will assume that the droplet is already at its boiling point so T w is equal to T b p and the mean conductivity and C p m of the considered phase are given. The heat of combustion of n hexane is 45.1 mega joules per kilogram and therefore, we want to calculate the initial burning rate and the burning time. Now, for any hydrocarbon reaction C m H n into m plus n by 4 into O 2 plus 3.76 n 2 gives the products. Therefore, our stoichiometric ratio R S T is equal to m plus n by 4 into molecular weight of O 2 divided by molecular weight of C m H n. So, in our present case m is equal to 6, n is also equal to 14 divided by this is 14 divided by 4 into 32 divided by molecular weight of this is 86 and that is what I have done and that would give you 3.535. So, if you look at this then over the b itself would be C p m into T infinity minus T boiling point because T w is equal to T boiling point omega O 2 infinity R S divided by R S T del H C divided by latent heat. This formula stems from the fact that H m can in defining H m I can say C p m into T minus T ref plus omega O 2 by R S T del H C. That means, I have associated the heat of combustion with oxygen rather than with fuel and therefore, I must divide this by R S T and that would give you the formula that C p m T infinity minus T b p omega O 2 infinity by R S T del H C divided by H F G and substituting for C p m 2.839 in kilojoules per kg 300 minus 342. So, remember the sensible part is negative, but the heat of combustion part is positive 0.232 divided by 3.535 into 45.1 into 10 raise to 3 divided by the heat latent heat that gives you 8.48. So, we get even a larger value in a burning problem compared to the evaporation problem which was 3.16 or so, here we get it as 8.48 and therefore, we expect the burning rate to be high and the burning time to be small. So, you can see now here I can evaluate burning rate, initial burning rate is 4.4 into pi into gamma H which is K m by C p m into R w i which is the initial radius 50 by 2 into 10 raise to minus 6 l n 1 plus 8.48 giving 4.12 into 10 raise to minus 8 kg per second. That will be the initial burning rate and the burning time would be rho l d w i square 8 into gamma h into l n 1 plus b h. So, substituting for these values you now see this is just 1.57 milliseconds the same droplet evaporating now it is burning and the time has reduced quite considerably to 1.57 seconds. So, what we have learnt from these sets of problems is that in general in air water evaporation problems the b is very very small, but when fuels evaporate b can be substance even bigger than 1 of the order of 2 or 3 or something of that kind. When the drop when the fuel burns it would be of the order of 8 to 10 or even little greater than 10 in variety of different liquid fuels. So, but all this in stagnant surroundings all this is done in stagnant surroundings. Our interest in convective mass transfer and that we will take up as you go along, but you will recall that even convective environment can be handled by treating it as a diffusion problem provided we multiply the multiply the effective G star of a Stefan problem by the Sherwood number. That is the kind of something we had done in our lecture on Stefan flow model at an earlier time where I showed you how evaporation of a liquid droplet in a moving surroundings can be handled.