 In lecture 32, we looked at Stefan flow model, which is essentially a diffusion model to be applied to stagnant surroundings. Therefore, momentum equation was not invoked, only the species transfer and the energy equations were invoked. We applied these equations to the three types of mass transfer or four types of mass transfer that we normally encounter. One is the inert mass transfer without heat transfer, inert mass transfer with heat transfer, then mass transfer with heat transfer and simple chemical reaction and then finally, we applied it to the case of mass transfer with arbitrary chemical reaction. All these four types are most describe the overall problem of mass transfer. Today, we are going to consider the Couette flow model and therefore, we shall be invoking both the momentum equation and continuity equation along with the species transfer and the energy equation. So, we shall apply the Couette flow model to mass transfer with wall suction and blowing. This is essentially a case of let us say air flowing over surface and air itself is being sucked or blown through the wall, which would be the simple case of suction and blowing that we considered even with similarity solutions. Then we will move to the general problem of convective mass transfer and in this we shall interpret the effective mass transfer coefficient, which we shall term as G star and by way of an example, we will calculate estimate the evaporation slash burning time of a liquid droplet. So, let us begin then with the momentum transfer with wall suction and blowing. Now, in the Couette flow model as you will recall, the velocity u is taken to be constant multiplied by y and all actual derivatives are set to 0 and therefore, a is constant and then under steady state transport equation would read like this d by dy of n psi y equal to d by dy of rho m v psi minus gamma plus gamma t, where gamma t is the turbulent exchange coefficient d psi by dy equal to s psi. Just by way of reminder, the Couette flow model is really let us say this is the surface and we assume that the velocity profile will be like so with u infinity here. All d by dx are 0 for all variables and u infinity remains constant along the plate and this is the direction y. So, this is really the Couette flow model. The mass transfer would be taking place in this direction. So, if you look at the meanings of psi, psi equal to 1 would imply simply continuity equation or mass conservation equation. Psi equal to u will imply momentum equation which gamma equal to mu and then if it is a mass transfer mass fraction omega k, this would be the species transfer equation and this would be the energy equation. We have ignored the radiation and other heat generation terms here m double dot y k is really the fixed law diffusion flux given by the fixed law and rho m v a will be m dot w is equal to constant which is mass flux would remain constant. So, let us consider momentum transfer. So, if psi is equal to u then the governing equation remember there is no source term here, but the pressure gradient is also 0 and therefore, the equation will be d by dy n w u minus mu plus mu t du by dy equal to 0. If I integrate this once and note that the boundary condition is u equal to 0 at y is equal to 0. So, shear stress is given by mu times du by dy y equal to 0 of course, mu t would be 0 at wall. The constant of integration c will be simply minus tau wall and hence the integrating from 0 to infinity would give me du divided by n w u tau over w equal to 0 to delta dy mu plus mu t which I am for the moment calling it as constant c 1 some integral value integrated value c 1 and the integration of this term would simply result in 1 over n w l n 1 plus n w u infinity by tau w, but what is n w u infinity by tau w we can interpret that remember n w is rho times v w because the same air same fluid is being blown into the boundary layer as is flowing over a plate rho v w u infinity divided by tau wall. If I multiply and divide this by u infinity then I would get v w divided by u infinity equal to rho u infinity square by tau w which is nothing but v w by u infinity c f x by 2 and as you will recall this is nothing but the blowing parameter which we had invoked during similarity solutions and integral solutions. So, essentially then n w would be 1 over n w l n 1 plus b f would equal some constant c 1. So, that is what I have written here l n 1 plus b f would equal c 1 times n w and c 1 rho u infinity b f over c f x by 2 because n w is rho v w which can be written as rho u infinity b f c f x by 2. Now, of course, as b f tends to 0 the c f x must tend to c f x at v w equal to 0 and then assuming that this integration that is integration 0 to delta mu plus mu t would remain the same whether there is mass transfer of the wall or no mass transfer of the wall which of course, would be exactly true if it was a laminar boundary layer but even in turbulent boundary layer. If we say that mu t essentially is a function of y and not affected by whether as you will recall from Prandtl's mixing length that it will not be too seriously affected. If we assume c 1 is independent of whether v w is finite or 0 we can show that it follows from this equation that c f x v w divided by c f x v w 0 would be l n 1 plus b f by b f. This equation is applicable to both laminar and turbulent flows and it is derived for d p d x equal to 0 but can be taken to be valid even for mild pressure gradients as was done during integral analysis of momentum equations in our previous analysis of fluid flow problems. Now