 In the last three lectures, we looked at the proxies of the boundary layer flow model, namely the Stefan flow model, the Couette flow model, and the Reynolds flow model, the last one being the algebraic model, whereas the Stefan and Couette flow models were one dimensional models. Now, we look at the complete two dimensional boundary layer flow model. So, as to recover some ideas as to why we expect the simpler models to mimic the boundary layer flow model. So, in this lecture, I will introduce some definitions and then of course, look at the governing equations. And then of course, we will convert them to conserved property forms for all types of mass transfer problems. And then the main important thing is, we will recover the boundary conditions from mass conservation principle and the energy conservation. And ultimately, we will show that N w is equal to G b for small and large mass transfer rates. So, the definition of the boundary layer flow model is as follows. Here is a surface and this is the considered phase and this is the neighboring phase as you know, this is the interface. And in the considered phase, there may be presence of chemical reactions, there may be presence of turbulence, concentration gradients, temperature gradients and so on so forth. Infinity state is one where all the gradients vanish. The total mass transfer that is both convective plus diffusive mass transfer takes place at the interface from the neighboring phase. And in the deep inside the neighboring phase, we define a T state which of course, where the concentration gradients and temperature gradients as again like infinity state. Here again in the T state, they are 0 or we say that the fluid properties are uniform. So, convective mass transfer takes place due to concentration gradients in the considered phase. And since the Reynolds flow model mimics the real flow, the interface mass flux is given is N w is N w equal to G b and N w and G have the same units. N w is required as a boundary condition because it to the mass transfer equation as well as the energy equation. And as well as the momentum equation, because N w brings with it a certain velocity at which the mass comes in. So, assuming steady state mass transfer, now you have d by dx of rho m u psi plus d by dy of rho m v psi plus d by dy of gamma psi d psi by dy plus a source term where psi is any property like when it is 1, we get bulk mass equation because that term is 0 and S w is specified as 0 and that being 1, you simply get d rho m u by dx plus d rho m v by dy equal to 0 and that is the bulk mass equation under steady state. If I say psi is equal to u, then gamma psi would be mu m effective assuming there is turbulent flow and S psi would be usually minus d p by dx which is the pressure gradient. If psi is equal to omega k, then gamma psi as you know would be rho m d effective plus a equal to and source would be R k which is the species transfer equation. Then there would be the element transfer equation for which the source is always 0 because element alpha is a conserved property and H m is the mixture enthalpy. That would have a source term comprising mass transfer due to diffusion flux M double dash y k. The source there can be other sources like d p, d t, q rad and many other sources these are all ignored. We also ignored the viscous dissipation here and all equations are coupled requiring numerical solutions. Simplifications of omega k and H m equations are possible under certain assumptions so that they are rendered to a conserved property equation. We have gone over all these processes of converting omega k and H m equations to conserved property forms in variety of mass transfer problems. So if I write in conserved property form, the equation is simply this with S phi equal to 0, N w equal to G b and b is equal to psi infinity minus psi w psi w minus psi t. Now in inert mass transfer without heat transfer as you will recall psi is simply omega of the vapor and gamma is rho m d the diffusivity. In inert mass transfer with heat transfer psi equal to omega b as N H m and it is energy equation and we make use of the assumption of Lewis number equal to 1 that is gamma m h equal to rho m d plus equal to rho m alpha m in mass transfer with simple chemical reaction means we choose psi appropriately as a combination of fuel and oxygen or fuel and product and so on so forth and H m and again make the Lewis number equal to 1 assumption and also say that the specific heats are equal that is what renders the energy equation in the conserved property form. And in mass transfer with arbitrary chemical reaction, psi is again formed from appropriate combinations of eta alpha and gamma m is rho m d. In each case we need boundary conditions at y equal to 0 which will come from that form of the Reynolds flow model. So, for the inert mass transfer considered mass conservation between t and w states as you will recall this is our w w state this is t state and this is the infinity state and we specify n w here n w here. Then you will see that the mass which is coming in here is n w omega k at t and what is going out is n w times omega k at w and minus the diffusion mass transfer which is rho m diffusivity omega k by d y at y equal to 0. So, this would be the mass balance because this is the multiplied by omega k will be the