for all four types of mass transfer I am not going to re-derive as I did in case of stiffen flow I will simply say that in each case we simply converted the applicable mass transfer and energy transfer equation to a conserved property equation with appropriately defined conserved property psi. So, for all types of mass transfer and an appropriately defined conserved property psi n w will be equal to n psi y equal to constant and hence for conserved property instead of psi I can also take psi minus psi w as a conserved property then n w into psi minus psi w gamma plus gamma t d by d y psi minus psi w is equal to 0 or n w into psi minus psi w minus gamma plus gamma t d psi by d y equal to c 1 some constant c 1 now remember d psi omega d psi w by d y is of course, always 0 so that is that does not appear here, but d psi by d y would certainly survive then if I write this equation in the w state then I would get c 1 equal to gamma d psi by d y at y equal to y equal to 0 and if I write it in the t state I will get n w psi minus psi t minus psi w and in the t state there are no variations of psi and therefore, this entire term would be 0 and again c 1 and therefore, equating the two equations because they are both equal to c 1 I would get c 1 equal to n w psi t minus psi w equal to minus gamma d psi by d y at w this is and at the w state of course, omega t gamma t is equal to 0 so that is what I have state here. We shall make use of this equation and substitute for c 1 here c 1 equal to n w psi t minus psi w I will substitute that here on the next slide and you will see that therefore, I would get n w psi minus psi t minus gamma plus gamma t d psi by d y equal to 0 if I integrate this equation from w state y equal to 0 to infinity state y equal to delta then I will get 1 over n w equal to 0 to infinity d psi by d y d psi by psi minus psi t equal to 0 to delta d y gamma plus gamma t where gamma in case of energy gamma is k by c p in case of species transfer it is rho m times diffusivity. So, let us say that integral like in the previous case we will say is equal to c 2 let us say then the integration of the left hand side would give me n w equal to 1 over c 2 l n 1 plus b psi because same as previous case b psi is equal to c psi infinity minus psi w divided by psi w minus psi t and n w will be from the previous side n w would be c 1 divided by psi t minus psi w and that would equal minus gamma d psi by d y at w psi t minus psi w. So, we have got one equation which is of this form and other equation which is of this form the first containing c 2 and the second containing c 1. So, now consistent with the theory of heat transfer we may write minus gamma d psi by d y at w is equal to g times psi w minus psi infinity where g is now the mass transfer coefficient. So, if I replace this quantity by this quantity in the previous expression here then I get c 1 would be I can get c 1 from there in terms of g and therefore, the final form would look like n w equal to g times psi infinity minus psi w divided by psi w minus psi t or simply g times b psi and g itself would be 1 over c 2 times l n 1 plus b psi divided by b psi. Now, if I let if I assume that as b psi tends to 0 g tends to g star the mass transfer coefficient let us say it tends to g star corresponding to b psi tending to 0 further and if I say that c 2 remains constant with or without mass transfer as we said earlier this statement is of course, perfectly true for laminar boundary layer, but even if I say it is true for turbulent boundary layer where 1 over where gamma t would be a function of y then it follows that g over g star would be simply l n 1 plus b psi by b psi. This is a very important result thus the fictitious g star flux is now given by n w g star l n 1 plus b psi where 1 over g star is equal to c 2 which is equal to 0 to delta d y over gamma plus gamma t. So, we can now view g star itself as the sum of layer by layer from 0 to delta resistances to mass transfer in the considered phase over the width delta because remember this is the this is diffusion coefficient. So, 1 over diffusion coefficient would simply be resistances resistance and we are simply saying that 1 over g star which itself is a kind of a resistance because g star is conductance then 1 over g star would be simply layer by layer addition of resistances to mass transfer in the considered phase. This interpretation of g star enables its evaluation from gamma y equal to gamma psi in a laminar boundary layer and from the known gamma t y from a turbulence model like a mixing length or for example, in a turbulent boundary layer. Thus the Couette flow model permits study of the property variations remember gamma itself could be a function of psi of temperature or the mass fraction itself. Gamma t on the other hand would be function of the turbulent characteristics of the boundary layer. In fact, if gamma is equal to constant and gamma t is equal to 0 which is the case of a laminar diffusion problem then g star would be simply gamma by delta which is same as the Stefan flow model in which g star was shown to be equal to gamma by l. We have recovered most of the features of the Stefan flow model for gamma equal to constant and gamma t equal to 0 and g star would then be gamma by l l n 1 plus b psi which is what we are shown in the Stefan flow model. Now, if we consider the case of pure heat transfer in the presence of suction and blowing something we have solved by similarity method then with psi is equal to H m equal to C p t because it is an inert situation just