convective transfer and diffusion transfer will be m dot k at w which is nothing but from Fick's law this is rho m diffusivity into omega k d y at y equal to 0. So, if I rearrange this equation take n w common out here then I get the form n w is equal to rho m d omega k d y at wall divided by omega k w minus omega k at t likewise for any other conserved property phi I have the same equation as you can see here and n w will be rho m d omega d phi by d y at w phi w minus phi t where phi would be as you recall omega f u minus omega over 2 divided by r s t equal to omega f u plus omega product by 1 plus r s a for a simple chemical reaction and phi can be any linear combination of eta alpha and you choose a alpha suitably for a given problem as we indicated last time. Now, these are the boundary conditions derived from mass conservation principle likewise we can do so for the energy equation. So, for example in the energy equation again if I consider this to be w w and this to be t t state then n w times h m t will be the flux coming in as well as q w will be another flux coming in whereas, what is going out is the convective flux n w h w plus m dot w diffusion of the w k h k sigma. So, if I equate the influxes with outgoing fluxes I would get the form which I have shown here. So, this is the incoming flux q w is also the incoming flux which would be on this side and n w is equal to h m w and this is the diffusion flux of species k multiplied by the enthalpy that goes with it of the species k. Now, q w is equal to k m d t d y because q w is shown inwards and that would be equal to c p m gamma h where gamma h is rho m alpha m d t by d y at the w and this enhance n w would written as sigma k gamma m d omega k by d y at w h k plus c p m gamma h d t by d y at w h m w minus h m t and this is the general energy equation conservation principle the final form of the numerator will depend on the mass transfer application at hand. So, let us see the part of the different forms that this numerator will take in different types of mass transfer problems. So, in a mass transfer with heat transfer if I make the assumption of Lewis number one as we usually do then gamma h will be simply equal to gamma m and hence c p m gamma h d t by d y would be gamma h into c p m is simply sigma omega k c p k into d t by d y and if I now absorb c p k inside this derivative then I get a gamma h into sigma omega k d h k by d y at w and hence you will say that the numerator now reads as gamma m into sigma k d omega k by d y at w h k plus gamma h into sigma k of omega k d h k by d y at w. But since we have assumed gamma h is equal to gamma m this is simply the differential these two terms are simply differentials of a product omega k h k d y at w summation of omega k h k is simply gamma m h d h m by d y at w over h m minus h m t h m w minus h m t. So, n w takes the same form as derived from the mass conservation principle, but with the variable enthalpy h m. Let us see the numerator when mass transfer with heat transfer and chemical simple chemical reaction. Now, here we take Lewis number equal to 1 and also say that c p k will be equal to c p m and for the moment let us say delta t stands for t minus t ref then we have enthalpy of the fuel would be c p m into delta t plus omega f u into delta h c that is I have associated the heat of combustion with the fuel and therefore, h o 2 and h product will be simply c p m delta t and hence summation of sigma gamma m d omega k by d y h k would simply amount to gamma m d f u by d by d y at the wall multiplied by d h c because the other quantity c p m delta t if I substitute c p m delta t for all the three species it will simply give me c p m delta t gamma h into summation of d omega k by d y at w equal to 0 because summation of sigma omega k is 1 and likewise c p m gamma h d t by d y would be gamma h d h m by d y at w from this relationship minus gamma h d h c by d f u by d y. So, again you see that the addition of the of this term with the summation of these two terms this one this term and this term would amount to this term will cancel with this term because gamma m is equal to gamma h and therefore, we get n w again is equal to gamma m h d h m by d y at w divided by h m w minus h m t. So, even in simple chemical reaction with these assumptions of equal Lewis number equal to 1 and equal specific heats we recover the same form as we had done earlier with inert mass transfer with heat transfer finally, let us look at single component convective mass transfer. Now, in this case omega k is equal to 1 therefore, this part of the numerator will be 0 and c p m gamma h d t by d y would be simply gamma h because I can multiply c p m is inside this and therefore, that will give me h m by d y at w and hence I get n w equal to d h m by d y w h m w minus h m t which is again what we had earlier. Single component convective heat transfer means supposing I have a boundary layer form with air I am injecting in it air itself from the wall at a given rate. So, the energy equation would take the form that I have shown here. Now, if I further make the assumption that the specific heats do not vary between w t and infinity states then of course, I can make the assumption I can show that n w will be equal to gamma h into d t by d y w divided by t w minus t t. Thus in all