consider phase as air flowing over it and suction and blown fluid is the same as the fluid in the considered phase it is temperature may be same or different that does not matter. But, therefore, we would get minus k d t d y equal to 0 which is the heat flux and that would equal g times C p into T w minus T infinity according to our model and that would equal heat transfer coefficient for suction and blowing into T w minus T infinity. So, that is what the heat transfer coefficient is and therefore, we deduce that the heat transfer coefficient for finite v w is nothing but g by C p and likewise H cof v w equal to 0 that is in the absence of mass transfer would be equal to g star by C p because g star is a value of g when v w is equal to 0 and therefore, we deduce that g over g star would also be equal to H cof v w divided by H cof v w 0 equal to stanton x at v w for v w divided by stanton x for v w equal to 0 and that would equal l n b h by b h where b h now is T infinity minus T w over T w minus T t because specific heats in all states are taken to be constant at the moment. So, now this relation was found to be applicable in real boundary layer flow in lecture 30 you will recall and therefore, the Couette flow model captures also the features of a real boundary layer flow. So, the Couette flow model on the one hand when gamma t is 0 captures the Stefan flow model features and it also captures the boundary layer flow features at least the when the species are the same it captures both the features very well. Now, we turn to the application of this to evaporation and burning in which we shall invoke the previous expression for can be used instantaneously to estimate the evaporation or burning time thus let us say if I have rho l d v by d t is equal to minus m dot w which is the rate of change of mass is equal to minus m dot w kg per second and that would be called a w n w and that would be called minus a w a g star l n 1 plus b psi where a w is the area of the surface through which the mass transfer is taking place then integrating this from time t equal to 0 to complete evaporation when the volume disappears gives us the relationship t evaporation or burn is equal to minus rho l l n 1 plus b psi equal to v initial to 0 d v by a w g star. Now, let us say if we are considering a liquid droplet and diffusion mass transfer that is gamma t is equal to 0 then a w will be equal to 4 pi r w square v will be 4 pi 3 pi r w square and g star will be gamma h by r w as you will recall then hence you will see that evaporation or burning time would be minus rho l divided by 1 over b psi r w i to 0 r w by gamma m h d r w and that will yield the so called d squared law rho l d w i squared h times gamma h l n 1 plus b psi and this expression is used extensively in designing dryers and so on and so forth because all it says is that if you reduce the smaller the diameter of the droplet faster will be the drying achieved because if you reduce the diameter by a factor of 2 the evaporation time will reduce by a factor of 4. In stagnant surroundings this would be considered as a guidance for atomization of fuels atomization of let us say milk which is to be dried into a powder and or in a cooling tower where you where you send hot cooling water for cooling purposes then in a shower then you want to reduce the diameter as small as possible so that you get very quick drying and a smaller dryer or a cooling tower. Now let us say the liquid droplet was in a convective environment as in inside a diesel engine where as you know that in a diesel engine the liquid droplets are injected in an atomized state and they come out like a cloud and when the piston is at the top net center the temperature is already very high in the surrounding and the droplets evaporate and then burn inside the cylinder. So we can assess such we can evaluate the evaporation or drying times for evaporation or burning times in such situations but the environment there is convective because the piston's head is modeled in such a way that there is a swirl inside the cylinder so the air movement and the particle movement and the droplet movement there is a relative velocity between them. So essentially you get now convective evaporation or convective drying same thing happens when you have a cooling tower where the droplet is falling down through let us say stagnant air but that sets up a resultant velocity between the droplet and the surrounding air which is stagnant. On the other hand in thermal power stations we have you will recall that we have cooling towers like that and air ingresses like this and on here you have showers of water where the cooling water from the condenser falls down and there is an air so you have a counter current of air and water and therefore the liquid droplets evaporate under counter flow so the relative velocity is additive of the air flow and the water flow droplet flow which is coming downstairs. In the process the water gets cooled and is sent back to condenser of course some amount of water is lost because the air moving upwards picks up some moisture which has to be topped up of course from time to time so that the condenser is not starved of cooled water. So in such situation we can use a shortcut method and thus we can say that m dot w that is mass transfer rate under convection divided by mass transfer rate under diffusion would be mass transfer rate under convection according to the to the Couette flow model is G star times A w which is 4 pi r w square ln 1 plus b and mass transfer rate according to diffusion