cases of mass transfer mass and energy conservation principles give identical formula for n w and combining with Reynolds slow model which claims to mimic the real boundary layer flow model we can say that n w will be equal to gamma psi d psi by d y w over psi w minus psi t which we have just now derived is equal to g times b and that must equal rho m v w. So, you can see now that the momentum equations which require this quantity are coupled with the mass fraction and energy equations through these quantity. What it shows is that the rate of mass transfer would be proportional to psi infinity minus psi w. It will also be proportional to the gradient of psi at the wall and it will be proportional to v w which is from that. This shows that even if gamma is uniform the psi equation is non-linear the gamma is the property. So, even if it was uniform through the considered phase the psi equation is non-linear because velocity field u v is a function of v w and psi infinity minus psi w. Now this kind of coupling through the boundary conditions of momentum and the mass trans scalar equations is akin to the natural convection problem in which u and v are coupled to the energy equation through the source terms we call the buoyancy term. So, in natural convection momentum and energy equations are coupled through boundary through buoyancy source terms whereas in mass transfer the coupling arises because of the boundary condition at the wall. This is the difference between a natural convection problem and a force convection mass transfer problem. So, the coupling between the momentum and psi equations can be ignored when n w which is proportional to v w tends to 0 which means if the mass transfer rate is very very small then we can say that g star which is the value of g for small mass transfer rates then n w divided by b psi n w tends to 0 would be minus gamma psi d omega by dy divided by w psi w minus psi infinity. This follows from the previous equation which I have shown here. So, I have simply cancelled psi w minus psi t psi w minus psi t and this is psi infinity minus psi w is taken to the denominator. So, you get g star is equal to this quantity. So, the g star now depends only on the psi profiles and not on the boundary condition because v w is tending to 0. This definition is analogous to that used to define the heat transfer coefficient. As you will recall we define the heat transfer coefficient as minus k dt dy p w minus t infinity. So, likewise we say that the g star which is the mass transfer coefficient at v w equal to 0 is minus gamma psi d psi by dy at w divided by psi w minus psi infinity. So, the two definitions are now analogous and now you can see why we have been calling g as the mass transfer coefficient. When n w is large the coupling between the momentum equations and the psi equation is strong and n w equal to g times b psi. Hence, g must be a function of b psi and g star equal to g when b psi tends to 0. Yes, g star is simply g with b psi siding to 0, but it will also g will also itself be a function of g and perhaps even of Reynolds number, Prandtl number and so on and so forth. But we shall shortly see what the state of affairs are. By analyzing experimental data on mass transfer with and without combustion, Spaulding showed that within experimental scatter g by g star is equal to n w by b divided by n w by b as at very very small mass transfer rates is in fact equal to f b only. This is this is worth noting f b only. In other words, the equation shows that g over g star is not influenced by Reynolds number, Prandtl number or Schmidt numbers. This is what the experimental data shows both in internal flows as well as external flows. Flow over cylinder, flow over flat plates and flow through the tubes and so on and so forth. It may be a problem of evaporation, it may be a problem of condensation, it may be a problem of combustion. All these problems for which experimental data were available, Spaulding showed that g over g star turns within experimental scatters say about plus minus 10 to 15 percent as g by g star is a function of b only, which is a remarkable result that this ratio should not be influenced by any other quantity other than the driving force b. So, it is independent of Reynolds, Prandtl and Schmidt numbers. So, all that we would require now is that is the value of g star evaluated from Hcoff Vw equal to 0 divided by Cpm and f b to obtain g. So, what are the forms of f b and that is what we want to find out. Incidentally, this book Spaulding db convective mass transfer published in 1963 is a pioneering book on mass transfer and particularly of great relevance to engineering community. So, now let us see what the form of f b should be. So, using computer simulations of the boundary layer equations as well as experimental data Spaulding showed that g over g star is equal to f b and that function f b is nothing but ln 1 plus b by b. Now, if you recall this is the relationship which we had also predicted using Stefan flow model as well as the Couette flow models. Of course, Stefan flow model was for diffusion mass transfer only whereas the Couette flow model was for included convection in its form and therefore, this relationship is unique it does not contain any Reynolds number Prandtl number or anything like that. We take up the issue as to whether this relationship can actually be predict can also be derived from the Reynolds flow model just as we are shown that it mathematically that the Stefan and Couette flow models yield that relationship. So, in order to do that let us reconsider of a t state w state and the infinity state with the considered phase shown here between w and infinity state. Now, here what I have done is let us consider an elemental strip here of thickness delta y the outer edge of that strip is y o the inner edge of that strip is y i and let us postulate g star star as a flux at the y o surface which is coming in and bringing with it the properties of the y o surface and likewise n w plus g star star is a outgoing flux which brings with it the properties of the y i state. So, this is what I have said here g star star crosses the y o surface carrying properties of the y o surface and similarly the Reynolds flux n g star star plus n w crosses y o surface carrying with it properties of the y i surface. The physical idea behind introduction of g star star is that the real flow process is like heat conduction mass diffusion turbulence etcetera do behave like the Reynolds flow, but on a much smaller scale delta y y o minus y i tending to 0. So, if I now write the mass conservation over y o and t states then n w psi t would be the incoming flux plus g star star into psi times y o would be the incoming flux from y o surface and that would equal g star star plus n w into psi y i and therefore this would rearrangement of this would give me n w divided by g star star is equal to psi y o minus psi y i divided by psi y i minus psi t or this difference is nothing but d psi y and this is psi y minus psi t. So, if we consider large numbers of delta y between 0 and delta of the considered phase or between infinity and w states then simply n w into summation of 1 over g star star would amount to integration of 0 to infinity d psi y divided by psi y minus psi t and the integration would naturally yield ln 1 plus psi y infinity minus psi w divided by psi w minus psi t which as you know is nothing but b psi ln 1 plus b psi. So, if I take this result further then as b psi tends to 0 n w into infinity 1 over g star star would simply tend to b psi itself and so as b psi tends to 0 w to infinity is g star star raise to minus 1 would stand to b psi by n w and therefore comparison of with the observation of slide 11 what was the observation we made on slide 11 g over g star is equal to n w b over n w by b n w 0 is equal to f b. So, comparison with that will show that as b psi tends to 0 the sum of g star star raise to minus 1 is nothing but the g star raise to minus 1 and hence n w equal to g b psi is equal to g star ln 1 plus b psi and g over g star is equal to f b and that is equal to 1 plus ln 1 plus b psi over b psi. So, this is a very interesting result that 1 over remember g is conductance and therefore 1 over g is resistance and therefore the total resistance between w and infinity states which we say is 1 over g star is nothing but sigma 1 over g star star where g star star is the resistance of the small element g delta y over delta y. So, we can interpret now that the 1 over g star the resistance is the sum of the resistances to mass transfer between w and infinity states and the inverse of that is the conductance of for mass transfer and in other words the ratio of the conductance at large mass transfer rates the conductance at small mass transfer rates or negligible mass transfer rates is simply a function of ln 1 plus b psi by b psi. This formula can be used for large mass transfer rates obtained in liquid fuel burning and in transpiration cooling as we shall see in later lectures where I will be considering problems. Small b psi usually occurs in combustion of solid fuels and in evaporative cooling or air conditioning and so on so forth. But where in these applications b psi would be a less than 0.1 in these applications combustion of solid fuel and evaporative cooling and in fact it will be of the order of 0.02, 0.03. Whereas in transpiration cooling and liquid fuel burning value of b psi can vary between 0.5 to nearly 10. So b psi can be really large in liquid fuel burning and in transpiration cooling. So we shall check how good is this formula from the analytical solutions we have derived so far for laminar flows as well as in turbulent flows and let us see how well this formula is so that we can confidently use it at different mass transfer rate. If the mass transfer rate is very low then of course there is no problem because g by g star will be simply 1 and you would have straight away there is no difficulty. If g by g star is simply is moderate then of course this formula will work. At very very large mass transfer rate remember the definition of b psi says that b psi is psi infinity minus psi w divided by psi w minus psi t. So when b psi is very large psi infinity minus psi w is also large and this means that when the property in the infinity state and the w state is large we would expect the properties to vary between the w and infinity state. So any departure from this formula if found in experiments would be largely due to property variations and therefore this formula is usually corrected for property variation and that is the matter for discussion in the next lecture.