a model which is a Stefan flow model is rho m d 4 times pi r w into ln 1 plus b and therefore cancelling the terms you will get 1 over 2 G star d w which is the diameter of the droplet d over rho m diffusivity and this quantity like Nusselt number is called the Sherwood number. In mass transfer the analog analogous to Nusselt number we have a Sherwood number which is essentially G star the mass transfer coefficient into the diameter divided by rho m d which is equivalent of the conductivity of the fluid. Now if I use analogy between heat and mass transfer for Lewis number 1 then you know that the for a flow over a sphere you know that the Nusselt number is 2 plus 0.6 Reynolds raise to 0.5 Prandtl raise to 1 third but I can now say Sherwood number would be 2 times 0.6 or Re raise to 0.5 Schmidt number raise to 1 third where Re is the is based on the relative velocity between the gas and the droplet the diameter of the droplet and the kinematic viscosity or this expression for example would now get changed to 2 times rho l ln 1 plus b psi R w i to 0 R w d R w divided by gamma as 2 plus 0.6 Re d w 0.5 Schmidt number raise to 1 by 3. Now since this also contains the radius R w where because d w is 2 times R w this expression integration requires numerical integration because close found solutions are not found. So, let us take for example if I took a problem let us say a water droplet d w i is 1 millimeter diameter at 25 degree centigrade evaporates in air whose relative humidity is let us say 25 percent then and temperature is 25 degree centigrade so that this is a case of mass transfer without heat transfer and no chemical reaction and let us assume that the relative velocity between the two is 5 meters per second. So, estimate the evaporation time and take Schmidt number equal to 0.6 which is quite typical of gaseous mixture. So, this is a mass inert mass transfer problem without heat transfer the mass fractions are in the infinity state corresponding to 25 percent R h and t equal to 25 will give you omega v infinity equal to 0.0078, omega v w corresponding to 100 percent relative humidity at the droplet surface and 25 degree centigrade will be 0.02, v m would be omega vapor in the infinity state minus omega vapor in the w state and omega vapor in the w state minus in the t state which is the transfer surface and that would be equal to 1. So, using this definition for omega v as we did in the in problems on diffusion mass transfer the v m would turn out to be 0.0124. At the mean conditions between the infinity and w states mixture density can be evaluated as 1.177 kilograms per meter cube liquid density would be 1000 kg per meter cube where is a water. Diffusivity from the lectures we say that the diffusivity of or water vapor through air is 2.376 into 10 to minus 5 and nu m will be diffusivity into Schmidt number which will be equal to 1.42 into 10 to minus 6. So, knowing nu m we can evaluate and you relative we can evaluate the Reynolds number that is required here in the Reynolds number expression and d m which is required here to calculate gamma h because this is rho m times d m. So, we can carry out the numerical integration with time step 0.01 second then evaporation time at r w equal to 0 would be 2.05 second and the radius would vary in this fashion sorry 0.5 mm to start with and then it drops down very gradually to 0 value here at about 2.05 seconds. Now, we can solve the same problem in in in stagnant surroundings by simply setting u w u relative equal to 0 which means that the Reynolds number here is 0 and therefore, this will be 2 times gamma h then you will see that the that the that the evaporation time becomes 4.66 seconds. So, clearly having a relative velocity between the between the gas and the droplet has reduced the the evaporation time and this is this is the principle that is used in cooling towers and in diesel engines for the purposes of reducing the size of in case of cooling towers and in in in diesel engines this is of great value because it enhances the rate of burning of the fuel and which in incidentally also reduces the cutoff ratio of a diesel engine which in turn improves the efficiency of the engine. So, finally, in summary then we can say that the Couette flow model with area equal to constant and u equal to constant times y and d psi by dx equal to 0 gives us the formula n w equal to g star ln 1 plus b psi where g over g star is equal to ln 1 plus b psi by b psi. We interpreted the g star flux as the sum of the layer by layer resistances to mass transfer in the considered phase over boundary layer width. In pure momentum and heat transfer in the presence of suction and blowing that is without my gradients of without gradients of species of any kind because the same species being sucked sucked or blown. We have shown that c f x in the presence of suction and blowing developed by c f x in the absence of it is simply ln 1 plus b f by b f and the heat transfer likewise would be ln 1 plus b h by b h. So, Couette flow model recovers essentially the results expected from a boundary layer flow model it also recovers the the results expected from the Stefan flow model and in turn gives us an opportunity to evaluate the effect the effect of property variations which we shall take up in subsequent lectures. So, in the next lectures in we will develop very similar results to what we have shown in in in Couette flow model via the algebraic Reynolds flow model and you will see what form N W and B relation has as per the Reynolds algebraic Reynolds flow model.